Difference between revisions of "2019 AMC 8 Problems/Problem 8"

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==Solution 1==
 
==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) <math>\boxed{\textbf{(E)}\ 54}</math>. of what she had in the beginning left.~heeeeeeeheeeeee
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After Gilda gives <math>20</math>% of the marbles to Pedro, she has <math>80</math>% of the marbles left. If she then gives <math>10</math>% of what's left to Ebony, she has <math>(0.8*0.9)</math> = <math>72</math>% of what she had at the beginning. Finally, she gives <math>25</math>% of what's left to her brother, so she has <math>(0.75*0.72)</math> <math>\boxed{\textbf{(E)}\ 54}</math>. of what she had in the beginning left.
  
 
==Solution 2==
 
==Solution 2==
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Then she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54.  
 
Then she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54.  
  
And \frac{54}{100} = 54% = <math>\boxed{\textbf{(E)}\ 54}</math>
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And <math>\frac{54}{100}</math>=54%=<math>\boxed{\textbf{(E)}\ 54}</math>
  
 
~phoenixfire
 
~phoenixfire
  
==See Also==
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==Solution 3==
{{AMC8 box|year=2019|num-b=7|num-a=8}}
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(Only if you have lots of time do it this way)
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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>.
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== Video Solution ==
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Solution detailing how to solve the problem: https://www.youtube.com/watch?v=WAmvVQzuzfc&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=9
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==See also==  
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{{AMC8 box|year=2019|num-b=7|num-a=9}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:48, 23 April 2021

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

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=WAmvVQzuzfc&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=9

See also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AJHSME/AMC 8 Problems and Solutions

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