Question

$(6ab-3abc+9abcd)÷(-\frac{1}{3}ab)$ is equal to

• A. $2+c+3cd$
• B. $-2+c-3cd$
• C. $-18+9c-27cd$
• D. $18-9c-27cd$

Answer 1

The correct option is C $-18+9c-27cd$

$\frac{6ab-3abc+9abcd}{-\frac{1}{3}ab}$

By separating denominators, we get

$=\frac{6ab}{-\frac{1}{3}ab}+\frac{(-3abc)}{-\frac{1}{3}ab}+\frac{9abcd}{-\frac{1}{3}ab}$

$=\frac{6ab}{ab}\times (-3)+\frac{(-3abc)}{ab}\times (-3)+\frac{\left(9abcd\right)}{ab}\times (-3)$

$=-18+9c-27cd$
Hence final value after given operation is $-18+9c-27cd$