# HCL Placement Paper | Quantitative Aptitude Set – 5

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

**The product of two numbers is 108 and the sum of their squares is 225. The difference of the number is:**.- 5
- 4
- 3
- None of these

**Answer:****57****Explanation:**Let the numbers be x and y.

Then xy = 108 and x^{2}+ y^{2}= 225

(x –y)^{2}= x^{2}+ y^{2}– 2xy

(x –y)^{2}= 225 – 216

(x –y)^{2}= 9

Therefore (x –y) = 3**Which of the following has the most number of divisors?**- 99
- 101
- 176
- 182

**Answer:****176****Explanation:**99 = 1 * 3 * 3 * 11

101 = 1 * 101

176 = 1 * 2 * 2 * 2 * 2 * 11

182 = 1 * 2 * 7 * 13

Clearly, 176 has the most number of divisors.**Ratio of two numbers is 3:2. If LCM of numbers is 60, then smaller number is?**- 20
- 30
- 40
- 50

**Answer:****20****Explanation:**say, 1st number =3x

2nd number =2x

LCM of numbers = 6x

given LCM = 60

=> x6 = 60

=>x = 10**6 men and 10 women were employed to make a road 360 km long. They were able to make 150 kilometres of road in 15 days by working 6 hours a day. After 15 days, two more men were employed and four women were removed. Also, the working hours were increased to 7 hours a day. If the daily working power of 2 men and 3 women are equal, find the total number of days required to complete the work.**- 19
- 35
- 34
- 50

**Answer:****34****Explanation:**We are given that the daily working power of 2 men and 3 women are equal.

=> 2 Em = 3 Ew

=> Em / Ew = 3/2, where ‘Em’ is the efficiency of 1 man and ‘Ew’ is the efficiency of 1 woman.

Therefore, the ratio of efficiency of man and woman = 3 : 2.

If ‘k’ is the constant of proportionality, Em = 3k and Ew = 2k.

Here, we need to apply the formula**=> (M**, where_{i}E_{i}) D_{1}H_{1}/ W_{1}= (M_{j}E_{j}) D_{2}H_{2}/ W_{2}

=> (M_{i}E_{i}) = (6 x 3k) + (10 x 2k)

=> (M_{j}E_{j}) = (8 x 3k) + (6 x 2k)

D_{1}= 15 days

D_{2}= Number of days after increasing men and reducing women

H_{1}= 6 hours

H_{2}= 7 hours

W_{1}= 150 km

W_{2}= 210 kmSo, we have

38k x 15 x 6 / 150 = 36k x D_{2}x 7 / 210

=> 38k x 6 = 12k x D_{2}

=> D_{2}= 19 days

Therefore, total days required to complete the work = 15 + 19 = 34 days**Two pipes A and B can fill a tank in 10 hours and 30 hours respectively. Due to a leak in the tank, it takes 2.5 hours more to fill the tank. How much time would the leak alone will take to empty the tank ?**- 20 hours
- 25 hours
- 30 hours
- 35 hours

**Answer:****30 hours****Explanation:**Let the capacity of the tank be LCM (10, 30) = 30 units

=> Efficiency of pipe A = 30 / 10 = 3 units / hour

=> Efficiency of pipe B = 30 / 30 = 1 units / hour

=> Combined efficiency of both pipes = 4 units / hour

Now, total time taken by A and B working together to fill the tank if there was no leak = 30 / 4 = 7.5 hours

=> Actual time taken = 7.5 + 2.5 = 10 hoursThe tank filled by A and B in these 2.5 hours is the extra work done to compensate the wastage by the leak in 10 hours.

=> 2.5 hours work of A and B together = 10 hours work of the leak

=> 2.5 x 4 = 10 x E, where ‘E’ is the efficiency of the leak.

=> E = 1 unit / hourTherefore, time taken by the leak alone to empty the full tank = 30 / 1 = 30 hours

**Max completes his journey at an average speed of 9 km/h. He covers the first 9 km at a speed of 6 km/h and he takes 1·5 hours to cover the remaining distance. Find out the speed at which he covered the remaining distance.**- 11 km/h
- 12 km/h
- 13 km/h
- 14 km/h

**Answer:****12 km/h****Explanation:**Let the required speed be x km/h.

Total time taken to finish his journey = (9/6 + 1·5) = 3 hours.

Total distance = 9 + 1·5x km.

Given, average speed = 9 km/h.

Therefore, (9 + 1·5x)/3 = 9

=> 9 + 1·5x = 27

=> 1·5x = 18

=> x = 12 km/h.**A train crosses a pole in 10 sec. If the length of train is 100 meters, what is the speed of the train in Kmph?**- 34
- 36
- 30
- 32

**Answer:****36****Explanation:**V = 100/10 = 10 m/s = 10*3600/1000 = 36Km/hr

**Jack and Robert appeared in an examination. Robert scored 9 marks less than Jack. Jack’s score was 56% of the sum of their scores added together. Calculate their individual scores.**- 22 and 33
- 41 and 35
- 40 and 35
- 42 and 33

**Answer:****42 and 33****Explanation:**Let Robert’s score be x. Then, Jack’s score = x+9.

Now, x+9 = 56% of [(x+9) + x]

=> x+9 = 14/25 × (2x + 9)

=> 25 × (x+9) = 14 × (2x+9)

=> 25x + 225 = 28x + 126

=> 3x = 99 => x = 33.

Therefore, Robert scored 33 marks and Jack scored 42 marks.**In a library, the ratio of the books on Computer, Physics and Mathematics is 5:7:8. If the collection of books is increased respectively by 40%, 50% and 75%, find out the new ratio:**- 3:9:5
- 7:5:3
- 2:3:4
- 2:5:4

**Answer:****2:3:4****Explanation:**40% increase will lead to a factor of 140 and similarly 150 and 175

so the new ratio is

(5*140):(7*150):(8*175)

on solving we get 2:3:4

**A person’s present age is one-third of the age of his mother. After 12 years, his age will be one half of the age of his mother. What is the present age of his mother?**- 30
- 34
- 38
- 36

**Answer:****36****Explanation:**Let the present ages of son and his mother are x years and 3x years.

Then (3x + 12) = 2( x + 12)

=> 3x + 12 = 2x + 24

=> x = 12

=> Present age of mother = 3x = 36 years**The average of 21 results is 20. Average of 1**.^{st}10 of them is 24 that of last 10 is 14. the result of 11’th is :- 42
- 44
- 46
- 40

**Answer:****40****Explanation:**11’th result = sum of 21 results – sum of 20 results

= 21 x 20 – (24 x 10 + 14 x 10)

= 420 – (240 + 140)

= 420- 380 = 40**A perfect number n is a number which is equal to the sum of its divisors. Which of the following is a perfect number?**- 6
- 9
- 15
- 21

**Answer:****9****Explanation:**6 is divisible by 1, 2 and 3.

And, 6 = 1 + 2 + 3.**Three numbers are in the ratio of 2 : 3 : 4 and their L.C.M. is 240. Their H.C.F. is:**- 40
- 20
- 30
- 10

**Answer:****20****Explanation:**Let the numbers be 2x, 3x and 4x

LCM = 12x

12x=240

=> x=20

H.C.F of 40, 60 and 80=20**A stadium was to be built in 1500 days. The contractor employed 200 men, 300 women and 750 robotic machines. After 600 days, 75% of the work was still to be done. Fearing delay, the contractor removed all women and 500 robotic machines. Also, he employed some more men having the same efficiency as earlier employed men. This led to a speedup in work and the stadium got built 50 days in advance. Find the additional number of men employed if in one day, six men, ten women and fifteen robotic machines have same work output.**- 1100
- 1340
- 1300
- 1140

**Answer:****1140****Explanation:**Let the total work be 4 units.

=> Work done in first 600 days = 25% of 4 = 1 unit

=> Work done in next 850 days = 75% of 4 = 3 unit

Also, we are given that the daily work output of 6 men, 10 women and 15 robotic machines are same.

=> 6 Em = 10 Ew = 15 Er

=> Em : Ew : Er = 5 : 3 : 2, where ‘Em’ is the efficiency of 1 man, ‘Ew’ is the efficiency of 1 woman and ‘Er’ is the efficiency of 1 robotic machine.

Therefore, ratio of efficiency of man, woman and robotic machine = 5:3:2.

If ‘k’ is the constant of proportionality, Em = 5k, Ew = 3k and Er = 2k.

Here, we need to apply the formula**=> (M**, where_{i}E_{i}) D_{1}H_{1}/ W_{1}=(M_{j}E_{j}) D_{2}H_{2}/ W_{2}

=> (M_{i}E_{i}) = (200 x 5k) + (300 x 3k) + (750 x 2k)

=> (M_{j}E_{j}) = (200 x 5k) + (m x 5k) + (250 x 2k), where ‘m’ is the additional men employed

D_{1}= 600 days

D_{2}= 850 days

H_{1}= H_{2}= Daily working hours

W_{1}= 1 unit

W_{2}= 3 units

So, we have

3400k x 600 / 1 = (1500 + 5m)k x 850 / 3

=> 3400k x 1800 = (1500 + 5m)k x 850

=> 1500 + 5m = 7200

=> 5m = 5700

=> m = 1140

Therefore, additional men employed = 1140**Two pipes A and B work alternatively with a third pipe C to fill a swimming pool. Working alone, A, B and C require 10, 20 and 15 hours respectively. Find the total time required to fill the pool.**- 7 hours 14 minutes
- 6 hours 54 minutes
- 5 hours 14 minutes
- 8 hours 54 minutes

**Answer:****66 hours 54 minutes****Explanation:**Let the total work be 3 units and additional men employed after 18 days be ‘x’.

=> Work done in first 18 days by 20 men working 8 hours a day = (1/3) x 3 = 1 unit

=> Work done in last 10 days by (20 + x) men working 9 hours a day = (2/3) x 3 = 2 unitHere, we need to apply the formula

M1 D1 H1 E1 / W1 = M2 D2 H2 E2 / W2,

where

M1 = 20 men

D1 = 18 days

H1 = 8 hours/day

W1 = 1 unit

E1 = E2 = Efficiency of each man

M2 = (20 + x) men

D2 = 10 days

H2 = 9 hours/day

W2 = 2 unitSo, we have

20 x 18 x 8 / 1 = (20 + x) x 10 x 9 / 2

=> x + 20 = 64

=> x = 44Therefore, number of additional men employed = 44

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