1988 AJHSME Problems/Problem 8

Problem

Betty used a calculator to find the product $0.075 \times 2.56$ . She forgot to enter the decimal points. The calculator showed $19200$ . If Betty had entered the decimal points correctly, the answer would have been

$\text{(A)}\ .0192 \qquad \text{(B)}\ .192 \qquad \text{(C)}\ 1.92 \qquad \text{(D)}\ 19.2 \qquad \text{(E)}\ 192$

Solution 1

The decimal point of 0.075 is three away from what Betty punched in, and that of 2.56 is two away. The decimal point is therefore $3+2=5$ units to the left of where it should be, so we would want $.19200\Rightarrow \mathrm{(B)}$ .

Solution 2

When you multiply one number with x number of decimal places with another number with y number of decimal places, the total number of decimal places is x+y. Using this we can can we the total number of decimal places should be $3+2=5$ . Then we take $19200.00$ and move the decimal place to the left 5 times, then we get $0.19200$ . Therefore the answer is $0.19200\Rightarrow \mathrm{(B)}$ .

~ algebraic_algorithmic

1988 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1988 AJHSME Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

See Also