1963 TMTA High School Algebra I Contest Problem 33

Problem

From the formula $A=\frac{1}{2} H(B+C),$ the value of $B$ in terms of $A, H,$ and $C$ is:

$\text{(A)} \quad \frac{2A-HC}{H} \quad \text{(B)} \quad (2A-C)H \quad \text{(C)} \quad \frac{H}{2A-HC}$

$\text{(D)} \quad \frac{HC-2A}{H} \quad \text{(E)} \quad H-2A+C$

Isolate $B$ step by step. $2A = H(B+C)$ $2A = HB+HC$ $2A-HC = HB$ $\frac{2A-HC}{H} = B$ The answer is $\boxed{\text{(A)}}.$

1963 TMTA High School Mathematics Contests (Problems)
Preceded by Problem 32	TMTA High School Mathematics Contest Past Problems/Solutions	Followed by Problem 34