Problem on LCM and HCF- Quant Aptitude

Problem on LCM and HCF

1. Find the least number which when increased by 4 is exactly divisible by 8, 16, 24, 30 and 32 ?

a)   480
b)   484
c)   476
d)   472
e)   None of these

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Explanation:

LCM of 8, 16, 24, 30 and 32 is 480

Therefore Required number is 480 - 4 = 476

2. What is the greatest number of five digits which when 3769 is added to it will be exactly divisible by 5, 6 , 10, 12, 15 and 18 ?

a)   4309
b)   99459
c)   100539
d)   99911
e)   None of These

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Explanation:

LCM of 5, 6, 10, 12 and 18 is 540

On dividing (99999 + 3769) by 540, the remainder is 88

therefore, The required number is 99999 - 88 = 99911

3. Find the minimum number of square tiles required to pave the floor of a room of 2.50m long and 1.50m broad ?

a)    50
b)   750
c)   45
d)   15
e)   None of these

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Explanation:

 HCF of 250 cm and 150 cm is 50 cm, which is the side of the tile 

Therefore The required number of tiles = (250 X 150) / (50X50) = 15

4. Five bells toll together at the intervals of 5, 6, 8, 12 and 20 seconds respectively. Find the number of times they toll together in one hour's time (Inclusive of the toll at the beginning)

a)   120
b)   31
c)   30
d)   5
e)   None of These

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Explanation: 

Time after which all the bells tol together is the LCM of 5, 6, 8, 12 and 20. i.e., 120 seconds = 20 minutes.

The number of times they toll together in one hour = 60/2 = 30 + 1 (beginning tone) 

therefore, the answer is 31

5. A milk man has three different kinds of milk 493liters, 551 liters and 435 liters. Find the minimum number of equal size containers required to store all the milk without mixing.

a)   29
b)   51
c)   58
d)   49
e)    None of these

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Explanation: 

As minimum number of containers are required, the size of the container should be maximum and the size is also equal. 

so size of the container will be HCF of 493, 551 and 435 i.e. 29 

therefore, required number of containers is = (493+551+435) / 29 = 51

6. The circumference of the front and back wheels of a vehicle are 6 3/14 m and 8 1/18 m respectively. At any given moment, a chalk mark is put on the point of contact of each wheel with the ground. Find the distance traveled by the vehicle so that both the chalk marks are again on the ground at the same time.

a)   217.5 m
b)   435 m
c)    412m
d)   419m
e)   None of these

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Explanation:

The required distance is the LCM of 6 3/14 and 8 1/18 

LCM of 6 3/14 and 8 1/18 = LCM (87/14, 145/18) 

 => LCM(87,145) / HCF (14,18) = 435/2 = 217.5m

7. The LCM of two numbers is 28 times of their HCF. The sum of their LCM and HCF is 1740. If one of the numbers is 420, the other number is

a)   150
b)   225
c)   180
d)   240
e)   None of these

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Explanation: 

LCM = 28 HCF 

LCM + HCF = 1740

 => 28 HCF + HCF = 1740 

 HCF = 1740/29 = 60 

therefore LCM = 28X60 

 if a and b are two numbers, then LCM of (a & b) X HCF of (a & b) = Product of number (aXb) 

 So, apply the above formula 28X60X60 = x X 420

 = > x = 240 

 So, the other number is 240

8. Two persons A and B walk around a circular track whose radius is 1.4 km. A walks at a speed of 176 meters per minute while B walks at a speed of 110 meters per minute. if they both start at the same time, from the same point and walk in the same direction, at what interval of time would they both be at the same starting point again? (in Hours)

a)   6 2/3
b)   2 1/3
c)   5 1/4
d)   3 2/3
e) None of these

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Explanation: 

Circumference of the track is 2X(pie)X(R) = 2 X (22/7) X 1400m = 8800 m 

Time taken by A to complete one round = (8800m) / (176m/min) = (8800X60)/(176) = 3000 sec 

Time taken by B to complete one round = (8800) / (110m/min) = (8800X60) / (110) = 4800 sec 

Time they meet together at the starting point is LCM of 3000 and 4800 sec i.e., 24000 sec = 6 2/3 hours 

therefore, they meet at the starting point after 6 2/3 hours

9. Find the least number which when divided by 8, 9, 15, 24, 32 and 36 leaves remainders 3, 4, 10, 19, 27 and 31 respectively?

a)   2880
b)   2885
c)   2974
d)   2875
e)   None of these

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Explanation:  

The difference between the numbers and their remainders is same i.e., 8-3 = 9-4 = 15-10 = 24-19 = 32-27 = 36-31 = 5 

So, required answer is LCM (8, 9, 15, 24, 32, 36) - 5 

 = > 2880 - 5 = 2875

10. Find the greatest number which when divide 357, 192 and 252 leaves same remainder in each case

a)   45
b)   1
c)   15
d)   Cant be determinde
e)    None of these

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Explanation:

Required answer is HCF (357-192, 192-252, 357-252)

 = > HCF (165, 60, 105) = 15

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