Difference between revisions of "2019 AMC 8 Problems/Problem 8"
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And \frac{54}{100} = 54% = <math>\boxed{\textbf{(E)}\ 54}</math> | And \frac{54}{100} = 54% = <math>\boxed{\textbf{(E)}\ 54}</math> | ||
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==See Also== | ==See Also== | ||
Revision as of 13:48, 20 November 2019
Contents
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 20% of the marbles to Pedro, she has 80% of the marbles left. If she then gives 10% of what's left to Ebony, she has (0.8*0.9) = 72% of what she had at the beginning. Finally, she gives 25% of what's left to her brother, so she has (0.75*0.72) 
. of what she had in the beginning left.~heeeeeeeheeeeee
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 \frac{54}{100} = 54% =
~phoenixfire
See Also
| 2019 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 8 | |
| 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. 
