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CAT 2021 Slot 2 — QA Question 16

ModulusEasy

For a real number x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if

Answer & solution

  • A

    6 < x < 11

  • B

    9 < x < 14

  • C

    10 < x < 15

  • 7 < x < 12

Solution

Easy

The two expressions inside the modulus, 3x−203x-20 and 3x−403x-40, are the distances of 3x3x from 2020 and from 4040. Their sum equals 2020 (the gap between 2020 and 4040) precisely when 3x3x lies between 2020 and 4040. Solve the three sign-cases to confirm, then find which option's interval lies entirely inside the solution set.

1

Case 3x≥403x\ge 40. Both terms non-negative.

(3x−20)+(3x−40)=20⇒ 6x=80 ⇒ x=403≈13.33(single point)\begin{aligned} &(3x-20)+(3x-40)=20\\ &\Rightarrow\ 6x=80\ \Rightarrow\ x=\tfrac{40}{3}\approx 13.33 \quad\text{(single point)} \end{aligned}
2

Case 20≤3x<4020\le 3x<40. First term non-negative, second negative.

(3x−20)−(3x−40)=20⇒ 20=20(true for the whole interval)\begin{aligned} &(3x-20)-(3x-40)=20\\ &\Rightarrow\ 20=20 \quad\text{(true for the whole interval)} \end{aligned}

So every xx with 203≤x≤403\tfrac{20}{3}\le x\le\tfrac{40}{3} satisfies the equation.

3

Case 3x≤203x\le 20 and combine. Both terms non-positive gives −6x=−40⇒x=203≈6.67-6x=-40\Rightarrow x=\tfrac{20}{3}\approx 6.67 (single point). The full solution set is the closed interval.

203≤x≤403 ≈ 6.67≤x≤13.33\begin{aligned} &\tfrac{20}{3}\le x\le \tfrac{40}{3}\ \approx\ 6.67\le x\le 13.33 \end{aligned}

The condition "necessarily holds" means we need an interval that is entirely a subset of this. Test the options: $7

7
CAT 2021 Slot 2 QA Q16: For a real number x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if — Solution | TheCATExam