Infosys Placement paper 1 & 2

Infosys Placement paper 1:

1. 125 small but identical cubes are put together to form a large cube. This large cube is now
painted on all six faces.
(i) How many of the smaller cubes have no face painted at all.
(a) 27
(b) 64
(c) 8
(d) 36
(ii) How many of the smaller cubes have exactly three faces painted?
(a) 98
(b) 100
(c) 96
(d) 95
(iii) How many of the smaller cubes have atleast one side painted?
(a) 4
(b) 8
(c) 9
(d) 27
Sol:
Side of larger cube is 125−−−√3
= 5
I) No face painted will be in the interior part of the cube.
Interior part will be a cube of side (5 – 2) = 3.
Hence no. of cubes with no face painted ll be 33
= 27
Ans : (a) 27
II) Cubes with 3 faces painted will be the vertices of the cube.
There will be 8 such cubes
Ans : 8 [Wrong options… 3rd options should come here]
III) Atleast 1 face painted ⇒
greater than or equal to 1
Cube with 1 face painted + cube with 2 side painted + cube with 3 side painted
Cube with 1 face painted will be the outermost layer of larger cube but not on the edges.
i.e. (5–2)2
= 9 cubes on 1 side
So totally 6 ×
9 = 54 cubes
Cube with 2 face painted ll be edges of the larger cube but
(5 – 2) = 3.
Since a cube has 12 edges, totally 12×
3 = 36 cubes
Cube with 3 sides painted = 8 cubes
Totally 54 + 36 + 8 = 98 cubes
Ans : 98

2. A train 110 m long is running with a speed of 60 km/hr. At what time will it pass a man who is running at 6 km/hr in the direction opposite to that in which the train is going?

a)5 sec
b)6 sec
c)7 sec
d)10 sec
Answer: Option b

3. 7528 : 5306 :: 4673 : ?
a) 2367
b) 2451
c) 2531
d) 2489
Sol:
Answer is 2451.
As there is a difference of 2222.
7528 – 2222 = 5306.
So 4673 – 2222 = 2451
4. x2–y2=16
and xy
= 15 so find out x + y ?
Sol:
x2–y2
= 16
(x+y)(x–y)
= 16
So 16 comes in following table
1 ×
16, 2 × 8, 4 ×
4
Using 2 x 8 equation
x+y=8
and x–y=2
So x = 5 or 3 and y = 3 or 5
So answer is 8.
5. Census population of a district in 1981 was 4.54 Lakhs, while in year 2001 it was 7.44 Lakhs.
What was the estimated mid-year population of that district in year 2009.
Sol:
1981 ⇒
4.54
2001 ⇒
7.44
Difference ( year ) = 20
Difference ( population ) = 2.9
So population per year = 2.920
= 0.145
2009 ⇒
x = ?
Hence x = 7.44 + 8×0.145
=8.6 Lakhs
6. Based on the statement in the question, mark the most logical pair of statement that follow
“Either he will shout or they will fire”.
(1) He shouted.
(2) He did not shout.
(3) They fired
(4) They did not fire
(a) 1,4
(b) 2,3
(c) 4,1
Sol:
Either or condition is true atleast one of the condition should happen. Answer is option C
because according to the given sentence.
“Either he will shout or they will fire”
One of the two must happen whether he shouting or they firing.
If one of them happens, the other will not happen.
So if he did not shout then the firing should happen,so they fired.
If they did not fire it means the first thing has happened, so he shouted.
7. Gautham passes through seven lane to reach his school. He finds that YELLOW lane is
between his house and KAMA lane. The third lane from his school is APPLE lane. PEACOCK
lane is immediately before the PARK lane. He passes ASH lane at the end. KAMA lane is
between YELLOW lane and PEACOCK lane. The sixth lane from his house is RAO lane.
I. How many lane are there between KAMA lane and RAO lane ?
a) one
b) two
c) three
d) four
II. After passing the park lane how many lane does Gautham cross to reach the school ?
a) 4
b) 3
c) 2
d) 1
III. After passing the YELLOW lane how many lane does Gautham cross to reach the school ?
a) 4
b) 6
c) 2
d) 1
IV. Which lane is between PARK lane and RAO lane ?
a) YELLOW lane
b) KAMA lane
c) APPLE lane
d) PEACOCK lane
V. If the house of Gautham,each lane and his school are equidistant and he takes 2 min to pass
one lane then how long will he take to reach school from his house ?
a) 18 min
b) 16 min
c) 14 min
d) 12 min
Sol:
1. 3 Lanes between KAMA lane and RAO lane
2. Answer is 2 because after passing the PARK lane Gautham cross 3 lane to reach the school.
3. After passing the YELLOW lane Gautham cross 6 lane to reach the school.
4. APPLE lane
5. 16 minutes
8. Find the maximum value of n such that 50! is perfectly divisible by 2520^n .
Sol:
2520 = 23×32×5×7
Here 7 is the Highest prime So find the number of 7’s in 50! only.
Number of 7’s in 50! = [507]+[5072]
= 7+1 = 8
For n(max) = 8, 50! is perfectly divisible by 25208
.
9. Find the no of ways in which 6 toffees can be distributed over 5 different people namely
A,B,C,D,E.
Sol:
We assume that all the toffees are similar. Then Number of ways are (n+r−1)Cr−1
. Here A + B + C + D + E = 6
Here r = 5, n = 6
Number of ways = 6+5−1C5−1 = 10C4 = 210.
If all the toffees are different, then each toffee can be distributed to any of the five. So total ways
are 56
.
10. A train covered a distance at a uniform speed .if the train had been 6 km/hr faster it would
have been 4 hour less than schedule time and if the train were slower by 6 km/hr it would have
been 6 hrs more.find the distance.
Sol:
Let t be the usual time taken by the train to cover the distance
Let d be the distance, s be the usual speed
Usual time taken→
d/s = t => d =t×s
ds+6
= t – 4
t×ss+6
= t – 4
ts = ts + 6t – 4s – 24
6t – 4s – 24 = 0 →
(1)
d/(s – 6) = t + 6
ts = ts – 6t + 6s – 36
– 6t + 6s – 36=0 →
(2)
Solving (1) and (2), v get
s = 30 km/h
t = 24 hrs
d = t×s
d = 30×24
= 720 km
Ans : 720 km
11. A girl leaves from her home. She first walks 30 metres in North-west direction and then 30
metres in South-west direction. Next, she walks 30 metres in South-east direction. Finally, she
turns towards her house. In which direction is she moving?
Option
A) North-east
B) North-west
C) South-east
D) South-west
E) None of these
Sol:
A.North-east
12. There are two containers on a table. A and B. A is half full of wine, while B,which is twice
A’s size, is one quarter full of wine. Both containers are filled with water and the contents are
poured into a third container C. What portion of container C’s mixture is wine?
Sol:
Let d size of container A is “x”
then B’s size will be “2x”
A is half full of wine ⇒x2
So remaining “x2
” of A contains water
B is quarter full of win ⇒2×4⇒x2
So remaining ⇒2x–x2=3×2
3×2
of B contains water
Totally C has A’s content + B’s Content = x + 2x = 3x
Wine portion in C = x2
of “A” + x2
of “B”
x portion of wine
Water portion in C = x2
of “A” + 3×2
of “B”
⇒4×2⇒2x
portion of water
So portion of wine in C is
x3x=13
portion of wine
if 1/3 expressed in %
13×100
= 33.33%
Ans : 33.33% of wine
13. Four persons A,B,C,D were there. All were of different weights. All Four gave a
statement.Among the four statements only the person who is lightest in weight of all others gave
a true statement.
A Says : B is heavier than D.
B Says : A is heavier than C.
C Says : I am heavier than D.
D Says : C is heavier than B.
Find the lightest and List the persons in ascending order according to their weights ?
Sol:
A says B > D
B says A > C
C says C > D
D says C > B
Since the person with lightest weight tells the truth
C lies ( If C tells the truth, then C is not the lightest and then C lies )

D > C is the true statement.
So D is also not the lightest person and D lies.
B > C
So from A and B only one is telling the truth and that is not B because
B > C, so B is not the lightest
A is the lightest
Ans: A
14. There is well of depth 30 m and frog is at bottom of the well. He jumps 3 m in one day and
falls back 2 m in the same day. How many days will it take for the frog to come out of the well?
Sol:
28 days
Frog jumps 3 m in day & falls back 2 m at night
so,frog will be 3 – 2 = 1 m up in a day.
Thus, in 27 days it will be 27 m up
On 28th day it will be at top i.e 27 + 3 = 30 m & will not fall down.
15. Find the next term in the given series
47, 94, 71, 142, 119, 238, _ ?
a.331
b.360
c.320
d.340
Sol:
Ans : 215, 430
(47, 94) (71, 142) (119, 238) (X, Y)
47×2
= 94
94 – 23 = 71
71×2
=142
142 – 23 = 119
119×2
= 238
238 – 23 = 215
215×2
= 430
So the next 2 terms are 215 , 430
16. A train leaves Meerut at 5 a.m. and reaches Delhi at 9 a.m. Another train leaves Delhi at 7
a.m. and reaches Meerut at 10.30 a.m. At what time do the two trains travel in order to cross each
other ?
Sol:
Let the total distance be x
So the speed of 1st train is x/4 and 2nd train x/3.5
In 2 hours 1st train covers half of the total distance .
So remaining is only half of the total distance(ie x/2).
Let t be the time taken
t×x4+t×x3.5=x2
t = 1415
i.e. 56 min
i.e. Total time taken= 2 hrs + 56 min
Time they cross each other is 7:56 am (5+2.56)
Answer 7:56 am
17. ‘A’ and ‘B’ started a business in partnership investing Rs 20000/- and Rs 15000/-
respectively. After six months ‘C’ jointed them with Rs 20000/-. What will be B’s share in the
total profit of Rs 25000/- earned at the end of two years from the starting of the business?
Sol:
A:B:C = (20000×24):(15000×24):(20000×18)
= 4 : 3 : 3
B’s Share = 3×250004+3+3
= 7500
18. b,x,e,u,h,_?
Sol:
We know that a = 1,b = 2, ……….., z = 26
Convert the alphabets into numbers.we get number series as follows
2, 24, 5, 21, 8
In these (2,5,8) belong to one group as they have common difference of 3
(24,21,_?)these are of one group as they have difference of –3.
So the next number is 21 – 3=18.
If we convert 18 into alphabet it is “r”.
Since r = 18.
19. 3,5,11,29,83,245, _ ?
Sol:
We have to find the differences between the given numbers and then by applying that number
with 3 we can get the result
5 – 3 = 2
See here the result is 2,then multiply it with 3
11 – 5 = 6
29 – 11 = 18
83 – 29 = 54
245 – 83 = 162
731– 245 = 486
5 – 3 = 2
11 – 5 = 6 (2×3
)
29 – 11 = 18 (6×3
)
83 – 29 = 54 (18×3
)
245 – 83 = 162 (54×3
)
731 – 245 = 486 (162×3
)
20. A Jar contains 18 balls. 3 blue balls are removed from the jar and not replaced.Now the
probability of getting a blue ball is 1/5 then how many blue balls jar contains initially ?
Sol:
x/15 = 1/5
x = 3
3 + 3 (removed 3 blue balls) = 6

Infosys Placement paper 2

1. The hour hand lies between 3 and 4. Tthe difference between hour and minute hand is 50 degree.What
are the two possible timings?
Sol:
The angle between the hour hand and minute hand at a given time H:MM is given by
θ = 30×H – 211×MM
The time after H hours, hour hand and minute hand are at
MM = | 211×((30×H)±θ) |
given H = 3, MM = 50
Substituting the above values in the formula
θ = 8011, 28011
2. Jack and Jill went up and down a hill. They started from the bottom and Jack met Jill again 20 miles
from the top while returning. Jack completed the race 1 min a head of Jill. If the hill is 440 miles high and
their speed while down journey is 1.5 times the up journey. How long it took for the Jack to complete the
race ?
Sol:
Assume that height of the hill is 440 miles.
Let speed of Jack when going up = x miles/minute
and speed of Jill when going up = y miles/minute
Then speed of Jack when going down = 1.5x miles/minute
and speed of Jill wen going up = 1.5y miles/minute
Case 1 :
Jack met jill 20 miles from the top. So Jill travelled 440 – 20 = 420 miles.
Time taken for Jack to travel 440 miles up and 20 miles down = Time taken for Jill to travel 420 miles up
440x+201.5x=420y
681.5x=420y
68y = 63x
y = 63×68 —(1)
Case 2 : Time taken for Jack to travel 440 miles up and 440 miles down = Time taken for Jill to travel 440
miles up and 440 miles down – 1
440x+4401.5x=440y+4401.5y – 1
440×53(1y−1x)=1—–(2)
Substitute (2) in (1) we get
x = 440×5×53×63
t = 440×53(1x)
t = 12.6min
3. Data Sufficiency question:
A, B, C, D have to stand in a queue in descending order of their heights. Who stands first?
I. D was not the last, A was not the first.
II. The first is not C and B was not the tallest.
Sol:
D because A is not first neither C and B is not the tallest person. The only person will be first is D.
So option (C). We can answer this question using both the statements together.
4. One of the longest sides of the triangle is 20 m. The other side is 10 m. Area of the triangle is
80 m2. What is the another side of the triangle?
Sol:
If a,b,c are the three sides of the triangle.
Then formula for Area = (s(s–a)×(s–b)×(s–c))−−−−−−−−−−−−−−−−−−−−√
Where s = (a+b+c)2=12×(30+c)
[Assume a = 20 ,b = 10]
Now,
Check the options.
5. Data Sufficiency Question:
a and b are two positive numbers. How many of them are odd?
I. Multiplication of b with an odd number gives an even number.
II.a2 – b is even.
Sol:
From the 1st statement b is even, as when multiplied by odd it gives even
a2 – b = even
⇒ a is even
Here none of a and b are odd
6. Mr. T has a wrong weighing pan. One arm is lengthier than other. 1 kilogram on left balances 8
melons on right, 1 kilogram on right balances 2 melons on left. If all melons are equal in weight, what is
the weight of a single melon.
Sol:
Let additional weight on left arm be x.
Weight of melon be m
x + 1 = 8 x m – – – – – – (1)
x + 2 x m = 1 – – – – – – (2)
Solving 1 & 2 we get.
Weight of a single Melon = 200 gm.
7. a, b, b, c, c, c, d, d, d, d, . . . . . . Find the 288th letter of this series.
Sol:
Observe that each letter appeared once, twice, thrice …. They form an arithmetic progression. 1+2+3……
We know that sum of first n natural numbers = n(n+1)2
So n(n+1)2 ≤ 288
For n = 23, we get 276. So for n = 24, the given series crosses 288.
Ans is X
8. If ABC =C3 and CAB = D3, Then find D3÷B3
Sol:
ABC = C3
So, look for a number, that has a 3 digit cube, and the last digit of the cube is same as the number
itself: 53 = 125
So, CAB = 512 = 83
D = 8 and B = 2
83÷23
Answer = 64.
9. There are three trucks A, B, C. A loads 10 kg/min. B loads 13 1/3 kg/min. C unloads 5 kg/min. If three
simultaneously works then what is the time taken to load 2.4 tones?
Sol:
Work done in 1 min =10 + 403 – 5= 553 kg/min
For 1 kg = 3/55 min
For 2.4 tonnes = 3/55 x 2.4 x 1000 = 130 mins = 2hrs 10min
10. If A = x3y2 and B=xy3, then find the HCF of A, B
Sol:
A=x3×y2
B = x×y3
To find the HCF of the above numbers, take minimum power of x and y in both the numbers.
HCF = Common terms from both A & B and minimum powers = x×y2
11. HERE = COMES – SHE, (Assume s = 8)
Find value of R + H + O
Sol:
HERE = COMES – SHE
HERE
+ SHE
————
COMES
————
E + E = S = 8 => E = 4
3 digit no. + 4 digit no. = 5 digit no. ⇒ C = 1 ,O = 0, H = 9 etc
So 9454 + 894 = 10348
10348
– 894
——–
9454
——-
R + H + O = 5 + 9 + 0 = 14
12. A person is 80 years old in 490 and only 70 years old in 500 in which year is he born?
a) 400
b) 550
c) 570
d) 440
Sol:
He must have born in BC 570
Hence in BC 500 he will be 70 years
And in BC 490 he will be 80 years
13. Lucia is a wonderful grandmother and her age is between 50 and 70. Each of her sons have as many
sons as they have brothers. Their combined ages give Lucia’s present age.what is the age?
Sol:
The question basically states that if Lucia were to have say 10 sons, then each son would have 9 sons
(Lucia’s grandsons – since each son has 9 brothers). So the total in this case would be 9×10 grandsons +
10 sons = 100.
Let us assume Lucia has got x sons. Now each son has (x – 1) sons. So total = x + (x – 1) x. For x = 8 we
get 64 which is in between 50 and 60. ( 7 x 8 grandsons + 8 sons = 64 )
14. A family X went for a vacation. Unfortunately it rained for 13 days when they were there. But
whenever it rained in the mornings, they had clear afternoons and vice versa. In all they enjoyed 11
mornings and 12 afternoons. How many days did they stay there totally?
Sol:
Clearly 11 mornings and 12 afternoons = 23 half days
since 13 days raining means 13 half days.
so 23 – 13 =10 half days ( not affected by rain )
so 10 half days = 5 full days
Total no. of days = 13 + 5 = 18 days.
15. Find the unit digit of product of the prime number up to 50 .
Sol:
Prime number up to 50 are
2,3,5,7,11,…,43,47
Product = 2×3×5×7×11×−−−×43×47
There’s a term 2×5=10
So unit digit of product = 0
16. HOW + MUCH = POWER Then P + O + W + E + R =
Sol:
HOW
+ MUCH
————-
POWER
————–
Here p = 1 and M = 9 because after adding carry bit it gives result 10. Hence O = 0,here three digits 0,1,9
have been used.
Now, put all remaining value in 3rd column and check which value is suitable for H,U and W and we get
H = 7,U = 8 and W = 5 and 1 carry which will be added in 4th column.
Now in first column we have W + H = R means 5 + 7 = 2 and 1 carry will add in 2nd column
in 2nd column, 0 + C = E,0 + 3 + 1 = 4 so C = 3,E = 4
Therefore,
9837
+ 705
———
10542
———
so P + O + W + E + R = 1 + 0 + 5 + 4 + 2 = 12
17. Complete the series..
2 2 12 12 30 30 ?
Sol:
Answer is 56.
It follows the series as:
1 x 2 = 2
2 x 1 = 2
3 x 4 = 12
4 x 3 = 12
5 x 6 = 30
6 x 5 = 30
7 x 8 = 56
This is the required number for the series.

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Infosys Placement Paper Syllabus

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