Difference between revisions of "2014 AMC 10B Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | If Leah has 1 more nickel, she has 14 total coins. Because she has the same number of nickels and pennies, she has 7 nickels and 7 pennies. This is after the nickel has been added, so we must subtract 1 nickel to get 6 nickels and 7 pennies. Therefore, Leah has <math>6\cdot5+7=\boxed{37 (\textbf{ | + | If Leah has 1 more nickel, she has 14 total coins. Because she has the same number of nickels and pennies, she has 7 nickels and 7 pennies. This is after the nickel has been added, so we must subtract 1 nickel to get 6 nickels and 7 pennies. Therefore, Leah has <math>6\cdot5+7=\boxed{37 (\textbf{C})}</math> cents. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|before=First Problem|num-a=2}} | {{AMC10 box|year=2014|ab=B|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:46, 20 February 2014
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 1 more nickel, she has 14 total coins. Because she has the same number of nickels and pennies, she has 7 nickels and 7 pennies. This is after the nickel has been added, so we must subtract 1 nickel to get 6 nickels and 7 pennies. Therefore, Leah has 
cents.
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 | ||
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