# Sample Aptitude Questions of Capgemini

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Aptitude questions form a part of the game-based section of Capgemini written test. These are based on the basic mathematical and reasoning concepts of puzzles, arithmetic operations, grids, etc. Aptitude section is a mix of moderate and slightly difficult questions. Here, we have given certain questions on fundamental topics, that will help you prepare for Capgemini aptitude test:
###### Capgemini Aptitude Questions
1. A 30% loss on cost price is what percent loss on selling price?
1. 30%
2. 20%
3. 15%
4. None of these
Let CP = 100 ; SP=70
Loss= 30/70 × 100 = 42.85%
2. A, B and C hire a taxi for Rs. 2400 for one day. A, B and C used the car for 6 hours, 8 hours and 10 hours respectively. How much did C pay?
1. Rs. 800
2. Rs. 1000
3. Rs. 600
4. Rs. 1200
Let total fair be = 2400 ;
Therefore c share =10/24 × 2400 = 1000
1. The ratio of investments of A and B is 8 : 7 and the ratio of their yearend profits is 20 : 21. If B invested for 12 months, then find the period of investment of A:
1. 6 months
2. 8 months
3. 10 months
4. 12 months
Let A invest for x months ; A = 8x months,
B = 7 × 12 = 84 months
8x/84 = 20/21
⇒ x = 10
1. What percent is 2 minutes 24 seconds of an hour?
1. 6%
2. 2%
3. 4%
4. 8%
%=144/60×60 = 4%
2. Evaluate: 3 cos 80° cosec 10° + 2 cos 59° cosec 31°
1. 1
2. 3
3. 2
4. 5
3 cos 80°. Cosec 10° + 2 cos 59° . cosec 31°
= 3 cos (90° - 10°). Cosec 10° + 2 cos (90° - 31°).Cosec 31°
=3sin10°.Cosec10° +2sin31°.cosec31°
=3+2=5
1. The total cost of 8 buckets and 5 mugs is Rs. 92 and the total cost of 5 buckets and 8 mugs is Rs. 77. Find the cost of 2 mugs and 3 buckets.
1. Rs. 35
2. Rs. 70
3. Rs. 30
4. Rs. 38
CP of 1 bucket = Rs. X
CP of 1 mug = Rs. Y
∴ 8x + 5y = 92....... (i)
5x + 8y = 77........(ii)
By equation (i) × 5 – equation (ii) × 8.
40x + 25y – 40x – 64y
= 460 – 616 ⇒ − 39y = - 156⇒ y = 4
From equation (i),
8x + 20 = 92 ⇒8x = 92 – 20 = 72 ⇒ x = 9
∴ CP of 2 mugs and 3 buckets
= 2 × 4 + 3 × 9 = 8 + 27 = Rs. 35
2. If 4x/3 + 2P = 12 for what value of P, x = 6?
1. 6
2. 4
3. 2
4. 1
When x = 6, (4 * 6)/3 + 2P = 12
⇒ 8 + 2P = 12
⇒ 2P = 12 – 8 = 4
⇒ P = 2
3. What number must be added to the expression 16a2 – 12a to make it a perfect square?
1. 9/4
2. 11/2
3. 13/2
4. 16
a2 - 2ab + b2 = (a-b)2
∴ 16a2 – 12a = (4a)2 - 2*4a*3/2
Hence, on adding (3/2)2 = 9/4, expression will be a perfect square.
4. The straight line 2x + 3y = 12 passes through:
1. 1st, 2nd and 3rd quadrant
2. 1st, 2nd and 4th quadrant
3. 2nd, 3rd and 4th quadrant
4. 1st, 3rd and 4th quadrant
The usual way to solve these type of questions is to put x = 0 once and find y coordinate. This would represent the point where the line cuts the Y axis.
Similarly put y = 0 once and find x coordinate. This would represent the point where the line cuts the X axis. Then join these points and you will get the graph of the line.
So when we put x = 0 we get y = 4.
When we put y = 0 we get x = 6.
So when we join these points we see that we get a line in 1st quadrant, which when extended both sides would go to 4th and 2nd quadrants. So option B.
5. In ΔABC, ∠A + ∠B = 65°, ∠B + ∠C = 140°, then find ∠B.
1. 40°
2. 25°
3. 35°
4. 20°