Difference between revisions of "1960 AHSME Problems/Problem 29"

(Solution to Problem 29)
 
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== Problem 29==
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== Problem ==
  
 
Five times <math>A</math>'s money added to <math>B</math>'s money is more than <math>\$51.00</math>. Three times <math>A</math>'s money minus <math>B</math>'s money is <math>\$21.00</math>.
 
Five times <math>A</math>'s money added to <math>B</math>'s money is more than <math>\$51.00</math>. Three times <math>A</math>'s money minus <math>B</math>'s money is <math>\$21.00</math>.

Revision as of 18:35, 12 May 2018

Problem

Five times $A$'s money added to $B$'s money is more than $$51.00$. Three times $A$'s money minus $B$'s money is $$21.00$. If $a$ represents $A$'s money in dollars and $b$ represents $B$'s money in dollars, then:

$\textbf{(A)}\ a>9, b>6 \qquad \textbf{(B)}\ a>9, b<6 \qquad \textbf{(C)}\ a>9, b=6\qquad \textbf{(D)}\ a>9, \text{but we can put no bounds on} \text{ } b\qquad \textbf{(E)}\ 2a=3b$

Solution

Use math symbols to write an equation and an inequality based on the conditions in the problem. \[5a+b>51\] \[3a-b=21\] From the second equation, $b = 3a-21$ and $a = 7 + b/3$.

Substituting $b$ in the inequality results in \[5a + 3a - 21 > 51\] \[8a > 72\] \[a > 9\]

Substituting $a$ in the inequality results in \[35 + \frac{5b}{3} + b > 51\] \[35 + \frac{8b}{3} > 51\] \[\frac{8b}{3} > 16\] \[b > 6\]

The answer is $\boxed{\textbf{(A)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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