Question

$16\frac{2}{3}\%$ of $600$gm $-33\frac{1}{3}\%$ of $180$gm

• A. $20$ gm
• B. $30$ gm
• C. $40$ gm
• D. $60$ gm

Answer 1

The correct option is C $40$ gm

$16\frac{2}{3}\%$ of $600gm-33\frac{1}{3}\%$ of $180gm$

$\Rightarrow [\frac{50}{3}\times \frac{1}{100}(600)-(\frac{100}{3}\times \frac{1}{100})(180\left)\right]gm$

$\Rightarrow [\frac{600}{6}-\frac{180}{3}]gm$

$\Rightarrow (100-60)gm$

$\Rightarrow 40gm$

$\therefore [162/3\%of600gm-331/3\%of180gm]=40gm$