Exercise 1.2(cbse)

 Question 1:

Express each number as product of its prime factors: 

Answer 1:

(i) 140=2x2x5x7=2²x5x7

(ii) 156=2x2x3x13=2²x3x13

(iii) 3825=3x3x5x5x17-3²x5²x17

(iv)5005= 5x7x11x13

(v)7429 =17x19x23

Question 2:

Find the LCM and HCF of the following pairs of integers and verify that

LCM X HCF = product of the two numbers.

Answer:2

(1) 26 and 91

26=2x13

91=7x13

HCF= 13 (highest common factor)

LCM=2x7x13=182 (every common number)

Product of the two numbers = 26x91=2366

HCFxLCM =13x182=2366

Hence, product of two numbers =HCF x LCM

Verified.

(ii)510 and 92

510= 2x3x5x17

92=2x2x23

HCF=2

LCM 2x2x3x5x17x23-23460

Product of the two numbers = 510x92=46920

HCFx LCM=2x23460 =46920

Hence, product of two numbers = HCF x LCM

Verified

(iii) 336 and 54

336= 2x2x2x2x3x7 336=2¹x3x7

54=2x3x3x3

54=2×3³

HCF=2×3=6

LCM=2¹x3¹x7=3024

Product of the two numbers 336x54=18144

HCF×LCM=6×3024= 18144

HENCE, HCF×LCM= Product of two numbers verified