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.

**The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 45. What is the difference between the two digits of that number?**.- 5
- 7
- 6
- None of these

**Answer:****5****Explanation:**Let the ten’s digit be x and unit’s digit be y

Then (10x + y) – (10y + x) = 45

9(x – y) = 45

x – y = 5**Two numbers are in the ratio of 5:7. If their LCM is 105, what is the difference between their squares ?**- 216
- 210
- 72
- 840

**Answer:****216****Explanation:**Let ‘h’ be the HCF of the two numbers.

=> The numbers are 5h and 7h.

We know that Product of Numbers = LCM x HCF

=> 5h x 7h = 105 x h

=> h = 3

So, the numbers are 15 and 21.

Therefore, difference of their squares = 21^{2}– 15^{2}= 441 – 225 = 216**Three people A, B and C working individually can finish a job in 10, 12 and 20 days respectively. They decided to work together but after 2 days, A left the work and after another one day, B also left work. If they got two lacs collectively for the entire work, find the difference of the highest and lowest share.**- 70000
- 60000
- 10000
- 20000

**Answer:****70000****Explanation:**Let the total work be LCM(10, 12, 20) = 60 units

=> Efficiency of A = 60/10 = 6 units / day

=> Efficiency of B = 60/12 = 5 units / day

=> Efficiency of C = 60/20 = 3 units / day

Since the number of working days are different for each person, the share of each will be calculated in the ratio of the units of work done.

Now, A works for 2 days and B works for 3 days.

=> Work done by A = 2 x 6 = 12 units

=> Work done by B = 3 x 5 = 15 units

=> Work done by C = 60 – 12 – 15 = 33 units

Therefore, ratio of work done = 12:15:33 = 4:5:11

So, A’s share = (4/20) x 2, 00, 000 = Rs 40, 000

B’s share = (5/20) x 2, 00, 000 = Rs 50, 000

C’s share = (11/20) x 2, 00, 000 = Rs 1, 10, 000

Therefore, difference of the highest and lowest share = Rs 1, 10, 000 – 40, 000 = Rs 70, 000**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 ?**- 132
- 35
- 12
- 36

**Answer:****132****Explanation:**

Product of numbers = LCM x HCF

=> 4235 = 11 x 385Let 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**Three pipes A, B and C are connected to a tank. Working alone, they require 10 hours, 20 hours and 30 hours respectively. After some time, A is closed and after another 2 hours, B is also closed. C works for another 14 hours so that the tank gets filled completely. Find the time (in hours) after which pipe A was closed.**- 1
- 1.5
- 2
- 3

**Answer:****2****Explanation:**Let the capacity of the tank be LCM (10, 20, 30) = 60

=> Efficiency of pipe A = 60 / 10 = 6 units / hour

=> Efficiency of pipe B = 60 / 20 = 3 units / hour

=> Efficiency of pipe C = 60 / 30 = 2 units / hour

Now, all three work for some time, say ‘t’ hours.

So, B and C work for 2 more hours after ‘t’ hours and then, C works for another 14 hours.

=> Combined efficiency of pipe A, pipe B and pipe C = 11 units/hour

=> Combined efficiency of pipe B and pipe C = 5 units/hour

So, we have 11 x t + 5 x 2 + 14 x 2 = 60

=> 11 t + 10 + 28 = 60

=> 11 t = 60 – 38

=> 11 t = 22

=> t = 2

Therefore, A was closed after 2 hours.**A policeman sees a thief at a distance of 100 meters and starts to chase him. The thief sees him and starts to run too. If the thief is running at the speed of 8 km/hr and the policeman is running at the speed of 10 km/hr, find out the distance covered by the thief before the policeman catches him.**- 250 meters
- 400 meters
- 450 meters
- 350 meters

**Answer:****400 meters****Explanation:**We can safely assume that the policeman is running in the same direction as the thief.

Speed of policeman w.r.t thief = (10 – 8) = 2 km/hr.

Time taken by policeman to cover the 100m distance between him and the thief = (100/1000) / 2 = 1/20 hr.

Therefore, the distance covered by thief in 1/20 hrs = 8 × 1/20 = 2/5 km = 400 meters.**A boat runs at the speed of 13 km/h in still water. If the speed of the stream is 4 km/h, how much time will it take to go 68 km downstream?**- 5 h
- 4 h
- 6 h
- 3 h

**Answer:**

**4 h****Explanation:**Speed of the boat downstream = 13 + 4 = 17 km/h.

Therefore, time taken to go 68 km downstream = (68/17) = 4 h.**The price of sugar is decreased by 10%. As a consequence, monthly sales is increased by 30%. Find out the percentage increase in monthly revenue.**- 17 %
- 19 %
- 18 %
- None of these

**Answer:****17 %****Explanation:**Let the price of sugar be Rs 100 and monthly sales be 100 units. Then,

total revenue = 100 × 100 = Rs 10000.

And, new revenue = 90 × 130 = Rs 11700.

Increase in revenue = 11700 – 10000 = Rs 1700.

Hence, percentage increase in revenue = (1700/10000) × 100% = 17%.**The present ages of A, B and C are in proportions 4:5:9. Nine years ago, sum of their ages was 45 years. Find their present ages in years**- 15, 20, 35
- 20, 24, 36
- 20, 25, 45
- 16, 20, 36

**Answer:****16, 20, 36****Explanation:**Let the current ages of A, B and C be ax years, 5x years and 9x respectively.

Then (4x-9) + (5x-9) + (9x-9) =45

=> 18x – 27 = 45

=> 18x = 72

=> x = 4

Present ages of A, B and C are 4x = 16, 5x = 20, 9x = 36 respectively.**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 Ashok’s present age is ?**- 40 years
- 28 years
- 32 years
- 36 years

**Answer:****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

=> Ashok’s present age = 4x = 40 years**A two-digit number is such that the product of the digits is 12. When 9 is subtracted from the number, the digits are reversed. The number is:**.- 34
- 62
- 43
- 26

**Answer:****43****Explanation:**Let the ten’s and unit’s digit be x and y.

Then 10x + – 9 = 10 x + x

10×2 + 12 -9x = 120 + x2

9×2 – 9x – 108 = 0

x2 –x – 12 = 0

x2 –4x + 3x – 12 = 0

(x – 4) (x + 3) = 0

Therefore x = 4

Hence the required no. is 43**Which is the largest number that divides 17, 23, 35, 59 to leave the same remainder in each case ?**- 2
- 3
- 6
- 12

**Answer:****6****Explanation:**Required Number = HCF (23-17, 35-23, 59-35, 59-17)

= HCF (6, 12, 24, 42)

= 6**Two numbers are in the ratio of 5:7. If their LCM is 105, what is the difference between their squares ?**- 261
- 210
- 72
- 840

**Answer:****216****Explanation:**Let ‘h’ be the HCF of the two numbers.

=> The numbers are 5h and 7h.

We know that Product of Numbers = LCM x HCF

=> 5h x 7h = 105 x h

=> h = 3

So, the numbers are 15 and 21.

Therefore, difference of their squares = 21^{2}– 15^{2}= 441 – 225 = 216**A alone and B alone can do a work in respectively 18 and 8 days more than both working together. Find the number of days required if both work together.**- 12
- 8
- 16
- 36

**Answer:****12****Explanation:**Let the time required to complete the work by A and B together = n days

=> Time required by A alone = n + 18 days

=> Time required by B alone = n + 8 days

Therefore, n^{2}= 18 x 8 = 144

=> n = 12

Hence, A and B require 12 days to complete the work if they work together.**Working alone, two pipes A and B require 9 hours and 6.25 hours more respectively to fill a pool than if they were working together. Find the total time taken to fill the pool if both were working together.**- 6
- 6.5
- 7
- 7.5

**Answer:****7.5****Explanation:**Let the time taken if both were working together be ‘n’ hours.

=> Time taken by A = n + 9

=> Time taken by B = n + 6.25

In such kind of problems, we apply the formula :

n^{2}= a x b, where ‘a’ and ‘b’ are the extra time taken if both work individually than if both work together.

Therefore, n^{2}= 9 x 6.25

=> n = 3 x 2.5 = 7.5

Thus, working together, pipes A and B require 7.5 hours.**Paul has to travel 24 km. After walking for 1 hour 40 minutes he sees that he has covered 5/7 of the distance left to cover. Find out Paul’s speed in meters per second.**- 5/3 m/s
- 7/5 m/s
- 2/3 m/s
- 8/5 m/s

**Answer:**

**5/3 m/s****Explanation:**Let the required speed be x km/hr.

Distance covered by Paul in 1 hr 40 min = x × 100/60 = 5x/3 km.

Remaining distance = (24 – 5x/3) km.

Therefore, 5x/3 = 5/7 × (24 – 5x/3)

=> 7/5 × 5x/3 = 24 – 5x/3

=> 7x/3 = (72 – 5x)/3

=> 7x = 72 – 5x

=> 12x = 72 => x = 6.

Paul’s speed in meters per second = 6 × 5/18 = 5/3 m/s.**Peter’s speedboat run at a speed of 9 km/h in still water. He rows to a place at a distance of 105 km and comes back to the starting point. If the speed of the stream is 1.5 km/h, find out the time taken by Peter.**- 24 h
- 21 h
- 23 h
- 22 h

**Answer:****24 h****Explanation:**Upstream speed = 9 – 1.5 = 7.5 km/h.

Downstream speed = 9 + 1.5 = 10.5 km/h.

Therefore, time taken = 105/7.5 + 105/10.5 = 14 + 10 = 24 h.**Jack consumes 75% of his salary. Later his salary is increased by 20% and he increases his expenditures by 10%. Find the percentage increase in his savings.**- 51 %
- 60 %
- 50 %
- 55 %

**Answer:****50 %****Explanation:**Let Jack’s original salary be Rs 100. Then,

his expenditure = Rs 75,

his savings = Rs 25.

Now, his new salary = Rs 120. So,

new expenditure = (110/100) × 75 = Rs 165/2,

new savings = 120 – 165/2 = Rs 75/2.

Increase in savings = 75/2 – 25 = Rs 25/2.

Therefore, percentage increase in savings = (25/2)/25 × 100% = 50%.

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