Difference between revisions of "UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1"

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This matches the expression we need to solve for, so we are done. <math>\boxed{14}</math>.
 
This matches the expression we need to solve for, so we are done. <math>\boxed{14}</math>.
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==Solution 2==
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Notice that BRYAN is simple BARRY rearranged, and SARAH is simple SHANA rearranged. Let the cost of rearranging SHANA be <math>x</math>. We have the following equation:
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<cmath>21+18=25+x</cmath>
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Solving for <math>x</math>, our answer is <math>\boxed{14}</math>.
  
 
==Credits==
 
==Credits==

Revision as of 22:43, 3 August 2020

Problem

Four siblings BRYAN, BARRY, SARAH and SHANA are having their names monogrammed on their towels. Different letters may cost different amounts to monogram. If it costs $\textdollar{21}$ to monogram BRYAN, $\textdollar{25}$ to monogram BARRY and $\textdollar{18}$ to monogram SARAH, how much does it cost to monogram SHANA?

Solution

We set up the following equations:

$b+r+y+a+n= 21$

$b+a+2r+y= 25$

$s+2a+r+h= 18$

We are asked to find the price of "Shana", or $s+h+2a+n$. We notice that this expression has no $b$ or $y$, so we subtract the first equation from the second to eliminate those variables:

$r-n=4$

$r=n+4$

Which we substitute into the third equation

$s+2a+n+4+h=18$.

$s+2a+n+h=14$.

This matches the expression we need to solve for, so we are done. $\boxed{14}$.


Solution 2

Notice that BRYAN is simple BARRY rearranged, and SARAH is simple SHANA rearranged. Let the cost of rearranging SHANA be $x$. We have the following equation: \[21+18=25+x\] Solving for $x$, our answer is $\boxed{14}$.

Credits

  • Credit to Football017 for writing the solution

See also

2014 UNM-PNM Contest II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10
All UNM-PNM Problems and Solutions
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