UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1

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. $\boxed{14}$ .

Slick Solution

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
Credit to yofro for writing the second solution

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

Page

Toolbox

Search

UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1

Contents

Problem

Solution

Slick Solution

Credits

See also