Difference between revisions of "1977 AHSME Problems/Problem 3"
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| + | == Problem 3 == | ||
| + | |||
| + | A man has \$2.73 in pennies, nickels, dimes, quarters and half dollars. If he has an equal number of coins of each kind, then the total number of coins he has is | ||
| + | |||
| + | <math>\text{(A)}\ 3 \qquad | ||
| + | \text{(B)}\ 5 \qquad | ||
| + | \text{(C)}\ 9 \qquad | ||
| + | \text{(D)}\ 10 \qquad | ||
| + | \text{(E)}\ 15 </math> | ||
| + | |||
| + | |||
==Solution== | ==Solution== | ||
Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
| − | Denote the number of pennies, nickels, dimes, quarters, and half-dollars by <math>x</math>. Then <math>x+5x+10x+25x+50x = 273</math>, and <math>x = 3</math>. The total amount of coins is <math>5x</math>, which is <math>\textbf{ | + | Denote the number of pennies, nickels, dimes, quarters, and half-dollars by <math>x</math>. Then <math>x+5x+10x+25x+50x = 273</math>, and <math>x = 3</math>. The total amount of coins is <math>5x</math>, which is <math>\textbf{\text{(E)}}\ 15</math>. |
Latest revision as of 11:25, 21 November 2016
Problem 3
A man has $2.73 in pennies, nickels, dimes, quarters and half dollars. If he has an equal number of coins of each kind, then the total number of coins he has is
Solution
Solution by e_power_pi_times_i
Denote the number of pennies, nickels, dimes, quarters, and half-dollars by 
. Then 
, and 
. The total amount of coins is 
, which is 
.
