Difference between revisions of "2019 AMC 8 Problems/Problem 8"
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Since she gave away 20% and 10% of what is left and then another 25% of what is actually left, we can do 20+10+25 or 55%. But it is actually going to be a bit more than 55% because 10% of what is left is not 10% of the total amount. So the only option that is greater than 100% - 55% is <math>\boxed{\textbf{(E)}\ 54}</math>. | Since she gave away 20% and 10% of what is left and then another 25% of what is actually left, we can do 20+10+25 or 55%. But it is actually going to be a bit more than 55% because 10% of what is left is not 10% of the total amount. So the only option that is greater than 100% - 55% is <math>\boxed{\textbf{(E)}\ 54}</math>. | ||
Revision as of 21:38, 3 November 2020
Problem 8
Gilda has a bag of marbles. She gives 
of them to her friend Pedro. Then Gilda gives 
of what is left to another friend, Ebony. Finally, Gilda gives 
of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?
Solution 1
After Gilda gives 
% of the marbles to Pedro, she has 
% of the marbles left. If she then gives 
% of what's left to Ebony, she has 
= 
% of what she had at the beginning. Finally, she gives 
% of what's left to her brother, so she has 
. of what she had in the beginning left.
Solution 2
Suppose Gilda has 100 marbles.
Then she gives Pedro 20% of 100 = 20, she remains with 80 marbles.
Out of 80 marbles she gives 10% of 80 = 8 to Ebony.
Thus she remains with 72 marbles.
Then she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54.
And 
=54%=
~phoenixfire
Solution 3
(Only if you have lots of time do it this way)
Since she gave away 20% and 10% of what is left and then another 25% of what is actually left, we can do 20+10+25 or 55%. But it is actually going to be a bit more than 55% because 10% of what is left is not 10% of the total amount. So the only option that is greater than 100% - 55% is .
See also
| 2019 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
