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> | ||
| − | + | ==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>. | 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>. | ||
| − | + | ==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 <math>\boxed{\textbf{(D) } 40}</math>. We can check that 40 blueberry and 20 cherry jelly beans indeed does work. | 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. | ||
| − | + | ==Solution 3== | |
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>. | 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>. | ||
Latest revision as of 17:21, 17 January 2021
Contents
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?
Solution 1
Denote the number of blueberry and cherry jelly beans as 
and 
respectively. Then 
and 
. Substituting, we have 
, so 
, 
.
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 
. We can check that 40 blueberry and 20 cherry jelly beans indeed does work.
Solution 3
Note that the number of jellybeans minus 
is a multiple of 
and positive. Therefore, the number of jellybeans is a multiple of and 
. The only answer choice satisfying this is .
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
