Difference between revisions of "2009 AMC 10A Problems/Problem 8"

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== Solution ==
 
== Solution ==
  
A senior ticket costs <dollar/><math>6.00</math>, so a regular ticket costs <math>6 \cdot \frac{1}{\frac{3}{4}}\:=\:6\cdot\frac{4}{3}\:=\:8</math> dollars. Therefore children's tickets cost half that, or <dollar/><math>4.00</math>, so we have:
+
A senior ticket costs <math>$6.00</math>, so a regular ticket costs <math>6 \cdot \frac{1}{\frac{3}{4}}\:=\:6\cdot\frac{4}{3}\:=\:8</math> dollars. Therefore children's tickets cost half that, or <math>$4.00</math>, so we have:
  
 
<math>2(6+8+4)\:=\:36</math>
 
<math>2(6+8+4)\:=\:36</math>

Revision as of 15:23, 30 December 2016

Problem 8

Three Generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a $50$% discount as children. The two members of the oldest generation receive a $25\%$ discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs <dollar/>$6.00$, is paying for everyone. How many dollars must he pay?

$\mathrm{(A)}\ 34 \qquad \mathrm{(B)}\ 36 \qquad \mathrm{(C)}\ 42 \qquad \mathrm{(D)}\ 46 \qquad \mathrm{(E)}\ 48$

Solution

A senior ticket costs $$6.00$, so a regular ticket costs $6 \cdot \frac{1}{\frac{3}{4}}\:=\:6\cdot\frac{4}{3}\:=\:8$ dollars. Therefore children's tickets cost half that, or $$4.00$, so we have:

$2(6+8+4)\:=\:36$

So Grandfather Wen pays $$36$, or $B$.

See Also

2009 AMC 10A (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 AMC 10 Problems and Solutions

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