Difference between revisions of "1990 AJHSME Problems/Problem 10"

Revision as of 23:05, 4 July 2013

Problem

On this monthly calendar, the date behind one of the letters is added to the date behind $\text{C}$ . If this sum equals the sum of the dates behind $\text{A}$ and $\text{B}$ , then the letter is

$[asy] unitsize(12); draw((1,1)--(23,1)); draw((0,5)--(23,5)); draw((0,9)--(23,9)); draw((0,13)--(23,13)); for(int a=0; a<6; ++a) { draw((4a+2,0)--(4a+2,14)); } label("Tues.",(4,14),N); label("Wed.",(8,14),N); label("Thurs.",(12,14),N); label("Fri.",(16,14),N); label("Sat.",(20,14),N); label("C",(12,10.3),N); label("$\textbf{A}$",(16,10.3),N); label("Q",(12,6.3),N); label("S",(4,2.3),N); label("$\textbf{B}$",(8,2.3),N); label("P",(12,2.3),N); label("T",(16,2.3),N); label("R",(20,2.3),N); [/asy]$

$\text{(A)}\ \text{P} \qquad \text{(B)}\ \text{Q} \qquad \text{(C)}\ \text{R} \qquad \text{(D)}\ \text{S} \qquad \text{(E)}\ \text{T}$

Solution

Looking at the positions of the letters, we see that if the date behind $\text{Q}$ is $q$ , then the date behind $\textbf{A}$ is $q-6$ and the date behind $\textbf{B}$ is $q+6$ . Thus, their sum is $2q$ .

The date behind $\text{C}$ is $q-7$ , so the desired letter is the one for which the date behind it is $2q-(q-7)=q+7$ . This is letter $\text{P}\rightarrow \boxed{\text{A}}$ .

1990 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AJHSME box|year=1990|num-b=9|num-a=11}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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Difference between revisions of "1990 AJHSME Problems/Problem 10"

Revision as of 23:05, 4 July 2013

Problem

Solution

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