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# IBM Placement Paper | Quantitative Analysis Set – 2

This is an IBM model paper for Quantitative Aptitude. This placement paper will cover aptitude that is asked in IBM placements and also strictly follows the pattern of questions asked in IBM papers. It is recommended to solve each one of the following questions to increase your chances of clearing the IBM placement.

1. If the average of 5 consecutive even numbers is 10, find the middle number?
1. 4
2. 8
3. 2
4. 10

```
10
```

Explanation:

Let the five consecutive even numbers be
x-4, x-2, x, x+2, x+4

Average of these = ((x-4)+(x-2)+(x)+(x+2)+(x+4))/5 = x
Given average = 10

Therefore x = 10

2. Two numbers are in the ratio of 2:9. If their H. C. F. is 19, numbers are:
1. 6, 27
2. 8, 36
3. 38, 171
4. 20, 90

```
38, 171
```

Explanation:

Let the numbers be 2X and 9X
Then their H.C.F. is X, so X = 19
=> Numbers are (2×19 and 9×19) i.e. 38 and 171

3. Three friends started running together on a circular track at 8:00:00 am. Time taken by them to complete one round of the track is 15 min, 20 min, 30 min respectively. If they run continuously without any halts, then at what time will they meet again at the starting point for the fourth time ?
1. 8:30:00 am
2. 9:00:00 pm
3. 12:00:00 pm
4. 12:00:00 am

```
12:00:00 pm
```

Explanation:

LCM (15, 20, 30) = 60
=> They meet at the starting point after every 60 min, i.e., after every 1 hour.
Therefore, they will meet at the starting point for the fourth time after 4 hours, i.e., at 12:00:00 pm.

4. Two friends A and B were employed to do work. The initial deadline was fixed at 24 days. Both started working together but after 20 days, A left the work and the whole work took 30 days to complete. In how much time can B alone can do the work?
1. 40
2. 50
3. 60
4. 70

```
60
```

Explanation:

Let the total work be 24 units. It is given that A and B together can do the work in 24 days.
=> Combined efficiency of A and B = 24/24 = 1 unit / day
=> Work done in 20 days = 20 units
=> Work left = 24 – 20 = 4 units
Now, this remaining 4 units of work was done by B alone in 10 days.
=> Efficiency of B = 4/10 = 0.4
Therefore, time required by B alone to do the work = 24/0.4 = 60 days

5. Two pipes A and B attached to a swimming pool can fill the pool in 20 minutes and 30 minutes respectively working alone. Both were opened together but due to malfunctioning of motor of pipe A, it had to be shut down after two minutes but B continued to work till the swimming pool was filled completely. Find the total time taken to fill the pool.
1. 20
2. 22
3. 25
4. 27

```
27
```

Explanation:

Let the capacity of the pool be LCM(20, 30) = 60 units.
=> Efficiency of pipe A = 60 / 20 = 3 units / minute
=> Efficiency of pipe B = 60 / 30 = 2 units / minute
=> Combined efficiency of pipe A and pipe B = 5 units / minute
Now, the pool is filled with the efficiency of 5 units / minute for two minutes.
=> Pool filled in two minutes = 10 units
=> Pool still empty = 60 – 10 = 50 units
This 50 units is filled by B alone.
=> Time required to fill these 50 units = 50 / 2 = 25 minutes
Â
Therefore, total time required to fill the pool = 2 + 25 = 27 minutes

6. Samuel covers the distance from his home to his office at a speed of 25 km/hr and comes back at a speed of 4 km/hr. He completes the whole journey within 5 hours 48 minutes. Find out the distance from his home to office:
1. 20 km
2. 18 km
3. 15 km
4. 25 km

```
20 km
```

Explanation:

Let the speed of travelling to office and back to home be x and y respectively.
So, his average speed is = 2xy / (x+y) = (2Â Ã— 25 Ã— 4) / (25 + 4) = 200/29 km/hr
He covers the whole journey in 5 hours 48 minutes = 5=> = 29/5 hrs
Therefore, total distance covered = (200/29Â Ã— 29/5) = 40 km
So, the distance from his home to office = 40/2 = 20 km

7. A boatman takes 3 hours 45 minutes to travel 15 km downstream and takes 2 hours 30 minutes to travel 5 km upstream of a river. What is the speed of the stream of the river in km/h?
1. 2 km/h
2. 1 km/h
3. 6 km/h
4. 4 km/h

```
1 km/h
```

Explanation:

Downstream:
Time taken = 3 + 45/60 = 3 + 3/4 = 15/4 h.
Distance covered = 15 km.
Downstream Speed = 15 / (15/4) = 4 km/h.
Upstream:
Time taken = 2 + 30/60 = 2 + 1/2 = 5/2 h.
Distance covered = 5 km.
Upstream Speed = 5 / (5/2) = 2 km/h.
We know, speed of stream
= 1/2 (Downstream Speed – Upstream Speed)
= 1/2 (4-2) = 1 km/h.

8. John earns 33.33% more than Peter. By what percentage is Peter’s earning less than that of John’s?
1. 22 %
2. 25 %
3. 26 %
4. 23 %

```
25 %
```

Explanation:

Let John’s income be j and Peter’s income be p. Then,
j = p + p Ã— 33.33% = p + p Ã— 100?3 % = p + p Ã— 1/3 = 4p/3
=> p = 3j/4 = (4 – 1)j/4 = j – j/4 = j – j Ã— 1/4 = j – j Ã— 100?4 % = j – j Ã— 25%.
Therefore, Peter’s earning is less than John’s earning by 25%.

9. Present age of Vinod and Ashok are in ratio of 3:4 respectively. After 5 years, the ratio of their ages becomes 7:9 respectively. What is Ashokâ€™s present age is ?
1. 40 years
2. 28 years
3. 32 years
4. 36 years

```
40 years
```

Explanation:

Let the present age of Vinod and Ashok be 3x years and 4x years respectively.
Then (3x+5) / (4x+5) = 7 / 9

=> 9(3x + 5) = 7(4x + 5)
=> 27x + 45 = 28x + 35
=> x = 10
=> Ashokâ€™s present age = 4x = 40 years

10. Sum of 4 children born at interval of 4 years is 36. What is the age of youngest child?
1. 2 years
2. 3 years
3. 4 years
4. 5 years

```
3 years
```

Explanation:

Let the ages of children be x, (x+4),
(x+8) and (x+12) years.

Then x + x + 4 + x + 8 + x +12 = 36
4x + 24 = 36
4x = 12
x = 3
Age of the youngest child = x = 3 years

11. If the sum of two numbers is 13 and the sum of their square is 85. Find the numbers?.
1. 6, 7
2. 5, 8
3. 4, 9
4. 3, 10

```
6, 7
```

Explanation:

Let the numbers be x and 13-x
Then x2 + (13 â€“ x)2 = 85
=> x2 + 169 + x2 â€“ 26x = 85
=> 2 x2 â€“ 26x + 84 = 0
=> x2 â€“ 13x + 42 = 0
=> (x-6)(x-7)=0

Hence numbers are 6 & 7

12. HCF of two numbers is 11 and their LCM is 385. If the numbers do not differ by more than 50, what is the sum of the two numbers ?
1. 132
2. 35
3. 12
4. 36

```
132
```

Explanation:

Product of numbers = LCM x HCF = 11 x 385 = 4235
Let the numbers be of the form 11m and 11n, such that ‘m’ and ‘n’ are co-primes.
=> 11m x 11n = 4235
=> m x n = 35
=> (m, n) can be either of (1, 35), (35, 1), (5, 7), (7, 5).
=> The numbers can be (11, 385), (385, 11), (55, 77), (77, 55).
But it is given that the numbers cannot differ by more than 50.
Hence, the numbers are 55 and 77.
Therefore, sum of the two numbers = 55 + 77 = 132

13. The LCM of two co-prime numbers is 117. What is the sum of squares of the numbers ?
1. 220
2. 1530
3. 250
4. 22

```
250
```

Explanation:

117 = 3 x 3 x 13
As the numbers are co-prime, HCF = 1.
So, the numbers have to be 9 and 13.
92 = 81
132 = 169

14. A and B took a job to be completed in 20 days. They started working together and after 12 days, C joined them and the whole job finished in 15 days. How much time would C require to complete the job if only C was hired?
1. 15
2. 12
3. 10
4. 8

```
12
```

Explanation:

Let the total job be 20 units. It is given that A and B took the job to be completed in 20 days.
=> Combined efficiency of A and B = 20/20 = 1 unit / day
Now, job done in 12 days = 12 units
=> Job Left = 8 units
Now, this remaining 8 units of job have been done by all A, B and C together.
Let the efficiency of C be ‘x’.
=> Combined efficiency of A, B and C = 1+x units/ day
Now, with this efficiency, the job got completed in 3 more days.
=> Job done in 3 days = 3 x (1+x) = 8 units
=> x = 5/3
Therefore, efficiency of C = x = 5/3 units / day
Hence, time required by C alone to do the job = 20/(5/3) = 12 days

15. Three pipes A, B and C were opened to fill a cistern. Working alone, A, B and C require 12, 15 and 20 minutes respectively. After 4 minutes of working together, A got blocked and after another 1 minute, B also got blocked. C continued to work till the end and the cistern got completely filled. What is the total time taken to fill the cistern ?
1. 6 minutes
2. 6 minutes 15 seconds
3. 6 minutes 40 seconds
4. 6 minutes 50 seconds

```
6 minutes 40 seconds
```

Explanation:

Let the capacity of the cistern be LCM(12, 15, 20) = 60 units.
=> Efficiency of pipe A = 60 / 12 = 5 units / minute
=> Efficiency of pipe B = 60 / 15 = 4 units / minute
=> Efficiency of pipe C = 60 / 20 = 3 units / minute
=> Combined efficiency of pipe A, pipe B and pipe C = 12 units / minute
Now, the cistern is filled with the efficiency of 12 units / minute for 4 minutes.
=> Pool filled in 4 minutes = 48 units
=> Pool still empty = 60 â€“ 48 = 12 units
Now, A stops working.
=> Combined efficiency of pipe B and pipe C = 7 units / minute
Now, the cistern is filled with the efficiency of 7 units / minute for 1 minute.
=> Pool filled in 1 minute = 7 units
=> Pool still empty = 12 â€“ 7 = 5 units
Now, B also stops working.
These remaining 5 units are filled by C alone.
=> Time required to fill these 5 units = 5 / 3 = 1 minute 40 seconds
Â
Therefore, total time required to fill the pool = 4 minutes + 1 minutes + 1 minute 40 seconds = 6 minutes 40 seconds

16. If John walks at the speed of 5 km/h, he reaches his office 7 minutes late. However, if he walks at a speed of 6 km/h, he reaches his office 5 minutes early. How far is his office from his home?
1. 9
2. 8
3. 10
4. 6

```
6
```

Explanation:

Let the distance of Johnâ€™s office from his home be x.
The time difference when covering the distance x at the two different speeds = 5 – (-7) = 12 min = 1/5 hr
=> x/5 – x/6 = 1/5
=> (6x – 5x)/30 = 1/5
=> x = 6.
So, his office is 6 km far from his home.

17. A speedboat runs 6 km upstream in a river and comes back to the starting point in 33 minutes. The stream of the river is running at 2 km/hr. What is the speed of speedboat in still water?
1. 25 km/h
2. 21 km/h
3. 26 km/h
4. 22 km/h

```
22 km/h
```

Explanation:

Let the speed of speedboat in still water be x km/h.
Then, speed downstream = (x + 2) km/h, speed upstream = (x – 2) km/h.
Since it goes 6 km upstream and comes back in 33 minutes, we have
6/(x+2) + 6/(x-2) = 33/60
=> 11xÂ² – 240x – 44 = 0
=> 11xÂ² – 242x + 2x – 44 = 0
=> (x – 22)(11x + 2) = 0
=> x = 22.
Therefore, the required speed = 22 km/h.

18. Mary’s salary is reduced by 10%. By what percentage must her new salary be increased in order to gain her old salary?
1. 137/9 %
2. 194/9 %
3. 100/9 %
4. 110/9 %

```