# H.C.F and L.C.M Aptitude Test Paper 4

16) There are two numbers. HCF of both the numbers is 11, and their LCM is 693. If the first number is 77, find the second number?

1. 89
2. 56
3. 78
4. 99

Explanation:

The product of two numbers = HCF * LCM

Let the required number is = x

So, 77 * x = 11 * 693

Hence, x=99.

17) Find the greatest number which divides 34, 90, and 104 and leaves the same remainder in each case.

1. 15
2. 17
3. 14
4. 18

Explanation:

Difference between numbers = 90-34 = 56, and 104- 90=14

HCF of 56 and 14 = 14

Hence, 14 is the largest number which divides the given number and leaves the same remainder in each case.

18) What is the least multiple of 7 which when divided by 6, 9, 15, and 18 respectively, leaves 4 as the remainder in each case?

1. 364
2. 255
3. 385
4. 266

Explanation:

Take LCM of 6, 9, 15, and 18. Their LCM = 90
As per given information, (90* x+4)/7

(90 *x)/ 7 = ((7 * 12 *x) + 6x + 4)/7
Or, (7*12x)/7 + (6x + 4)/7

x=4 satisfy the condition.
Hence, the required number is 90 * 4 + 4 = 364.

19) Which least number should be subtracted from 1936 so that the resulting number, when divided by 9, 10, and 15, will leave the same remainder 8 in each case?

1. 39
2. 38
3. 37
4. 36

Explanation:

Take LCM of 9, 10, and 15. Their LCM is = 90

If the number 1936 is divided by 90, we get the remainder 46, but we need 8 as remainder.

Hence, the required number is 46 - 8 = 38.

20) Find the least number which when divided by 5, 6, 7, and 8 leaves remainder 3, but when divided by 9 leaves remainder 0.

1. 843
2. 1683
3. 2523
4. 3363

Explanation:

Take LCM of 5, 6, 7, and 8. It will be = 840.

Let the number 840*x + 3 is completely divisible by 9 if the sum of digits is divisible by 9.

Let x= 2, the number will be 1683, and the sum of its digits is equals to 18 which is divisible by 9. So, the required number is 1683.

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