2006 AMC 10B Problems/Problem 22

Revision as of 16:18, 26 July 2019 by Snorelaxrules (talk | contribs) (Solution 1)

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$ (Error compiling LaTeX. ! Missing $ inserted.)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$and the third and fourth pair have no positive integer solutions for$B$and$J$.

So,$ (Error compiling LaTeX. ! Missing $ inserted.)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$$ (Error compiling LaTeX. ! Missing $ inserted.)= $1.65 \Rightarrow 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 $3\cdot11\cdot{0.05}\implies 1.65 \implies\boxed{D}$.

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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. AMC logo.png

Invalid username
Login to AoPS