Question

If $-\frac{20}{7}<-3z+6<-\frac{11}{5}$, find the greatest possible integer value of $9z-18$.

• A. $6$
• B. $7$
• C. $8$
• D. $9$

Answer 1

Answer: (C)

$8$

Given, $-\frac{20}{7}<-3z+6<-\frac{11}{5}$

Multiply by $-3$ all sides, we get

$-\frac{20}{7}\times (-3)>(-3)(-3z+6)>-\frac{11}{5}\times (-3)$

$\Rightarrow \frac{60}{7}>9z-18>\frac{33}{5}$

$\Rightarrow 8.57>9z-18>6.6$

Then $9z-18<8.57$

So the greatest integer value of $9z-18$ is $8$.