Find the product by suitable rearrangement

Board CBSE
Textbook NCERT
Class  Class 6
Subject Maths
Chapter  Chapter 2 Whole Numbers
Exercise Exercise 2.2

Question – Find the product by suitable rearrangement:

(a) 2 × 1768 × 50   
(b) 4 × 166 × 25      
(c) 8 × 291 × 125
(d) 625 × 279 × 16 
(e) 285 × 5 × 60       
(f) 125 × 40 × 8 × 25

Solution

(a) 2 × 1768 × 50
On rearranging, we get
= 1768 × (2 × 50)
Multiplying the numbers in the parentheses, we get
= 1,768 × 100
= 1,76,800

(b) 4 × 166 × 25
On rearranging, we get
= 166 × (4 × 25)
Multiplying the numbers in the parentheses, we get
= 166 × 100
= 16,600

(c) 8 × 291 × 125
On rearranging, we get
= 291 × (8 × 125)
Multiplying the numbers in the parentheses, we get
= 291 × 1000
= 2,91,000

(d) 625 × 279 × 16
On rearranging, we get
= 279 × (625 × 16)
Multiplying the numbers in the parentheses, we get
= 279 × 10,000
= 27,90,000

(e) 285 × 5 × 60
On rearranging, we get
= 285 × (5 × 60)
Multiplying the numbers in the parentheses, we get
= 285 × 300
= 85,500

(f) 125 × 40 × 8 × 25
On rearranging, we get
= (125 × 8) × (40 × 25)
Multiplying the numbers in the parentheses, we get
= 1000 × 1000
= 10,00,000

Hence, the product by suitable rearrangement of 2 × 1768 × 50 is 176800, 4 × 166 × 25 is 16600, 8 × 291 × 125 is 291000, 625 × 279 × 16 is 2790000, 285 × 5 × 60 is 85500 and 125 × 40 × 8 × 25 is 1000000.

 

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