Difference between revisions of "2018 AMC 10A Problems/Problem 8"
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x &= 6. | x &= 6. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Joe has <math> | + | Joe has six <math>5</math>-cent coins, nine <math>10</math>-cent coins, and eight <math>25</math>-cent coins. Thus, our answer is |
<math>8-6 = \boxed{\textbf{(C) } 2}.</math> | <math>8-6 = \boxed{\textbf{(C) } 2}.</math> | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Video Solution== | + | ==Video Solution (HOW TO THINK CREATIVELY!)== |
+ | https://youtu.be/zbcnOfDJmQI | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | ==Video Solutions== | ||
https://youtu.be/ZiZVIMmo260 | https://youtu.be/ZiZVIMmo260 | ||
+ | |||
https://youtu.be/BLTrtkVOZGE | https://youtu.be/BLTrtkVOZGE | ||
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~savannahsolver | ~savannahsolver | ||
− | == Video Solution == | + | ==Video Solution by OmegaLearn== |
https://youtu.be/HISL2-N5NVg?t=1861 | https://youtu.be/HISL2-N5NVg?t=1861 | ||
− | ~ pi_is_3.14 | + | ~pi_is_3.14 |
== See Also == | == See Also == |
Latest revision as of 15:14, 3 July 2023
Contents
Problem
Joe has a collection of coins, consisting of
-cent coins,
-cent coins, and
-cent coins. He has
more
-cent coins than
-cent coins, and the total value of his collection is
cents. How many more
-cent coins does Joe have than
-cent coins?
Solution 1 (One Variable)
Let be the number of
-cent coins that Joe has. Therefore, he must have
-cent coins and
-cent coins. Since the total value of his collection is
cents, we can write
Joe has six
-cent coins, nine
-cent coins, and eight
-cent coins. Thus, our answer is
~Nivek
Solution 2 (Two Variables)
Let the number of -cent coins be
the number of
-cent coins be
and the number of
-cent coins be
Set up the following two equations with the information given in the problem:
From there, multiply the second equation by to get
Subtract the first equation from the multiplied second equation to get or
Substitute in for
into one of the equations to get
Finally, the answer is
- mutinykids
Solution 3 (Three Variables)
Let and
be the numbers of
-cent coins,
-cent coins, and
-cent coins in Joe's collection, respectively. We are given that
Substituting
into each of
and
and then simplifying, we have
Subtracting
from
gives
from which
Substituting this into either
or
produces
Finally, the answer is
~MRENTHUSIASM
Video Solution (HOW TO THINK CREATIVELY!)
~Education, the Study of Everything
Video Solutions
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=1861
~pi_is_3.14
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.