Question

5. The product of the H.C.F. and the L.C.M. of two numbers is 1152. If one number is 48, find
the other one.
6. () Find the smallest number that is completely divisible by 28 and 42.​

Answer 1

5) L.C.M. × H.C.F. = first no. × second no.

1152 = 48 × y

y = 1152/48

y = $\huge\red{24}$

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Answer 2

5. The product of the H.C.F. and the L.C.M. of two numbers is 1152. If one number is 48, findthe other one.

Solution:

LCM ×HCF =first number ×second number

1152 = 48 × x

x = 1152/ 48

x = 24

Therefore, the other number is 24 respectively.

6. Find the smallest number that is completely divisible by 28 and 42.

Solution:

Factors of 28 = 2 × 2× 7

Factors of 42 = 2× 3× 7

Common factors = 2× 7

Uncommon factors = 2 × 3

Total= 2×2×3×7

= 84

Therefore, 84 is the smallest number that is completely divisible by 28 and 42.

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