Difference between revisions of "1988 AJHSME Problems/Problem 8"
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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== | ||
Latest revision as of 15:27, 1 January 2025
Contents
Problem
Betty used a calculator to find the product 
. She forgot to enter the decimal points. The calculator showed 
. If Betty had entered the decimal points correctly, the answer would have been
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 
units to the left of where it should be, so we would want 
.
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 . Then we take 
and move the decimal place to the left 5 times, then we get 
. Therefore the answer is 
.
~ algebraic_algorithmic
See Also
| 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. 
