2019 AMC 8 Problems/Problem 8

Problem 8

Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ 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?

$\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$

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 $(0.75*0.72)$ $\boxed{\textbf{(E)}\ 54}$. 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 $\frac{54}{100}$=54%=$\boxed{\textbf{(E)}\ 54}$

~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 $\boxed{\textbf{(E)}\ 54}$.

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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

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