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Difference between revisions of "2017 AMC 10B Problems/Problem 5"

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<math>\textbf{(A)}\ 10\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50</math>
 
<math>\textbf{(A)}\ 10\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50</math>
  
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==Solution==
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==Solution 1==
  
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===Solution 1===
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Denote the number of blueberry and cherry jelly beans as <math>b</math> and <math>c</math> respectively. Then <math>b = 2c</math> and <math>b-10 = 3(c-10)</math>. Substituting, we have <math>2c-10 = 3c-30</math>, so <math>c=20</math>, <math>b=\boxed{\textbf{(D) } 40}</math>.
  
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Denote the number of blueberry and cherry jelly beans as <math>b</math> and <math>c</math> respectively. Then <math>b = 2c</math> and <math>b-10 = 3(c-10)</math>. Substituting, we have <math>2c-10 = 3c-30</math>, so <math>c=20</math>, <math>b=\boxed{\textbf{(D) } 40}</math>.
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==Solution 2==
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From the problem, we see that 10 less than one of the answer choices must be a multiple of 3 and positive. The only answer choice satisfying this is <math>\boxed{\textbf{(D) } 40}</math>. We can check that 40 blueberry and 20 cherry jelly beans indeed does work.
  
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===Solution 2===
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==Solution 3==
  
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From the problem, we see that 10 less than one of the answer choices must be a multiple of 3 and positive. The only answer choice satisfying this is <math>\boxed{\textbf{(D) } 40}</math>. We can check that 40 blueberry and 20 cherry jelly beans indeed does work.  
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Note that the number of jellybeans minus <math>2 \cdot 10 = 20</math> is a multiple of <math>4</math> and positive. Therefore, the number of jellybeans is a multiple of <math>4</math> and <math>>20</math>. The only answer choice satisfying this is <math>\boxed{\textbf{(D) } 40}</math>.
  
 
==Video Solution==
 
==Video Solution==
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==Video Solution by TheBeautyofMath==
 
==Video Solution by TheBeautyofMath==
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With new whiteboard at home: https://www.youtube.com/watch?v=zTGuz6EoBWY?t=774
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With new whiteboard at home: https://youtu.be/zTGuz6EoBWY?t=774
  
 
~IceMatrix
 
~IceMatrix

Latest revision as of 18:21, 17 January 2021

Problem

Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating 10 pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50$

Solution 1

Denote the number of blueberry and cherry jelly beans as $b$ and $c$ respectively. Then $b = 2c$ and $b-10 = 3(c-10)$. Substituting, we have $2c-10 = 3c-30$, so $c=20$, $b=\boxed{\textbf{(D) } 40}$.

Solution 2

From the problem, we see that 10 less than one of the answer choices must be a multiple of 3 and positive. The only answer choice satisfying this is $\boxed{\textbf{(D) } 40}$. We can check that 40 blueberry and 20 cherry jelly beans indeed does work.

Solution 3

Note that the number of jellybeans minus $2 \cdot 10 = 20$ is a multiple of $4$ and positive. Therefore, the number of jellybeans is a multiple of $4$ and $>20$. The only answer choice satisfying this is $\boxed{\textbf{(D) } 40}$.

Video Solution

https://youtu.be/QrCSysVHhPU

~savannahsolver

Video Solution by TheBeautyofMath

With new whiteboard at home: https://youtu.be/zTGuz6EoBWY?t=774

~IceMatrix

See also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 AMC 10 Problems and Solutions

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