# HCL Placement Paper | Quantitative Aptitude Set – 4

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.

**Find a positive number which when increased by 16 is equal to 80 times the reciprocal of the number**.- 20
- -4
- -10
- 4

**Answer:****4****Explanation:**Let the number be x.

Then x + 16 = 80 * (1/x)

x

^{2}+ 16x â€“ 80 = 0

x^{2}+ 20x â€“ 4x â€“ 80 =0

(x + 20) (x -4)

Therefore x = 4**The LCM. of two numbers is 30 and their HCF. is 15. If one of the numbers is 30, then what is the other number?**- 30
- 25
- 15
- 20

**Answer:****15****Explanation:**Say another number = x

Product of two numbers = product of HCF and LCM

=> x.30 = 15*30

=> x=15**Which of the following is the largest of all ?**- 7/8
- 15/16
- 23/24
- 31/32

**Answer:****31/32****Explanation:**LCM (8, 16, 24, 32) = 96

7/8 = 84/96

15/16 = 90/96

23/24 = 92/96

31/32 = 93/96

Hence, 31/32 is the largest of all.**Three friends A, B and C are employed to make pastries in a bakery. Working individually, they can make 60, 30 and 40 pastries respectively in an hour. They decided to work together but due to lack of resources, they had to work on shifts of 30 minutes. Find the time taken to make 185 pastries.**- 4 hours
- 3 hours 45 minutes
- 4 hours 15 minutes
- 5 hours

**Answer:****4 hours 15 minutes****Explanation:**It is given that A, B and C make 60, 30 and 40 pastries respectively in an hour.

=> In 30 minutes, they will make 30, 15 and 20 pastries respectively.

So, in one cycle of 1 hour 30 minutes where each works for 30 minutes, pastries made = 30 + 15 + 20 = 65

Now, in 2 cycles (3 hours), 130 pastries would be made.

In the next 30 minutes, A would make 30 pastries.

So, total time elapsed = 3 hours 30 minutes and pastries made = 130 + 30 = 160

In the next 30 minutes, B would make 15 pastries.

So, total time elapsed = 4 hours and pastries made = 160 + 15 = 175

In the next 15 minutes, C would make 10 pastries.

So, total time elapsed = 4 hours 15 minutes and pastries made = 175 + 10 = 185

Therefore, total time taken = 4 hours 15 minutes**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. Another pipe D, which is a waste pipe, can empty the filled tank in 30 minutes working alone. What is the total time (in minutes) taken to fill the cistern if all the pipes are simultaneously opened ?**- 5
- 6
- 7
- 8

**Answer:****6****Explanation:**Let the capacity of the cistern be LCM(12, 15, 20, 30) = 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

=> Efficiency of pipe D = 60 / 30 = 2 units / minute

=> Combined efficiency of pipe A, pipe B, pipe C and pipe D = 10 units / minute

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Therefore, time required to fill the cistern if all the pipes are opened simultaneously = 60 / 10 = 6 minutes**The ratio of the speed of two trains is 7:8. If the second train covers 400 km in 4 h, find out the speed of the first train.**- 69.4 km/h
- 78.6 km/h
- 87.5 km/h
- 40.5 km/h

**Answer:****87.5 km/h****Explanation:**Let the speed of the two trains be 7x and 8x.

Then, 8x = 400 / 4

=> 8x = 100 => x = 12.5 km/h.

Hence, speed of the first train = 7x = 7 Ă— 12.5 = 87.5 km/h.**A motorboat crosses a certain distance in 1 hour and comes back in 1Â˝ hours. If the stream is running at 3 km/h, find out the speed of motorboat in still water.**- 10 km/h
- 15 km/h
- 12 km/h
- None of these

**Answer:****15 km/h****Explanation:**Let the speed of motorboat in still water be x km/h. Then,

Downstream speed = (x + 3) km/h.

Upstream speed = (x – 3) km/h.

Then, (x + 3) Ă— 1 = (x – 3) Ă— 3/2

=> 2x + 6 = 3x – 9

=> x = 15.

So, the speed of motorboat in still water is 15 km/h.**Barack spends Rs 6650 to buy some goods and gets a rebate of 6% on it. After this, he pays a sales tax of 10%. What is his total expenditure?**- Rs 6870.10
- Rs 6876.10
- Rs 6865.10
- Rs 6776.10

**Answer:****Rs 6876.10****Explanation:**Rebate received by Barack = 6% of Rs 6650 = 6/100Â Ă— 6650 = 3/5 Ă— 665 = Rs 399.

Sales Tax paid by Barack = 10% of Rs (6650-399) = 10% of Rs 6251 = Rs 625.10.

Therefore, Barack’s total expenditure = Rs (6251 + 625.10) = Rs 6876.10.**In a box, there are 10p, 25p and 50p coins in the ratio 4:9:5 with the total sum of Rs 206. How many coins of each kind does the box have?**- 200, 360, 160
- 135, 250, 150
- 90, 60, 110
- Cannot be determined

**Answer:****200, 360, 160****Explanation:**Let the number of 10p, 25p, 50p coins be 4x, 9x, 5x respectively. Then,

4x/10 + 9x/4 + 5x/2 = 206 (Since, 10p = Rs 0.1, 25p = Rs 0.25, 50p = Rs 0.5)

=> 8x + 45x + 50x = 4120 (Multiplying both sides by 20 which is the LCM of 10, 4, 2)

=> 103x = 4120

=> x = 40.

Therefore,

No. of 10p coins = 4 x 40 = 160 (= Rs 16)

No. of 25p coins = 9 x 40 = 360 (= Rs 90)

No. of 50p coins = 5 x 40 = 200 (= Rs 100)**At present, the ratio between the ages of Ram and Shyam is 6:5 respectively. After 7 years, Shyamâ€™s age will be 32 years. What is the present age of Ram?**- 32
- 40
- 30
- 36

**Answer:****30****Explanation:**Let the present age of Ram and Shyam be 6x years and 5x years respectively.

Then 5x + 7 = 32

=> 5x = 25

=> x = 5

=> Present age of Ram = 6x = 30 years**What is the sum of two consecutive odd numbers, the difference of whose squares is 56?**.- 30
- 28
- 34
- 32

**Answer:****28****Explanation:**Let the no. be x and (x +2).

Then (x +2)2 â€“ x2 = 56

4x + 4 = 56

x + 1 = 14

x = 13

Sum of numbers = x + (x +2) = 28**Express 252 as a product of primes.**- 2 * 2 * 3 * 3 * 7
- 3* 3 * 3 * 3 * 7
- 2 * 2 * 2* 3 * 7
- 2 * 3 * 3 * 3 * 7

**Answer:****2 * 2 * 3 * 3 * 7****Two numbers are in the ratio 3 : 5. If their L.C.M. is 75. what is sum of the numbers?**- 25
- 45
- 40
- 50

**Answer:****40****Explanation:**1st number = 3x

2nd number =5x

LCM of 3x and 5x is 15x

=> 15x = 75

=> x = 5

sum = 15+25 =40**A person employed a group of 20 men for a construction job. These 20 men working 8 hours a day can complete the job in 28 days. The work started on time but after 18 days, it was observed that two-thirds of the work was still pending. To avoid penalty and complete the work on time, the employer had to employ more men and also increase the working hours to 9 hours a day. Find the additional number of men employed if the efficiency of all men is the same.**- 40
- 44
- 64
- 80

**Answer:****44****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 unit

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

M_{1}= 20 men

D_{1}= 18 days

H_{1}= 8 hours/day

W_{1}= 1 unit

E_{1}= E_{2}= Efficiency of each man

M_{2}= (20 + x) men

D_{2}= 10 days

H_{2}= 9 hours/day

W_{2}= 2 unit

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So, we have

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

=> x + 20 = 64

=> x = 44

Therefore, additional men employed = 44**Three pipes A, B and C were opened to fill a tank. Working alone, A, B and C require 10, 15 and 20 hours respectively. A was opened at 7 AM, B at 8 AM and C at 9 AM. At what time the tank would be completely filled, given that pipe C can only work for 3 hours at a stretch, and needs 1-hour standing time to work again.**- 12 : 00 PM
- 12 : 30 PM
- 1 : 00 PM
- 1 : 30 PM

**Answer:****12 : 30 PM****Explanation:**Let the capacity of the tank be LCM (10, 15, 20) = 60

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

=> Efficiency of pipe B = 60 / 15 = 4 units / hour

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

=> Combined efficiency of all three pipes = 13 units / hour

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Till 9 AM, A works for 2 hours and B work for 1 hour.

=> Tank filled in 2 hours by A = 12 units

=> Tank filled in 1 hour by B = 4 units

=> Tank filled till 9 AM = 16 units

=> Tank still empty = 60 – 16 = 44 units

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Now, all three pipes work for 3 hours with the efficiency of 13 units / hour.

=> Tank filled in 3 more hours = 39 units

=> Tank filled till 12 PM = 16 + 39 units = 55 units

=> Tank empty = 60 – 55 = 5 units

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Now, C is closed for 1 hour and these remaining 5 units would be filled by A and B working together with the efficiency 10 units/hour.

=> Time taken to fill these remaining 5 units = 5 / 10 = 0.5 hours

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Therefore, time at which the tank will be completely filled = 12 PM + 0.5 hours = 12 : 30 PM

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