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Difference between revisions of "1990 AJHSME Problems/Problem 13"

(Created page with '==Problem== One proposal for new postage rates for a letter was <math>30</math> cents for the first ounce and <math>22</math> cents for each additional ounce (or fraction of an …')
 
 
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==Solution==
 
==Solution==
  
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After the first ounce, there are <math>3.5</math> ounces left.  Thus, it is required to pay for <math>3</math> ounces and then the extra half.
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After the first ounce, there are <math>3.5</math> ounces left.  Since each additional ounce or fraction of an ounce adds <math>22</math> cents to the total cost, we need to add <math>4\times 22</math> to the cost for the first ounce.
  
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The total price is <math>30+4\times 22=118</math> cents, which is choice <math>\boxed{\text{C}}</math>.
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So, the total price is <math>30+4\times 22=118</math> cents. The answer is choice <math>\boxed{\textbf{(C)}~\mathbf{1.18}~\textbf{dollars}}</math>.
  
 
==See Also==
 
==See Also==
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{{AJHSME box|year=1990|num-b=12|num-a=14}}
 
{{AJHSME box|year=1990|num-b=12|num-a=14}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 17:01, 8 August 2018

Problem

One proposal for new postage rates for a letter was $30$ cents for the first ounce and $22$ cents for each additional ounce (or fraction of an ounce). The postage for a letter weighing $4.5$ ounces was

$\text{(A)}\ \text{96 cents} \qquad \text{(B)}\ \text{1.07 dollars} \qquad \text{(C)}\ \text{1.18 dollars} \qquad \text{(D)}\ \text{1.20 dollars} \qquad \text{(E)}\ \text{1.40 dollars}$

Solution

After the first ounce, there are $3.5$ ounces left. Since each additional ounce or fraction of an ounce adds $22$ cents to the total cost, we need to add $4\times 22$ to the cost for the first ounce.

So, the total price is $30+4\times 22=118$ cents. The answer is choice $\boxed{\textbf{(C)}~\mathbf{1.18}~\textbf{dollars}}$.

See Also

1990 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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