Question

If $|r-6|=11$ and $|2q-12|=8$ then, the minimum value of $\frac{q}{r}$

• A. $-2$
• B. $\frac{17}{10}$
• C. $\frac{-1}{5}$
• D. $\frac{2}{5}$

Answer 1

Answer: (A)

$-2$

For the first equation, $|r-6|$
Case 1, when $|r-6|>0$, then $|r-6|=r-6$
So, $r-6=11=>r=17$
Case 2, when $|r-6|<0$, then $|r-6|=-(r-6)$
So, $-r+6=11=>r=-5$
For the second equation, $|2q-12|$
Case 1, when $|2q-12|>0$, then $|2q-12|=2q-12$
So, $2q-12=8=>q=10$
Case 2, when $|2q-12|<0$, then $|2q-12|=-2q+12$
So, $-2q+12=8=>q=2$
So, Minimum value of $\frac{q}{r}=\frac{Maxofq}{Minofr}=\frac{10}{-5}=-2$[For negetive values]