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Difference between revisions of "2006 AMC 12B Problems/Problem 14"
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== Problem ==
 
== Problem ==
 +
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> cents per glob and <math>J</math> blobs of jam at <math>5</math> cents per glob. 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 all positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches?
  
 +
<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>
 
== Solution ==
 
== Solution ==
 +
From the given, we know that
 +
 +
<math>253=N(4B+5J)</math>
 +
(The numbers are in cents)
 +
 +
since <math>253=11\cdot23</math>, and since <math>N</math> is an integer, then <math>4B+5J=11</math> or <math>23</math>. It is easily deduced that <math>11</math> is impossible to make with <math>B</math> and <math>J</math> integers, so <math>N=11</math> and <math>4B+5J=23</math>. Then, it can be guessed and checked quite simply that if <math>B=2</math> and <math>J=3</math>, then <math>4B+5J=4(2)+5(3)=23</math>. The problem asks for the total cost of jam, or <math>N(5J)=11(15)=165</math> cents, or <math>1.65\implies\mathrm{(D)}</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2006|ab=B|num-b=13|num-a=15}}
 +
{{MAA Notice}}

Latest revision as of 16:21, 8 December 2013

Problem

Elmo makes $N$ sandwiches for a fundraiser. For each sandwich he uses $B$ globs of peanut butter at $4$ cents per glob and $J$ blobs of jam at $5$ cents per glob. The cost of the peanut butter and jam to make all the sandwiches is $2.53$. Assume that $B$, $J$ and $N$ are all 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

From the given, we know that

$253=N(4B+5J)$ (The numbers are in cents)

since $253=11\cdot23$, and since $N$ is an integer, then $4B+5J=11$ or $23$. It is easily deduced that $11$ is impossible to make with $B$ and $J$ integers, so $N=11$ and $4B+5J=23$. Then, it can be guessed and checked quite simply that if $B=2$ and $J=3$, then $4B+5J=4(2)+5(3)=23$. The problem asks for the total cost of jam, or $N(5J)=11(15)=165$ cents, or $1.65\implies\mathrm{(D)}$

See also

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 12 Problems and Solutions

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