Art of Problem Solving

1988 AJHSME Problems/Problem 8: Difference between revisions

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<math>\text{(A)}\ .0192 \qquad \text{(B)}\ .192 \qquad \text{(C)}\ 1.92 \qquad \text{(D)}\ 19.2 \qquad \text{(E)}\ 192</math>
<math>\text{(A)}\ .0192 \qquad \text{(B)}\ .192 \qquad \text{(C)}\ 1.92 \qquad \text{(D)}\ 19.2 \qquad \text{(E)}\ 192</math>


==Solution==
==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 <math>3+2=5</math> units to the left of where it should be, so we would want <math>.19200\Rightarrow \mathrm{(B)}</math>.
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 <math>3+2=5</math> units to the left of where it should be, so we would want <math>.19200\Rightarrow \mathrm{(B)}</math>.
==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 <math>3+2=5</math>.  Then we take <math>19200.00</math> and move the decimal place to the left 5 times, then we get <math>0.19200</math>. Therefore the answer is <math>0.19200\Rightarrow \mathrm{(B)}</math>.
~ algebraic_algorithmic


==See Also==
==See Also==
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{{AJHSME box|year=1988|num-b=7|num-a=9}}
{{AJHSME box|year=1988|num-b=7|num-a=9}}
[[Category:Introductory Algebra Problems]]
[[Category:Introductory Algebra Problems]]
{{MAA Notice}}

Latest revision as of 15:27, 1 January 2025

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

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
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

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