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

Latest revision as of 17:21, 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

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.

@@ Line 5: / Line 5: @@
 <math>\textbf{(A)}\ 10\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50</math>
-==Solution==
+==Solution 1==
-===Solution 1===
+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.
-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 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>.
+==Video Solution==
+https://youtu.be/QrCSysVHhPU
+~savannahsolver
-===Solution 2===
+==Video Solution by TheBeautyofMath==
+With new whiteboard at home: https://youtu.be/zTGuz6EoBWY?t=774
-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.
+~IceMatrix
+==See also==
 {{AMC10 box|year=2017|ab=B|num-b=4|num-a=6}}
 {{MAA Notice}}

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