Page 45 - ICSE Math 5

Example 2: Multiply 2,169 by 57.

Solution: 2,169 × 57

Multiplicand: 2,169 Multiplier: 57 = 5 tens + 7 ones
Step 1: Multiply the multiplicand by the digit at the TTh Th H T O
ones place of the multiplier. 1 4 6
2 1 6 9
2,169 × 7 ones = 2,169 × 7 = 15,183
× 7

1 5 1 8 3

Step 2: Multiply the multiplicand by the digit at the L TTh Th H T O
tens place of the multiplier. 3 4
2 1 6 9
2,169 × 5 tens = 2,169 × 50 = 1,08,450
× 5 0

1 0 8 4 5 0

Step 3: Add the products obtained in Step 1 and Step 2.
15,183 + 1,08,450 = 1,23,633

Alternatively,

L TTh Th H T O
3 4  Carry over of 2,169 × 50
1 4 6  Carry over of 2,169 × 7
2 1 6 9  Multiplicand
So, 2,169 × 57 = 1,23,633.
× 5 7  Multiplier
It is read as one lakh
1 1
1 5 1 8 3 (2,169 × 7) twenty-three thousand six
hundred thirty-three.
+ 1 0 8 4 5 0 (2,169 × 50)
1 2 3 6 3 3  Product

Example 3: Multiply 4,396 by 158.

Solution: 4,396 × 158
1 4 3 Carry over of 4,396 × 50
3 7 4 Carry over of 4,396 × 8
4 3 9 6 Multiplicand
× 1 5 8 Multiplier So, 4,396 × 158 = 6,94,568.
It is read as six lakh ninety-
2 1
3 5 1 6 8 (4,396 × 8) four thousand five hundred
sixty-eight.
2 1 9 8 0 0  (4,396 × 50)
+ 4 3 9 6 0 0  (4,396 × 100)

6 9 4 5 6 8 Product

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