Question 14

# For a real number x the condition $$\mid3x-20\mid+\mid3x-40\mid=20$$ necessarily holds if

Solution

Case 1 : $$x\ge\frac{40}{3}$$
we get 3x-20 +3x-40 =20
6x=80
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 7<x<12.