capegemini interview questions and answers paper 410 - skillgunNote: Paper virtual numbers may be different from actual paper numbers . In the page numbers section website displaying virtual numbers .
Find the probability that a number selected at random from the set of two digit numbers will not be a multiple of 12 and not a multiple of 4?
The two digit numbers are 10,11,12,...,99
There are 90 two digit numbers.
Since the set of multiples of 4 contains multiple of 12 then it's enough to find the number of multiples of 4.
The first two digit number that is divisible by 4 is 12 and every 4th number from 13 is multiple of 4.
Here, 22 (integral value of 90/4) numbers of multiples of 4 & 12 together.
Then the number of two-digit non-multiple of 12 and non-multiple of 4 is 90-22 = 68.
Hence the required ratio is 68/90 = 34/45.
Every day, a man spends 1/4 of his time to work, 3/8 of his time to sleep and the rest of the time in family. Then, during a week how many hours did he spend with his family?
Time taken to work and sleep = 1/4 + 3/8 = 5/8 part
Then the remaining part = 1 - 5/8 = 3/8
Remaining part of job 3/8 spend with his family.
1 part of a day means he takes 24 hour
Then 3/8 part of a day takes = 3 x 24/8 = 9 hours
And, for a week he spends (9 x 7) = 63 hours with his family.
Assume that 1/6, 1/10 and 1/3 part of a human body's weight is made of bones, skins and muscles respectively. Also assume that the remaining weight is due to water and fluids. If a man's weight is 60 kg find the weight of water and fluids in his body.
Given that, 1/6 part made of bones,
1/10 part made of skin and 1/3 part made of muscles.
Then the remaining part (i.e., weight of water and fluids) = 1 - (1/6 + 1/10 + 1/3)
1 - (10 + 6 + 20)/60 = 1 - 36/60 = 2/5
Given that the man's weight is 60 kg.
Then water and fluid weight = 2 x 60/5 = 24 kg.
Hence the required answer is 24 kg.
A college student spends 2/5 of his time for maths, 3/10 of his time for science and 1/8 of his study time for arts. If he spends remaining 30 minutes for playing with friends then find the total time taken by him to learn science and arts.
2 1/7 hours
3 4/9 hours
1 3/10 hours
Part of the time left (remaining time) = 1 - (2/5 + 3/10 + 1/8)
1 - 33/40 = 7/40 Assume that the total time taken by him to study all subjects be x hours. We have just found that he had 7/40 remaining time after studying maths, science and arts.
From the question, we can say that he spends this remaining 7/40 of his time for playing and this time equals 30 minutes.
Then 7/40 of x = 1/2 hour
7x / 40 = 1/2
x = 20/7 hours.
Time taken to learn science = 3/10 of x = 3/10 x 20/7 = 6/7 hours
Time taken to learn arts = 1/8 of X = 1/8 x 20/7 = 5/14 hours.
Time required to learn science and arts = (1) + (2) = 6/7 + 5/14 = 17/14 hours = 1 3/14 hours.
Find the speed of boat in still water if it takes 5 hours to cover a certain distance upstream and takes just 2 and half hours to cover the same distance downstream. The rate of stream is 10 kmph.
If the speed downstream is a km/hr and the speed upstream is b km/hr, then:
Speed in still water = (a + b)/2 km/hr.
Rate of stream = (a - b)/2 km/hr.
Let boat's upstream speed be x km/hr and downstream be y km/hr
Distance covered in upstream for 5 hours = Distance covered in downstream for 2 1/2 or 5/2 hours.
By using distance = speed x time, we get
x * 5 = y * 5/2
2(5x) = 5y
2x = y
y - 2x = 0
Given that, the speed of stream is 10 km/hr
i.e., (y-x)/2 = 10
y - x = 20 ...eqn 2
Solving the above eqn, we get
x = 20 and y = 40
Speed of Boat in still water = (x + y)/2
(20 + 40)/2 = 30.
The rate of current is 5km/hr and the speed of boat in still water is 19km/hr. The ratio between the distance travelled downstream in 15 minutes to the distance travelled upstream in 30 minutes is:
If the speed of boat in still water is u km/hr and the speed of the stream is v km/hr, then:
Speed downstream = (u + v) km/hr.
Speed upstream = (u - v) km/hr.
Given that the speed of boat in still water is 19 km/hr and the rate of current is 5 km/hr.
Downstream speed = (19 + 5) km/hr = 24km/hr
Distance travelled downstream in 15 minutes with a speed of 24km/hr = 24 x (15/60) = 6 km
Upstream speed = (19 - 5) km/hr = 14km/hr
Distance travelled upstream in 30 minutes with a speed of 14km/hr = 14 x(30/60) = 7 km
Then the required ratio = 6:7 = 3 : 3.5
Hence the answer is 3 : 3.5
A man takes certain time to row 8 km downstream. But when he swims upstream for the same time, he can cover 6 km. If he rows a distance of 96 km upstream and comes back in 28 hours, then find the rate of the stream ?
Assume x hours be the time taken to move 8 km downstream. then x will be the time taken to cover 6 km upstream.
Total distance travelled = 96 km
Downstream speed = 8/x km/hr
Time taken to row 96 km downstream = 96 / (8/x) = 12x hrs (time = speed / distance )
Upstream Speed = 6/x km/hr
Time taken to row 96 km upstream = 96 / (6/x) = 16x hrs
Time taken to return back is 28 hrs
Total Time taken = Time taken to row downstream + Time taken to row upstream
28 = 12x + 16x
28 = 28x
x = 1
Substituting x = 1 we get,
Downstream Speed = 8 km/hr and Upstream Speed = 6 Km/hr.
Rate of stream = 8 - 6 / 2 = 1 Km/hr
Two bag contains different kinds of rice. The first bag contains 40% of worst quality and the rest good quality. The second bag contains 60% of worst quality. How much rice should be mixed from each of the container so as to get 63 kg of rice such that the ratio of worst quality to good quality is 3:4 ?
33 kg,30 kg
50 kg,13 kg
54 kg,9 kg
40 kg,23 kg
How many litres of milk worth Rs.12 per litre must be mixed with 60 litres of milk worth Rs.8 per litre so as to gain 20% by selling the mixture at Rs.10.80 per litre?
S.P of 1 litre of mixture = Rs.10.80 and Gain 20%.
Therefore, C.P of 1 litre of mixture = Rs.100/120 x 10.80 = Rs.9 ...(1)
C.P of 1 litre milk of 1st kind = d = Rs.12
C.p of 1 litre milk of 2nd kind = c = Rs.8
Mean Price = m = Rs.9 (from equation 1)
Then d - m = 3 , m - c = 1
Applying above values to our allegations formula (refer introduction before question)
Ratio of quantities of 1st and 2nd kind = 1:3 .Assume x litres of milk of 1st be mixed with 60litres of 2nd kind.
Then, 1 : 3 = x : 60
1/3 = x/60
x = 60/3 = 20.
Thus 20 litres milk of 1st kind to be mixed with 2nd.
A Milk can labelled P is filled with milk and water is mixed in the ratio 11:4 and another labelled Q has mixture in ratio 3:2. In what ratio these two mixtures be mixed together to get a new one labelled R containing milk and water in the ratio 2:1?
The c.p of milk be Re.1 per litre. ...(A)
Now, milk in 1 litre mixture of P = 11/15 ...(1)
Then c.p of 1 litre mixture of P = (11/15) x 1 = Re.11/15 (Based on value in previous equation and our assumption A)
Milk in 1 litre mixture of Q = 3/5 ...(2)
Then c.p of 1 litre mixture of Q = (3/5) x 1 = Re.3/5.As given in question, milk and water in final mixture are in the ratio 2:1. So Milk in 1 litre mixture of R = 2/(1 + 2) x 1 = 2/3. ...(3)
Then the c.p of 1 litre mixture of R = Re.2/3
Here, d = 11/15, c = 3/5 and m = 2/3
Then d - m = 11/15 - 2/3 = (11-10)/15 = 1/15
and m - c = 2/3 - 3/5 = (10-9)/15 = 1/15
Required ratio (as per our formula) = d-m : m-c = 1/15 : 1/15 = 1:1
15625 6250 2500 1000 ? 160
80 370 ? 1550 2440 3530
Here the difference between the two numbers are more than 200.i.e 290,490,690,890 and 1090
Among the following which is not an appropriate order of delivered reports?
M, A, N, G, O, L, D
M, D, N, G, L, O, A
M, N, A, L, D, O, G
M, N, A, O, D, L, G
Here Order is like Each of the first mates delivered their report exactly after his or her captain.
In case L speaks after A, and A is the third of the first mates to speak, then among the following statements which would be untrue?
O spoke immediately after G.
The order of the first four speakers was M, G, N, D.
O's first mate was present.
A was the fourth speaker after M.
Here 4th option because if A was the fourth speaker the given Question Condition is wrong
Among the following statements which statement must be true?
In case the second speaker was a captain, the seventh speaker was a first mate.
In case the second speaker was a first mate, the seventh speaker was a captain.
In case the third speaker was a first mate, the seventh speaker was a captain
In case the third speaker was a captain, the seventh speaker was a first mate.
In case A spoke immediately after L and immediately before O, and O was not the last speaker, L spoke
Using given information L Spoke fourth
Statements : All puppets are dolls
All dolls are toys
Conclusions : I. Some toys are puppets
II. All toys are puppets
if only conclusion I follows
if only conclusion II follows;
if either I or II follows
if neither I or II follows
By using The information some toys are puppets
Statements : All apples are oranges
Some oranges are papayas
Conclusions : I. Some apples are papayas
II. Some papayas are apples
if either I or II follows
We are not getting the answer
Statements :Some players are singers
All singers are tall
Conclusions : I. Some players are tall
II. All players are tall
By using I we get the answer
Statements : All coins are crows
Some crows are pens
Conclusions : I. No pen is coin
II. Some coins are pens
By using the given iformation we cannot find the answer
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