Question

$A=1\frac{1}{2};B=\frac{7}{8};C=1\frac{1}{3}$

$\text{The fastest approach to calculate}A\times B\times C\text{is}$

• A. Approach 1
• B. Approach 2
• C. Approach 3

Answer 1

The correct option is B Approach 2

$\text{Converting the mixed numbers to}\text{fractions, we get:}$

$A=1\frac{1}{2}=\frac{(2\times 1)+1}{2}=\frac{3}{2}$

$C=1\frac{1}{3}=\frac{(3\times 1)+1}{3}=\frac{4}{3}$

$\text{Also,}$

$B=\frac{7}{8}$

$\text{Calculating}A\times B\times C\text{through each}\text{approach, we find:}$

$Approach1:$

$(A\times B)\times C$

$=(\frac{3}{2}\times \frac{7}{8})\times \frac{4}{3}$

$=\left(\frac{3\times 7}{2\times 8}\right)\times \frac{4}{3}$

$=\frac{21}{16}\times \frac{4}{3}$

$=\frac{21\times 4}{16\times 3}$

$=\frac{7\times 1}{4\times 1}$

$=\frac{7}{4}$

$Approach2:$

$(A\times C)\times B$

$(\frac{3}{2}\times \frac{4}{3})\times \frac{7}{8}$

$=\frac{3\times 4}{2\times 3}\times \frac{7}{8}$

$=2\times \frac{7}{8}$

$=\frac{2\times 7}{1\times 8}$

$=\frac{2\times 7}{1\times 8}$

$\frac{1\times 7}{1\times 4}$

$=\frac{7}{4}$

$Approach3:$

$(B\times C)\times A$

$(\frac{7}{8}\times \frac{4}{3})\times \frac{3}{2}$

$=\frac{7\times 4}{8\times 3}\times \frac{3}{2}$

$=\frac{7\times 1}{2\times 3}\times \frac{3}{2}$

$=\frac{7}{6}\times \frac{3}{2}$

$=\frac{7\times 3}{6\times 2}$

$=\frac{7\times 1}{2\times 2}$

$=\frac{7}{4}$

$\text{In}approach2\text{the numerator and the}\text{denominator cancel out easily.}$

$\therefore \text{The fastest approach to calculate the}\text{product is}approach2.$