Difference between revisions of "2006 AMC 10B Problems/Problem 22"
m (→Solution 2) |
|||
(13 intermediate revisions by 10 users not shown) | |||
Line 4: | Line 4: | ||
<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> | <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 1 == |
− | The peanut butter and jam for each sandwich costs <math>4B+5J\cent</math>, so the peanut butter and jam for <math>N</math> sandwiches costs <math>N(4B+5J)\cent</math>. | + | 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>. |
Setting this equal to <math>253\cent</math>: | Setting this equal to <math>253\cent</math>: | ||
Line 13: | Line 13: | ||
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> | ||
− | The first pair violates <math>N>1</math> and the third and fourth | + | 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>. |
− | So, <math>N=11</math> and <math>4B+5J=23</math> | + | So, <math>N=11</math> and <math>4B+5J=23</math>. |
− | The only integer solutions for <math>B</math> and <math>J</math> are <math>B=2</math> and <math>J=3</math> | + | The only integer solutions for <math>B</math> and <math>J</math> are <math>B=2</math> and <math>J=3</math>. |
− | Therefore the cost of the jam Elmo uses to make the sandwiches is <math>3\cdot5\cdot11=165\cent</math> <math>= \$1.65 \Rightarrow D </math> | + | 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> |
==Solution 2== | ==Solution 2== | ||
− | Note as above, you get the equation <math>N(0.04B+0.05J=2.53 | + | Note as above, you get the equation <math>N(0.04B+0.05J)=2.53</math> |
− | Notice that we can multiply by <math>100</math> on both sides to get whole numbers. Hence <math>\implies N(4B+5J=253 | + | Notice that we can multiply by <math>100</math> on both sides to get whole numbers. Hence <math>\implies N(4B+5J)=253</math> |
Note that the prime factorization of <math>253=11\cdot23</math>. | Note that the prime factorization of <math>253=11\cdot23</math>. | ||
Line 41: | Line 41: | ||
Hence, <math>B=2,J=3</math>. | Hence, <math>B=2,J=3</math>. | ||
− | Hence, the price of the | + | Hence, the price of the jam is <math>3\cdot11\cdot{0.05}\implies 1.65 \implies\boxed{D}</math>. |
+ | |||
+ | == Video Solutions == | ||
+ | Video solution: https://www.youtube.com/watch?v=7248rEcCSfM | ||
+ | |||
+ | https://www.youtube.com/watch?v=dIaBGFwwZEc ~David | ||
== See Also == | == See Also == |
Latest revision as of 10:55, 24 June 2024
Problem
Elmo makes sandwiches for a fundraiser. For each sandwich he uses
globs of peanut butter at
per glob and
blobs of jam at
per blob. The cost of the peanut butter and jam to make all the sandwiches is
. Assume that
,
, and
are positive integers with
. What is the cost of the jam Elmo uses to make the sandwiches?
Solution 1
The peanut butter and jam for each sandwich costs , so the peanut butter and jam for
sandwiches costs
.
Setting this equal to :
The only possible positive integer pairs whose product is
are:
The first pair violates and the third and fourth pairs have no positive integer solutions for
and
.
So, and
.
The only integer solutions for and
are
and
.
Therefore, the cost of the jam Elmo uses to make the sandwiches is
Solution 2
Note as above, you get the equation
Notice that we can multiply by on both sides to get whole numbers. Hence
Note that the prime factorization of .
Hence, we want or
Now, we have two cases to test.
Case 1:
Notice that we want or
Taking
Hence, .
Hence, the price of the jam is .
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 | |
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.