INFOSYS WHOLE TEST PAPER 3
INFOSYS WHOLE TEST PAPER 3
1. 4, 6, 10, 14, 22, 26, 34, 38, 46, _ ? what is next term in
the series.
Sol:
Divide each number by 2. Then we get 2, 3, 5, 7, 11, 13,
......, 23. This is a prime number
series. So next number will be 2 x 29 =
58
2. y, _?, q, m, i
1. w
2. u
3. t
4. l
Sol:
Difference of 3 is in between two alphabets i + 3 = m
(j,k,l)
m + 3 = q (n,o,p)
q + 3 = u (r,s,t) u is the answer.
3. What is the next number in the series 3,7,13,19....
Sol:
Prime numbers from 3 on wards are 3, 5, 7, 11, 13, 17,
19, 23, 29 . . . Write alternate primes numbers starting from 3.
3, 7, 13, 19, 29
Answer is 29
4. Data Sufficiency Question: Is w a Whole number?
Statement
1: 3w is an Odd number. Statemet 2: 2w is an
Even number Sol:
Statement 2 is enough to solve
this
3w is Odd means w may be Odd or Fraction like 5/3 we
can not guess what w is from the first statement. 2w is Even.
So must and should w either Odd r Even i.e which is
whole number. No Fraction will give Even output.
5. Joe's age, Joe's sister's age and Joe’s fathers age sums up
to a century. When son is as old as his
father, Joe's
sister will be twice as old as now. When Joe is as old as his father then his
father is twice as old as when his sister was as old as her father.Age of her
father ?
Sol:
Joe + sister + father = 100
After x years let us consider
Joe's age is equal to his father
Joe + x = father
Therefore,
sister + x = 2 x sister
⇒
sister = x
Joe + sister = father
Therefore, 2 × father = 100 Hence, Father = 50
6. The sum of
series represented as
1/(1×5)+1/(5×9)+1/(9×13)+−−−−+1/(221×225)
is
a) 28/221
b) 56/221
c) 56/225
d) None of these Sol:
11×5
+ 15×9
+ 19×13
+ - - - - 1221×225 = 14×
[(5–1)1×5 + (9–5)5×9 + (13–9)9×13 + - - - - (225–221)221×225
]
= 14×[(1–15)+(15–19)+(19–113)+...(1221–1225)] = 14×(1–1225) = 14×224225 = 56225
7. What are the next three terms in the series 3, 6, 7, 12, 13,
18, 19, 24, _ _ _?
Sol:
This is a mixed series. 6, 12,
18, 24, . . . . form a series. Adding 1 to 6, 12, 18, forms another series. So
next three terms are 25, 30, 31.
8. What is the next number in the series. a, b, d, h, _?
Sol: a = 1 b = 2 d = 4
h = 8 This is a 2n series starting with n = 0, 1, 2, ...
24
= 16 which is p. Ans = p
9. Find the letter that comes in the place of
" - " b, _, d, d, e, d, f, g, d.
Sol:
[b c] d [d e] d [f g] d [h i]
d....
The series follow above
manner. Answer will be c.
10. The number of
zeros at the end of the product of all prime numbers between 1 and 1111 is?
Sol:
Prime numbers between 1 &
1111 are 2,3,5,7,11,...
There is no other prime no.
ending with 5 as unit digit, except one '5'
2 x 5 = 10 gives only one zero in
the product of all prime numbers
So, number of zeros at the end of
the product = 1
11.
A train goes from
stations A to B. One day there is a technical problem at the very beginning of
the journey & hence the train travels at 3/5 of it's original speed and so
it arrives 2 hours late. Had the problem occurred after 50 miles had been
covered, the train would have arrived 40 min earlier(i.e., only 120-40 = 80 min
late). What is the distance between the 2 stations?
Sol:
For 1 mile the train is late by
40 / 50 min or 4/5 minutes. Or it is
late by 1 minute for every 5/4 miles. For 120 minutes late it has to travel 120
x 5/4 = 150 miles.
12.
Due to some defect in
our elevator, I was climbing down the staircase. I’d climbed down just 7 steps when I saw a
man on the ground floor. Continuing to walk down, I greeted the man and I was
surprised to see that when I was yet to get down 4 steps to reach the ground
floor, the man had already finished climbing the staircase. He perhaps climbed
up 2 steps for every 1 of mine.
How many steps
did the staircase have?
Sol:
Let us consider x be the number
of steps
7 + x + 4 = 2x
As old man takes 2 steps for every one steps he takes
and he has to complete 4 steps, So x = 11 and total steps = 2x = 22
39. A card board of size 34 ×
14 has to be attached to a wooden
box and a total of 35 pins are to be used on the each side of the card box.
find the total number of pins used.
Sol:
Total 35 pins are there and 4
sides of card board.
So 35 x 4 = 140
Now in the rectangle 4 vertices have 4 pins which is
common to the sides. So 140 – 4 = 136.
13. In the Garbar
Jhala, Ahmadabad a shopkeeper first raises the price of Jewellery by x% then he
decreases the new price by x%. After one such up down cycle, the price of a
Jewellery decreased by Rs. 21025. After a second updown cycle the jewellery was
sold for Rs. 484416.
What was the
original price of the jewellery.
Sol:
Let the original price be
"p":
I cycle:
Up by x% means new price is p
+ px100
Down by x% on current price means
new price is (p
+ px100) – (p
+ px100)
×x100
Price after one up down cycle is
(p – 21025)
Thus, (p
+ px100) – (p
+ px100)
×x100
= $(p – 21025) = p' - - - - - (1) II cycle:
Up by x% means new price is p′
+ p′x100
Down by x% on current price means
new price is
(p′
+ p′x100)
– (p′ + p′x100)
×x100
Price after second up down cycle
is 484416.
Thus,(p′
+ p′x100)
– (p′ + p′x100)
×x100
= 484416 - - - - - - -
(2)
Putting value of p' = p – 20125
in equation (2) and dividing (1) & (2) to eliminate x.
We get a quadratic equation in p:
p2–526466p–(21025)2=0
The equation has real roots in
the form 525625, 841.
14. Three
football teams are there. Given below is the group table. Fill in the x's
P - Played
W - Won
L - Lost
D - Draw
F - Goals For
A - Goals Against
P
W L D
F A
A 2 2
x x x 1
B 2 x x
1 2 4
C 2 x
x x 3 7
Sol:
P W L D F A
A
2 2 0 0 7 1
B
2 0 1 1 2 4
C
2 0 1 1 3 7
Total goals for = Total goals
against
1 + 4 + 7 = 3 + 2 + x
x = 7
A has
played two and won 2 therefore lost = 0, draw = 0
B has
played 2 but one is draw as A has 0 draw, it should be against C i.e C draw = 1
C played 2 draw = 1 therefore lost = 1 because A has won both matches played
against them. So we can conclude that each team has played a match with every
other team.
15.
A dog takes 4 leaps for
every 5 leaps of hare but 3 leaps of dog is equal to 4 leaps of hare compare
speed?
Sol:
Dog and hare speeds according to
the number of leaps = 4 : 5
But their leap lengths are in the
ratio = 4 : 3 (3 x D = 4 x H )
Multiplying number of leaps and
leap lengths we get their speeds as = 4 x 4 : 5 x 3 = 16 : 15 Answer = 16 : 15
16.
A bird keeper has got P
pigeons, M mynas and S sparrows. The
keeper goes for lunch leaving his assistant to watch the birds. Suppose p = 10, m = 5, s = 8 when the bird
keeper comes back, the assistant informs the x birds have escaped. The bird
keeper exclaims: "Oh no! All my sparrows are gone."
How many birds
flew away?
When the bird
keeper comes back, the assistant told him that x birds have escaped. The keeper
realized that atleast 2 sparrows have escaped.
What is minimum
no of birds that can escape?
Sol:
This question can be solved using
the pigeonhole principle.
I guess the answer for the first
question is 23 (10 + 5 + 8 ).
Since if all the birds are
escaped, then only he can be sure that all sparrows are gone.
And for the second one, answer is
17 (10 p + 5 m + 2 s ).
If 17 birds escaped then best
case such that least number of sparrows escaped will be like 10 pigeon, 5 myna
and 2 sparrows escaped.
17.
3,4,7,10,13,16,19,22, . . . Find 10th
term in series Sol:
3
3×1 = 3 + 1 = 4
3×2 = 6 + 1 = 7
3×3 = 9 + 1 = 10 3×4
= 12 + 1 = 13
3×5
= 15 + 1 = 16
3×6 3×7 3×8 3×9
= 18 + 1 = 19
= 21 + 1 = 22
= 24 + 1 = 25
= 27 + 1 = 28
10th term = 28
18. a,d,i,p,?
what is next term
a) q
b) r
c) s
d) t Sol: a = 1×1 d = 2×2 i = 3×3
p = 4×4
Next will be 5×5
= 25 = Y
19. Marbles are
to be distributed. Ann gets 1,Mary gets 2, Rose gets 3 and Lisa gets 4. John
Brown gets as
much as his sister.Tim Smith gets 2 times as much as his sister. Neil Johnson
gets 3 times as much as his sister. Sam Paul gets 4 times as much as his
sister.Find the surnames of Ann,Mary,Rose and Lisa ?
Sol:
Ann's brother is Neil John (1×3=3
).
Mary 's brother Sam paul ( 2×4=8
).
Rose's brother John Brown (3×1=3
).
Lisa's Brother is Tim Smith (4
\times 2 = 8$).
1 + 2 + 3 + 4 + 3 + 8 + 3 + 8 =
32
20. A shop
has 4 shelf, 3 wardrobes, 2 chairs and 7 tables for sell. You have to buy a. 1 shelf
b. 1 wardrobe
c. either 1 chair or 1 table How
many selection can be made?
Sol:
The way to answer this question
4C1×3C1×2C1+4C1×3C1×7C1
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