CAT 2021

Slot 2

QA

Question & Solution

Case 1 :
$x\ge\frac{40}{3}$
we get 3x-20 +3x-40 =20 
$\Rightarrow$ 6x=80
$\Rightarrow$x=$\frac{80}{6}$=$\frac{40}{3}$=13.33

Case 2 :
$\frac{20}{3}\le\ x<\frac{40}{3}$
we get 3x-20+40-3x =20
we get 20=20
So we get x$\in\ \left[\frac{20}{3},\frac{40}{3}\right]$

Case 3 :
$x<\frac{20}{3}$
we get 20-3x+40-3x =20
40 = 6x
x=$\frac{20}{3}$
but this is not possible
so we get from case 1, 2 and 3
$\frac{20}{3}\le\ x\le\frac{40}{3}$
Now looking at options
we can say only option C satisfies for all x.
Hence, the answer is 7 $\lt x \lt$ 12.