Question

There are three boxes. Box A is $8\frac{1}{2}$ in shorter than Box B. Box B is $2\frac{1}{3}$ in taller than Box C. Height of Box C is $26\frac{5}{6}$ in. What are the heights of Box A and Box B?

• A. $A=20\frac{4}{6}$ in, $B=29\frac{1}{6}$ in,
• B. $A=29\frac{4}{6}$ in, $B=20\frac{1}{6}$ in,
• C. $A=29\frac{1}{6}$ in, $B=20\frac{4}{6}$ in,
• D. $A=29\frac{5}{6}$ in, $B=20\frac{1}{6}$ in,

Answer 1

The correct option is A $A=20\frac{4}{6}$ in, $B=29\frac{1}{6}$ in,

The height of Box $C=26\frac{5}{6}$ in
Box B is $2\frac{1}{3}$ in taller than Box C.
So, the height of Box $B=26\frac{5}{6}$ in $+2\frac{1}{3}$ in. $=\frac{26\times 6+5}{6}$ in $+\frac{3\times 2+1}{3}$ in $=\frac{161}{6}$ in $+\frac{7}{3}$ in $=\frac{161}{6}$ in $+\frac{7\times 2}{3\times 2}$ in $=\frac{161}{6}$ in $+\frac{14}{6}$ in $=\frac{175}{6}$ in $=29\frac{1}{6}$ in
Box A is $8\frac{8}{2}$ in shorter than Box B.
So, the height of Box $A=\frac{175}{6}$ in $-8\frac{1}{2}$ in $=\frac{175}{6}$ in $-\frac{175}{6}$ in $-\frac{17\times 3}{2\times 3}$ in $=\frac{175}{6}$ in $-\frac{51}{6}$ in $=\frac{175-51}{6}$ in $=\frac{124}{6}$ in $=20\frac{4}{6}$ in
Height of Box $A=20\frac{4}{6}$ in and height of Box $B=29\frac{1}{6}$ in
Hence, Option A is correct.