Question

$\text{If}A\text{and}B\text{are fractions and}A\times B<1,\text{then the value of}A\text{and}Bcannot\text{be}$

• A. A = 1/2 ; B = 1/3
• B. A = 3/2 ; B = 4/3
• C. A = 3/2 ; B = 1/4

Answer 1

The correct option is B A = 3/2 ; B = 4/3

$\text{We}\text{check the validity of each case by}\text{giving}A\text{and}B\text{suitable values.}$

$Case1:A=\frac{3}{2};B=\frac{1}{4}⟹A\times B=\frac{3}{2}\times \frac{1}{4}=\frac{3\times 1}{2\times 4}=\frac{3}{8}$


$\therefore \text{For}A\times B<1,\text{it is possible to have}A=\frac{3}{2};B=\frac{1}{4}$

$Case2:A=\frac{3}{2};B=\frac{4}{3}⟹A\times B=\frac{3}{2}\times \frac{4}{3}=\frac{3\times 4}{2\times 3}=\frac{12}{6}=2$


$\therefore \text{For}A\times B<1,\text{it is}impossible\text{to}\text{have}A=\frac{3}{2};B=\frac{4}{3}$

$Case3:A=\frac{1}{2};B=\frac{1}{3}⟹A\times B=\frac{1}{2}\times \frac{1}{3}=\frac{1\times 1}{2\times 3}=\frac{1}{6}$


$\therefore \text{For}A\times B<1,\text{it is possible to have}A=\frac{1}{2};B=\frac{1}{3}$

$\text{Hence, for}A\times B<1\text{the values of}A\text{and}Bcannot\text{be}A=\frac{3}{2};B=\frac{4}{3}$