**Problem 1 : **

Find the HCF of :

(2^{3}^{ }⋅ 3^{2}^{ }⋅ 5 ⋅ 7^{4}), (2^{2}^{ }⋅ 3^{5 }⋅ 5^{2 }⋅ 7^{6}) and (2^{3 }⋅ 5^{3 }⋅ 7^{2})

**Problem 2 : **

Find the LCM of :

(2^{3}^{ }⋅ 3^{3}^{ }⋅ 5), (2^{3}^{ }⋅ 3^{2 }⋅ 5^{2}) and (2^{ }⋅ 3^{ }⋅ 5^{2})

**Problem 3 : **

Find the LCM of :

0.63, 1.05 and 2.1

**Problem 4 : **

Two number are in the ratio 15 : 11 and their HCF is 13. Find the numbers.

**Problem 5 : **

The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other number.

**Problem 6 : **

Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.

**Problem 7 : **

Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

**Problem 8 : **

Find the largest number which divides 62, 132 and 237 leaves the same remainder in each case.

**Problem 9 : **

Find the greatest number of four digits which is divisible by 15, 25, 40 and 75.

**Problem 10 :**

Two numbers are in the ratio 6 : 7. If their LCM is 84, then find the two numbers.

**Problem 1 : **

Find the HCF of :

(2^{3}^{ }⋅ 3^{2}^{ }⋅ 5 ⋅ 7^{4}), (2^{2}^{ }⋅ 3^{5 }⋅ 5^{2 }⋅ 7^{6}) and (2^{3 }⋅ 5^{3 }⋅ 7^{2})

**Solution : **

In the above given numbers, we find 2, 5 and 7 in common.

Take 2, 5 and 7 with minimum power.

That is 2^{2}, 5 and 7^{2}.

The HCF of the given numbers is

= 2^{2}^{ }⋅ 5 ⋅ 7^{2}

= 4^{ }⋅ 5 ⋅ 49

= 980

So, the HCF of the given numbers is 980.

**Problem 2 : **

Find the LCM of :

(2^{3}^{ }⋅ 3^{3}^{ }⋅ 5), (2^{3}^{ }⋅ 3^{2 }⋅ 5^{2}) and (2^{ }⋅ 3^{ }⋅ 5^{2})

**Solution : **

In the above given numbers, each base is a prime number.

Take all the different prime numbers with maximum powers.

They are 2^{3}, 3^{3} and 5^{2}.

The LCM of the given numbers is

= 2^{3}^{ }⋅ 3^{3} ⋅ 5^{2}

= 8^{ }⋅ 27 ⋅ 25

= 5400

So, the LCM of the given numbers is 5400.

**Problem 3 : **

Find the LCM of :

0.63, 1.05 and 2.1

**Solution : **

We we look at the given numbers 0.63, 1.05, and 2.1, maximum number of digits after the decimal is 2.

So, let us multiply each number by 100 to get rid of the decimal.

When the given numbers are multiplied by 100, we get 63, 105 and 210.

Now, decompose 63, 105 and 210 into prime factors.

63 = 3^{2 }⋅ 7

105 = 5 ⋅ 3 ⋅ 7

210 = 2 ⋅ 5 ⋅ 3 ⋅ 7

Take all the different prime factors with maximum powers.

They are 2, 3^{2}, 5 and 7.

The LCM of (63, 105 and 210 ) is

= 2^{ }⋅ 3^{2} ⋅ 5 ⋅ 7

= 2^{ }⋅ 9 ⋅ 5 ⋅ 7

= 630

To get L.C.M of (0.63, 1.05, 2.1), divide 630 by 100.

630 / 100 = 6.3

So, the L.C.M of (0.63, 1.05, 2.1) is 6.3

**Problem 4 : **

Two number are in the ratio 15 : 11 and their HCF is 13. Find the numbers.

**Solution : **

From the given ratio, the two numbers can be assumed as 15x and 11x.

H.C.F of 15x and 11x is x. **Given :** H.C.F of two numbers is 13.

Then, we have

x = 13

Therefore, we have

15x = 15 ⋅ 13 = 195

11x = 11 ⋅ 13 = 143

So, the two numbers are 195 and 143.

**Problem 5 : **

The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other number.

**Solution : **

Let x be the other number.

Product of two numbers = H.C.F ⋅ L.C.M

Substitute.

77 ⋅ x = 11 ⋅ 693

Divide each side 77.

x = 99

So, the other number is 99.

**Problem 6 : **

Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.

**Solution : **

Required greatest possible length is

= H.C.F of (495, 900, 1665)

To find HCF, decompose 495, 900 and 1665 into prime factors.

495 = 3^{2 }⋅ 5 ⋅ 11

900 = 2^{2 }⋅ 3^{2 }⋅ 5^{2}

1665 = 3^{2 }⋅ 5 ⋅ 37

In the prime factors of (495, 900, 1665), we find 3 and 5 in common.

Take 3 and 5 with minimum power.

They are 3^{2} and 5.

Then the H.C.F is

= 3^{2 }⋅ 5

= 9 ⋅ 5

= 45

So, the required greatest possible length is 45 cm.

**Problem 7 : **

Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

**Solution : **

Required greatest number is

= H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of (1651, 2032)

= 127

So, the required number is 127.

**Problem 8 : **

Find the largest number which divides 62, 132 and 237 leaves the same remainder in each case.

**Solution : **

Required greatest number is

= H.C.F. of (132 - 62), (237 - 132) and (237 - 62)

= H.C.F. of (70, 105 and 175)

= 35

So, the required number is 35.

**Problem 9 : **

Find the greatest number of four digits which is divisible by 15, 25, 40 and 75.

**Solution : **

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

If a four digit number is divisible by 15, 25, 40 and 75, then the same four digit number should also be divisible by their L.C.M 600.

On dividing 9999 by 600, the remainder is 399.

To find the greatest four digit number exactly divisible by 600, we need to subtract the remainder 399 from 9999.

9999 - 399 = 9600

Therefore, 9600 is exactly divisible by 600.

Because, 9600 is exactly divisible by 600, the it should also be divisible by 15, 25, 40 and 75.

So, the greatest number of four digits which is divisible by 15, 25, 40 and 75 is 9600.

**Problem 10 :**

Two numbers are in the ratio 6 : 7. If their LCM is 84, then find the two numbers.

**Solution : **

From the given ratio, the numbers can be assumed as 6x and 7x.

We can find LCM of 6x and 7x using synthetic division as given below.** **

Therefore, LCM of (6x, 7x) is

= x ⋅ 6 ⋅ 7

= 42x

**Given :** LCM of the two numbers is 84.

Then, we have

42x = 84

Divide each side by 42.

x = 2

Substitute 2 for x in 6x and 7x to find the two numbers.

6x = 6 ⋅ 2 = 12

7x = 7 ⋅ 2 = 14

So, the two numbers are 12 and 14.

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