Solution - Long multiplication
Step-by-step explanation
1. Rewrite the numbers from top to bottom aligned to the right
Place value | tens | ones | . | tenths |
1 | 0 | . | 5 | |
× | 3 | |||
Ignore the decimal points and multiply as if these are whole numbers (as if each most right digit is the ones digit):
In this case we removed 1 decimal place(s). So once calculated, the result will be reduced by the factor of 10.
Place value | hundreds | tens | ones |
1 | 0 | 5 | |
× | 3 | ||
2. Multiply the numbers using long multiplication method
Start by multiplying the ones digit (3) of the multiplier 3 by each digit of the multiplicand 105, from right to left.
Multiply the ones digit (3) of the multiplicator by the number in the ones place value:
3×5=15
Write 5 in the ones place.
Because the result is greater than 9, carry the 1 to the tens place.
Place value | hundreds | tens | ones |
1 | |||
1 | 0 | 5 | |
× | 3 | ||
5 |
Multiply the ones digit (3) of the multiplicator by the number in the tens place value and add the carried number (1):
3×0+1=1
Write 1 in the tens place.
Place value | hundreds | tens | ones |
1 | |||
1 | 0 | 5 | |
× | 3 | ||
1 | 5 |
Multiply the ones digit (3) of the multiplicator by the number in the hundreds place value:
3×1=3
Write 3 in the hundreds place.
Place value | hundreds | tens | ones |
1 | |||
1 | 0 | 5 | |
× | 3 | ||
3 | 1 | 5 |
Because we have 1 digit(s) to the right of the decimal point in the numbers that are being multiplied, we move the decimal point 1 time(s) to the left (reducing the result by the factor of 10) to get the final result:
The solution is: 31.5
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