HCF and LCM Class 7 Practice Set 14 Answers Solutions Chapter 3
Solution & Step-by-Step Answer:
(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
Solution & Step-by-Step Answer:
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
Solution & Step-by-Step Answer:
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
Solution & Step-by-Step Answer:
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.
Solution & Step-by-Step Answer:
i.
ii.
iii.
Solution & Step-by-Step Answer:
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 = ∴ The other number is 144.
Solution & Step-by-Step Answer:
Here, HCF = 3, Product of the given numbers = 765 Now, HCF x LCM = Product of the given numbers ∴ 3 x LCM = 765 ∴ LCM = = 255 ∴ The LCM of the two two-digit numbers is 255.
Solution & Step-by-Step Answer:
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.
Solution & Step-by-Step Answer:
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.