# HCF and LCM Question and Answer Set 1

Hi students, welcome to **Amans Maths Blogs (AMB)**. On this post, you will get the HCF and LCM Question and Answer Set 1 is the questions with solution for SSC CGL CHSL CAT and other competative exams like NTSE NSEJS NMTC etc. It will help you to practice the questions on the topics of maths as HCF and LCM.

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**HCF and LCM Questions and Answer Set 1: Ques No 1**

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainderĀ in each case

**Options:**

A. 4

B. 7

C. 9

D. 13

**HCF and LCM Questions and Answer Set 1: Ques No 2**

Find the greatest number of six digits which on beingĀ divided by 6, 7, 8, 9 and 10 leaves 4, 5, 6, 7 and 8 as remainders respectively.

**Options:**

A. 999999

B. 997999

C. 997918

D. 997789

**HCF and LCM Questions and Answer Set 1: Ques No 3**

The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:

**Options:**

A. 15 cm

B. 35 cm

C. 25 cm

D. 52 cm

**HCF and LCM Questions and Answer Set 1: Ques No 4**

Two numbers are in the ratio of 15:11. If their H.C.F. is 13, find the numbers.

**Options:**

A. 185,153

B. 195,153

C. 153,195

D. 195,143

**HCF and LCM Questions and Answer Set 1: Ques No 5**

In finding the HCF of two numbers, the last divisors is 37 and the quotients are 7, 5 and 2. Find the numbers.

**Options:**

A. 507, 3716

B. 42, 337

C. 407, 2923

D. None of these

**HCF and LCM Question and Answer Set 1: Ques No 6**

The HCF of three prime numbers will be.

**Options:**

A. Greater than 1

B. Always 1

C. An Odd Number

D. Impossible

**HCF and LCM Question and Answer Set 1: Ques No 7**

While finding the HCF of two coprime numbers by division method it was found that four successive quotients are 3, 5, 7 and 2. Find the greater number.

**Options:**

A. 297

B. 139

C. 246

D. 548

**HCF and LCM Question and Answer Set 1: Ques No 8**

Find the HCF of 111111 and 111111111?

**Options:**

A. 111

B. 1369

C. 333

D. None of these

**HCF and LCM Question and Answer Set 1: Ques No 9**

HCF of two numbers is 26 and their product is 8112. How many pairs of such numbers are possible?

**Options:**

A. 2

B. 3

C. 4

D. 5

**HCF and LCM Question and Answer Set 1: Ques No 10**

There are two positive integers X and Y such that their product is 69972 and their HCF is 7. How many pairs of (X, Y) are possible?

**Options:**

A. 16

B. 12

C. 8

D. 64

# HCF and LCM Question and Answer Set 1 : Solutions

**Answer 1: A**

**Answer 2: C**

**Answer 3: B**

**Answer 4: D**

**Answer 5: C**

Since last divisor = 37, quotient = 2, so dividend = 37 x 2 = 74

Divisor = 74, Quotient = 5 and Remainder = 37

Dividend = 74 x 5 + 37 = 407

Now, divisor = 407, quotient = 7 and remainder = 74.

So the dividend = 407 x 7 + 74 = 2923.

Thus, the numbers are 407 and 2923.

**Answer 6: B**

HCF of any number of prime numbers is always 1.

**Answer 7: C**

Since HCF of two coprime numbers = 1, So a = 2

The two coprime numbers are (36a + 5) and (115a + 16).

36a + 5 = 36×2 + 5 = 77

115a + 16 = 115×2 + 16 = 246

Greater Number = 246

**Answer 8: A**

111111 and 111111111 are both repunits having 3m digits. Hence both are divisible by 3 and 37. Hence (3 x 37) is necessarily a factor of both of the numbers. Thus HCF = 3 x 37 = 111

**Answer 9: A**

Let two numbers X and Y. Since 26 is HCF of X and Y.

So, X = 26a, Y = 26b where a and b are prime to each other.

XY = 8112

So, (26a)(26b) = 8112

So, 676ab = 8112

So, ab = 12.

Now, their product if 12 and they are prime to each other.

These are (1, 12) and (3, 4). Hence, only two pairs are possible.

**Answer 10: C**

XY = 69972

Let X = 7a and Y = 7b, where a and b are coprimes. H = 7. Now, we need to find the possible pairs of (a, b) whose product is equal to

(69972/49) = 1428 = 2^{2} x 3 x 7 x 17

There are four different prime factors in 1428.

Thus, the number of possible pairs of (a, b) is 2^{4-1} = 8