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Difference between revisions of "2006 AMC 10B Problems/Problem 22"

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== Problem ==
 
== Problem ==
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Elmo makes <math>N</math> sandwiches for a fundraiser. For each sandwich he uses <math>B</math> globs of peanut butter at <math>4</math>¢ per glob and <math>J</math> blobs of jam at <math>5</math>¢ per blob. The cost of the peanut butter and jam to make all the sandwiches is <math>#2.53</math>. Assume that <math>B</math>, <math>J</math>, and <math>N</math> are positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches?
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Elmo makes <math>N</math> sandwiches for a fundraiser. For each sandwich he uses <math>B</math> globs of peanut butter at <math>4\cent</math> per glob and <math>J</math> blobs of jam at <math>5\cent</math> per blob. The cost of the peanut butter and jam to make all the sandwiches is <math> \$ 2.53</math>. Assume that <math>B</math>, <math>J</math>, and <math>N</math> are positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches?
  
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<math> \mathrm{(A) \ } &#1.05\qquad \mathrm{(B) \ } &#1.25\qquad \mathrm{(C) \ } &#1.45\qquad \mathrm{(D) \ } &#1.65\qquad \mathrm{(E) \ } &#1.85 </math>
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<math> \mathrm{(A) \ } \$ 1.05\qquad \mathrm{(B) \ } \$ 1.25\qquad \mathrm{(C) \ } \$ 1.45\qquad \mathrm{(D) \ } \$ 1.65\qquad \mathrm{(E) \ } \$ 1.85 </math>
  
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== Solution ==
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== Solution 1 ==
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The peanut butter and jam for each sandwich costs <math>4B+5J</math>¢, so the peanut butter and jam for <math>N</math> sandwiches costs <math>N(4B+5J)</math>¢.  
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The peanut butter and jam for each sandwich costs <math>4B\cent+5J\cent</math>, so the peanut butter and jam for <math>N</math> sandwiches costs <math>N(4B+5J)\cent</math>.  
  
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Setting this equal to <math>253</math>¢:
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Setting this equal to <math>253\cent</math>:
  
 
<math>N(4B+5J)=253=11\cdot23</math>
 
<math>N(4B+5J)=253=11\cdot23</math>
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The only possible positive integer pairs <math>(N , 4B+5J)</math> whose product is <math>253</math> are: <math> (1,253) ; (11,23) ; (23,11) ; (253,1) </math>
 
The only possible positive integer pairs <math>(N , 4B+5J)</math> whose product is <math>253</math> are: <math> (1,253) ; (11,23) ; (23,11) ; (253,1) </math>
  
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The first pair violates <math>N>1</math> and the third and fourth pair have no positive integer solutions for <math>B</math> and <math>J</math>.  
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The first pair violates <math>N>1</math> and the third and fourth pairs have no positive integer solutions for <math>B</math> and <math>J</math>.  
  
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So, <math>N=11</math> and <math>4B+5J=23</math>
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So, <math>N=11</math> and <math>4B+5J=23</math>.
  
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The only integer solutions for <math>B</math> and <math>J</math> are <math>B=2</math> and <math>J=3</math>
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The only integer solutions for <math>B</math> and <math>J</math> are <math>B=2</math> and <math>J=3</math>.
  
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Therefore the cost of the jam Elmo uses to make the sandwiches is <math>3\cdot5\cdot11=165</math>¢ <math> = </math> <math>1.65 \Rightarrow D </math>
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Therefore, the cost of the jam Elmo uses to make the sandwiches is <math>3\cdot5\cdot11=165\cent</math>  <math>=  \$1.65 \Rightarrow \boxed{\mathrm{D}}</math>
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==Solution 2==
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Note as above, you get the equation <math>N(0.04B+0.05J)=2.53</math>
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Notice that we can multiply by <math>100</math> on both sides to get whole numbers. Hence <math>\implies N(4B+5J)=253</math>
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Note that the prime factorization of <math>253=11\cdot23</math>.
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Hence, we want <math>4B+5J=23</math> or <math>11</math>
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Now, we have two cases to test.
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Case 1: <math>4B+5J=23</math>
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Notice that we want <math>B\le5</math> or <math>J\le4</math>
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Taking <math>\pmod{5}\implies 4B\equiv3\pmod{5}\implies B=2</math>
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Hence, <math>B=2,J=3</math>.
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Hence, the price of the jam is <math>3\cdot11\cdot{0.05}\implies 1.65 \implies\boxed{D}</math>.
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== Video Solutions ==
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Video solution: https://www.youtube.com/watch?v=7248rEcCSfM
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 +
https://www.youtube.com/watch?v=dIaBGFwwZEc  ~David
  
 
== See Also ==
 
== See Also ==

Latest revision as of 09:55, 24 June 2024

Problem

Elmo makes $N$ sandwiches for a fundraiser. For each sandwich he uses $B$ globs of peanut butter at $4\cent$ per glob and $J$ blobs of jam at $5\cent$ per blob. The cost of the peanut butter and jam to make all the sandwiches is $$ 2.53$. Assume that $B$, $J$, and $N$ are positive integers with $N>1$. What is the cost of the jam Elmo uses to make the sandwiches?

$\mathrm{(A) \ } $ 1.05\qquad \mathrm{(B) \ } $ 1.25\qquad \mathrm{(C) \ } $ 1.45\qquad \mathrm{(D) \ } $ 1.65\qquad \mathrm{(E) \ } $ 1.85$

Solution 1

The peanut butter and jam for each sandwich costs $4B\cent+5J\cent$, so the peanut butter and jam for $N$ sandwiches costs $N(4B+5J)\cent$.

Setting this equal to $253\cent$:

$N(4B+5J)=253=11\cdot23$

The only possible positive integer pairs $(N , 4B+5J)$ whose product is $253$ are: $(1,253) ; (11,23) ; (23,11) ; (253,1)$

The first pair violates $N>1$ and the third and fourth pairs have no positive integer solutions for $B$ and $J$.

So, $N=11$ and $4B+5J=23$.

The only integer solutions for $B$ and $J$ are $B=2$ and $J=3$.

Therefore, the cost of the jam Elmo uses to make the sandwiches is $3\cdot5\cdot11=165\cent$ $= $1.65 \Rightarrow \boxed{\mathrm{D}}$

Solution 2

Note as above, you get the equation $N(0.04B+0.05J)=2.53$

Notice that we can multiply by $100$ on both sides to get whole numbers. Hence $\implies N(4B+5J)=253$

Note that the prime factorization of $253=11\cdot23$.

Hence, we want $4B+5J=23$ or $11$

Now, we have two cases to test.

Case 1: $4B+5J=23$

Notice that we want $B\le5$ or $J\le4$

Taking $\pmod{5}\implies 4B\equiv3\pmod{5}\implies B=2$

Hence, $B=2,J=3$.

Hence, the price of the jam is $3\cdot11\cdot{0.05}\implies 1.65 \implies\boxed{D}$.

Video Solutions

Video solution: https://www.youtube.com/watch?v=7248rEcCSfM

https://www.youtube.com/watch?v=dIaBGFwwZEc ~David

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AMC 10 Problems and Solutions

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