TCS APTITUDE PLACEMENT PAPER
TCS APTITUDE PLACEMENT PAPER
Q 1.2/3rd of the balls in a bag are
blue, the rest are pink. if 5/9th of the blue balls and 7/8th of the pink balls
are defective, find the total number of balls in the bag given that the number
of non defective balls is 146?
a) 216
b) 649
c) 432
d) 578
Solution- let total
no of balls =x
blue=2x/3 pink=x/3 total no of defective balls =
10x/27 +7x/24
=143x/216 non defective balls=x-143x/216=146 x=432
Q 2. Find no of ways in which 4
particular persons a,b,c,d and 6 more persons can stand in a queue so that A
always stand before B. B always stand before C, And C always stand before D.
a)6!
b)7!
c)1006*6!
d)10046!
Solution - a,b,c,d are grouped ie consider them as
one and remaining as 6. total 6+1 = 7! Ways
Q 3. 100 students appeared for two
different examinations 60 passed the first,50 the second and 30 both the
examinations.Find the probability that a student selected at random failed in
both the examination
a)5/6
b)1/5
c)1/7
d)5/7
Solution-60+50-30=80
100-80=20 20/100=1/5. so B is the answer.
Q 4. There are 10 points on a
straight line AB and 8 on another straight line AC none of them being point A.
how many triangles can be formed with these points as vertices?
Option
a. 680
b. 720
c. 816
d. 640
Solution-
To form a triangle we need 3 points select 2 points
from the 10 points of line AB & 1 from the 8 on AC
= (10C2)*(8C1)
select 2 points from the 8 points of line AC & 1 from the 10 on AB=
(8C2)*(10C1)
total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640
d.640
Q 5. From a bag containing 8 green and 5 red
balls,three are drawn one after the other .the probability of all three balls
beings green if the balls drawn are replace before the next ball pick and the
balls drawn are not replaced , are respectively. a)512/2197,336/2197
b) 512/2197,336/1716
c) 336/2197,512/2197
d) 336/1716,512/1716
Solution-THE
PROBABILITIES OF GETTING WITH REPLACEMENT IS=8/13*8/13*8/13=512/2197
THE PROBABILITIES OF
GETTING WITHOUT REPLACEMENT =8/13*7/12*6/11=336/2197
Q 6. find the greatest number that
will divide 148 246 and 623 leaving remainders 4 6 and 11 respectively.
a) 20 b) 12 c)6 d)48
Solution-Hcf ( (148-4), (246-6),
(623-11))=12
Q 7. a mother her little daughter and her just born infant
boy together stood on a weighing machine which shows 74kgs.how much does the
daughter weigh if the mother weighs 46kg more than the combined weight of
daughter and the infant and the infant weighs 60% less than the daughter. a)9
b)11
c)cannot be determined
d)10
Solution-daughter weight is x infant weight is 60%
less than daughter i.e., 0.4x
mother weight is (x+0.4x+46) total weight = x + 0.4x
+ (x+0.4X+46) = 74 solving the eq. x =
10 option d is correct
Q 8. find the number of ways a
batsman can score a double century only in terms of 4's & 6's?
a)
15
b)
16
c) 17
d)
18
Solution-4's and 6's
50 0
47 2
44 4
41 6
38 8
35 10
32 12
29 14
26 16
23 18
20 20
17 22
14 24
11 26
8 28
5 30
2 32
So total 17 ways but
here it is 4's & 6's both so don't consider 1st one
Final ans : 16 ways
Q 9. 98. Thomas takes 7 days to
paint a house completely whereas Raj would require 9 days to paint the same
house completely. How many days will take to paint the house if both them work
together. (Give answers to the nearest integer)?
A.
4
B.
2
C. 5
D. 3
Solution-work done by
thomas in day=1/7
work done by other in a day=1/9 work done by both in
a day=1/7 +1/9 =16/63 days required by both = 63/16
Answer-A
Q 10. how many positive integers
less than 4300 of digits 0-4.
a) 560 b)565 c)575
d)625 Solution-Ans is 575 one digit no =4 (0 is not a positive integer) two
digit no=4*5=20 three digit no=4*5*5=100
four digit no=3*5*5*5=375(the
possibility for 1,2,3 will come in the first position)four digit no=1*3*5*5(the
possibility of 4is fixed in the first position and then 0,1,2 is comes in
second position)and the last digit is 4300 we include this number also
Ans is 4+20+100+375+75+1=575
Q 11. A person travels from Chennai
to Pondicherry in cycle at 7.5 Kmph. Another person travels the same distance
in train at a speed of 30 Kmph and reached 30 mins earlier. Find the
distance.
A)5
Km
B)10
Km
C)15km
D)20km
Solution-Let, time taken by ϲ =t
//ly, time taken by train ω =t-30
We know that.....speed=distance
/time
Speed of bicycle ϲ, 7.5=d/t....(1)
Speed of ω,
30=d/(t-30/60)....(2)
Sol 1&2,we get t=0.666
By sub and value in equal(1)
We d=4.999~5km
Q 12. A bag contains 8 white balls,
and 3 blue balls. Another bag contains 7 white, and 4 blue balls. What is the
probability of getting blue ball?
a)3/7 b)7/22 c)7/25 d)7/15
Solution-First we have to select a
bag and then we will draw a ball.
Probability of selection of both
bags is equal =1/2
Now probability of
blue ball taken from first bag = ( 1/2) x (3/11) and probability of blue ball
taken from second bag = (1/2) x (4/11) So probability of blue ball = ( 1/2) x
(3/11) + (1/2) x (4/11) = 7/22
Q 13. In a 3*3 square grid
comprising 9 tiles each tile can be painted in red or blue color. When the tile
is rotated by 180 degree, there is no difference which can be spotted. How many
such possibilities are there?
a) 16 b)32 c)64 d)256
Solution-ans is 32 .
ans grid has to be rotated at 180
degrees.
hence, 11=33 12=32
13=31
21=23
22=22
31=13
32=12 33=11
cell 11-33 can be
altered in 2 ways (as thr are 2 colours) cell 12-32 can be altered in 2 ways cell
13-31 can be altered in 2 ways cell 21-23 can be altered in 2 ways and , cell
22 can be altered in 2 ways
, so
2X2X2X2X2 = 32.
Q 14. Jake can dig a well in 16 days. Paul
can dig the same well in 24 days. Jake, Paul and Hari together dig the well in
8 days. Hari alone can dig the well in option A)48
B)96
C)24
D)32
Solution-1/x = 1/8 - (1/16 + 1/24)
so x=48
ans = 48 days
Q 15. For any two numbers we define
an operation $ yielding another number, X $ Y such that following condition
holds:
•
X $ X = 0 for all X
•
X $ (Y $ Z) = X $ Y + Z
Find the Value of 2012 $ 0 + 2012 $
1912
Options
a)
2112
b)
100
c)
5936
d)
Can not be determined
Solution-here $
represent the - operator so X-X=0 first condition x$(y$z)=x-(y-z)=x-y+z it can
be written as x$y+z
2012-0+2012-1912=2112 so ans is
2112
Q 16. On a toss of two dice, A
throws a total of 5. Then the probability that he will throw another 5 before
he throws 7 is
a)40% b) 45% c)50% d)60%
Solution-ans: 40%
Explanation:
total probabilities for getting 5 =
4/36 total probabilities for getting 7 = 6/36 Total Probability = 10/36
We need only 5, hence prob of
getting only 5 is (4/36)/(10/36)
=40%
Q 17.
1,2,2,3,3,3,4,4,4,4,1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,…………………………………..
Then what is the 2320 position of
the number in the sequence?
a) 2 b) 1c) 3 d) 4
Solution-answer is b)1
1,2,3,4(1-1time 2- 2times 3-3
4-4)=10 terms /completes cycle and starts from 1
1,2,3,4(1-2 2-4 3-6 4-8)= 20 terms
/completes cycle and starts from 1
(1,2,3,4)each digit 3 time to its
value =30 terms/completes cycle and starts from 1
10+20+30+40+50+.......=x
x is nearer value to 2320 solving n(n+1)/2
10.(21.22)/2=
2310 analysing it 2310 completes cycle
and starts from 1 again now it 22 times
(1-22 times 2-44 times ......)
2320 position will occupied by 1
Q 18. In 2003, there are 28 days in February and there are
365 days in the year. In
2004, there are 29 days in February and there are 366 days in the year.
If the date March 11, 2003 is a Tuesday, then which one of the following would
be the date March 11, 2004 be?
A.
Wednesday
B.
Tuesday
C.
Thursday
D.
Monday
Solution-Every year day is
increased by 1 odd day. Or in leap year it is increased by 2 odd days.
so 11 march 2003 is tuesday,
11 march 2004 is thursday
C. Thursday
Q 19. How many 6 digit even numbers
can be formed from digits 1 2 3 4 5 6 7 so that the digit should not repeat and
the second last digit is even?
a)6480 b)320
c)2160 d)720 solution-given 6th digit even number , so last digit 2 or 4 or
6-> 3 ways " 5th digit should be even...so there will be 2 ways(rep.
not allowed) so,therefore we get 5*4*3*2*2*3=720 ways
Q 20. There are 5 letters and 5
addressed envelopes. If the letters are put at random in the envelops, the
probability that all the letters may be placed in wrongly addressed envelopes
is.
a)119 b)44 c)53 d)40
Solution-If there is one letter and
one envelope then no way you can put it wrong(S1).
If there are 2 letters and 2
envelopes then you can put them wrong in 1 way(S2).
If there are 3 letters and 3
envelopes then you can put them wrong in 2 ways(S3).
If there are 4 then you can put
them wrong in 9 ways(S4).
If there are 5 then you can put
them wrong in 44 ways(S5).
If you observe you can find a
pattern.
S3=(S1+S2)*2
S4=(S2+S3)*3
S5=(S3+S4)*4
S6=(S4+S5)*5
In general, Sn=(Sn-2 + Sn-1)*(n-1)
So, if there are 5 letters then
S5=(S3+S4)*4=(2+9)*4=44
Q 21. How many liters of a 90% of
concentrated acid needs to be mixed with a 75% solution of concentrated acid to
get a 30 liter solution of 78% concentrated acid?
a) 8 b)9 c) 7 d)6
Solution-the concentration is given
which is wrt 100, hence we can take x lt of 90% and (30-x)of 75%
x*90 + (30-x)*75 = 30*78 hence the
ans is 6 ltr...
Q 22. Average marks of a,b,c is 48.
When d joins average becomes 47. E has 3 more marks than d. Average marks of
b,c,d,e is 48. What is the marks of a?
a) 42 b) 43 c) 53 d)56 Solution- a+b+c=144
a+b+c+d=188 d=44 e=47 b+c+d+e=192 b+c=101 b=43 ans.
Q 23. On a certain assembly line,
the rejection rate for Hyundai i10s production was 4 percent, for Hyundai i20s
production 8 percent and for the 2 cars combined 7 percent. What was the ratio
of Hyundais i10 production?
option
a) 3/1
b) 2/1
c) 1/1
d) ½
Solution-let the
no of i10 cars be x and i20 be y now th rejcted i10 cars are 4x/100 and i20
cars are 8y/100
and it is given
that 4x/100 +8y/100 = 7(x+y)/100 so we get y = 3x so the ratio is 3:1
Q 24. For a car there are 5 tyres
including one spare tyre(4+1). All tyres are equally used. If the total
distance travelled by the car is 40000km then what is the average distance
travelled by the each tyre?
Option
a) 10000
b) 40000
c) 32000
d) 8000
Solution-total
distance travelled by the car=40000km so total distance travelled by 4 wheels=4*40000=160000
as all tyres(4+1) are equally used so average distance travelled by the each
tyre=160000/5=32000 option(c)
Q 25. If A=x3 y2 and B=xy3, then
find the HCF of A,B.
a)x4y5 b)xy2 c)xy d)x3
Solution-if
A=x^3*y^3 B=xy3 then hcf=xy2
bcoz hcf is always equal to the
highest common powers between the expressions,
i.e x1 is common in both and also
y2 is common in both.
Q 26. In a clock the long hand is
of 8cm and the short hand is of 7cm. if the clock runs for 4 days find out the
total distance covered by both the hands
a)
1824 π cm
b)1648π cm
c)1724π cm d)2028π cm
Solution-Explanation: Short Hand ( Hour hand)
Hour hand makes a full rotation in
12 hours.
One full rotation in 12 hours
=>2πr=14π cm traversed every 12 hours.
For one day(24 hours ), we have 28π
cm, twice that of a 12 hour period.
For 3 days, we then have 4 x 28
π=112π cm traversed.
Long Hand ( Minute hand)
One full rotation in 1 hour
=>2πr=16π cm traversed every hour.
For one day, we have 24 x 16 π=384π cm.
For 4 days, we then have 4 x
384π=1536π cm traversed.
Total Distance
For the total, we have
112π+1536π=1648π cm.
Q 27. A can finish a work in 18
days and B can do the same work in 15 days. B worked for 10 days and left the
job. In how many days A alone can finish the remaining work?
a) 7 b) 6 c) 5 d) 10
Solution-A can finish 118118th of
the total work in a day and B can finish 115115th of the total work in a day.
After working 10 days by B alone,the work remains is 1−1015=131−1015=13th of
the total work.
It can be finished by A
in13÷118=1×183×1=613÷118=1×183×1=6days.
Q 28. In how many way possible ways
can you write 1800 as a product of 3 positive integers a,b,c
a)350 b)360 c)380 d)450
Solution-1800=2^3*3^2*5^2 then 5c2*4c2*4c2=360 ans Q 29. In ahorse racing
competition there were 18 numbered 1 to 18.The organizers assigned a
probability of winning the race to each horse baesd on horses health and
training the probability that horse one would win is 1/7, that 2 would win is
1/8, and that 3 would win is 1/7.Assuming that tie is imposible Find the chance
that one of these three will win the race?
a)22/392 b)1/392 c)23/56 d)391/392
Solution-HORSE 1: 1/7 WINNING
PROBABILITY
HORSE 2: 1/8 WINNING PROBABILITY
HORSE 3: 1/7 WINNING PROBABILITY
ONE OF THESE WIN THE RACE:
=> 1/7 + 1/8 + 1/7
=> 8/56 +7/56 + 8/56 (TAKING
LCM)
=> 23/56
Q 30. Apple costs L rupees per
kilogram for first 30kgs and Q rupees per kilogram for each additional
kilogram. If the price of 33 kilograms is 11.67and for 36kgs of Apples is 12.48
then the cost of first 10 kegs of Apples is
a) 3.50 b) 10.53 c) 1.17 d)2.8
Solution-30L+3Q=11.67
30L+6Q=12.48
------------
3Q=.81 Q= .27 from
that L=0.362 cost of 10 kg apple is 10*.362=3.6 a) 3.50
Q 31. How many vehicle registration
plate numbers can be formed with digits 1,2,3,4,5(no digits being repeated)if
it is given that registration number can have 1 to 5 digits ?
a) 205 b) 100 c) 325 d) 105
Solution-you can have registration
plates of 5,4,3,2 or 1 digits
So, it's 5*4*3*2*1 + 5*4*3*2 +
5*4*3 +5*4 + 5
=120 + 120 + 60 + 20 +5
=> 325
Q 32. Jake and Paul each walk 10
km. The speed of jack is 1.5 faster than paul speed . what is the Jake's speed
?
a)4 b)6 c)7 d)8
Solution-Let pauls speed be x kmph
Then jacks speed is (x+1.5 ) kmph
(10/x)-(10/1.5+x)=1.5
X=2.5
Jacks speed=2.5+1.5=4kmph
Q33. In this question A^B means A raised to the power B. If
f(x)=a*x^4-b*x^2+x+5.
f(-3)=2. Then f(3)=?
a)3 b) 7 c) 8 d) 6
Solution-Given
that f(x)=a*x^4-b*x^2+x+5 and f(-3)=2 so => a*(-3)^4-b*(-3)^2+(-3)+5=2
=> 81a-9b+2=2
=> 81a-9b=0
-------equation(1) now f(3)=a*(3)^4-b*(3)^2+3+5
=81a-9b+8
and from equation
(1) 81a-9b=0 so f(3)=0+8=8
Q 34. Of a set of 30 numbers,
average of first 10 numbers = average of last 20 numbers. Then the sum of the
last 20 numbers is ?
(a) 2
X sum of first ten numbers
(b)
2 X sum of last ten numbers
(c)Sum of first ten numbers
(d)Cannot be
determined with given data Solution-since, average=(sum of n no.s)/(total no)
therefore, (sum of first 10 no.s)/10 ==(sum of last 20 no)/20 hence. (sum of
last 20 no.s) = 2*(sum of first 10 no.s)
Q 35. Mother, daughter and infant
total weight is 74 kg. Mother's weight is 46 kg more than daughter and infant's
weight. Infant's weight is 60% less than daughter's weight. Find daughter's
weight.
a)10 b)9 c)8 d) 7
Solution-Total Age
is M+D+I=74 given that M-D-I=46 solving above 2 eq's we get Mother age = 60 now
remaing age=14 which is sum of daughter and Infant age.
Given that Infant age is 60% lass
than Daughter.
i.e If daughter
age is 100 then infant age is 40. So ages ratio Of D and I is 100 : 40 i.e ...
5 : 2 So (5+2)=7 parts equal to 14. then 5 parts equal to 10.
2 parts equal to
4. daughter age is =10 and infant age is = 4
Answer a) 10
Q 36. In a certain city, 60% of the
registered voters are congress supporters and the rest are BJP supporters. In
an assembly election, if 75% of the registered congress supporters and 20% of
the registered BJP supporters are expected to vote for candidate A, what
percent of the registered voters are expected to vote for candidate A?
a) 53 b) 20 c) 60 d) 75
Solution-let the people in the city
be x
60% are
congress=60% of x 40% are BJP=40% of x out of 60%,75% voted for
congress=75%(60% of x)=18x/40 out of 40%,20% voted for congress=20%(40% of
x)=8x/100 total=18x/40+8x/100=106x/200 total percent=106x/200*100=53% of x
Q 37. In a particular year the
month of january had exactly 4 thursdays and 4 sundays , on which day of the
weekk, jan 1 occurs?
a) Monday b) Tuesday c) Thursday d)
Wednesday
Solution-as there are 4 fulls weeks
i.e 28 days.. so..every day occurs min 4 times.
den remaining 3 days
(as jan has 31 days) will be monday tuesday wednesday. so on 31st jan comes
wednesday.
so 1st jan ll be MONDAY
Q 38. A number when divided
successively by 4 and 5 leaves remainder 1 and 4 respectively. when it is
successively divided by 5 and 4, then the respective remainders will be;
a). 1,2
b). 2,3
c). 3,2
d). 4,1
Solution-When dividing a positive
integer nn by another positive integer DD (divider), we obtain a quotient QQ,
which is a non-negative integer and a remainder R, which is an integer such
that 0≤R<D0≤R<D. We can write n=DQ+R.n=DQ+R.
When dividing our number nn by 4 we
obtain a remainder of 1, so, if the quotient is some integer QQ, we can write
n=4Q+1.n=4Q+1.
Now, dividing QQ by 5, we obtain
another quotient say qq and remainder 4, thus we can write Q=5q+4.Q=5q+4.
It follows that
n=4(5q+4)+1=20q+17.n=4(5q+4)+1=20q+17.
Since
n=20q+17=5(4q+3)+2n=20q+17=5(4q+3)+2, it means that when dividing nn by 5
first, we get a quotient 4q+34q+3 and remainder 2.
Then dividing 4q+34q+3 by 4 we
obviously obtain a remainder of 3.
Q 39. Average salary of 17 teachers
is 45000.3 teachers left and the average salary dropped by 2500.What is the sum
of salaries of 3 teachers who left?
a)173000 b) 176000 c) 170000 d) 85000
Solution- Total Initial Salary :
17*45000 = 765000
Average Salary After removal of 3
Teachers = 45000-2500 = 42500
Total Final Salary : 14*42500 =
595000
Sum of Salaries of 3 teachers who
left : 765000 - 595000 = 170000
Q 40. There are 20 persons sitting
in a circle. In that, there are 18 men and 2 sisters. How many arrangements are
possible, in which the two sisters are always separated by a man?
a.
18!*2
b.
17!
c.
17!*2
d.
12
Solution- 18!*2 Consider 1 man
along with two sisters as one group.. so they can be arranged in 17! ways as
circular.. The one man in between the two sisters can be out of any 18 men..
so,17!*18.. and the two sisters can be arranged in 2 ways..so 18!*2
Q 41. a number plate can be formed
with two alphabets followed by two digits with no repetition. then how many
possible combinations can we get?
a) 58500 b) 67600 c) 57850 d) 58761
Solution- no.of alphabets=26 (a-z),
no.of digits=10(0-9). ways of arranging
two alphabets with out repetition=26*25; ways of forming two digits without
repetition=10*9 no.of combinations of forming the number on number plate=
26*25*10*9=58500 Q 42. The letters in the word "PLACES" are permuted
in all possible ways and arranged in the alphabetical order.Find the word at 48
position.
a)AESPCL
b)ALCEPS
c)ALSCEP
d)AESPLC
Solution- ans =
(d) for words AC**** possible ways for arranging * will be 4!=24 now next seq
in alphabatical order will be AE**** so....
same way for AE**** possible ways
for arranging * will be 4!=24
.. ..
Thus, 48th element will be last
element in AE**** that means AE followd by reverse alphabatical order! which is
AESPLC
Q 43. A alone can do 1/4th of the
work in 2 days. B alone can do 2/3th of the work in 4 days. If all the three
work together, they can complete it in 3 days so what part of the work will be
completed by C in 2 days?
A. 1/12
B. 1/8
C. 1/16
D. 1/20
Solution- A can do the total work
in 8 days, and B can do it in 6 days.
Let the total work be 24 units. Now
capacities are
A =
24/8 = 3,
B =
24/6 = 4,
A + B + C = 24/3 = 8
So Capacity of C = 1 unit.
In two days C will do 2 units which
is 2/24th part of the total work. So 1/12th part.
Q 44. A sum is sufficient to pay
either George age for 15 days or marks wage to 10dayshow long together?
a)5 b) 6 c) 8 d) 9
Solution- George one day work is
(1/15)
Marks one day work is (1/10)
G+M=(1/15)+(1/10)=(1/6)
6 is the answer
Q 45. In how many ways a team of 11
must be selected a team 5 men and 11 women such that the team must comprise of
not more than 3 men.
a) 1565 b) 2256 c) 2456 d) 1243
Solution- Maximum 3 men can be
played which means there can be 0, 1, 2, 3 men in the team.
(5C0×11C11)+(5C1×11C10)+(5C2×11C9)+(5C3×11C8)=2256
Q 47. how many prime numbers less
that 100 and greater than 3 are of the form:4x+1 , 5y-1
a) 11 b)12 c)7 d)None of the above
Solution- all the prime numbers between 3
and 100 are:5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73, 79, 83, 89
and 97 numbers must end with 4 or 9 to follow 5y-1 condition.
so the numbers
are:29,89 so there are two numbers.
Q 48. Three dice are rolled. What
is the probability of getting a sum of 13?
a) 19/216 b) 21/216 c) 17/216 d)
23/216
Solution- Just count the number of
ways to get 13. We just need to count possibilities for two dice because the
third dice value is fixed. For two dice, the sum can be anywhere from 7 to 12
and that would be 6 + 5 + 4 + 3 + 2 + 1 = 21. So, the probability is
21/216
Q 49. On a 26-question test, five
points were deducted for each wrong answer and eight points were added for each
correct answer. If all the questions were answered, how many were correct if
the score was zero?
a). 10
b). 11
c). 12
d). 13
Solution- if x no. of question is
wrong and y no. of question is correct then
-5*x+8*y=0 and
x+y=26 by solving we get x=16 & y=10
Q 50.Two alloys A and B are composed of two
basic elements. The ratios of the compositions of the two basic elements in the
two alloys are 5:3 and 1:2, respectively. A new alloy X is formed by mixing the
two alloys A and B in the ratio 4:3. What is the ratio of the composition of
the two basic elements in alloy X? a)1:1 b) 2:3 c) 5:2 d) 4:32
Solution- Let the actual amount of
A be 4x and actual amount of B be 3x since A and B are in the ratio 4/3
amount of first basic element in
the new alloy T (5/8)*4x + (1/3)*3x = (7x)/2
amount of second basic element in
the new alloy T (3/8)*4x + (2/3)*3x= (7x)/2
so ratio of first basic element to
second basic element: [(7x)/2 ] / [ 7x/2] = 1/1 = 1:1 ( Answer A)
Q 51. Babla alone can do a piece of
work in 10 days ashu alone can do it in 15 days. The total wages for the work
in Rs.5000 .how much should babla be paid of they work together for entire
duration of the work
a)4000 b)3000 c)5000 d)2000
Solution- babla 10
day------ now 30/10=3 unit per day L.C.M = 30(total work they have to perform)
ashu 15 day------- 30/15=2 unit per day
so their ratios of work is 3:2 so
babla will get 3/5*5000=3000 Q 52. Average of 3 numbers ABC is given as 48.
Average of A,B,C,D is 46. Its given that E is having 3 more than D, then
Average of B,C,D,E is 45. What is the score of A?
a) 46 b) 48 c) 49 d) 47
Solution- average
A+B+C= 48; A+B+C=3*48=144; average A+B+C + D= 46; A+B+C+D =4*46=184;
D=40; GIVEN E=D+3=40+3=43;
B+C+D+E=45*4= B+C+40+43=45*4
B+C=180-40-43
B+C=97
A+B+C=144
A+97=144
A=47
Q 53. Raj travels a part of journey
by taxi paying 15 per km and rest by train paying 21per km. If he travels a
total of 450 Km and pay Rs.8130 then the distance travelled by raj in train?
a) 230 b) 260 c) 190 d) 180
Solution- let the
distance traveled by train =x therefore distance travelled by taxi = (450-x)
now, 15*(450-x) + 21*x=8130 or, x=230
Q54. A sum is sufficient to pay
either George age for 15 days or marks wage to 10dayshow long together?
a) 9 b) 5 c) 6 d) 8
Solution- George one day work is
(1/15)
Marks one day work is (1/10)
G+M=(1/15)+(1/10)=(1/6)
6 is the answer
Q 55. Radius of the bigger circle
is 1. Which area will be greater?
a) 5 b) 4 c) cannot be determined
d) none of these
Solution- If the radius of the
bigger circle is 1, then diameter = 2 units.
Which in turn equals to diagonal of square.
Let the side of the square be x. Then x2 + x2 = 22 ⇒ 2x2 = 4 ⇒ x = √22
Now diameter of the inner circle =
side of the square. So radius of the
inner circle = √22=1√222=12
Areas marked by 1, 2, 3, 4 = (Area
of the circle - area of the square)/4 = π(1)2−(√2)24π(1)2−(2)24 = π−24π−24 =
0.285
Areas marked by 5, 6, 7, 8 = (Area
of the square - area of the inner circle)/4 = (√2)2−π(1√2)24(2)2−π(12)24 =
2−π242−π24 = 0.1075
So Area marked by 4 is bigger.
Q 56. University of Vikram has
enrolled nine Phd candidates.Babu Chitra,Dheeraj , Eesha,Farooq,Gowri ,
Hameed,Iqbal,Jacob.
-Farooq and Iqbal were enrolled on
the same day as each other, and no one else was enrolled that day.
-Chitra and gowri were enrolled on
the same day as each other, and no one else was enrolled that day.
-On each of the other days of
hiring , exactly one candidate was enrolled.
-Eesha was enrolled before Babu.
-Hameed was enrolled before Dheeraj
-Dheeraj was enrolled after Iqbal
but before Eesha
- Gowri was enrolled after both
Jacob and Babu
-Babu was enrolled before Jacob
Who were the last two candidates to
be enrolled?
a)Gowri and Chitra
b) Babu and Chitra c) Babu and Gowri d) Eesha and Jacob Solution- Given that
1.
Easha < Babu
2.
Hameed < Dheeraj
3.
Iqbal < Dheeraj < Easha
4. Jacob/Babu
< Gowri 5. Babu < Jacob
from 1 and 5, Easha was before Babu
and Jacob so she cannot be in the last two. Option B ruled out
from 4 and 5, babu is before Jacob and
Gowri so he cannot be in the last two. Options a, c ruled out. So option d is
correct.
Q 56. A boy buys 18 sharpeners
(brown or white) for Rs. 100. For every white sharpener, he pays one rupee more
than the brown sharpener. What is the cost of white sharpener & how much
did he buy?
a) 5,13 b)5,10 c)6,10 d) None of
these
Solution- If he bought x white sharpeners
@ Rs (y+1) and (18-x) brown sharpeneres @ Rs y per sharpener, then x*(y+1)+x*(18-x)=100 x= 100-18y
Only integral
value of x less than 18 will be 10. then x=10, y=5
so he bought 10 white sharpeners @
Rs 6 per sharpeners and 8 brown sharpeners @ Rs 5 per sharpener.
c)6,10
Q 57. If M is 30% of Q, Q is 20% of
P, N is 50% of P, Then M/N = ?
a) 4/3 b) 3/25 c) 6/5 d) 3/250
Solution-
m=3/10q,q=1/5p,n=1/2p here we can find d value of p and put into d second to
get value of q so m/n will be 3/25
Q 58. In a staircase, there are 10
steps. A child is attempting to climb the staircase. Each time, she can either
make 1 step or 2 steps. In how many different ways can she climb the stair
case?
a). 10 b). 21 c). 36 d). None of
these.
Solution- she can go by
1 steps-1
way
that is choosing 1 two-step in 9
moves:
9C1 : 9 ways//
2 two-steps:
choosing 2 two-steps in 8 moves:
8C2 = 28 ways//
3 two-steps
7C3 = 35 ways//
4 two-steps//
6C4 = 15 ways//
5 two-steps//
which covers all the 10 stairs.. that means only one way
2 2 2 2 2 = 1 way//
Adding all the ways:
1 + 9 + 28 + 35 + 15 + 1 = 89
ways//
Q 59. Eesha bought two varities of
rice costing 50Rs per kg and 60 Rs per kg and mixed them in some ratio.Then she
sold that mixture at 70 Rs per kg making a profit of 20 % What was the ratio of
the mixture?
a)1:10 b) 3:8 c)
1:5 d) 2:7 solution- let x be the price of the rice after mixing,then
120*x/100=70
x=175/3 by eligation method the required ratio is(60-175/3):(175/3-50)=1:5
Q 60.Find the 32nd wordin the list
where the word MONOS IS permuted in all posiblities ways and arranged in
alphabetical order.
a) OSMON b) OSNOM c) OSMNO d) ONMSO
Solution-
Arranging in alphabetical order
MNOOS
M_ _ _ _ CAN BE ARRANGED IN
4!/2!=12 WAYS
N_ _ _ _ CAN BE ARRANGED IN
4!/2!=12 WAYS (12+12=24)
O_ _ _ _ CAN BE ARRANGED IN
4!/2!=12 WAYS (12+12+12=36) OUT OF BOUND
OM_ _ _CAN BE ARRANGED IN 3!=6
WAYS(12+12+6=30)
30TH WORD IS OMSON
31ST WORD IS ONMOS
32ND WORD IS ONMSO
Q 61. One card is lost out of 52
cards. two cards are drawn randomly. They are spade. What is the probability
that the lost card is also spade?
a)1/52 b)1/13 c) 1/4 d) 11/50
Solution- (13 - 2) / (52 - 2) = 11
/ 50
Q 62. What is the value of
44444445*88888885*44444442*44444438/44444448^2
a)88888883
b)88888884 c)88888888 d)44444443 Solution- take x=44444444 that equ.can be
written as- (x+1) (2x-3) (x-2)+(x-6) / x^2 solving equation we get 2x-5
substitute x=44444444 in above equation we get- 88888883
Q 63.Complete the series 4,20,35,49,62,74,?
a)76 b)79 c) 78 d) 85
Solution- 4+16=20
20+15=35
35+14=49
49+13=62
62+12=74
74+11=85
Q 64. Walking at 3/4 of his normal
speed, Mike is 16 minutes late in reaching his office. The usual time taken by
him to cover the distance between his home and his office is
a). 42 minutes
b). 48 minutes
c). 60 minutes
d). 62 minutes
Solution- Let s =
his normal speed t = his normal time
Then
D = (3/4)s * (t+16)
Since the distance is the same we can equate
this to his regular day which is D = s*t s*t = (3/4)s * (t+16) t=48
Q 65. The sum of 5 numbers in AP is
30 and the sum of their squares is 190. Which of the following is the third
term?
a)5 b) 6 c) 8 d) 9
Solution- consider the
5 numbers in AP as a-2d,a-d,a,a+d,a+2d; given,a-2d+a-d+a+a+d+a+2d=30;
5a=30==>a=6
here a is the 3 rd term so..third
term is 6.
Q 66. A, B and C can do a piece of
work in 20, 30 and 60 days respectively. In how many days can A do the work if
he is assisted by B and C on every third day?
a). 12 days
b). 15 days
c). 16 days
d). 18 days
solution-
A's 2 day's work =2 * (1/20)= 1/10
(A
+ B + C)'s 1 day's work =((1/20)+(1/30)+(1/60))= 1/10
Work done in 3 days =(1/10)+(1/10)=
1/5
Now, (1/5) work is done in 3 days.
Whole work will be done in (3 x 5)
= 15 days.
Q 67. The least number which when
divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders
respectively, is:
a) 4562
b) 15110
c) 2135
d) 7589
Solution- Here (48 - 38 ) = ( 60 -
50 ) = ( 72 - 62 ) = ( 108 - 98 ) = ( 140 - 130 ) = 10 in every case
Leat number will be LCM of ( 48,
60, 72, 108, 140 ) - 10
LCM = 15120
So, required number = 15120 - 10 =
15110
Q 68. A sum of money is borrowed
and paid back in two annual instalments of Rs.882 each allowing 5% C.I.The sum
borrowed was:
a) Rs.1680
b) Rs.1142 c)
Rs.640 d) Rs.1640 Solution- amount=p*(1+r/100)^n borrowed
amount= 882 / (1+5/100) + 882 / (1+5/100)^2
= (882*20)/21 + (882*20*20)/(21*21)
= 1640
Q 69. how many parallelograms are
formed by a set of 4parallel lines intersecting an other set of 7 parallel
lines.
a) 125 b)126 c) 127 d) 128
Solution- Let there be 4 horizontal
set of parallel lines and 7 vertical set of parallel line.(U can also consider
vice-versa)
Now for a parallelogram, u need 2
horizontally parallel and 2 vertically parallel lines ie. we need to choose 2
lines from each set.
So the solution will be 7C2*4C2=126
Q 70. A completes 80% of a work in
20 days. Then B also joins and A and B together finish the remaining work in 3
days. How long does it need for B if he alone completes the work?
a) 37 ½ b) 32 c)32 ½ d) 37
Solution- let total work is 100 unit .
A can do 80 unit in 20 days ....so
he can do 4 unit in 1 days .
now A has finished 80 units so
remaining work is 20 unit
so , 20 unit work takes 3 days time
to complete in which A will do 12 unit
(since A do 4 unit/days )
thus ,B do 8 unit in 3 days
and total work is 100 unit so B
will take (3/8)*100 = 37.5 days to complete total work
Q 71. a person starts writing all 4
digits numbers.how many times had he written the digit 2?
a) 4200 b) 4700 c) 3700 d) 3200
Solution-1)when 2 is at unit
place=9*10*10*1
2)when 2 is at tenth
place=9*10*1*10
3)when 2 is at
hundred place=9*1*10*10 4)when 2 is at thousand place=1*10*10*10 so total no.
of 2s=900+900+900+1000=3700
Q 72. There is a tank,and two pipes
A and B.A can fill the tank in 25 minutes and B can empty the tank in 20
minutes.If both pipes are opened at same time.how much time required the tank
can fill?
a) 15 min b) 18 min c) 13 min d)
Never be filled
Solution-We have pipe A filling the
tank completely in 25 minutes.
Ie, it can fill 100% tank in 25
minutes.
Hence, in 1 min, the tank is filled
= 100/25 = 4%
We have pipe B which can empty the
tank in 20 minutes.
Hence, in 1 min, tank is empties =
100/20 = 5%.
Since we have the rate
of emptying the tank more than the rate of filling the tank, We can say that
the tank will never get filled.
Q 73. An Old man and a Young man
are working together in an office and staying together in a near by apartment.
The Old Man takes 30 minutes and the Young 20 minutes to walk from apartment to
office. If one day the old man started at 10:00AM and the young man at 10:05AM
from the apartment to office, when will they meet?
a) 10:15 b) 10:30 c) 10:45 d) 10:00
Solution-Ratio of old man speed to
young man speed = 2:3
The distance covered by old man in
5 min = 10
The 10 unit is
covered with relative speed=10/(3-2)=10 min so, they will meet at 10:15 am.
Q 74. the shopkeeper charged 12 rupees for a
bunch of chocolate. but i bargained to shopkeeper and got two extra ones, and
that made them cost one rupee for dozen less then first asking price . how many
chocolates i recieved in 12 rupees ??
a) 10
b)16
c) 14
d)18
Solution-let no.
of chocolates=x=x/12dozen price of x/12 dozen is rs 12 so price of 1 dozen is
144/x
2nd condition
144/(x+2) +1=144/x go through option x=16
Q 75. there are 16 teams divided in
4 groups.Every team from each group will play with each other once. The top 2
teams will go to the next round and so on the top two teams will play the final
match. Minimum hw many matches will be played in that tournament?
a) 43 b) 40 c) 14 d) 50
Solution-for first
round 4*4C2=24 second round 2*4C2=12
1*4C2=6
FINAL=1
TOTAL=43
Q 76. A sealed envelope contains a card with
a single digit written on it. Three of the following statements are true and
one is false. I. The digit is 1.
II.
The digit is not 2.
III.
The digit is not 9.
IV. The
digit is 8.
Which one of the
following must necessarily be correct? a) II is false
b) III
is true
c) IV
is false
d) The
digit is even.
e) I is
true
Solution-III is
true since when 3 is wrong then there are 2 possibilities that 1 or 8 which is not possible
Q 77. Tickets are numbered from
1,2....1100 and one card is drawn randomly what is the probability of having 2
as a digit?
a) 29/110 b) 32/110 c) 30/110 d)
22/110
Solution-for every 100 probability
of having 2 as digit is 19 and 2 is used 20 times for every 100.according to
qstn, there r 10 100's having 19 digits as 2 n 200 to 299 has 100 digits having
2. hence,
((10*19)+100)/1100=29/110
Q 78. How many 2’s are there
between the terms 112 to 375?
a) 313 b) 159 c)
156 d) 315 Solution-156 from 112 to 199....(19) from 200 to 299....(100+20)
from 300 to 375....(18) so 100 + 19 + 18 + 20 -1=156
-1 bcz 112 shouldn't be taken
Q 79. Ram and Shakil run a race of 2000 meters. First, Ram
gives Shakil a start of
200 meters and beats him by one
minute. If , Ram gives Shakil a start of 6 minutes Ram is beaten by 1000
meters. Find the time in minutes in which Ram and Shakil can run the races
separately.
a) 12,18 b) 10,12 c) 11,18 d) 8,10
Solution-Let x and y are the speeds
of Ram and Shakil.. Then by problem we got following equation 2000/x=(1800/y)-1
1000/X=(2000/y)-6 solve equation 1 and 2, we get x=250 and y=200
Therefore, Time taken by Ram and
Shakilk to complete a race of 2000m is 8 min and 10 Min
d) 8,10
Q80. The average temperature of
June, July and August was 31 degrees. The average temperature of July, August
and September was 30 degrees. If the temperature of June was 30 degrees, find
the temperature of September (in degrees).
a) 25
b) 26
c) 27
d) 28
Solution-c.27
June+July+August=3*31=93.......(1)
July+august+Sept=3*30=90........(2)June=30
Hence, putting
june =30 in equ 1 July+August=63
putting in 2nd
equ. 63+Sept=90 hence, Sept=27
Q 81.Three generous friends, each
with some money, redistribute the money as follows: Sandra gives enough money
to David and Mary to double the amount of money each has. David then gives
enough to Sandra and Mary to double their amounts. Finally, Mary gives enough
to Sandra and David to double their amounts. If Mary had 11 rupees at the
beginning and 17 rupees at the end, what is the total amount that all three
friends have?
a) 105 b) 60 c) 88 d) 71
Solution-let sandra,
david and mary each has s, d and 11(given) respectively. After first
distribution
David has d+d=2d,
marry has 11+11=22 and sandra has s-d-11. After second distribution, sandra has
2*(s-d-11) , mary has 2*22=44 and david has 2d-(s-d-11)-22=3d-s-11. After third
distribution,
sandra has 2*2(s-d-11), david has
2*(3d-s-11) and mary has 44-2(s-d-11)-(3d-s-11)=77-s-d
It is given that finally Mary has
17 rs. So,
77-s-d=17
=>s+d=60
=> s+d+11(Mary's
money)=60+11=71.
Q 82. George walks 36 kms partly at
a speed of 4 kms per hour and partly at 3 km per hour If he had walked at a
speed of 3km per hour when he had walked at 4 and 4 km per when he had walked
at 3 he would have walked only 34 kms. The time (in hours) spent by George in
walking was).
a) 8 b) 12 c) 5 d) 10
Solution-the question can be solved
with the help of two equations...
4x+3y=36----equ(1)
3x+4y=34----equ(2) by solving these two equations x=6 and y=4 so total time is
10hours.
Q 83. The sum of the four
consecutive two digit odd number when divided by 10 becomes a perfect square,
which of the following can be one of these four numbers?
a) 21 b) 25 c) 41 d) 67
Solution-Let the
four 2-digit odd numbers be n-3 n-1 n+1
n+3
Sum of the 4
numbers ==> 4n acc to qn, when the
sum is divided by 10 we get a perfect
square... perfect squares
include==>1,4,9,16,25,36,49,.....
Possible values of
4n/10 ==> 4, 16,36... If 4n/10=4 n=10
Hence, the corresponding nos are
7,9,11,13(all of which are NOT 2-digit nos)
If 4n/10=16 n=40
Hence the
corresponding nos are 37, 39, 41, 43 If
4n/10=36 n=90
Hence the corresponding nos are 87,
89, 91, 93
The answer to the ques therefore is
Option C
Q 84. Consider the sequence of
numbers 0, 2, 2, 4,... Where for n > 2 the nth term of the sequence is the
unit digit of the sum of the previous two terms.
Let sn denote the sum of the first
n terms of this sequence. what is the smallest value of n for which sn>2771?
a) 692 b) 693 c) 694 d) 700
Solution-0 2 2 4 6 0 6 6 2 8 0 8 8 6 4
0 4 4 8 2 .0 2 2...this series will be repeated after every 20 terms sum of 20 terms of series=80
2771/80
quotient=34 remainder=51 sum of 13 terms =52 number of terms is 693 since
34*20=680
680+13 =693 sum of
693 terms will be > 2771
Q 85. a number plate can be formed
with two alphabets followed by two digits with no repetition. then how many
possible combinations can we get?
a) 52500 b) 58500 c) 56500 d) 56800
Solution-no.of alphabets=26 (a-z),
no.of digits=10(0-9). ways of arranging
two alphabets with out repetition=26*25; ways of forming two digits without
repetition=10*9
no.of combinations of forming the
number on number plate= 26*25*10*9=58500
Q 86. I bought a certain number of
marbles at rate of 59 marbles for rupees 2 times M, where M is an integer. I
divided these marbles into two parts of equal numbers, one part of which I sold
at the rate of 29 marbles for Rs. M, and the other at a rate 30 marbles for Rs.
M. I spent and received an integral number of rupees but bought the least
possible number of marbles. How many did I buy?
a) 870 b) 102660 c) 1770 d) 1740
Solution-Let he bought x
marbles.
59 marbles costs = Rs. 2M
Therefore, x marbles costs = Rs.
(2M * x) / 59
Since the marble is divided into 2
equal parts so the number x should be an even number.
For first x/2 marbles,
29 marbles
s.p. is = Rs. M
Therefore, x/2 marbles s.p. = Rs.
(M * x) / 58
For other x/2 marbles,
30 marbles
s.p. is = Rs. M
Therefore, x/2 marbles s.p. = Rs.
(M * x) / 60
Now we can't equate like [(M * x) /
58] + [(M * x) / 60] = (2M * x) / 59
because (M *x ) will get cancel
each side and of course 1/58 + 1/60 is not equal to 2/59
So here we don't need M and we can
cancel it. After that we have,
CP = 2x/59
For first x/2 marbles,
SP = x/58
And for other x/2 marbles,
SP = x/60
Now this CP and SP must be an
integer (as per question). So we have to find a number x which will be divisible
simultaneously by 59, 58 and 60. So we have to find the LCM of 59, 58, 60 which
will turn out minimum value as 102660 and it is even as well. So the value of x
will be 102660 minimum
Q 87. Cara, a blue whale participated in a
weight loss program at the biggest office. At the end of every month, the
decrease in weight from original weight was measured and noted as 1, 2, 6, 21,
86, 445, 2676. While Cara made a steadfast effort, the weighing machine showed
an erroneous weight once. What was that.
a) 2676
b) 2
c) 445
d) 86
Solution-1*1+1=2
2*2+2=6
6*3+3=21
21*4+4=88 and not 86
Q 88. How many different integers
can be expressed as the sum of three distinct numbers from the set
{3,8,13,18,23,28,33,38,43,48}
a) 421 b) 20 c) 10 d) 22
Solution-minimum sum
of 3 no.=24 maximum sum =129
now the series
24,29,34,....................,129
no of terms ={(129-24)/5}+1=22
Q 89. Aman walking at the speed of
4 km/h crosses a square field diagonally in 3 minutes. The area of the field
(in m2) is:
a) 20000 b) 21000 c) 25000 d) 26000
Solution-convert speed into m/s i.e
4*5/18 = 2/9 this
is distance covered in 1 sec.
so distance covered in 3 min i.e
180 sec = 2/9*180
= 200m
this is the lenth
of diagonal and are of square = 1/2 (daigonal)*(diagonal)
1/2*200*200
20,000 m2
Q 90. A owes B Rs 50. He agrees to pay
B over a number of consecutive day starting on a Monday, paying single note of
Rs 10 or Rs 20 on each day. In how many different ways can A repay B. (Two ways
are said to be different if at least one day, a note of a different
denomination is given) a) 8
b) 7
c) 6
d) 5
Solution- 10,20,20=3!/2!==3 ways
10,10,10,20=4!/3!==4 ways
10,10,10,10,10=5!/5!==1 way so.total=3+4+1==8 ways.
Q 91. The HCF of 2472, 1284 and a third number ‘N’ is 12. If
their LCM is 2^3 × 3^2
× 5 × 103 × 107, then
the number ‘N’ is:
a) 2^2
× 3^2 × 7
b) 2^2
× 3^3 × 103
c) 2^2
× 3 × 5
d) None
of these
Solution-
2472=8*3*103 1284=4*3*107
so n = 4*3*x
dividing by hcf 2472/12=2*103 1284/12=107 lcm/12=2*3*5*103*107
x=lcm/first*second x=2*3*5*103*107/2*103*107=3*5=15 number is = 15*12=180
ans. d)
Q 92. A certain function f
satisfies the equation f(x)+2*f(6-x) = x for all real numbers x. The value of
f(1) is
a) 2
b) can't
determine
c) 1
d) 3
Solution-
f(1)+2*f(6-1)=1......... (1) f(5)+2*f(6-5)=5......... (2)
substituting we have (2) in (1) we
have :-
-3f(1)=-9,
hence answer f(1)=3
Q 93. what is the value of
77!*(77!-2*54!)^3/(77!+54!)^3+54!*(2*77!-54!)^3/(77!+54!)^3
a)2*77!+2*54! b) 77!-54! c) 77!+54!
d) 2*77!-2*54!
Solution-77!-54!
take 77!=a, 54!=b then we get in
simple (a-2b)^3/a+b solve it u get a-b as answer
Q 94.Find sum of the series
1-2+3-4+....-98+99
a)-49 b) 0 c) 50 d)-50
Solution-(1+3+.....+99)-(2+4+....+98)ie
50 terms+49 terms resp. using formla s=n/2[a+l] we get, 50/2[1+99]-49/2[2+98]
den 2500-2450=50.
Q 95. In a city
there are few engineering, MBA and CA candidates. Sum of four times the
engineering, three times the MBA and 5 times CA candidates is 3650. Also three
times CA is equal to two times MBA and three times engineering is equal to two
times CA. In total how many MBA candidates are there in the city? a) 200 b) 300
c) 450 d) 400
Solution-e = no. of engineering
students, m = no. of MBA students and c = no. of CA students
4e + 3m + 5c = 3650, therefore e =
(3650-3m-5c)/4 --------(1)
3c = 2m , therefore c = 2m/3
3e = 2c replacing e with (1) we
get,
3(3650- 3m - 5c)/4 =
2c.....replacing c with 2m/3 and solving this equation we get m = 450
So the number of MBA students is
450
Q 96. Find the sum of angles 1,2,3,4,5
in a star.
a) 180 b) 300 c) 360 d) 400
Solution-et an
angle of 5 star is a,b,c,d,e then (180-a)+(180-b)+(180-c)+(180-d)+(180-e)=720
now a+b+c+d+e=900-720=>180
Q 97. Consider a triangle drawn on
the X-Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each
vertex being represented by its (X, Y) coordinates. The number of points with
integer coordinates inside the triangle (excluding all the points on the
boundary) is
(1) 780
(2) 800
(3) 820
(4) 741
Solution-draw the triangle and view
it carefully
The number of points with integer
coordinates inside the triangle are
(1,39)
(1,38),(2,38)
(1,37),(2,37),(3,37)
.......................
...........................
(1,2),(2,2),(3,2),........(38,2)
(1,1),(2,1),(3,1),(4,1)...(38,1),(39,1)
so total no of points
are=1+2+3+...+38+39
=39*(39+1)/2=39*20=780
ANS(1)
Q 98. the marked price of coat was
40%less than the suggested retail price. Eesha purchased the coat for half of
the marked price at the 15th anniversary sale. What percent less than the
suggested retail price did eesha pay?
a)60%b)20% c)70% d)30%
Solution-suppose
retail price = 100 so the market price will be = 60 as given coat purchased =
half f d market price = 30 so its clear isha paid 70% less than retail price.
Q 99. There is a school were 60%
are girls and 35% of the girls are poor.
Students are selected at random,
what is the probability of selecting a poor girl out of total strength.
a) 21 b) 27 c) 28 d) 29
Solution- '21 will be the
probability of poor girl out of total strength....
let 100 be the
total strength 60% out of 100= 60 number of girls now 35% of gals are poor
..i.e. (35/60)*100 are poor girl=21
so 21% of total strength are poo...
i.e. the probability is .21
Q 100. If m+n is divided by 12
leaves a reminder 8,If m-n is divided by 12 leaves a reminder 6,then If mn is
divided by 6 what is the reminder?
a) 4 b) 3 c) 2 d) 1
Solution- let m=19
and n=1; m+n=20, gives the remainder=8 m-n=18, remainder=6 so mn=19*1=19 after
division by 6 it wud giv the remainder 1...:)
Q 101. there is conical tent in
which 10 persons can stand. Each person need 6m square to stand and 60m cube
air to breath. what is the height of tent?
a) 60 b) 30 c) 20 d) 45
Solution-Amount of area reqd. by 1
person to stand = 6 sq.metre
Amount of area
reqd. by 10 person to stand = 6x10 sq.metre i.e 60 sq.metre So,
Base Area, pi*r*r=60
=>r*r = 60/pi ---------------(1)
Now,
Volume of air reqd. to breathe 1
person= 60 cu.metre
Volume of air reqd. by 10 person to
breathe = 60x10=600 cu.metre
So,
Volume, 1/3*pi*r*r*h = 600
---------------------(2)
Putting the value of eqn 1 in eqn
2, we get
1/3*pi*60/pi*h = 600
==> 1/3*60*h = 600
==> 20*h = 600
==> h = 600/20 = 30
Therefore height = 30 metres
Q 102. In a triangle ABC, the lengths
of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on
the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then
what is the radius (in cm) of the circle circumscribing the triangle ABC?
a) 17.05 b) 27.85 c) 22.45 d) 26.25
Solution-circum radius = R
Area of the triangle ABC = Δ = ½
(BC)(AD) = ½ (a)(3) = 3a/2
The formula here is : R = abc/4Δ =
abc / 4(3a/2) = bc/6 = (17.5)(9)/6 = 26.25
Q 103. A rectangle is divided into
four rectangles with area 70,36,20 and X. The value of X is:
a) 350/90 b) 350/7 c) 350/11 d)
350/13
Solution-70/x = 36/20
=> x*36= 20*70
=> x= 20*70/36 = 350/9
Q 104. the ratio of radii of
cylinder to that of cone is 1:2.heights are qual.find ratio between volume.
a) 3:4 b) 1:2 c) 1:4 d) 4:1
Solution-Let ratio of radius of
cylinder to cone=...r1/r2=1/2
Volume of cylinder=π*r1^2*h
Volume of cone=π*r2^2*h/3
Ratio of volumes=3*(r1/r2)^2= 3:4
Q 105. A hollow pipe has
circumference 14 cm. A bug is on its wall at a distance of 48 cm from top. A
drop of honey is on the wall at 24 cm from top but diametrically opposite to
bug. Find the shortest distance bug has to travel to reach honey.
a) 25 cm b)
39 cm c) 21 cm d) 24 cm
Solution-Total Vertical Distance
bug has to through = 24 cm Circumference of pipe= 14 cmSince bug has to move to
diametrically opposite side, it has to cover half of the circumference = 14/2 =
7 cmSO by Pythagoras, diagonal distance would be = sqrt(7^2+24^2) = sqrt(625) =
25 cm
Q 106. if a ladder is 100m long, and
distance b/w bottom of ladder and wall is 60. top side of bottom and wall is
joint. what is the maximum size of cube that place b/t them.
a) 34.28 b) 24.28 c) 21.42 d) 28.56
Solution- using trigonometry we
have
(80/60)=tan(y)....1
(x/60-x)=tan(y)....2
solve it... u will have 35 so a) 34.28
Q 107. what is the next three
numbers for the given series? 11 23 47 83 131
a)145 b) 178 c) 176 d) 191
Solution- there are multiples of
12:
11+12=23
23+(12*2)=47 and so on .....answer is 191 as 131+(12*5)=191
Q 108. A series of book was
published at 7 year intervals.When the 7th book was issued the sum of
publication year is 13524.When was the 1st book published ?
a) 1911 b) 1910 c) 2002 d) 1932
Solution- sum of
A.P. series=(n/2)(2*a+(n-1)*d) here sum=13524,n=7,d=7 substituting these values
in above eq. n solving we get a=1911
which was the 1st book published year.
Q 109. There are 14 digits of
credit card number to be filled. Each of the below three boxes contains
continuous digits of 18 as sum. Given: 4th digit is 7 and 7th digit is x. Then
what is the value of x?
a) 1 b) 7 c) 4 d) 2
Solution- a+b+c=18 and b+c+d=18
given d=7 ,so b+c=11
then we get b=5
and c=6 c+d+e=18, so e=18-(c+d)=5 since d+e+f=18 we get f=7 ie x=7
Q 110. Crusoe hatched from a mysterious
egg discovered by Angus, was growing at a fast pace that Angus had to move it
from home to the lake. Given the weights of Crusoe in its first weeks of birth
as 5, 15, 30,135, 405, 1215, 3645. Find the odd weight out.
a) 3645
b) 135
c) 30
d) 15
Solution- 30.
5
5*3=15
15*3=45(here it is given as 30)
45*3=135
135*3=405
405*3=1215
1215*3=3645
Q 111. Arun makes a popular brand
of ice-cream in a rectangular shaped bar 6 cm long, 5 cm wide and 2 cm thick.
To cut costs, the company had decided to reduce the volume of the bar by 19%.
The thickness will remain the same, but the length and width will be decreased
by the same percentage. The new width will be.
a) 4.5cm b) 5.5cm
c) 6.5 cm d) 7.5cm Solution- volume v=l*b*t new volume is 81% of old v.
therefore new vol = 81%(v)=l'*b'*t
=> 81% (l*b*t)
= l'*b'*t => 81% (l*b) = l' * b'
since the change
in new l and b is the same. hence consider change to be x. => 81% (l*b) =
xl*xb
81%=x^2 x=9%.
9% change in 5cm
width is 0.45 therefore new width is 5-0.45=4.55cm
Q 112. A can complete a piece of
work in 8 hours, B can complete in 10 hours and C in 12 hours. If A,B, C start
the work together but A laves after 2 hours. Find the time taken by B and C to
complete the remaining work.
a) 2
(1⁄11) hours
b) 4
(1⁄11) hours
c) 2
(6⁄11) hours
d) 2
hours
Solution- A,B,C 1 hour work
is=1/8+1/10+12= 37/120
A,B,C work together
for 2 hours, so,A,B,C 2 hours work is: (37/120)*2=37/60 remaining
work=1-37/60=23/60 (23/60 work is done by B and C together)
B,C 1 hour work is= 1/10+1/12=11/60
23/60 hours work done by B,C=
(11/60)* (60/23)=11/23
so... ans is: 2(1/11)
Q 113. What is the greatest
possible positive integer n if 8^n divides (44)^44 without leaving a remainder?
a)14 b)28 c)29 d)15
Solution- 44^44 = 2^88 *11^44
= 8^29 * 2 * 11^44
so ans is 29
Q 114. A tree of height 36m is on
one edge of a road of width 12m. It falls such that the top of the tree touches
the other edge of the road. Find the height at which the tree breaks.
a) 16 b) 24 c) 12 d) 18
Solution- let the
height at which it broke be Xmt let the length of other piece be Ymt which
touches ground
X+Y=36;
Y^2 - X^2 = 144
then on solving x=16mt
Q 115. How many 6 digit even
numbers can be formed from digits 1 2 3 4 5 6 7 so that the digit should not
repeat and the second last digit is even?
a)6480 b)320 c)2160 d)720
Solution- given 6th digit even
number , so last digit 2 or 4 or 6-> 3 ways
" 5th digit
should be even...so there will be 2 ways(rep. not allowed) so,therefore we get
5*4*3*2*2*3=720 ways
Q 116. At the end of 1994 rohit was
half an old as his grand mother.The sum of years in which they were born is
3844. How old rohit was at the end of 1999. a) 48 b) 55 c) 49 d) 53
Solution- let at
the end of 1994 grand mother's age is x and rohit's age x/2 then we can
say....birth year of GM is =(1994-x) and rohit is = (1994 - x/2) sum of years
is 3844
i.e (1994 -x) + (1994 - x/2) = 3884
=> x = 96
i.e GM age is 96
so rohit age will be 96/2 = 48 years in 1994 age is 48
1995 49
1996 50
1997 51
1998 52
1999 53
so ans should be
53 years.......
Q 117. Find the number of divisors
of 1728.?
a) 28 b) 21 c) 24 d) 18
Solution- 1728= 2^6 * 3^3
Hence the Number of factors = (6+1)
x (3+1) = 7 x 4 = 28.
We know that if a number
represented in standard form (a^m *b^n) , then the number of factors Is given
by (m+1)(n+1).
Answer is 28
Q 118. 17 x 8 m rectangular ground
is surrounded by 1.5 m width path. Depth of the path is 12 cm. Gravel is filled
and find the quantity of gravel required.
a)
5.5
b)
7.5
c)
6.05
d)
10.08
Solution- area of
the rectangular ground=(17*8)m^2=136 m^2 taking into account the path: total
area=[(17+(2*1.5))*(8+(2*1.5))]=220 m^2 area of the path=220-136=84m^2 gravel
required=[84*(12/100)]=10.08m
Q 119. Ashok, Eesha, Farookh, and
Gowri ran a race. Ashok said, “I did not finish 1st or 4th “.
Eesha said, “I did not finish 4th
”. Farookh said, “I finished 1st ”. Gowri said, “I finished 4th ”. There were
no ties in the competition, and exactly three of the children told the truth.
Who finished 4th? explain
a)
Farookh
b)
Eesha
c)
Gowri
d)
Ashok
Solution- Gowri finished with 4th
place:
According to the qsn :
exactly three of
the children told the truth. then 4conditions may arise i.e.
TTTF,TTFT,TFTT,FTTT
For 1st condition :
Let Gowri lies that means she nver be
in 4th Place and other 3 tell the truth then nobody is in 4th place so from the
above we conclude that Gowri finished at the 4th
Q 120. A circle has 29 points
arranged in a clockwise manner numbered from 0 to 28, as shown in the figure
below. A bug moves clockwise around the circle according to the following rule.
If it is at a point i on the circle, it moves clockwise in 1 second by ( 1 + r
) places, where r is the reminder ( possibly 0 ) when i is divided by 11. Thus
if it is at position 5, it moves clockwise in one second by ( 1 + 5 ) places to
point 11. Similarly if it is at position 28 it moves ( 1 + 6 ) or 7 places to
point 6 in one second. If it starts at point 23, at what point will it be after
2012 seconds?
a) 1 b) 7 c) 15 d) 20
Solution- after 1st second :
(1+23%11 = 1) = 2 places [25] after 2nd second : (1+25%11 = 3) = 4 places [0]
after 3rd second : (1+0%11 = 0) = 1 place [1] after 4th second : (1+1%11 = 1) =
2 places [3] after 5th second : (1+3%11 = 3) = 4 places [7]
after 6th second :
(1+7%11 = 7) = 8 places [15] after 7th second : (1+15%11 = 4) = 5 places [20]
after 8th second : (1+20%11 = 9) = 10 places [1]
now,for the same pattern from 4th
sec to 8th sec will repeat itself (5 sec intervals)..
total time = 2012
secs first 3 secs out of pattern...so time left 2012 - 3 =2009 secs
now no. of
repetitions in the leftover time = 2009/5 = 401....remainder = 4 for the next 4
iterations following the similar pattern
the position will be 20..
Q 121. A team won 80% of the games
it played. It played 5 more games of which it won 3 and lost 2. Its loss
percentage changed to 25%. How many games did it play overall?
a) 20
b) 14
c) 16
d) 25
Solution- ans= 20
if game played=x then lost game=x/5 now they played 5 more games in which they
lost 2 so, (x+5)*25/100=x/5 + 2 x=15 so total game he played=15+5 = 20
Q 122. find the sum of the series
given below 1(1!)+2(2!)+3(3!)+.........+2012(2012!)
a) 2013!+1 b) 2013!-1 c) 2012!+1 d)
2013!-1
Solution- let x =
1(1!)+2(2!)+3(3!)+.........+2012(2012!) & y = 1!+2!+3!+.........+2012! x+y
= 2(1!)+3(2!)+....+2013(2012!) = 2!+3!+....+2012!+2013! x+y+1 =
1!+2!+3!+.........+2012!+2013! = y+2013!
x = 2013! -1
Q 123. there is a circle which
circumscribes three unit circle which are tangential to each other.what is the
circumference of bigger circle.?
a) pi(4+2sqrt3)/sqrt3 b)
pi(6+2sqrt3)/sqrt3 c) pi(3+2sqrt3)/sqrt3 d) pi(6+2sqrt3)/sqrt3
Solution- Just
draw according to question, now, join the center of the smaller circles. radius of bigger circle= 1+ radius of circle
circumscribing the equilateral triangle.
Hence radius of
bigger circle = 2/root 3 + 1 Hence circumferance = 2* pie * R ie. pi(4+2sqrt3)/sqrt3
Q 124. a man starts work on monday
and works for 8 days and works for every ninth day as his holiday. His 12th day
will be on which day
a) Monday b) Wednesday c) Thursday
d) Tuesday
Solution- his work cycle includes 9
days(Mon-Mon=8days and 1 day Tue is holiday) so to get 12 holiday there should
be 12 cycle=12*9=108 days get remainder by dividing it by 7(108/7 i.e 3)
hence from monday 3rd day is
wensday
Q 125. the value of a scooter
depreciates in such a way that at the end of each year, is ¾ of its value at
the beginning of same year. If the initial value of the scooter is rs40,000.
What is the value at the end of 3yrs?
a)23125 b)19000 c)13435 d)16875
Solution- As, it
is given that , the cost becomes 3/4 at the end of year. so, after 3 years ,the
price of Scooter=40000*3/4*3/4*3/4= Rs.16875
Q 126. At 12:00 hours jake starts
to walk from his house at 6kms an hour. At 13:30 hours, paul follows him from
jake's house on his bicycle at 8 kmph. When will jake be 3 kms behind paul?
a) 19:00 b) 18:30 c) 20:00 d) 19:30
Solution- upto
13:30 jake covered 6*1.5=9km then paul need to be ahead of 3 km.so he go 12km.
their relative speed is 2km per hour.(same direction). so 12km divided by 2 is
6hrs. hence from 13:30hrs ,6hrs is added.so answer is 19:30 hours.
ANS:19:30 hrs
Q 127. There are five boxes in a
cargo hold. The weight of the first box is 200 kg and theweight of the second
box is 20% higher than the weight of the third box, whose weight is 25% higher
than the first box’s weight. The fourth box at 350 kg is 30% lighter than the
fifth box. Find the difference in the average weight of the four heaviest boxes
and the four lightest boxes.
a) 80
kg
b) 75
kg
c) 37.5
kg
d) 116.8
kg
Solution- weight
of 1st box=200 kg so weight of 3rd box=250 kg
weight of 2nd box = 300 kg nd 4th box= 350 thus 5th box=500 kg avg weight of four
heaviest boxes= 1400/4=350 nd lightest
box=1100/4=275 diff=75 kg
Q 128. A rectangle of height 100 squares and
width 200 squares. Squares is drawn on a graph paper. It is colored square by
square from top left corner and moving across in a spiral turning right.
Whenever a side of this rectangle or a colored square is reached. Which square
is colored last (give its row and column numbers). The bottom right square is
on row 100, column 200? a) 51,150
b) 51,
50
c) 50,
150
d) 50,
50
Solution- For (4, 8) rectangle,
ends at (3, 2)
Likewise,
For (2, 4) rectangle, ends at (2,
1)
For (3, 6) rectangle, ends at (2,
5)
For (4, 8) rectangle, ends at (3,
2)
For (5, 10) rectangle, ends at (3,
8)
For (6, 12) rectangle, ends at (4,
3)
For (7, 14) rectangle, ends at (4,
11)
For (8, 16) rectangle, ends at (5,
4)
For (9, 18) rectangle, ends at (5,
14)
For (10, 20) rectangle, ends at (6,
5)
Analyzing the above 10 points:
We can get some idea that,
For Even number of rows, End points
column increases 1 by previous column.
For Odd number of rows, End points
column difference of the present column and the previous even number column.
For End point row, the value
repeats two times from row 2.
In our problem, (100, 200) i.e.
Even number row,
So End point column value must be
50 for 200 columns and End point row value must be 51 for 100 rows.
For even number of rows and
columns,
Condition: Column =2*Row
Endpoint (R, C) =
{((Row/2) +1), Column/4} so last square to be coloured is (51, 50)
Q 129. A bag contains six sticks of
the following lengths 1 cm, 3 cm, 7 cm, 11 cm and 13 cm. three sticks are drawn
at random from the bag. What is the probability that we can form a triangle
with those sticks?
a)
11/20
b)
1
c)
1/4
d)
2/5
Solution- A bag contains six sticks of the
following lengths 1 cm, 3 cm, 7 cm, 5cm,11 cm and 13 cm. three sticks are drawn
at random from the bag. What is the probability that we can form a triangle
with those sticks total possibilities=6c3=20
now for making a triangle,, the sum
of any 2 sides must be greater than 3rd side so
3 5 7
3 11 13
5 7 11
7 11 13
5 11 13
5/20=1/4
Q 130. 3 mangoes and 4 apples costs
Rs 85. 5 apples and 6 peaches costs Rs. 122. 6 mangoes and 2 peaches cost
Rs.114. what is the combined price of 1 apple, 1 peach and 1 mango?
a) 37
Rs
b) 39
Rs
c) 35
Rs
d) 36
Rs
Solution- 3m+4a=85...(i)
5a+6p=122....(ii)
6m+2p=114....(iii)
Adding (i) (ii) (iii) we get
9a+9m+8p= 321
9a+9m+9p= 321+p
a+m+p=(321+p)/9------(iv) It must
have to be a integer number.
To make (iv) as an integer p must
have to be either 3 or (3+9) or (3+18)or.....
Let check it out..
if p=3 then
5a+18=122 5a=104
a!= an integer so
p=3 is false If a=12, then
by(ii)
5a=122-72 a=10
by(i)
3m=85=40
a=15 a+m+p=
15+10+12=37.
Q 131. how many number x (x being
an integer) with 10<=x<=99 are 18 more than sum of their digits?
a) 12
b)9
c) 18
d)10
Solution- Let the
no. be 10y+z. then,
10y+z= y+z+18
=> Y=2
So the no. are 20, 21, 22, 23, 24,
25, 26, 27, 28, 29. Total nos. 10
Q 132. F, G, H, K, L, M and N are 8
people. They need to group into 2 with the condition.
* F
and J must be same group
* G
and N must be in different groups
* H
and L must be in same group
* M
and G are not in the same group Correct ordering of groups
a) FJ,KL,MN,GH
b) FH,JL,MN,GK
c) FJ,HL,MN,GK
d) FJ,Hl,MN,GH
Solution- ans is c because c
satisfies all the conditions specified
Q 133. In how many ways can the
digit of the number 2233558888 be arranged so that the odd digits are placed in
the even positions?
a) 900 b) 450 c) 225 d) 360
Solution- odd place combination =
5!/2!*2!=30
even place
combination= 6!/2!*4!=15 to total=30*15=450
Q 134. Find the probability that a
leap year chosen at random will contain 53 sundays.
a) 2/7
b) 3/7
c) 1/49
d) 1/7
Solution- In leap year,we have 366
days.
366/7=52Weeks+2days
we have already 52 sundays,mondays,etc.
then we have 2 days may be
{sunday,monday} , {monday,tuesday} , {tuesday,wednesday}
,....{saturday,sunday}
we have 2 possible chances out of 7. hence ans:2/7.
Q 135. A shop sells chocolates It
is used to sell chocolates for Rs.2 each but there were no sales at that
price.When it reduced the price all the chocolates sold out enabling the
shopkeeper to realize Rs 164.90 from the chocolates alone If the new price was
not less than half the original price quoted How many chocolates were sold?
a) 39 b) 97 c) 37 d) 71
Solution- just go on trial and
error method until u get u a integer. 164.90/1.7=97
Q 136. The Mean of three numbers is 10
more than than the least number and 15 less than the highest. Their median is
5. Find the sum of the three numbers. a) 5 b) 20 c) 30 d) 25
Solution- given:
x+y+z/3=m y=5;
m=x+10;x=m-10;
z=m+15
sub in above give
eq u will get m= 10; sub in x, y, z =0,5,25=total=30
Q 136. what is the remainder of
(16937^30)/31
a) 1 b) 2 c) 3 d) 6
Solution- 16937=16926+11,now 16926
is completely divisible.. So Wat remains is (11^30)/31
Which is (11^6)^5./31
11^6 gives 4 as remainder..so
4^5/31 is remaining...which gives 1 as remainder.
Q 137.
8+88+888+.....+8888........8888. There are 21 “8” digits in the last term of
the series.
Find the last three digits of the
sum.
a) 458 b) 648 c) 658 d) 568
Solution- 21*8=168 carry 16
20*8=160+16=176
carry 17 19*8=152+17=169
the value is 968
Q 138. If x^y denotes x raised to
the power y, find last two digits of (1941 ^ 3843)+ (1961 ^4181).
a) 02 b) 82 c) 42 d) 22
Solution- we should use power cycle
method
take last 2 digit of 1941 take last
2 digits of 1961
41^1=41 r=1 61^1=61 r=1
41^2=81(take last 2 digits of
ans)r=2 61^2=21 r=2
41^3=21 r=3 61^3=81 r=3
41^4=61 r=4 61^4=41 r=4
41^5=01 r=0
61^5=01 r=0 41^6=41 61^6=61
hence power cycle
is repeating and it is 5 hence for 1961 it is 5 r-remainder for 1961
powercycle-5 for 1941 powercycle-5 4181mod5=rem1=61
3843mod5=rem 3=21
therefore 21+61=82
Q 139. in a g6 summit being held in
London a French, a german, an Italian, a british, a Spanish and a polish
diplomat represent their respective countries and participate in a round table
conference to strengthen the co operation between these countries. There are
exactly 6chairs evenly spaced around a circular table. The chairs are numbered
1 through 6, with successively numbered chairs next to each other and chair
number 1 next to chair no6. Each chair is occupied by one of the diplomats. The
following conditions apply. • Polish sits
immediately next to british
•
German sits immediately next to Italian
•
French doesn’t sit immediatedly next to Italian
•
If Spanish sits immediatedly next to polish,
Spanish doesn’t sit next to
Italian
Which of the following seating
arrangement of the 6 diploamts in chair 1-6 would not violate the given
conditions?
A)
French, polish, british, Italian, Spanish,
german
B)
French, german, Italian, polish, british,
Spanish
C)
French, german, Italian, Spanish, polish,
british
D)
French, Spanish, polish, british, german,
Italian
a)
C b) D c)
A d) B
Solution- from the above inference
its clear that german sits beside italian but spanish must sit beside polish
and polish sits nxt to british. So D matches n hence option b)D
Q 140. Figure shows an equilateral
triangle of side of length 5 which is divided into several unit triangles. A
valid path is a path from the triangle in the top row to the middle triangle in
the bottom row such that the adjacent triangles in our path share a common edge
and the path never travels up (from a lower row to a higher row) or revisits a
triangle. An example is given below. How many such valid paths are there?
a) 120 b) 16 c) 23 d) 24
Solution- given side length=5
.....so (n-1)!=(5-1)!=4!
ans is 24
Q 141. A sum of rs 3000 is
distributed amongst A, B, and C. A gets 2/3 of what B and C got together and c
gets 1/3 of what A and B got together, C's share is?
a) 1200 b) 2250 c) 750 d) 1050
Solution- A+B+C=3000 from data it
is clear that A=2/3(B+C) and C=1/3(A+B)
==> 3C=A+B so replace A+B can be
replaced by 3c.....
===> 4c=3000 and now
c=3000/4===>750... So c gets 750
Q 142. Joke is faster than Paul, Joke and Paul each walk 24
KM. The sum of their speed is 7 Km per hour. And the sum of times taken by them
is 14 hours. Then,
Jake speed is
a) 7 kmph b) 3
kmph c) 5 kmph d) 4 kmph Solution- peed=TimedistanceSpeed=Timedistance
let the speed of joke x then speed
of paul will be 7-x
24x+247−x=1424x+247−x=14
Try to plugin the values from the
options. If Jake speed is 4 the paul is 3.
Q 143. A child is looking for his
father. He went 90 meters in the east before turning to his right. He went 20
meters before turning to is right again to look for his father at his uncle's
place 30 meters from this point. His father was not there. From there, he went
100 meters to his north before meeting his father in a street. How far did the
son meet his father from starting point ?
a) 80
metre
b) 90
metre
c) 100
metre
d) 110
metre
Solution- Clearly, the child moves
from A to B 90 metres eastwards upto B, then turns right and moves 20 metre
upto C, then turns right and moves upto 30 metre upto D. Finally he turns right
and moves upto 100 metre upto E.
So AB = 90 metre, BF = CD = 30
metre,
So, AF = AB - BF = 60 metre
Also DE = 100 metre, DF = BC = 20
metre
So, EF = DE - DF = 80 metre
as we can see in image that
triangle AFE is a right angled triangle and we are having two sides, need to
calculate third one, so we can apply Pythagoras theorem here
A = AE = sqrt{AF^2 + EF^2}
= sqrt{(60)^2+(80)^2}
= sqrt{3600+6400}
= sqrt{10000} = 100
So from starting point his father
was 100 metre away.
Q 144. assume that f(1)=0 and
f(m+n)=f(m)+f(n)+4(9mn-1).for all natural no(integer>0)m and n.what is the
value of f(17)?
a) 5436 b) 4831 c) 5508 d) 4832
Solution- 4832...
calculate upto f(5)by putting
values in the given expression i.e. f(1)=0, f(2)=32, f(3)=100, f(4)=204,
f(5)=344. then cal f(7)=f(5+2)orf(3+4)=732 and f(10)=f(5+5)=1584, and now
f(17)=f(10+7)=f(10)+f(7)+4(9*15*7 -1)..........
f(17)=1584+732+2516=4832
Q 145. The numbers 272738 and
232342, when divided by n, a 2 digit number, leave a remainder of 13 and 17
respectively. Find the sum of the digits of n?
a)
7
b)
8
c)
5
d)
4
Solution- as remainder is 13 &
17 so dividend must be greater than 17
272738 = n*a+13
=> 272725=n*a ---(1) 232342 = n*b+17 => 232325=n*b ---(2)
last two digit of (1)&(2) is 25
so n must be 25, no other two digit no. greater than 25 satisfies this
272725=25*10909
232325=25*9293
n=25, sum of digits of n = 2+5 = 7
Q 146. 60 48 38 28 24 20 18 choose
odd one.
a) 28 b) 38 c) 60 d) 18
Solution- 28 is
odd one.because coming in reverse order 18+2=20
20+4=24
24+6=30
30+8=38
38+10=48
48+12=60
Q 147. The perimeter of a
equilateral triangle and regular hexagon are equal.Find out the ratio of their
areas?
a) 3:2
b) 2:3
c) 1:6
d) 6:1
Solution- Given that perimeter of
equ.triang and hexagon are equal.consider length of triang
as 'x' and length of hex as 'y'.so
the relation is x=2y.Hexagon is made of six equ traingles and formula for area
of equ triang is sqrt(3)/4*x^2 and using this we get ratio of areas as 2:3
Q 148. what is the
remainder of (32^31^301) when it is divided by 9? a) 3 b) 5 c) 2 d) 1
Solution-
32^31^301 when 31 divided by 9 gives remainder 5 5 5^2 5^3 all gives the same
unit digit 5 so 31^32 gives unit digit 5 same rule applicable to 31^301 when 31
divided by 9 gives remainder 4
4 4^2 4^3 4^4 =4 6 4 6 unit place
repeats for every 2 times i,e for even power its uint place is 6 and for odd
its 4 as 301 is odd its unit place is 4
so 31^32^301=31^4=5^4=5 is the ans
Q 149. A take 12 hrs to make publication
B take 10 hrs to make publication. find the time taken by them to make two
publications working independently?
a) 12
hours b) 11 hours
b) 22
hours d) 11 hours 40 minutes
Q 150. Which of the following
numbers must be added to 5678 to give a reminder 35 when divided by 460?
a) 980 b) 797 c) 955 d) 618
Solution- ans is 797
If you need of remainder 35, while
dividing a number by 460. The Unit place of that number must be 5. So 5678+797=
6475
6475/460 gives reminder 35
Q 151. If a number is divided by
357 the remainder is 5, what will be the remainder if the number is divided by
17?
a) 9 b) 3 c) 5 d) 7
solution- ANS:5 ...to get the
original number add the 357 with remainder 5, you will get 362.then divide it
by 17 you will get 5 as remainder.
Q 152. A girl entered a store and
bought x flowers for y dollars (x and y are integers). When she was about to
leave, the clerk said, “If you buy 10 more flowers I will give you all for $2,
and you will save 80 cents a dozen”. The values of x and y are:
a) (15,1)
b) (10,1)
c) (5,1)
d) Cannot
be determined from the given information.
Solution- initially price/dozen= 12y/x;aftr purchasing 10
more--new dozen prices=
2*12/x+10...now.. 12y/x- 24/x+10
must be equal to 80/100 dollars...keep substituting the options.. (5,1) fits in
Q 153. In how many possible ways
can write 3240 as a product of 3 positive integers a,b and c
a) 450 b) 420 c)
350 d) 320 Solution- 3240=2*2*2*5*3*3*3*3 so, no. of ways=8!/(3!*4!)=420......
Q 154. a and b starts from their house at 10am. they
travel from their house on MG road at 20kmph and 40kmph. they meet at T
junction at 12:00 pm B reaches the T junction earlier and turns right .Both of
them continue travelling till 2:00pm what is distance between a& b at
2:00pm?
a) 160km b) 120km c) 140km d) 150km
Solution- a
reaches the junction at 12 after travelling 40 km but b reaches junction at 11
after travelling 40 km.
at 2 am. a will
travel 40 km from junction at 2 am. b will travel for 3 hrs that is 120 km from
junction so it travel 10 km.
distance between them is 120
+40=160 km
Q 155. On door A - It leads to
freedom
On door B - It leads to Ghost house
On door C - door B leads to Ghost
house
The statement written on one of the
doors is wrong.
Identify which door leads to
freedom.
a)
A
b)
B
c)
C
d)
None
Solution- lets consider door A is
wrong then A leads to ghost on door B written is it leads to ghost and on door
c it is written door B leads to ghost house,henc C leads to freedom in the same
way assume B is wrong,means lead to freedm but on door c written door B leads
to ghost house,it is contradiction last case if we assume C is wrong,the
written on door B is also becoming wrong
so our ansr is C lead to freedm
which comes from 1st case where we assume A is wrong
Q 156. The savings of employee
equals income minus expenditure.If the income of A,B,C are in the ratio 1:2:3
and their expense ratio 3:2:1 thenwhat is the order of the employees in
increasing order of their size of their savings?
a) A>C>B b)
B>A>C c) B>C>A d) C>B>A Solution-C >B>A salary of C is
highest among all n expenses are least salary of A is lowest among all n
expenses are high
Q 157. A manufacturer of chocolates
makes 6 different flavors of chocolates. The chocolates are sold in boxes of
10. How many different boxes of chocolates can be made?
(NOTE: A box is considered
different from another only if, regardless of the order, the box contains a
different number of chocolates of at least one type)
a) 3003
b) 10^6
c) 3000
d) 6^10
Solution- f n similar articles are
to be distributed to r persons, x1+x2+x3......xr=n each person is eligible to
take any number of articles then the total ways are n+r−1Cr−1
In this case
x1+x2+x3......x6=10 in such a case the formula for non negative integral
solutions is n+r−1Cr−1 Here n =10 and r=6. So total ways are 10+6−1C6−1 = 3003
Q 158. 4 men throw a die each
simultaneously. Find the probability that at least 2 people get the same
number.
a) 5/18 b)13/18 c) 1/36 d) ½
Solution- 13/18 as 1-(probability
of not getting same no.)=1-((6*5*4*3)/(6*6*6*6))
Q 159. How many positive integers
less than 500 can be formed using the numbers 1,2,3,and 5 for digits, each
digit being used only once.
a) 52 b) 68 c) 66 d) 34
Solution- 3 digits no formed= 3*3*2
(5 cant come at place,no should be less than 500)
=18
2 digit no formed=4*3
=12
1 digit no formed=4 total positive
integers=18+12+4 =34
Q 160. In a rectangular region of 300X400 foot, there are 3
ants per square inch.
how many
ants(approximate value) are there in the square region??? [Gave the hint:
1foot=12inches]
a) 500million
b) 50million'
c) 500000
d) 5million
Solution- 00*400 (foot)
300*400*12 (inches)
300*400*12*3
4320000 approximately 50 million ants
Q 161. The letters in the word
ROADIE are permuted in all possible ways and arranged in alphabetic order. Find
the word in the 44th rank.
a) AERIOD b) AERDOI c) AERODI d)
AEODRI
Solution- A----- => 5!=120
AD---- => 4!=24
AED--- => 3!=6
AEI--- => 3!=6
AEO--- => 3!=6
24+6+6+6=42
AERDIO => 43th
AERDOI => 44th
Q 162. There is a pool of radius X and
there is a path way around the pool with a width of 4 feet. Find the radius of
the pool if the path area / pool area = 11/25. a) 12
b) 5
c) 25
d) 29
Solution-
(PI.(X+4)^2-PI.X^2)/PI.X^2=11/25
(X+4)^2/X^2-1=11/25
(X+4)^2/X^2=36/25
(X+4)/X=6/5
X=20
Q 163. a certain organization has
three committees.only two persons are members of all committees,but every pair
of committees have three members in common.what is the least possible no of
member of members on any one committee?
a) 4 b) 5 c) 6 d) none of these.
Solution- the least possible no of
member of members on any one committee = 4 .. option a)
In all 3 committees, say X,Y,Z, 2
persons say A and B are common.
C is
common between X and Y.
D is
common between Y and Z.
E
is common between X and Z.
so X committee has ABCE. Y
committee has ABCD. Z committee has ABDE.
Q 164. In how many ways can 2310 be
expressed as product of three factors??
1) 41
2) 56
3) 23
4) 46
Solution-
(3^(n-1)+1)/2 2310=2*3*5*7*11 n=5
do this
ans===41
Q 165. In an office, at various
times during the day, the boss gives secretary a letter to type , each time
putting the letter on top of the secretary's inbox. when there is time, the
secretary takes the top letter off the pile and type it. if there are 5 letters
in all and boss delivers them in the order 12345, which of the following could
not be the order in which secretary types them?
a) 24351
b) 45231
c) 32415
d) 12345
Solution- B) 45231 cant be the
order of typing.....
Suppose boss put first time 4
letters in order 1,2,3,4
Then definitely she will strt
typing the top one tht is 4th..after completing 4th letter...boss again put the
5th letter on the top....sho she again choose the top most tht is 5th one and
type it...after tht queue will have order 1,2,3
So she should shoosed the top most
one tht is 3rd card...but in option it is given 2 which is wrong...check like
this in all the options.
Q 166. 6 task and 6 persons. P1 and
P2 does not do task T1. T2 is assigned to P3 or P4;. Each person should be
assigned with at least 1 task. In how many ways the task can be assigned.
a)
192
b)
360
c)
144
d)
180
Solution- 192
As T1 can be done
in 4 ways(p3,p4,p5,p6) T2 in tw0 ways (p3,p4) and the remaining in 4! ways
4*2*4!=192
Q 167. How many lattice points are
there between (2,0) and (16,203)?
a) 8 b) 10 c) 14 d) 15
Solution- draw a
line from origin(0,0) to (14,203) this line is parallel to the line
joining(2,0) and (16,203).
A lattice point is defined as the positive
values for x and y coordinates along the line in a graph. after drawing this
line you will come across 8 such points.
so the answer is 8
Q 168. It takes 52 days to complete
an agreement deal by a certain number of men.After 17 days,300 men are added
and 21 days are reduced.how many men were working initially?
a) 250 b) 150 c) 200 d) None of the
above
Soultion- Let x no people works
initially.
So the left work after 17 days can
be done by x people is in 35, but the same work can be done by x+300 is in
35-21=14 days.
so,
35*(x/(x+300))=14 x=200.
initially 200 people worked
Q 169. four parallel lines are
drawn parallel to one side of an equilateral triangle such that it cuts the
other two sides at equal intervals.the area of the largest segment thus formed
is 27msqr.find the area of the triangle?
a) 100 b) 75 c) 81 d) 54
Solution- Area of Trapezoid=
((a+b)*h))/2(where a and b are length of parallel sides)
Now Since side is divided in 5
parts therefore,length of one interval=a/5
Now use Sin60 and to calculate the height of
trapezoid( ie. a root 3/10). and cos60 to calculate the base of the triangle
whose hypotenuse is a/5..(i.e. will be a/10); We get b= a-(a/10+a/10) b= 4a/5
now equate the area of trapezoid with 27..
Hence area of equilateral triangle
is 75 sqm
Q 170. In the sample subtraction
problem below, single digits are replaced by letters,
Find the values of 3*A + &*B +
4*C *D = ?
A5C1
3U79
------
397D
a) 80
b) 99
c) 89
d) 96
Solution- I think the answer is 96,
let me explain...
from the above question it is clear
that there is a printing mistake, for now let us assume b=u;
after solving the subtraction we get
the value A=7,d=2,u=5,c=5; now if we suppose &=a then the ans will be
96(Which is option d) if &=d then the ans will be 71 if &=c then the
ans will be 86
so, from all the
assumptions only the first one matches with the options given, so option d will be our answer, 96
Q 171. two dice are thrown. find
the probability of getting a multiple of 3 of 4 as the sum.
a) 5/9 b) 4/9 c) 2/9 d) 1/9
Solution- 20/36=5/9
(1,2),(1,3),(1,5),(2,1)(2,2)(2,4)(2,6)(3,1)(3,3)(3,5)(3,6)(4,2)(4,4)(4,5)(5,1)(5,3)(5,4)(6,2)(6
,3)(6,6) as its some is multiple of
3 or 4
Q 172. 2 gears. one with 12 teeth
and the other one with 14 teeth are engaged with each other. One tooth in
smaller gear and one tooth in bigger gear are marked and initially those 2
marked teeth are in contact with each other. After how many rotations of the
smaller gear with the marked teeth in the other gear will again come into
contact for the first time?
a)
7
b)
12
c)
Data Insufficient
d)
84
Solution- Assume the distance
between the teeth is 1 cm. Then the circumference of first gear is 12 cm and
the second is 14 cm.
Now LCM (12, 14) = 84.
So to cover 84 cm, the first gear
has to rotate 84/12 = 7 rounds
(the second gear rotates 84 / 14 =
6 rounds as it is bigger)
Q 173. there are 4 couples who go for honeymoon together.at
one of the places they all have to cross the river but only one boat is
available.wives are jealous that they dont want their husbands to travel with
another women and husbands are also possessive that they dont want their wives
to travel with some other men.the no of minimum possible ways in which they
will cross the river are. a) 16 b) 17 c) 18 d) 19
Solution- we have 4 couple
1)first two husbands goes to
another side
2)then one husband comes back
3)then one couple will go
4)then second husband comes back
so in this 4 ways
1 couple reach to another side so for 4 couples there are 16 ways
a)16 is ans
Q 174. Professor absentminded has a
very peculiar problem, in that he cannot remember numbers larger than 15.
However, he tells his wife, I can remember any number up to 100 by remembering
the three numbers obtained as remainders when the number is divided by 3, 5 and
7 respectively. For example (2,2,3) is 17. Professor remembers that he had
(1,1,6) rupees in the purse, and he paid (2,0,6) rupees to the servant. How
much money is left in the purse?
option
a) 59
b) 61
c) 49
d) 56
Solution- He had 76
rupees,according to questions, he remember any no. larger than 15 as remainder
when divided by 3,5 and 7 respectively.
Divide 76 by 3,5 and 7, we will get
(1,1,6)as remainder.
He paid 20 rupees bcoz when 20
divided by 3,5 & 7 , we will get (2,0,6) as remainder.
Now money left=76-20
=56 (Ans)
Q 175. A and B completed a work
together in 5 days. Had A worked at twice the speed and B at half the speed, it
would have taken them four days to complete the job. How much time would it
take for A alone to do the work?
a) 5days b) 20days
c) 10days d) 25days Solution- a nd b's 1 day work, a+b=1/5 with twice d speed
of a nd half of b completes work in 4 days,
so, 2a+b/2=1/4.
on solving both d
eq. we get a=1/10 so a will complete d whole work in 10 days.
Q 176. 0>a>b>c>d. Which
is largest
a)
(b+d)/(a+c)
b)
(a+d)/(b+c)
c)
(b+c)/(a+d)
d)
(c+d)/(a+d)
Solution- given
statement is 0>a>b>c>d that means all the values of a,b,c,d are
less than ZERO so lets consider a=-1 b=-2, c=-3 and d=-4 so that
0>a>b>c>d will satisfy by
solving the options we get the values as follows
a.(b+d)/(a+c)= 1.5
b. 1
c.1
d.1.4 among all of
these option a is highest so the ans is option a.
Q 177. An engineer undertakes a
project to build a road of 15km in 300 days and employs 45 men for the purpose
.After 100 days,he finds only 2.5km of the road has been completed.find the
(approx)number of extra men must employs to finish the work in time.
option
a) 43
b) 45
c) 55
d) 68
Solution- 45 workers working
already
Let x be the total men required to
finish the task in next 200 days
2.5 km done hence remaining is 12.5
km
Also, work has to
be completed in next 200 days (300 - 100 = 200) We know that, proportion of men
to distance is direct proportion and,
proportion of men to days is inverse proportion
Hence, X = (45 * 12.5 * 100) / (2.5 * 200) thus, X = 112.5 that is
approximately 113
Thus, more men needed to finish the
task = 113-45=68
Q 178. there are 5 sweets - jamun,
kulfi, peda. laddu and jilebi that i wis t eat on 5 consecutive days, monday
throufh friday, one sweet a day, based on following self imposed
constraints:
1) laddu
is not eaten on monday
2) if
jamun is eaten on monady , then laddu must be eaten on friday
3) if
jamun is eaten on tuesday, kulfi should be eaten on monday
4) peda
is eaten the day following the day eating jelabi based on the above , peda can
be eaten on any day expect??
Solution-monday bcoz the day before
jalebi is required to be taken bt its sunday and no sweet is eaten on Sunday
Q 179. A circular swimming pool is
surrounded by a concrete wall 4ft wide. If the area of the wall is 11/25 of the
area of the pool then radius of the pool in feet is a) 20
b) 8
c) 16
d) 30
Solution- i thnk ans is 20.Bcoz let
r be the radius of pool.
(11/25)(pi*r^2)=(pi(r+4)^2)-(pi*r^2)
By solving this we get ans as 20
Q 180. How many 5’s will be there
in the number 121122123... till 356?
a) 51 b) 54 c) 50 d) 49
Solution- 121122123... till 356
121,122,123,124... till 356
121 to 200 => 125,135,145, from
150 to 159 => 11 5's , 165,175,185,195
[from 150 to 159, each no. has 1
no. of 5 except 155, 155 has 2 no. of 5]
total=18
201 to 300 =>
205,215,225,235,245, 250 to 259 => 11 5's , 265,275,285,295
=> total=20
301 to 356 =>
305,315,325,335,345,350,351,352,353,354,355(2 5's),356
=> total=13
total 5's = 18+20+13 = 51
Q 181. Car A leaves city C at 5 pm
and drives at a speed of 40 kmph. 2 hours later another car B leaves city C and
drives in the same direction as car A. In how much time will car B be 9 km
ahead of car A. Speed of car B is 60 kmph.
option
a) 4.25
hrs
b) 4.17
hrs
c) 4.30
hrs
d) 4.45
hrs
Solution-we take
the relative speeds of the 2 cars. relative speed=(60-40)=20km/h the car a has
travelled (40*2)km in 2 hrs. thence car b has to travel 80km+9km ahead to get
ahead by 9 km. so time= 89/20=4hr 45 min.
Q 182. The rupee/coin changing
machine at a bank has a flaw. It gives 10 ten rupee notes if you put a 100
rupee note and 10 one rupee coins if you insert a 10 rupee note but gives 10
hundred rupee notes when you put a one rupee coin!
Sivaji, after being ruined by his
rivals in business is left with a one rupee coin and discovers the flaw in the
machine by accident. By using the machine repeatedly, which of the following
amounts is a valid amount that Sivaji can have when he gets tired and stops at
some stage (assume that the machine has an infinite supply of notes and coins):
a) 26975
b) 53947
c) 18980
d) 33966
Solution-initially sivaji had only
one coin so he earns only 1000/-
to earn more he must convert one 100
note to 10 notes and then one ten note to ten coins now he has Rs.990 and 10 coins after converting 10 coins he has 10990 after
converting another 10 coins he has 20980 after converting another 10 coins he
has 30970 after converting another 10 coins he has 40960 after converting
another 10 coins he has 50950 after converting another 10 rupee note to coins
and using only 3 he has seven coins and 53940 so sum is 53940+7=53947
Q 183. 2/3rd of the balls in a bag
are blue, the rest are pink. if 5/9th of the blue balls and 7/8th of the pink
balls are defective, find the total number of balls in the bag given that the
number of non defective balls is 146.
a) 216 b) 649 c) 432 d) 578
Solution-let total no of balls =x blue=2x/3 pink=x/3
total no of defective balls =
10x/27 +7x/24
=143x/216 non
defective balls=x-143x/216=146 ,x=432
Q 184. One day Eesha started 30 min
late from home and reached her office 50 min late while driving 25% slower than
her usual speed. How much time in min does eesha usually take to reach her
office from home?
a) 80min b) 50min
c) 60min d) 70min Solution-let time taken by eesha daily = x and usual speed =
y
(x+20)*(.75*y) = x*y extra time
taken = 50-30 = 20min
.75x+15 = x .25x =
15 x = 60 min
Q 185. 7,17,19,43,91,131...find the
odd term.
a) 17 b) 43 c) 91 d) 131
Solution-except 91 all other
are prime no.
so 91 odd one
Q 186. find the no. of zeros in the
product of 1^1*2^2*3^3....*49^49?
a) 250 b) 225 c) 545 d) 135
Solution-the number of zero's can
be found by finding number of 2's and 5's.
since in this number, number of
occurrences 2's will always be greater than number of occurrences of 5's.
therefore we need to find number of 5's only.
number of
occurrences of 5's in 5^5 = 5
10^10=10
15^15=15
20^20=20
25^25=25
30^30=30
35^35=35
40^40=40 45^45=45
sum=225. therefore 225 5's can be paired with 225 2's.
total number of zeros = 225
Q 187. a merchant buys 20kg of
wheat at 30rs/kg and 40k at 25rs/kg.he mixes and sells 1/3 rd of mixture at
26rs/kg.price at whch merchant should sell remaining mixture so that a profit
of 25% on whole outlay is?
a) Rs30 b) Rs40 c) Rs360 d) Rs 37
Solution-cp of
total mixture=(20*30)+(25*40)=1600rs as he needs 25% profit, so he needs to
earn 1600+(1600*(25/100))=2000rs.
he has alredy got rs by selling 1/3
part=1/3 of 60 kg..=20 kg
=20*26=520rs.. now
he needs 2000-520=1480rs.
so he must sell remaining 40 kg
at=1480/40=37rs/kg
Q 188. the addition of
641+852+973=2456 is incorrect. What is the largest digit that can be changed to
make the addition correct?
a)
5 b) 6 c)
4 d) 7
Solution-641
852
973
2466
Given Sum is 2456
So if we change
10's digit 7 to 6 then the sum will be correct . so 7 is the largest digit .
Q 189. a, b, c are non negative
integers such that 28a+30b+31c=365. Then a+b+c is?
a)13 b)> 13 c) =13 d) 12 Solution-Here, a + b +
c = 12
Explanation : a =
1, because in a month of twelve only Feb has 28 days b = 4, there are four
months namely April,June,Sep and Nov having 30 days
c = 7, there are 7 months namely
Jan,Mar,May,Jul,Aug,Oct and Dec having 31 days
So, a + b + c = 1 + 4 + 7 = 12.
Q 190. 26ab5 is a four-digit number
divisible by 25. If the number formed from the two digits ab is a multiple of
13, then ab =?
a) 52
b) 65
c) 10
d) 25
Solution-Any number last two digits
are 25 then that should be divisible by 25.
When ab=52, the number is 26525 and
52/13=4,
So ab=52
Q 191. A owes B Rs 50. He agrees to
pay B over a number of consecutive day starting on a Monday, paying single note
of Rs 10 or Rs 20 on each day. In how many different ways can A repay B. (Two
ways are said to be different if at least one day, a note of a different
denomination is given) a) 8
b) 7
c) 6
d) 5
Solution-10,20,20=3!/2!==3 ways
10,10,10,20=4!/3!==4
ways 10,10,10,10,10=5!/5!==1 way so.total=3+4+1==8 ways.
Q 192. After 6 years Raju's fathers
age will twice of the Raju's age 2 years ago.His mothers age was twice that of
Raju's age. Sum of the age of their parents.
a) 4
less than four times Raju’s age
b) 2
more than four times Raju’s age
c) 4
more than four times Raju’s age
d) 2
less than four times Raju’s age
Solution-F+6=2(R+6)
F= 2R+6
M-2=2(R-2)
M= 2R-2
Therefore the sum of Raju’s
Parent’s age is
F+M=2R+6+2R-2
F+M=4R+4
4 more than four times Raju’s age
Q 193. Length, Breadth and Height
of a 3D figure is in the ratio 3:2:1. If the length is doubled and Breadth
& Height are halved, then what is the % decrease in the volume of the
solid?
a) Decreased by 15% b) Decreased by
18% c) Decreased by 30%
d) Decreased by 50%
Solution-50% decrease
If original length, breadth &
height are 3x, 2x and x respectively, then volume=3x*2x*x=6x^3 With length
doubled, breadth & height halved, new dimensions are 6x, x and x/2
respectively and volume=6x*x*x/2=3x^3
So % decrease in
volume=100*(6x^3-3x^3)/6x^3=50
Q 194. 12 divides, ab313ab (in
decimal notation, where a,b are digits>0, the smallest value of a+b is
a)
7
b)
6
c)
2
d)
4
Solution-If a
number is divisible by 12 then it should be divisible by 4&3 for divisible
by 4 last [2 digit]no's should be divisible by 4 so last no's are
12,15,18,,,,,,and soon..
now least is 12
it,. 3 which is not the option,, no other least is 16 it's sum 1+6=7..
so 7 is option :)
Q 195. In a telecom assembly
factory,there are 250men and 150 women.The average productivity of all works is
12 units per day.The average productivity of men is 15 units per day..what is
the average productivity of women per day? a) 6 b) 9 c) 7 d) 8
Solution-250M + 150W = 400
Thus, 400*12 = 4800
Now, 250*15 = 3750
Now, 4800-3750 = 1050 Thus, for
150W avg prod is : 1050/150 = 7
Q 196. If a lemon and an apple
together cost Rs. 12.00, a tomato and a lemon cost Rs. 4.00 and an apple cost
Rs.8.00 more than a tomato or a lemon then which of the following can be.
a) Rs 2 b) Rs 4 c)
Rs 1 d) Rs 3 Solution-lemon+apple=12rs tomato+lemon=4
then apple=8+t or 8+l by solving we
get l+(8+l)=12
2l=4
lemon=2
so a=8+2=10; so a)
Q 197. George, Paul and Hari start
a business by contributing Rs. 30000/-, Rs. 40000/- and Rs. 50000/-
respectively. After ½ a year George withdraws half his contribution. At the end
of the year the business showed a profit of Rs 90000 which was divided amongst
the 3 men proportionate the to amount and duration of their investment in the
enterprise. Paul got,
a)
Rs 25000/-
b)
Rs 18000/-
c)
Rs 32000/-
d)
Rs 24000/
Solution-G,P & h ratio will be
(30,000*6+15000*6):(40000*12):(50000*12)
=27:48:60 hence
135x=90000 x=2000/3 paul got x*48 means 2000/3*48=32000
Q 198. A drinks machine offers
three solutions – Tea, Coffee or Random but the machine has been wired up
wrongly so that each button does not give what it claims. If each drink costs
Rs. 50, what is the mimimum amount of money that must be spent to determine
with certainity the correct labeling of the buttons?
a)
Rs. 100
b)
Can not be determined from the given information
c)
Rs 150
d)
Rs 50
Solution-You have to put just
50rs.
Put 50rs and push the button for
Random. There are only 2
possibilities. It will give either
Tea or Coffee.
If
it gives Tea, then the button named Random is for Tea. The button named Coffee
is for Random selection. And the
button named Tea is for
Coffee.
If
it gives Coffee, then the button named Random is for Coffee. The button named
Tea is for Random selection. And the button named Coffee is for Tea.
Thus, you can make
out which button is for what by putting just 50rs and pressing Random selection
first.
Q 199. P,Q,R,S are distinct
integers numbered from 1 to 12.What is the possible smallest value for (P/Q)+(R/S)=?
a) 1/12+2/11 b) 1/11+9/10 c)
1/11+2/12 d) 1/10+1/11
Solution-he given options for the
above question are, a.1/12+2/11 b.1/11+9/10 c.1/11+1/12 d.1/10+1/11
In mathematics,
two things are called distinct if they are not equal. so p,q,r,s are not equal.
Ans:- 1/12 + 2/11
Q 200. If ab64ab is
divisible by 12, then what is the least possible value of a+b ? a) 4 b) 5 c) 6
d) 7
Solution-1.Condition for
divisibility by 3--
The sum of digits of number must be
divisible by 3.
i.e a+b+6+4+a+b = 0 mod 3
=> 10 + 2(a+b) = 0 mod 3
=> 1 + 2(a+b) = 0 mod 3 (10 = 1
mod 3)
=> 2(a+b) = -1 mod 3
=> 2(a+b) = 2 mod 3
=> a+b = 1 mod 3
So, a+b must be of the form 3n + 1
for some positive integer n.
Max value of (a+b) is 18 as a and b
can take values upto 9 since they are digits.
The possible values are
1,4,7,10,13,16.
2. Condition for divisibility by
4--
For divisibility by 4, Last two
digits must be divisible by 4.
i.e 10a+b = 0 mod 4
=> 2a + b = 0 mod 4 (Since, 8a = 0 mod 4)
So, 2a + b can be
4,8,12,16,20,24,28. Combining these,we get
a + b = 3n +1
2a + b = 4k
The smallest values which satisfies
these two equation is a = 4 ,b = 0.
So, the minimum sum of a + b
is 4.
Q 201. There are 100 in a class and
they attend a test. 20 students are failed in both the subjects. 50 students
pass in subject A. 60 students passed in subject B. How many students passed in
subject A only.
a) 20 b) 30 c) 15 d) 25
Solution-LET x be the total no. of
students who passed in both test.
Then, no. of students who passed in A only= total no. of
students who passed in
A - total no. of
students who passed for both A & B =50 - x similarly, no. of students who
got pass only in B = 60 - x now, total no. students who got pass = 80
80= (50 - x) +(60-x)+x then we'll
get on solving, x=30
students who passed in A only = 50
-x= 50-30=20
Q 202. Aravind can do a job in 24 days. Mani can dig the same
well in 36 days.
Aravind, Mani, and Hari can do the
work together in 8 days. How long does it take Hari to do the work alone?
a) 12 days b) 18 days c) 16 days d)
24 days
Solution-This can be solved with
the help of efficiency.Efficiency of (aravind +mani+hari) =total efficiency
Here,
No. Of Days taken by Aravind =24
Efficiency of aravind =1/24
No.of days taken by mani=36
Efficiency of mani=1/36
No. Of days taken when all the three
work together =8
Efficiency of all thr three
together=1/8
Let the no. Of days taken by hari
to complete the job be x.
Then efficiency of hari =1/x Now,
from the above formula
1/24 + 1/36 + 1/x = 1/8
x=18
Q 203. Truck A and Truck B move grain into a box at the rate
of 20 kilos/min and
13 1/3 kilos a min respectively while truck
C removes grain from the box at the rate of 10 kilos/min. If the capacity of
the box is 2.4 tons, and Truck A, Truck B and Truck C are working
simultaneously Then the box will be filled in? a) 1 1/2 hrs
b) 3/5
hrs
c) 1
5/7 hrs
d) 2
1/8 hrs
Solution-otal
filled in i min=(20+40/3-10)=70/3 to fill 70/3 time req-----=1min to fill 2400
time req-----=(3/70)*(2400/60)hr
=15/7
Q 204. If 5 + 3 + 2 = 151022, 9 + 2
+ 4 = 183652, 8 + 6 + 3 = 482466 and 5 + 4 + 5 = 202541, then 7 + 2 + 5 = ?
a) 143547 b) 132234
c) 2577224 d) 112321
Solution-5+3+2=(5*3)(5*2)(5*3+5*2-3(ie
second term))=15,10,22
9+2+4=(9*2)(9*4)(9*2+9*4-2(ie
second term))=18,36,52 8+6+3=(8*6)(8*3)(8*6+8*3-6(ie second term))=48,24,66 so
answer is 143547
Q 205. Two full tanks one shaped
like the cylinder and the other like a cone contain liquid fuel the cylindrical
tank held 500 lts more then the conolical tank After 200 lts of fuel is pumped
out from each tank the cylindrical tank now contains twice the amount of fuel
in the canonical tank How many lts of fuel did the cylindrical tank have when
it was full?
a) 1200 b) 1000 c) 700 d) 1100
Solution-let area of cylinder=A
& area of
cone=B acc. to ques,
A=B+500 .....(1)
A-200=2(B-200) .....(2) from (1) &(2) A=1200
volume of fuel when cylinder is
full=1200 lts
Q 206. Asha and Eesha – Eesha lies
on Monday, Tuesday and Wednesday. Asha lies on Thursday, Friday and Saturday.
Other days they will say the truth. Professor forgot and asked them what day it
is. Both of them said yesterday I was lying and then professor got the day.
What day it is?
a) Tuesday b) Thursday c) Friday d)
Cannot be determined
Solution-Thursday. Asha lies on
thursday and today she is lying that she was lying yesterday. Eesha is not
lying today but she was lying yesterday.
Q 207. Three sisters are identical
triplets. The oldest by is Aasha, and she always tells anyone the truth. The
next oldest is Usha, and Usha always will tell anyone a lie. Eesha is the
youngest of the three. She sometimes lies and sometimes tells the truth.
Mukund an old friend of the family,
visited them one day and was able to recognize who was who, so he asked each of
them one question.
Mukund asked the sister that was
sitting on the left. “Which sister is in the middle of you three ?” and the
answer he received was, “Oh,that’s Aasha.”
Mukund then asked the sister in the
middle. “What is your name?” The response given was, “I’m Eesha.”
Mukund turned to the sister on the
right,then asked. “Who is that in the middle ?” The sister then replied, “She
is Usha.”
This confused
Mukund. Who was in the middle ?
a) Aasha b) Eesha c) Usha d) Cannot
be determined
Solution-its USHA ! as the left one
said the middle one is aasha.. if it was aasha she would have admitted! so the
left one and the middle one could be eesha usha both. as eesha sometimes says
truth and sometimes lie. and usha always lie. so the right one would be aasha
who will say truth and she said the middle one is USHA,so option (c) Usha is a
right answer !
Q 208. If 4 examiners can examine a certain
number of answer books in 8 days by working 5 hours a day, for how many hours a
day would 2 examiners have to work in order to examine twice the number of
answer books in 20 days. a) 6 hours
b) 7
hours
c) 15/2
hours
d) 8
hours
Solution-4
examiner working for 8 days(5 hours a day) thus,
1 examiner=8*5=40
hrs/person
Now,
2 examiner
working for 20 days(x hrs a day) to have twice the work. thus,2*x hrs spent
each day
We need 160 hrs.Working 2 hrs the
duration will be:
160/2*x=20
=>x=4 hrs
Twice work is done,thus,time=4*2=8
hrs.
Q 209. A circle has 11 points
arranged in a clockwise manner from 0 to 10. A bug moves clockwise on the
circle according to following rule. If it is at a point i on the circle, it
moves clockwise in 1 sec by (1 + r) places, where r is the remainder (possibly
0) when i is divided by 2. If it starts at 4th position, at what position will
it be after 2012 seconds?
a) 7 b) 9 c) 5 d)
1 solution-(D) 1
cycle= 1 +7n after
this cycle always is 5th position. now after 2008 sec point is 5th position.
now next 4 sec 1st position.
or
Trick:- starting position - 3 =
present position. maximum cases.
Q 210. in a group of 5, Anooj
said"one of us are lying", Pooja said"Exactly two are
lying", Bitto said "Exactly three are lying", Billa
said"Exactly four are lying", Chitra said,"Exactly five of us
are lying". Which one said the Truth?
a) Billa b) Anooj c) Chitra d)
Pooja
Solution-Billa because 4 of them
are lying and one is saying the truth....
1/5+2/5+3/5+4/5+5/5
15/5=3 So Billa
Q 211. A travels at 40kmph. B
travels at 60kmph. They are travelling towards each other. BY the time they
meet ,B would have travelled 120 km more than A. Find the total distance.
a)600km b) 720km c) 400km d) 540km
Solution- dist.
traveeld by A = 40x (let x be the time taken to meet) dist travelld by B = 60x
difference in dist = 20x = 120; which gives x=6hours so relative speed of A and
B is 100km/hr. distance travlld = 6*100 = 600km.
Q 212. If all the numbers between
11 and 100 are written on a piece of paper. How many times will the number 4 be
used?
a) 20 b) 19 c) 9 d) none of these
Solution- 19, every word can only
be appeared 20 times from 1 to 100. nd it is starting from 11, so 19 times
used..
Q 213. 10 people are there, they are
shaking hands together, how many hand shakes possible, if they are in no pair
of cyclic sequence. a) 45
b) 9
c) 12
d) 10
Solution- 10c2= n!/(n-r)!.r!
=10!/8!.2!
=10*9/2
=45
Q 214. What should be the value of
a,in the polynomials x^2-11x+a and x2-14x+2a so that these two polynomials have
common factors.
a) 24 b) 1 c) -1 d) ½
Solution- a=24,
then polynomials will become x^2-11x+24 and x^2-14x+48 which can be written as
(x-8)*(x-3) and (x-8)*(x-6)
(x-8) is common factor.
Q 215. a man cannot remember the
number larger than 15. However he remember any number upto 100 by remembering
the three numbers obtained as remainders when the numbers is divided by 3,7 and
11 respectively. he remembers that he had (2,4,8) rupees in his purse. and he
paid (2,5,4)rupees to the servant. How much money is left in the purse?
a) 40 b) 48 c) 46 d) 52
Solution- the number that satisfies
the all remainders of 2,4,8 by numbers 3,7,11 respectively is 74.
and the the number that satisfies
the all remainders of 2,5,4 by numbers 3,7,11 is 26.
hence the remaining money
is=74-26=48
ans is 48
Q 216. Next number in the given
series 1, 7, 8, 49, 50, 56, 57, 343
a) 344 b) 350 c) 2401 d) Cannot be
determined
Solution- 344 because
1*7=7
7+1-8
7*7=49
49+1=50
8*7=56
56+1=57
49*7=343 343+1=344
ans= 344
Q 217. A man sold 12 candies in 10$
had loss of b% then again sold 12 candies at 12$ had profit of b% find the
value of b.
a) 9 b) 9.09 c) 10 d) 11
Solution- Let the CP be 'x' then:
(x-10)*100/x = b% ----------->1
(12- x) * 100 /x = b% ------->2
(x-10)*100 / x =
(12-x)*100 / x x-10 = 12-x x = 11
So by Submitting
value of x into one of above question (1 or 2) we get b% which is 9.09 or 9
Approx
Q 218. Two breakers are kept on a table. the
capacity of the first breaker is x liters and that of the second breaker is 2x.
Two thirds of the 1st breaker and one fourth of the 2nd breaker is filled with
wine. The remaining space in both breakers is filled with water. If the content
in these breakers are mixed in a larger breaker of volume 3x, What is the
proportion of wine in the breaker? a)
11/12
b) 11/36
c) 7/6
d) 7/18
Solution- 2/3+1/2(1/4th of 2x =1/2
of x)=7/6. when pored in breaker of vol 3x then d proportion will be 7/6
*1/3=7/18
Q 219. When all possible six-letter
arrangements of the letters of the word "MASTER" are sorted in
alphabetical order, what will be the 49th word?
a) AREMST b) ARMEST c) AMERST d)
ARMSET
Solution- ARRANGING THE LETTERS IN
MASTER ALPHABETICALLY WE GET.... AEMRST
NOW A E _ _ _ _
can be filled in 4! ways A M _ _ _ _ can be filled in 4! ways so words upto 4!+4! =48 words are of ae
series and am series
So 49th word can be A R E M S T
Q 220. find a number such that when
it is added to 7249 will be perfectly divisible by 12,14,21,33 and 54
a) 8136 b) 9123 c)
8727 d) 9383 Solution- lcm of 12,14,21,33,54=8316 now checking through options,
on adding 7249+8136/8316=not perfetily divisible similarly on checking through
options,(9383+7249)/8316=2 answer=9383
Q 221.60 men can complete a piece
of work in 40 days.60 men start the work but after every 5 days 5 people
leave.in how many days will the work be completed?
a) 60 b) 80 c) 120 d) None of these
Solution- (60*40) men complete the
work in 1 day
1man's 1 day work
is 1/60*40 60 men's 5 day work is 5*(1/40)=1/8 remaining work (1-1/8)=7/8 as 5
men leave
55 men's 1 day
work=55*(1/60*40)=11/60*8
now
11/60*8 work done by them 1day
7/8 work done by them
[(60*8)/11]*(7/8)=420/11=38.18
so they complete in (38.18+5)
days{5 day added bcoz alrdy 5dy work had been done}
43.18 day total work will be
completed.
Q 222. A person walks at 4km/hr for a
particular duration T1 and 3km/hr for another duration T2 and covers a total
distance of 36km. If he walks at 4km/hr for the duration T2 and at 3Km/hr for
the duration T1, then he covers only 34km. What willl be the time taken by him
to cover the one of the legs? a) 4hrs
b) 7hrs
c) 10hrs
d) 6hrs
Solution- 4T1 + 3T2 =36...(i)
3T1 + 4T2 =34...(ii)
4*(i)-3*(ii),we get
T1=6hrs and T2=4hrs
Q 223. If ABERSU are in sorted in
alphabetical order, if 24 sortings are req for ABUSRE, 25 - AEBRSU, 49- ARBESU,
den how many for AEUSRB.
a) 45 b) 48 c) 47 d) 46
Solution- AEUSRB
ALPHA OREDER ABERSU
A B _ _ _ _ =4! =24
A E B _ _ _ =3!=6
A E R _ _ _ =3!=6
A E S _ _ _ =3! =6
A E U B _ _ =2! =2
A E U R _ _=2!=2
A E U S B _=1
A E U S R B=1
24+6+6+6+2+2+1+1=48
Q 224. a,band c can do some work in
36 days a and b together do twice as much work as c.a and c together do thrice
as much work as b.eind time taken by c alone to do complete work
a) 72
c) 120
c) 96
d) 108
Solution- 1 day work for (A+B+C) =
1/36 ............(ii)
Now, (A+B) = 2C
..........................(i)
From (i)and(ii): 3C = 1/36
C = 1/108
Thus, C takes 108 days to complete
the whole work alone.
(Remaining data are not required in
this case)
Q 225. A sum of rs 3000 is
distributed amongst A, B, and C. A gets 2/3 of what B and C got together and c
gets 1/3 of what A and B got together, C's share is?
a) 1200 b) 2250 c) 750 d) 1050
Solution- A+B+C=3000 from data it
is clear that A=2/3(B+C) and C=1/3(A+B)
==> 3C=A+B so replace A+B can be
replaced by 3c.....
===> 4c=3000 and now
c=3000/4===>750... So c gets 750
Q 226. raj writes a number. He sees
that the first no of 2digits exceeds 4times the sum of its digits by 3. If the
number is increased by 18, the result is the same as the number formed by
reversing its digits. Find the number
a) 35 b) 57 c) 42 d) 49
Solution- 35-4*8=3
35+18=53 answer is 35
Q 227. P ( x ) = ( x2012 + x2011 +
x2010 + . . . . + x + 1)2 – x2012
Q ( x ) = x2011 + x 2010 + . . . .
+ x + 1
The reminder when P ( x ) is
divided by Q ( x ) is
a) 1 b) 0 c) x+1 d) x-1 Solution-
substitute 1 in x
then it will become
((2013)^2-1)/2012 (2013+1)(2013-1)/2012
2014 is quotient
So, remainder is '0'.
Q 228. If YWUSQ is 25 - 23 - 21 -
19 - 17 then MKIGF is what?
a) 13-11-9-7-6 b) 1-2-3-5-7 c)
9-8-7-6-5 d) 7-8-4-5-3
Solution- YWUSQ - 25-23-19-17
THEN Y=25,W=23,U=21.....
THEN MKIGF = 13-11-9-7-6
Q 229. when numbers are written in
base b, we have 12x25=333. The value of b is
a)
6 b) 8 c) 7 d) none of these
Solution- when a no. is written in
base b
(1*b^1+2*b^0)*(2*b^1+5*b^0)=(3*b^2+3*b^1+3*b^0)
(b+2)*(2b+5)=(3b^2+3b+3)
solve it further a quadratic eq. is formed b^2-6b-7 after solving it can be
written as
(b-7)(b-1)
therefore b=7 b=7
Q 230. How many polynomial
functions f of degree >=1 satisfy f(X2) = (f(X)) 2 =f (f(X)).
a) 1 b) 0 c) 2 d) more than 2
Solution- :Let
f(x)=x^2 f(x^2)=x^4 [f(x)]^2=x^4 f(f(x))=x^4
Only 1 sol.
Q 231. A cow and a horse are bought
for Rs.200000. The cow is sold at a profit of20% and the horse is sold at a
loss of 10%. The overall gain is Rs. 4000.The cost pric of the cow is:
a) Rs
130000
b) Rs
80000
c) Rs
70000
d) Rs
120000
Solution- let cost
of cow be x let cost of horse be 200000-x so, (120%)x-(90%)(200000-x)=204000
solve the above equation vil get x=80000 thus cost price of cow is 80000/-
Q 232. Raj drives slowly along the
perimeter of a rectangular park at 24kmph and completes one full round in
4mins. The ratio of length is to breadth is 3:2. What are its dimensions?
a) 450m*300m
b) 150m*100m
c) 480m*320m
d) 100m*100m
Solution- The ratio of length is to
breadth is 3:2
So,
Length is 3x and breadth is 2x
Now, The perimeter of the
rectangular park is; 2( Length + Breadth)
2 ( 3x+ 2x)
2*5x
=10x
Raj drives slowly along the
perimeter of a rectangular park that's mean he covered the total distance 10x.
So, the total distance is = 10x
Total time 4mins, so 4/60 hrs.
Speed 24kmph
We know that,
Distance = Time * Speed
Distance = 4/60 * 24
Distance= 1.6
Now, we can write;
10x= 1.6 x =
1.6/10 x = 0.16
Length is (3*0.16) = .48km or 480 m
Breath (2*0.16) = .32km or 320 m
Ans. Its dimensions are 480m * 320
m
Q 233. or which of the following n
is the number 2^74 + 2^2058 + 2^2n is a perfect square.
a) 2010 b) 2018 c) 2012 d) 2020
Solution- 2^74 + 2^2058 + 2^2n
=(2^37)^2 + (2^1029)^2 + (2^n)^2
now, if we put 2^37 = a ; 2^1029 =
b; Then for the above expression to be perfect square 2^2n must be equal to
(2*a*b)= 2*(2^37)*(2^1029);
==> 2^2n =
2^(1067) ==> 2n = 1067 ,
but this case is not possible since
R.H.S is an odd integer whereas L.H.S is an even integer.
So , the above mentioned case can't
hold.
Now,if we put 2^37 = a; 2^n = b ;
So, for the given expression to be perfect square 2^2058 = (2*a*b)=
2*(2^37)*(2^n) = 2^(n+38);
So, 2058 = (n+38)
=> n = 2020
So,The answer is n = 2020
Q 234. Father is 5 times faster
than son. Father completes work in 40days before son.If both of them work
together when will work get complete?
a) 8 days b) 50/6 days c) 10 days
d) 20 days
Solution- et takes
x days TO COMPLETE THE WORK than father will take x-40 as we know time is inversely propotional to
speed so time=1/speed
fatherstime/sonstime
x-40/x and given
that speed ration btwn father and son is 5:1 so x-40/x=5/1 x-40=5x 4x=40 x=10
so if father and son work together than
1/10+1/50
25/3 days is ans
Q 235. a beaker contains 180 litres
of alcohol. On day 1, 60 litres of alcohol is replaced with water. On 2nd and
3rd days 60 litres of the mixture in the beaker is replaced with water. What
will be the quantity of alcohol in the beaker after 3rd day?
a) 40 liters b) 80 liters c) 53.33
liters d) 100 liters
Solution-
1st time 180-60=120
2nd time 120-60*120/180=80lit
3rd time 80-60*80/180= 160/3 lit
So c) 53.33 liters
Q 236. A man asks 5 people to make
a guess about the amount of money in his pocket which is less than 50.
A guess
that the amt is a multiple of 10..
B guess
that the amt is a multiple of 12..
C guess
that the amt is a multiple of 15..
D guess
that the amt is a multiple of 18...
E guess
that the amt is a multiple of 30 Which of the following guesses are correct?
a)AE
b) AB
c) BC
d) DE
Solution- AE:30 is multiple of both
10 and 30 which is less than 50. ... possible answer.
AB: there is no multiple of 10 and
12 less than 50. ..... not possible answer.
BC: Same logic as that of AB. .....
not possible answer.
DE: Same logic as that of AB. .....
not possible answer.
Hence answer is AE.
Q 237. Letters of alphabets no from
1 to 26 are consecutively with 1 assigned to A and 26 to Z. By 27th letter we mean A, 28th B. In general
26m+n, m and n negative intezers is same as the letters numbered n.
Let P = 6, strange country military
general sends this secret message according ot the following codification
scheme. In codifying a sentence, the 1st
time a letter occurs it is replaced by the pth letter from it. 2nd time if occurred it is replaced by P^2
letter from it. 3rd time it occurred it
is replaced by p^3 letter from it. What is the code word for ABBATIAL
a) GHNNZOOR
b) GHKJZOHR
c) GHHGZOGR
d) GHLKZOIR
Solution-A should be coded as 1+6 =
G (it occurred for first time)
B should be coded as 2+6 = H (it
occurred for first time)
B Should be coded as 2
+ 36 = 38 - 26 = 12 = L (it occurred for second time) Option d)
Q 238. 30L + 3Q = 1167 ,30L + 6Q =
1284
Find L?
a) 30 b) 35 c) 40 d) 45
Solution- 30L+3Q=1167 ----(i)
30L+6Q=1284 ----(ii)
Solving (i) & (ii),Q=39, L=35
Q 239. The price of a book in four
different shops and the successive discounts offered for the books is given
below. Select the option in which the price of the book is the least.
(a) 10%, 5%, and 5% discount on
Rs.195
(c) 12.5%
and 12.5% discounts on Rs.205
(b)25%, discount on Rs.200
(d) 10%,
and 15% discounts on a marked price of Rs.190
Solution- we know that there is a formula on discount of m%
and n% respectively which is(m+n-mn/100)%.According to this for first
case,total discount is
18.775%,so after discount the price
of the book is Rs.158.38.For second case total discount is 23.4375%,so the
price of the book after discount is Rs.156.95.For third case after discount the
pice of the book is Rs.150.For last case total discount is 23.5% and the price
of the book after discount is Rs.145.35.
So ans is option d in which the
price is least.
Q 240. Divide 50 into two parts
such that the sum of their reciprocals is 1/12. the numbers are
a) (24,26) b) (28,22) c) (27,23) d) (20,30)
let one number be
x the other number will be 50-x
1/x + 1/(50-x)= 1/12
50-x+x =
x(50-x)/12 600 = 50x-x^2 x^2-50x+600=0
x^2-30x-20x+600=0
x(x-30)-20(x-30)=0 (x-30)(x-20)=0 x= 20or 30 the numbers are 20 and 30
Q 241. John told mark that if mark
gives 1/3 rd of his money to him, he will have rs. 75. Mark told john that if
john gives half of hi money to him, he will have rs. 75. How much money did
they totally have?
a) 105 b) 125 c) 150 d) 75
Solution- let mark be y nd john be x
x+y/3=75 y+x/2=75 by solving x=60 y=45
total=105
Q 242. A city in the U S has a
basketball league with three basketball teams, the Arêtes, the Braves and the
Deities. A sports writer notices that the tallest player of the Arêtes is
Shortest than the shortest player of the Braves. The shortest of the Deities is
shortest of the Arêtes, while the tallest of the braves is taller than the
tallest of the Arêtes. Which of the following can be judged with certainty? X)
Paul, a brave is taller than David, an Arêtes.
Y) David, a Deities, is shorter
than Edward, an Arêtes.
a) X
only
b) Both
X and Y
c) Neither
X and Y
d) Y
only
Solution- its clearly written that
the tallest player of the Arêtes is Shortest than the shortest player of the
Braves.
=> all players of braves are
taller than all players of Arates.
=> X is true
=> ans = (a)
Q 243. Sum of the digits in the
equation (16^100)*(125^135) is
a) 2 b) 5 c) 3 d) 8
Solution- 16^100 * 125^135 =>
(2^4)^100 * (5^3)^135 => 2^400 * 5^405 = 2^400 * 5^400 * 5^5
= (2*5)^400 *3125
=10^400 * 3125
sum of 10^400= 1+0
+ 0+ 0+ ...... = 1 sum of 3125 = 11 therefore= 1 * 11 = 11 therefore = 1 + 1= 2
ans is 2
Q 244. what is the
average of the first 200 terms of the series 1,-2,3,-4,5,-6,7.... a) -0.5 b)
-50.5 c) 0.5 d) 50
Solution- average of the first 200
terms of the series 1,-2,3,-4,5,-6,7...
=1+(-2)+3+(-4)+...+199+(-200)/200
=(1-2)+(3-4)+(5-6)+...+(199-200)/200
=(-1)+(-1)+(-1)+...100times/200
=-100/200
=-0.5
Q 245. A sequence x1,x2,x3…. Is
said to be in harmonic progression if the reciprocals 1/x1,1/x2,1/x3… are in
arithmetic progression. The 5th and 7th term for harmonic progression are 30
and 50 respectively what is difference between 6th and 4th term ?
a) 16 b) 14.5 c)
13.4 d) 12.5
Solution- hp
series is
a,a/1+d,a/1+2d,.........
so applying this 5th term is 30=a/1+4d
and 7th term is 50=a/1+6d so from this two eq. we get a=50/3 and d=-1/9 putting this two for 4th
and 6th term we get 4th term =25 ..6th term=37.5 difference=12.5
Q 246. Each of A,B
and C need a certain unique time to do certain work. C needs 1 hour less than A
to complete the work. Working together they require 30 minutes to complete 50%
of the work. The work also gets completed if A and B start working together and
A leaves after 1 hour and B works further
3 hours. How much work does C do
per hour?
a) 16.66%
b) 66.66%
c) 50%
d) 33.33%
Solution- 50% suppose A do work in
x hrs ,B in y, then c would do in x-1 hrs..then 1/x+1/x-1+1/y=1 in 1 hrs..also
A&B 1 hrs work 1/x+1/y then work remaining xy-x-y/xy which is done by B in
3 hrs so (xy-x-y)/xy=3/y it results y=4x/x-1 putting value we get x=3 so y=6 so
A CAN DO WORK in 3 hrs B in 6 hrs & C IN 2 hrs that is 50%
Q 247. In an year N, the 320th day
of the year is Thursday. In the year N+1 the 206th day of the year is also
Thursday. What is the 168th day of In the year N-1?
a)
Friday
b)
Thursday
c)
Tuesday
d)
Saturday
Solution- riday..If the Nth year is
a non-leap year, then 320th day of year N to 206th day of year N+1 is [(365
– 320) + 206] = 251 days i.e. 35
weeks + 6 days. But this will not make the two days to be
Thursday.
Thus, the Nth year has to be a leap
year. In this case 320th day of year N to 206th day of year
N+1 is [(366 – 320) + 206) = 252 days
i.e. 36 weeks. Hence both days will be same day of the week i.e. Thursday as
given by the data.
168th day of N-1 year to 320th day
of N year is [(365 – 168) + 320) = 517 days i.e. 73 weeks +
6 days. Thus, if the 320th day is
Thursday, then the 168th day of year will be Friday, option
Q 248. What is the
remainder when 6^17+17^6 is divided by 7? a) 1
b) 6
c) 0
d) 3
Solution- 6^17+17^6 = (7-1)^17 +
(21-4)^6 = (7-1)^17 + (7*3 -4)^6....[eqn1]
if the [eqn1] is expanded then
every term of the expansion except [(-1)^17 + (-4)^6] will have 7 as one of its
factors.
Just think a little bit about the
binomial expansion of both [(7-1)^17] and [(7*3 -4)^6] ,then u can readily
point out that only the last term of both the expansions , i.e , [(-1)^17]
& [(-4)^6] respectively,will don't have 7 as one of its factors.
So, we have to
calculate the remainder when [(-1)^17 + (-4)^6] is divided by 7. Now, clearly
(-1)^17 = -1 and, (-4)^6 = 4^6 = 2^12 = (2^3)^4 = (7+1)^4.
Now, a same reasoning related to
binomial expansion mentioned previously explains why, when (7+1)^4 is divided
by 7 will leave a remainder 1.
So, (7+1)^4 will be of the for (7*A
+ 1); where A is some +ve integer, to know whose value isn't important in this
case.
So, when
[(-1)^17 + (-4)^6]
will be divided by 7 or, [-1 + (7+1)^4]
will be divided by 7 or, when [-1 + 7*A +1] will be divided by 7
or, when [ 7*A ] will be divided by
7 , clearly therefore the remainder will be zero, i.e, 0.
So, the answer is OPTION 3)0.
Q 249. George and Mark can paint
720 boxes in 20 days. Mark and Harry in 24 days and Harry and George in 15
days. George works for 4 days, Mark for 8 days and Harry for 8 days. The total
number of boxes painted by them is
a) 252 b) 516 c) 348 d) 492
Solution- Capacity of G + M = 720 /
20 = 36
M + H = 720 / 24 = 30
H + G = 720 / 15 = 48
Combined capacity = 2 (G + H + M) =
114
G + H + M = 114 / 2 = 57
Now capacity of G = (G+H+M) - (H +
M) = 57 - 30 = 27
M = (G+H+M) - (H + G) = 57 - 48 = 9
H = (G+H+M) - (G + M) = 57 - 36 =
21
Given that G
worked for 4 days, and mark for 8 and harry for 8 days So total work by them =
4 x 27 + 8 x 9 + 8 x 21 = 348
Q 250. Megha drives along the
perimeter of square field of side 10kms.She drives along the first side at
10kmph.along second side 20 kmph along 3rd side 30kmph and along the forth side
at 40 kmph.Her avearse speed is ?
a) 19.2kmph b) 18kmp c) 30kmph d)
20kmph
Solution- Average
speed : total distance/total time Total distance=10*4=40 km total time =10/40 +
10/20 + 10/30 + 10/40
So, Average Speed = 19.2 Km/h
Q 251. For real number x, int(x)
denotes integer part of x. int(x) is the largest integer less than or equal to
x.int(1,2)=1, int(-2,4)= -3. Find the value of
int(1/2)+int(1/2+100)+int(1/2+2/100)+....+int(1/2+99/100)
a) 50 b) 49 c) 51 d) 48
Soluiton- if it is int(1/2 + 100)
the answer is 150 lest if it is int(1/2 + 1/100) the answer is 50.
from int(1/2 + 49/100)= 0 and from
int(1/2 + 50/100)= 1 and int(1/2 + 99/100) = 1 so from 50 to 99 it is equal to
50
Q 252. 10 years ago 10 people age
was33. After 3 years a person of age 40 dies. After another 3 years anothe
person of 40years dies.After another 3 years another person of 27years dies.
Find the present average age?
a) 43 b) 44 c) 35 d) 40
Solution- 10 year ago 10 people= 33
10 year ago total
age=330 after 3 year 1 person with age 40 died = ie take his age as 37 before 3
years
lly for next 2 persons ; consider as
34 (40-6) and as 18(27-9) in 6 and 9 years ago( ie 37+34+18=89) 10 years ago
age of 7 people =330-89=241 nw consider
present age 7*10=70+241=311 nw avg= 311/7=44.43(ans)
Q 253. Raj invested in Indigo, HUL
and SBI shares at Rs. 300,
Rs. 200 and Rs. 5 per share, 100
shares for Rs. 10000. The number of Indigo and HUL shares he bought are
a) 15, 25 b) 23, 17
c) 17, 23 d) 17, 60
Solution- ans will be 17,23 coz..
left sbi share
will be 100-(17+23)=60 then indigo=300*17=5100 and HUL=200*23=4600 sbi=5*60=300
............
total=5100+4600+300=10000
Q 254. Raju can do a piece of work
in 10 days..vicky 12days,tinku 15 days..day all start the work together,but
raju leaves aftr 2 days,vicky leaves 3 days before the work is completed..how
many days work is completed?
a) 7 b) 5 c) 9 d) 6
Solution-
raju+vicky+tinku one day work = (1/10)+(1/12)+(1/15)= 1/4 now for 2 days
together work done= 2*(1/4)= 1/2
......now work done
by tinku in last 3 days alone=(3/15)=(1/5) remaining work= 1-(1/2 + 1/5)= 3/10
....now
vicky+tinku one day work=(1/12)+(1/15)=9/60 therefore 3/10 work will be done by
both of them in= (60/9)*(3/10)= 2 days
now answer= 2days(as got from
above; it is days required to do rem. work)+ 3days(when vicky
leaves)+2days(raju leaves)=7 days ans.
Q 255. Box of
fruits can be loaded into a truck in 9 minutes by a worker. And 8 boxes will
fill a truck completely. How many trucks can be loaded in 1 ½ hours if there
are 16 men together. a) 1
b) 2
c) 3
d) 0
Solution- 1 worker can in 9 mins
load 1 box
1 worker can in 90 mins load
(1*90/9)=10boxes
16 worker can in 90 mins load
(10*16)=160 boxes
So no. of trucks needed=160/8=20 i
think...
Q 256. the first 44 positive
integers are written in order to form the largest number
N=12345678910111213..424344. what will be the he remainder when N is diveded by
45?
a) 4 b) 9 c) 14 d) 18
Solution- no will
be div by 45 when it is divided y both 9 and 5 a no is div by 9 if sum of
digits is div by 9 now sum of digits of given no is 270 so the no is div by 9
and since last digit is 4 when div by 5 it leaves rem 4 let us say numbers such
as 9,54... we see they are div by 9 but leaves rem 4 when div by 5..
so when div by 45
they leave rem 9 so req ans=9
Q 257. The sum of 3 consecutive
numbers of the four numbers A, B, C, D are 4613,4961,5010,5099 then what is the
largest number among A,B,C,D ?
a) 1948 b) 1463 c) 1601 d) 1550
Solution- Let, S1,S2,S3,S4 be the
sums of A,B,C,D taking 3 of them at a time.
So, by the given data in the
problem,it follows that :
S1+S2+S3+S4 =
4613+4961+5010+5099
=>S1+S2+S3+S4 = 19683....[eqn 1]
Now, Without the loss of generality
we can assume
S1= A+B+C
S2= B+C+D
S3= C+D+A
S4= D+A+B
-------------------------------------------------------
adding the above terms we get,
S1+S2+S3+S4 = 3(A+B+C+D)
=>19683 = 3(A+B+C+D); [From [eqn
1] ]
=>(A+B+C+D)=(19683/3)
=>(A+B+C+D)=6561
So,the greatest term among them
=sum of all four - sum of lowest three terms
So,the greatest term among
them=6561-4613 = 1948.
So,the greatest term among them =
1948
So, the correct answer is : OPTION
a)1948
Q 258. Three cars A, B and C are
participating in a race.A is twice as likely as B to win and B is thrice as
likely as C to win. What is the probability that B will win, if only one of
them can win the race?
a) ½ b) 2/5 c) 3/10 d) 1/10
Solution- p(A)=2*p(B)
P(B)=3*p(c)
P(A) +P(B)+P(C) =1
2 * P(B) + P(B)+P(B)/3 =1
10*p(B) /3=1
P(B)=3/10
Q 259. 3 white chips, 7 blue chips,
16 green chips, 2 chips drawn from the box in succession what is the
probability that one is blue and other is white?
a) 7/50 b) 8/30 c) 7/25 d) 21/25*13
Solution- no of ways
that 2 chips selected from total of 26 is 26C2=26*25/2=13*25 no of ways that 1
white and 1 blue is selected is 3C1*7C1=21 total probability=21/13*25
Q 260. George is 2/3 rd as efficient as
smith and smith is ¾ th as efficient as John. George working alone is what
fraction of All of them working together. a) 2/3 b) 2/9 c) 4/9 d) 1/3 Solution-
g:s=2/3:1=2:3 s:j=3/4:1=3:4 g:s:j=2:3:4 so ans is 2/9
Q 261. A man can load one box in 9
minutes. A truck can contain 8 boxes. If 16 men load for one and a half hour,
how many trucks will be loaded?
a) 20 b) 10 c) 15 d) 40
Solution- 1 person in 9 min can
load 1 box
1 person in 1 min can load 1/9
boxes
16 person in 1 min can load 16/9
boxes
16 person in 90 min can load
16*90/9=160 boxes
now
8 boxes in 1 truck so
160 boxes comes
under 20 trucks so answer is 20 trucks
loaded
Q 262. In the simple subtraction problem
Below , sum single digits (not necessarily distinct) are replace by letters ,
find the value of 7*A + 7*B +6*C*D A 7 C 2
- 4 B 6 8
————–
5 4 3 D
a) 77 b) 95 c) 84 d) 70
Solution- A7C2
-4B68
_________
543D
So after solving the equation we
get A=9,B=2,C=0,D=4(because no are not distinct so it may be repeated).
so Ans of
7A+7B+6CD=7*7+7*2+6*0*4=77
Q 263. There are 5 distinct
integers a,b,c,d,e in ascending order.
(68-a)(68-b)(68-c)(68-d)(68-e)=725.
then what is the value of a+b+c+d?
a) 34 b) 136 c)
306 d) 238 Solution- 725 = -5*-1*1*5*29
so we can write,
68-e = -5
=> e = 73
68-d = -1
=> d = 69
68-c = 1
=> c = 67
68 - b = 5 => b
= 63
68-a= 29
=> a = 39
so a+b+c+d = 39+63+67+69 = 238
Q 264. A circle has 29 points
arranged in a clockwise manner numbered from 0 to 28, as shown in the figure
below. A bug moves clockwise around the circle according to the following rule.
If it is at a point i on the circle, it moves clockwise in 1 second by ( 1 + r
) places, where r is the reminder ( possibly 0 ) when i is divided by 11. Thus
if it is at position 5, it moves clockwise in one second by ( 1 + 5 ) places to
point 11. Similarly if it is at position 28 it moves ( 1 + 6 ) or 7 places to
point 6 in one second. If it starts at point 23, at what point will it be after
2012 seconds?
a) 1 b) 7 d) 15 d) 20
Solution- after 1st second :
(1+23%11 = 1) = 2 places [25] after 2nd second : (1+25%11 = 3) = 4 places [0] after
3rd second : (1+0%11 = 0) = 1 place [1] after 4th second : (1+1%11 = 1) = 2
places [3] after 5th second : (1+3%11 = 3) = 4 places [7]
after 6th second :
(1+7%11 = 7) = 8 places [15] after 7th second : (1+15%11 = 4) = 5 places [20]
after 8th second : (1+20%11 = 9) = 10 places [1]
now,for the same pattern from 4th
sec to 8th sec will repeat itself (5 sec intervals)..
total time = 2012
secs first 3 secs out of pattern...so time left 2012 - 3 =2009 secs
now no. of
repetitions in the leftover time = 2009/5 = 401....remainder = 4 for the next 4
iterations following the similar pattern
the position will be 20..
Q 265. Total income of Eesha in the
years 2003,2004,2005 was $36,400.Her income increased by 20% each year.What was
her income in 2005?
a) 14500 b) 14000 c) 14200 d) 14100
Solution- let in 2003 income=x.so
in 2004 it will be (x+x*20%)=6x/5.and in 2005 it will be(6x/5+6x/5*20%)=36x/25.
x+6x/5+36x/25=36400 then x=10000.in
2004 income=12000 and in 2005 income=14400
Q 266. Length, Breadth and Height
of a cuboid is in the ratio 1:3:27. Volume of the cuboid is 27 m3. If the
length is doubled and Breadth & Height are halved, then what is the change
in the volume of the cuboid?
a) Decreased by 15% b) Decreased by
18% c) Decreased by 30%
d) Decreased by 50%
Solution- given l:b:h=1:3:27
so now volume will
be lbh=1*3*27=81 now given l is doubled so l= 2l b and h are halved so
b=3b/2,h=27h/2 now new volume 2l*b/2*h/2 so substitute values then u get
lbh=2*(3/2)*(27/2)=40.5 so the change is 81-40.5=40.5 so half of the original
volume
Q 267. Two people, Ranbir and
Katrina decide to meet at a a beach between 1 pm to 2 pm, given that both will
surely turn up once in the given time frame. If Ranbir arrives, he waits for 15
minutes and then leaves feeling betrayed and similarly Katrina waits for 15
minutes after she arrives. So what's the probability that they meet?
a) 0.35 b) 0.45 c) 0.25 d) 0.15
Solution- waiting time of both
katrina and runbir is 15 min. Means they can meet only meet each other in 15
Min. , if they 1 more min. than they can't meet each other .. so chances of
meeting in only(Favourable event )=15 min
Total no. of min.(Sample space)= 60
min
Probability of Meeting is
=(Favourable event)/(Sample space)
15/60=0.25 ans.
Q 268. If (3a+6b)/(5a+12b)=12/23
determine the value of 3a2+5b2/ab
a) 19/2 b) 32/3 c) 9 d) 31/3
Solution- consider
(3a+6b)/(5a+12b)=12/23---->eq 1 sol eq--->1 the v vil get a/b=2/3 div 'ab
'to the num and den for 3a2+5b2/ab therfore, 3*(a/b)+5*(ba) sub a/b=2/3 in ab
eqtn v get 19/2
Q 269. A series of books was
published at 10years intervals when the 10th book was issued the sum of
publication years was 19,560when was the 1st book published
a)1910 b)1914 c)1911 d)1909
Solution- 19560=10/2[a+a+(10-1)*10]
19560=5[2a+90]
19560-450=10a a=19110/10 a=1911
Q 270. given the digits 1,3,6,9
find the probability that a 3 digit number formed by using them with no digit
repeated is divisible by 4
a) 1/12 b) 1/4 c) 1/6 d) none
Solution- divisibility rule for
four is last 2 digit must be divisible by 4. so last 2 digit must be 16,36,96.
(1) -- -- 1 6,(2) -- -- 3 6 , (3)
-- -- 9 6
mow you can put in 1 st case 9,3 yoy have 2*1 choice, similar
in 2nd and 3rd case
..so total number
of choice is =6 and total no. odf case = 4!=24 p(divisible 4)= 6/24=1/4
Q 271. The diagonal of a square is
twice the side of equilateral triangle the ratio of Area of the Triangle to the
Area of Square is?
a) √3:8 b) √2:5 c)
√3:6 d) √2:4 Solution- diagonal of a square is 2a area of the
square=1/2*(diagonal)^2 i.e 2a^2 side of the equi triangle is a area of equi
triangle=(3/4)a^2
3/4a^2:2a^2
3:8
Q 272. The number of different non
congruent triangles with integer side and perimeter 15 is
a) 9 b) 7 c) 10 d) 6
Solution- B)7, Let ,1st side of
triangle =a, 2nd side =b & 3rd side =c
A/C to law of triangle a+b>c so
(a+b) must be 8 or 9 or 10 or 11 or 12 or 13 or 14 but not be 15,
Q 273. The number of committes of
size 10 that could be formed from 10 men & 10 women such that committee has
atleast 6 women is
a)60626 b) 210 c) 10210 d) none
Solution- Option: (a)
10C6x10C4 + 10C7X10C3 + 10C8X10C2 +
10C9X10C1 + 10C10X10C0
= 44100+14400+2025+100+1
= 60626
Q 274. How many positive multiples
of 10 that are less than 1000 are the sum of 4 consecutive integers
a) 51 b) 50 c) 49 d) none
Solution- in the question what they
given is we have to find the sum of the 4 consecutive numbers that sum should
be multiple of like 10 ie..10,20.....and should be less than 1000. so that we
can conclude tat last 10 multiple before 1000 is 990 then w.k.t 1+2+3+4=10
6+7+8+9=30--->2
11+12+13+14=50
16+17+18+19=70
like wise it will be repeat,
based upon the
given we need to find upto 990 how many time it occurs is so that use this formula to find last number
in AP sn=a+(n-1)d we knw sn=990=>last number a=10=>initial number
d=20==>diff b/w two consecutive numbers
990=10+(n-1)20
990=10+20n-20
990=20n-10
20n=1000 n=50
Q 275. Inabhi’s class of 44
students,28 students speak malayalam,26 students speak tamil,9 students speak
none of the two languages.How many students speak both tamil& Malayalam
a) 54 b) 38 c) 19 d) 10
Solution- (44-9)=28+26-n(A$B)
35=54-n(A$B)
n(A$B)=19
Q 276. The sum of two numbers is
2016 and their product is 32 the sum of their reciprocals is
a) 63 b) 9 c) 32+
2014 d) 32- 2014 Solution- let two no are x,y x+y=2016 ......(1) xy=32
......(2)
therefore
1/x+1/y=x+y/xy=2016/32=63
Q 277. The sum of two numbers is
45. The sum of their quotient and is reciprocal is 2.05, the product of the
numbers is?
a) 450 b) 205 c)
400 d) 500 Solution- a + b = 45 a/b+b/a = 2.05
=>(a^2+b^2)/ab = 2.05
=>((a+b)^2−2ab)/ab=2.05
=>(a+b)^2 = 2.05ab + 2ab =
4.05ab
=> ab = 45^2 / 4.05 = 500
Q 278. Find the number of divisors
of 1728.?
a) 28 b) 20 c) 30 d) 18
Solution- 1728= 2^6 * 3^3
Hence the Number of factors = (6+1)
x (3+1) = 7 x 4 = 28.
We know that if a number represented
in standard form (a^m *b^n) , then the number of factors Is given by
(m+1)(n+1).
Answer is 28
Q 279. In 4 years, Rajs father will
be double Rajs age then. Two years ago, while his mother was twice his age that
time. If Raj is going to be 32 years old 8 years from now, then what is the sum
of his parents age now?
a) 96 b) 100 c) 102 d) 98
Solution- R.F=RAJS FATHER AGE
R=RAJS AGE
R.M=RAJS MOTHER AGE
(R.F+4)=2*(R+4)
(R.M-2)=2*(R-2)
R+8=32
therefore, R=24
therefore, R.M=46 therefore, R.F=52
R.F + R.M = 98
Q 280. when a no. divided by 406 it
gives 115 as a remainder. when it divided by 29 what is the remainder.???
a) 27 b) 7 c) 28 d) 19
Solution- let x is
divided by 406 & gives quotient y & remainder 115 then x = 406 * y +
115 now, x / 29 = (406 * y + 115)/29 =
406 * y/29 + 115/29
=> rem = 115/29 = 28
Q 281. Raj divided 50 into two
parts such that the sum of their reciprocal is 1/12, we get the parts as
a)
28, 22
b)
24, 36
c)
36, 14
d)
20, 30
Solution- (1/20)+(1/30)=5/60=1/12
so ans is (d) 20,30
Q 282. p/q - q/p
=21/10 then
4p/q + 4q/p = ?
a) 58/10 b) 113/10
c) 58/5 d) 121/110 Solution- 58/5 let us take p/q=x and q/p=1/x x-1/x=21/10 then
by solving we get a quadratic equation like 10x^2-10-21x=0
10x^2-25x+4x-10=0 then by taking
factors we get two values for x,x=-2/5 and x=5/2 by substituting x=5/2
4(5/2)+4(2/5)=58/5
Q 283. what is the
2015th no. of problemsolvingproblemsolving............................ a) g b)
p c) n d) o
Solution-
"problemsolving"it is a 14 letters word total no=2015th
"g" is repeat=2015/14=143
remainder=13
13th words is "N"
Q 284. In the sixth, seventh,
eighth, and ninth basketball games of the season, a player scored 23,14, 11,
and 20 points, respectively. Her points-per-game average was higher after nine
games than it was after the first five games. If her average after ten games
was greater than 18, what is the least number of points she could have scored
in the tenth game?
a) 29 b) 30 c) 28 d) 27
Solution- The sum of the scores for
games 6 through 9 is 68. The average in these four games is \frac{68}{4} = 17
The total points in all ten games
is greater than $10\cdot 18 = 180. Thus, it must be at least 181.
There are at least 181 - 68 = 113
points in the other six games: games 1-5and game 10.Games 1-5 must have an
average of less than 17. Thus we cannot put more than $16 + 17 + 17 + 17 + 17 =
84 points in those five games.
Thus, the tenth game must have at
least $113 - 84 = 29 points
Q 285. 2/3rd of the balls in a bag
are blue, the rest are pink. if 5/9th of the blue balls and 7/8th of the pink
balls are defective, find the total number of balls in the bag given that the
number of non defective balls is 146.
a) 216 b) 649 c) 432 d) 578
Solution- let the
total balls be x 2/3x=blue & 1/3x=pink non defective balls=146 total no. of
balls=defective+non defective hence defective=total-non defective(146)
(2/3x)*5/9+(1/3x)*7/8=t-146
0.66203x=t-146
146=0.3303x x=432
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