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

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==Problem 8==
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==Problem==
 
Gilda has a bag of marbles. She gives <math>20\%</math> of them to her friend Pedro. Then Gilda gives <math>10\%</math> of what is left to another friend, Ebony. Finally, Gilda gives <math>25\%</math> 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?
 
Gilda has a bag of marbles. She gives <math>20\%</math> of them to her friend Pedro. Then Gilda gives <math>10\%</math> of what is left to another friend, Ebony. Finally, Gilda gives <math>25\%</math> 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?
  
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==Solution 1==
 
==Solution 1==
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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.~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==
 
Suppose Gilda has 100 marbles.  
 
Suppose Gilda has 100 marbles.  
  
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Then she gives Pedro 20% of 100 = 20, she remains with 80 marbles.
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Then, she gives Pedro 20% of 100 = 20, she remains with 80 marbles.
  
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Out of 80 marbles she gives 10% of 80 = 8 to Ebony.  
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Out of 80 marbles, she gives 10% of 80 = 8 to Ebony.  
  
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Thus she remains with 72 marbles.  
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Thus, she remains with 72 marbles.  
  
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Then she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54.  
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Then, she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54.  
  
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And <math>\frac{54}{100}</math>=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
  
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==See Also==
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==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 <math>\boxed{\textbf{(E)}\ 54}</math>.
 +
 
 +
== Video Solution ==
 +
 
 +
==Video Solution by Math-X (Do percentage questions like this!!!)==
 +
https://youtu.be/IgpayYB48C4?si=_eGyqshtrNAg3tAm&t=2362
 +
 
 +
~Math-X
 +
 
 +
The Learning Royal : https://youtu.be/8njQzoztDGc
 +
== Video Solution 2 ==
 +
 
 +
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=WAmvVQzuzfc&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=9
 +
 
 +
==Video Solution 3==
 +
https://youtu.be/Cwnq8XRxJ6Y
 +
Soo, DRMS, NM
 +
 
 +
==Video Solution 4==
 +
https://youtu.be/eApJGrvDT5U
 +
 
 +
~savannahsolver
 +
 
 +
== Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/yAdewhYZ3Uc
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 
 +
~Hayabusa1
 +
 
 +
==See also==  
 
{{AMC8 box|year=2019|num-b=7|num-a=9}}
 
{{AMC8 box|year=2019|num-b=7|num-a=9}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:30, 9 November 2024

Problem

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

Video Solution by Math-X (Do percentage questions like this!!!)

https://youtu.be/IgpayYB48C4?si=_eGyqshtrNAg3tAm&t=2362

~Math-X

The Learning Royal : https://youtu.be/8njQzoztDGc

Video Solution 2

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

Video Solution 3

https://youtu.be/Cwnq8XRxJ6Y Soo, DRMS, NM

Video Solution 4

https://youtu.be/eApJGrvDT5U

~savannahsolver

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/yAdewhYZ3Uc

~Education, the Study of Everything

Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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

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