Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 14 Answers Solutions Chapter 3 HCF and LCM.

## HCF and LCM Class 7 Practice Set 14 Answers Solutions Chapter 3

Question 1.

Choose the right option.

i. The HCF of 120 and 150 is __

(A) 30

(B) 45

(C) 20

(D) 120

Solution:

(A) 30

Hint:

120 = 2 x 2 x 2 x 3 x 5

150 = 2 x 3 x 5 x 5

∴ HCF of 120 and 150 = 2 x 3 x 5 = 30

ii. The HCF of this pair of numbers is not 1.

(A) 13,17

(B) 29,20

(C) 40, 20

(D) 14, 15

Solution:

(C) 40, 20

Hint:

40 = 2 x 2 x 2 x 5

20 = 2 x 2 x 5

∴ HCF of 40 and 20 = 2 x 5 = 10

Question 2.

Find the HCF and LCM.

i. 14,28

ii. 32,16

iii. 17,102,170

iv. 23,69

v. 21,49,84

Solution:

i. 14 = 2 x 7

28 = 2 x 14

= 2 x 2 x 7

∴ HCF of 14 and 28 = 2 x 7

= 14

LCM of 14 and 28 = 2 x 2 x 7

= 28

ii. 32 = 2 x 16

= 2 x 2 x 8

= 2 x 2 x 2 x 4

= 2 x 2 x 2 x 2 x 2

16 = 2 x 8

= 2 x 2 x 4

= 2 x 2 x 2 x 2

∴ HCF of 32 and 16 = 2 x 2 x 2 x 2

= 16

∴ LCM of 32 and 16 = 2 x 2 x 2 x 2 x 2

= 32

iii. 17 = 17 x 1

102 = 2 x 51

= 2 x 3 x 17

170 = 2 x 85

= 2 x 5 x 17

∴ HCF of 17, 102 and 170 = 17

∴ LCM of 17, 102 and 170 = 17 x 2 x 3 x 5

= 510

iv. 23 = 23 x 1

69 = 3 x 23

∴ HCF of 23 and 69 = 23

∴ LCM of 23 and 69 = 23 x 3

= 69

v. 21 = 3 x 7

49 = 7 x 7

84 = 2 x 42

= 2 x 2 x 21

= 2 x 2 x 3 x 7

∴ HCF of 21, 49 and 84 = 7

∴ LCM of 21, 49 and 84 = 7 x 3 x 7 x 2 x 2

= 588

Question 3.

Find the LCM.

i. 36, 42

ii. 15, 25, 30

iii. 18, 42, 48

iv. 4, 12, 20

v. 24, 40, 80, 120

Solution:

i. 36, 42

∴ LCM of 36 and 42 = 2 x 3 x 2 x 3 x 7

= 252

ii. 15, 25, 30

∴ LCM of 15, 25 and 30 = 5 x 3 x 5 x 2

= 150

iii. 18, 42, 48

∴ LCM of 18,42 and 48 = 2 x 3 x 2 x 2 x 3 x 7 x 2

= 1008

iv. 4, 12, 20

∴ LCM of 4, 12 and 20 = 2 x 2 x 3 x 5

= 60

v. 24, 40, 80, 120

∴ LCM of 24, 40, 80 and 120 = 2 x 2 x 2 x 5 x 3 x 2

= 240

Question 4.

Find the smallest number which when divided by 8,9,10,15,20 gives a remainder of 5 every time.

Solution:

Here, the smallest number for division is LCM of 8, 9, 10,15 and 20.

8 = 2 x 2 x 2

9 = 3 x 3

10 = 2 x 5

15 = 3 x 5

20 = 2 x 2 x 5

LCM of given numbers = 2 x 2 x 2 x 3 x 3 x 5 = 360

∴ Required, smallest number = LCM + Remainder

= 360 + 5

= 365

∴ The required smallest number is 365.

Question 5.

Reduce the fractions \(\frac{348}{319}, \frac{221}{247}, \frac{437}{551}\) to the lowest terms.

Solution:

i.

ii.

iii.

Question 6.

The LCM and HCF of two numbers are 432 and 72 respectively. If one of the numbers is 216, what is the other?

Solution:

Here, LCM = 432, HCF = 72, First number = 216

First number x Second number = LCM x HCF

∴ 216 x Second number = 432 x 72

∴ Second number = \(\frac{432 \times 72}{216}=432 \times \frac{72}{216}=432 \times \frac{1}{3}=144\)

∴ The other number is 144.

Question 7.

The product of two two-digit numbers is 765 and their HCF is 3. What is their LCM?

Solution:

Here, HCF = 3, Product of the given numbers = 765

Now, HCF x LCM = Product of the given numbers

∴ 3 x LCM = 765

∴ LCM = \(\frac { 765 }{ 3 }\) = 255

∴ The LCM of the two two-digit numbers is 255.

Question 8.

A trader has three bundles of string 392 m, 308 m and 490 m long. What is the greatest length of string that the bundles can be cut up into without any left over string?

Solution:

The required greatest length of the string is the highest common factor (HCF) of 392, 308 and 490.

∴ 392 = 2 x 2 x 2 x 7 x 7

308 = 2 x 2 x 7 x 11

490 = 2 x 7 x 7 x 5

∴ HCF of 392, 308 and 490 = 2 x 7

= 14

∴ The required greatest length of the string is 14 m.

Question 9.

Which two consecutive even numbers have an LCM of 180?

Solution:

LCM of two consecutive even numbers = 180

But, HCF of two consecutive even numbers = 2

Now, product of the given number = HCF x LCM

= 2 x 180

= 360

To find the two consecutive even numbers, we have to factorize 360.

360 = 2 x 2 x 2 x 3 x 3 x 5

360 = (2 x 3 x 3) x (2 x 2 x 5)

= 18 x 20

∴ The two consecutive even numbers whose LCM is 180 are 18 and 20.