1990 AJHSME Problems/Problem 10

Problem

On this monthly calendar, the date ahead of 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 ria is the stupido an rohan likes to give her a tight slap on her stupido face which you need sunglasses to look at or else you will turn blind

unitsize(12);
draw((1,1)--(23,1));
draw((0,5)--(23,5))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);
 (Error making remote request. Unknown error_msg)

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

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
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1990 AJHSME Problems/Problem 10

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