1990 AJHSME Problems/Problem 10

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

Let the date behind $C$ be $x$. Now the date behind $A$ is $x+1$, and after looking at the calendar, the date behind $B$ is $x+13$. Now we have $x+1+x+13=x+y$ for some date $y$, and we desire for $y$ to be $x+14$. Now we find that $y$ is the date behind $P$, so the answer is $\boxed{(\text{A})}$ ~motorfinn

See Also

1990 AJHSME (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS