Difference between revisions of "2014 AMC 10B Problems/Problem 1"
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==Solution== | ==Solution== | ||
If Leah has <math>1</math> more nickel, she has <math>14</math> total coins. Because she has the same number of nickels and pennies, she has <math>7</math> nickels and <math>7</math> pennies. This is after the nickel has been added, so we must subtract <math>1</math> nickel to get <math>6</math> nickels and <math>7</math> pennies. Therefore, Leah has <math>6\cdot5+7=\boxed{37 (\textbf{C})}</math> cents. | If Leah has <math>1</math> more nickel, she has <math>14</math> total coins. Because she has the same number of nickels and pennies, she has <math>7</math> nickels and <math>7</math> pennies. This is after the nickel has been added, so we must subtract <math>1</math> nickel to get <math>6</math> nickels and <math>7</math> pennies. Therefore, Leah has <math>6\cdot5+7=\boxed{37 (\textbf{C})}</math> cents. | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/0IBJKYzefMU | ||
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| + | ~Education, the Study of Everything | ||
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| + | |||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 11:06, 2 July 2023
Problem
Leah has 
coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?
Solution
If Leah has 
more nickel, she has 
total coins. Because she has the same number of nickels and pennies, she has 
nickels and pennies. This is after the nickel has been added, so we must subtract nickel to get 
nickels and pennies. Therefore, Leah has 
cents.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
