| Board | CBSE |
| Textbook | NCERT |
| Class | Class 6 |
| Subject | Maths |
| Chapter | Chapter 2 Whole Numbers |
Question – Find the value of the following
a 297 × 17 + 297 × 3
b. 54279 × 92 + 8 × 54279
c. 81265 × 169 – 81265 × 69
d. 3845 × 5 × 782 + 769 × 25 × 218
Solution
(a) 297 × 17 + 297 × 3
= 297 × (17 + 3) [By distributive property i.e., (a+ b)× c = a× c + c× b]
Adding the numbers in the parentheses, we get
= 297 × 20
= 5,940
(b) 54279 × 92 + 8 × 54279
= 54279 × (92 + 8)
Adding the numbers in the parentheses, we get
= 54279 × 100
= 54,27,900
(c) 81265 × 169 – 81265 × 69
= 81265 × (169 – 69)
Subtracting the numbers in the parentheses, we get
= 81265 × 100
= 81,26,500
(d) 3845 × 5 × 782 + 769 × 25 × 218
= 3845 × 5 × 782 + (769 × 5) × 5 × 218
Multiplying the numbers in the parentheses, we get
= 3845 × 5 × 782 + 3845 × 5 × 218
= 3845 × 5 × (782 + 218)
Adding the numbers in the parentheses, we get
= 3845 × 5 × 1000
= 3845 × 5000
= 1,92,25,000
Related Questions –
- Find the sum by suitable rearrangement: (a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647
- Find the product by suitable rearrangement
- Find the value of the following a. 297 × 17 + 297 × 3 b. 54279 × 92 + 8 × 54279 c. 81265 × 169 – 81265 × 69 d. 3845 × 5 × 782 + 769 × 25 × 218
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