1963 TMTA High School Algebra I Contest Problem 33

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Problem

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

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

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

Solution

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

See Also

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