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Placement Papers with Solutions

(#M40000618) TCS QUESTION Keep an EYE

In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
option
(a) 1 hour
(b) 2 hours
(c) 3 hours
(d) 4 hours

Asked In TCS Sumandeep Ghosh (14 years ago)

Solved mohd anas Read Solution (2)

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(#M40000612) TCS QUESTION Keep an EYE

A Roman was born the first day of the 35th year before Christ and died the first day of the 35th year after Christ. How many years did he live?
option
(a) 70
(b) 69
(c) 71
(d) 72

Solution

A Roman was born the first day of the 35th year before Christ means before the start of year 0, 35 years has been past and died the first day of the 35th year after Christ means 35th year is about to start at his death so only 34 years has been completed so total years = 69


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Asked In TCS Sumandeep Ghosh (14 years ago)

Solved Read Solution (2)

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(#M40000606) TCS QUESTION Keep an EYE

Leena cut small cubes of 10 cm dimension each. She joined it to make a cuboid of length 100 cm, width 50 cm and depth 50 cm. How many more cubes does she need to make a perfect cube?
a) 500
b) 250
c) 750
d) 650

Solution

volume of given cube = 100*50*50 = 250000
required volume = 100*100*100 = 1000000
so extra volume required = 750000
so cubes required = 750000/10*10*10 = 750

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Asked In TCS Sumandeep Ghosh (14 years ago)

Solved Read Solution (4)

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(#M40000605) TCS QUESTION Keep an EYE

1! + 2! +......... + 50!=?
a)3.1035*10^64
b)2.1021*10^65
c)3.1035*10^63
d)3.1035*10^62

Solution

solve 50! + 49! + 48!
others are lower terms
this can be done by calculator only

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Asked In TCS Sumandeep Ghosh (14 years ago)

Solved SANTOSH AGRASINGH Read Solution (2)

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(#M40000604) TCS QUESTION Keep an EYE

Given a collection of points P in the plane , a 1-set is a point in P that can be separated from the rest by a line, .i.e the point lies on one side of the line while the others lie on the other side. The number of 1-sets of P is denoted by n1(P). The minimum value of n1(P) over all configurations P of 5 points in the plane in general position (.i.e no three points in P lie on a line) is
a) 3
b) 5
c) 2
d) 8

Solution

The meaning of this question is little complicated. In easy language the question should be:


There are some points in the plane. We have to make a separation between these points by forming a line in such a way that one point could be separate from the others. No 3 points are in a line. Maximum no of methods for plotting the line and minimum no of methods for plotting the line will be (arrangement of points is your choice)

Now Answer Is:

For Maximize: The number of methods, Points should be drawn in the circumference of the CIRCLE So answer would be the number of total point in the plane.

For minimize: The number of methods we should draw 3 points in a TRIANGLE and all other point inside this triangle in such a way that no 3 points could be in a line.

If it will be allowed to draw the points in a line then minimum possibilities will be 2 only, because if we take all points in a line then you can separate only the corner points from the others.

For example
If we have 10 points then maximum possibilities = 10
and minimum = 3

similarly for 5 points
Max = 5, Min = 3

for 19
max= 19 Min = 3


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Asked In TCS Sumandeep Ghosh (14 years ago)

Solved Read Solution (5)

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(#M40000603) TCS QUESTION Keep an EYE

Six friends decide to share a big cake. Since all of them like the cake, they begin quarreling who gets to first cut and have a piece of the cake. One friend suggests that they have a blindfold friend choose from well shuffled set of cards numbered one to six. You check and find that this method works as it should simulating a fair throw of a die. You check by performing multiple simultaneous trials of picking the cards blindfold and throwing a die. You note that the number shown by the method of picking up a card and throwing a real world die, sums to a number between 2 and 12. Which total would be likely to appear more often 8,9 or 10?
a) 8
b) All are equally likely
c) 9
d) 10

Solution

the best solution will be 7, and there will be total 6 cases for this choice (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
but this one is not in the choices so out of the 4 choices givem 8 will be the best choice. For this choice there are 5 methods (2,6), (3,5), (4,4), (5,3), (6,2)
similarly for 9 there are 4 choices (3,6), (4,5), (5,4), (6,3)
for 10 there are 3 choices (4,6), (5,5), (4,6)
Answer is (a)

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Asked In TCS Sumandeep Ghosh (14 years ago)

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(#M40000602) TCS QUESTION Keep an EYE

There is a bacteria which has the probability of die 1/3 of its total number or it may tripled. Find out the probability
A. P=1/3+(2/3*p^3)
B. P=2/3+(2/3*p^3)
C. P=2/3+(1/3*p^3)
D P=2/3+(2/3*p^3)

Asked In TCS Sumandeep Ghosh (14 years ago)

Solved khushboo mishra Read Solution (10)

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(#M40000601) TCS QUESTION Keep an EYE

The bacteria has the probability of split into 3 and probability to die is 1/3rd of the total bacteria.Let the probability is P.Some of them survived with probability 1/5.Then which among the following relation is true?
a)P=1/3+1/5*3
b)P=1/5*(1/8-3)

Asked In TCS Sumandeep Ghosh (14 years ago)

Solved marudhupandian .p Read Solution (10)

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(#M40000600) TCS QUESTION Keep an EYE

A man goes north 37km.turns left goes 2km.turns right goes 17km.turns right goes 2km. find distance b/w starting ending point.
a) 54
b) 27
c) 81
d) 67

Solution

Since man moves 2 km east and 2 km west so distance between initial and final points will be (37+17)km = 54km

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Asked In TCS Sumandeep Ghosh (14 years ago)

Solved Pradeepthi Read Solution (8)

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(#M40000599) TCS QUESTION Keep an EYE

In T.Nagar the building were numbered from 1 to 100.Then how many 0’s will be present in the numbers?
a) 18
b) 19
c) 20
d) 11

Solution

4, 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 64, 74, 84, 94
total 20 times 4 is there


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Asked In TCS Sumandeep Ghosh (14 years ago)

Solved Read Solution (7)

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