## Aptitude questions and answers on H.C.F and L.C.M concepts

1) What is the HCF of 1095 and 1168?

1. 37
2. 73
3. 43
4. 83

Explanation:

By Division Method

Therefore the HCF is 73

2) Find the HCF of 210, 385, and 735.

1. 7
2. 14
3. 21
4. 35

Explanation:

210 = 2 x 5 x 3 x 7 x 1
385 = 5 x 7 x 11 x 1
735 = 5 x 7 x 3 x 7 x 1
HCF = 5 x 7 = 35

3) What will be the HCF of 608, 544; 638, 783; and 425, 476 respectively?

1. 32, 29, 17
2. 17, 32, 29
3. 29, 32, 17
4. 32, 17, 29

4)

Explanation:
Formula:

LCM:

LCM = 2 x 2 x 5 = 20

HCF:

3=3 x 1
6=3 x 2 x 1
9=3 x 3 x 1
27=3 x 3 x 3 x 1
HCF =3
Formula:

5)If the HCF of two numbers is 27, and their sum is 216, find these numbers.

1. 27, 189
2. 154, 162
3. 108, 108
4. 81, 189

Explanation:

Let the numbers be 27x and 27y.

Then, 27x + 27y = 216
27 (x+ y) = 216
(x + y) = 216/27
X + y = 8

co-prime numbers with sum 8 are (1, 7) and (3, 5).

Therefore we can say that the required numbers are (27 x1, 27 x 7) and (27 x 3, 27x 5), i.e. (27, 189) and (81, 135).
Out of the above two answers, the given one in the answer is the pair (27, 189).

6) Two numbers are in the ratio of 15:11. If the HCF of numbers is 13, find the numbers.

1. 75, 55
2. 105, 77
3. 15, 11
4. 195, 143

Explanation:

Let the two numbers are 15x and 11x
HCF = x
x = 13
Therefore, the numbers are (15 x 13 and 11 x 13), i.e. 195 and 143.

7) Find the greatest integer that divides 358, 376, and 334 and leaves the same remainder in each case.

1. 6
2. 7
3. 8
4. 9

Explanation:

Take the difference between two numbers

376 - 358 = 18;
376 - 334 = 42;
358 - 334 = 24;

By taking the HCF of 18, 42, 24.

18 = 2 x3 x 3
42 = 2 x 3 x 7
24 = 2 x 2 x 2 x 3
HCF = 2 x 3 = 6

Hence the required number is 6.

8) Three bells toll at intervals of 36 sec, 40 sec, and 48 sec respectively. They start singing together at a particular time. When will they toll next together?

1. 6 minutes
2. 12 minutes
3. 18 minutes
4. 24 minutes

Explanation:

Firstly find the LCM of 36, 40, and 48

LCM = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 720 seconds

Hence, after 720 seconds or 12 minutes, three bells will toll together.

9) The LCM of the two numbers is 7700, and their HCF is 11. If one of these numbers is 275, what is the other number?

1. 279
2. 283
3. 308
4. 318

Explanation:

Apply Formula: HCF x LCM = first number x Second number 11 x 7700 = 275 x second number

10) 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.

11) 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.

12) 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.

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

14) The LCM of two prime numbers x and y is 161. If x>y, find the value of 15y-7x.

1. -32
2. -56
3. 38
4. 74

Explanation:

Note: The HCF of two prime numbers = 1

LCM/HCF =161/1 =161
161=7 * 23
The ratio of numbers: 7: 23
Hence, the expression gives15 * 7 - 7 * 23 = -56

15) Find the smallest digit that will leave the same remainder 4 in each case when divided by 12, 15, 20, and 54.

1. 454
2. 540
3. 544
4. 450

Explanation:

Take LCM of 12, 15, 20, and 54. It will be = 540.

ATQ, the remainder in each case = 4.

Hence, the required number = 540+4 = 544.

16) What is the number nearest to 10000 which is exactly divisible by 3, 4, 5, 6, 7, and 8?

1. 9956
2. 10080
3. 10096
4. 9240

Explanation:

Take LCM of 3, 4, 5, 6, 7, and 8. It will be = 840.

Now, 10000/840 = 760 (remainder).

840 - 760 = 80. If we add 80 in the given number, the number is exactly divisible by 840.

So, the required number is 10080.

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