Difference between revisions of "UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1"
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== Solution == | == Solution == | ||
+ | We set up the following equations: | ||
+ | |||
+ | <math>b+r+y+a+n= 21</math> | ||
+ | |||
+ | <math>b+a+2r+y= 25</math> | ||
+ | |||
+ | <math>s+2a+r+h= 18</math> | ||
+ | |||
+ | We are asked to find the price of "Shana", or <math>s+h+2a+n</math>. We notice that this expression has no <math>b</math> or <math>y</math>, so we subtract the first equation from the second to eliminate those variables: | ||
+ | |||
+ | <math>r-n=4</math> | ||
+ | |||
+ | <math>r=n+4</math> | ||
+ | |||
+ | Which we substitute into the third equation | ||
+ | |||
+ | <math>s+2a+n+4+h=18</math>. | ||
+ | |||
+ | <math>s+2a+n+h=14</math>. | ||
+ | |||
+ | This matches the expression we need to solve for, so we are done. <math>\boxed{14}</math>. | ||
+ | |||
+ | |||
+ | ==Slick Solution== | ||
+ | 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: | ||
+ | <cmath>21+18=25+x</cmath> | ||
+ | Solving for <math>x</math>, our answer is <math>\boxed{14}</math>. | ||
+ | |||
+ | ==Credits== | ||
+ | *Credit to Football017 for writing the solution | ||
+ | *Credit to yofro for writing the second solution | ||
== See also == | == See also == |
Latest revision as of 23:44, 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 to monogram BRYAN,
to monogram BARRY and
to monogram SARAH,
how much does it cost to monogram SHANA?
Solution
We set up the following equations:
We are asked to find the price of "Shana", or . We notice that this expression has no
or
, so we subtract the first equation from the second to eliminate those variables:
Which we substitute into the third equation
.
.
This matches the expression we need to solve for, so we are done. .
Slick Solution
Notice that BRYAN is simple BARRY rearranged, and SARAH is simple SHANA rearranged. Let the cost of rearranging SHANA be . We have the following equation:
Solving for
, our answer is
.
Credits
- Credit to Football017 for writing the solution
- Credit to yofro for writing the second solution
See also
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 |