GATE 2022 IN GA Q.8: solution and answer

GATE 2022 · INGA Q.8 · 2 marks · MCQ

Consider the following inequalities.

(i) $3 p - q < 4$

(ii) $3 q - p < 12$

Which one of the following expressions below satisfies the above two inequalities?

(A)
$p + q < 8$
(B)
$p + q = 8$
(C)
$8 \leq p + q < 16$
(D)
$p + q \geq 16$

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Answer and solution

ANSWER

(A) $p + q < 8$ . Adding the two inequalities gives $2 p + 2 q < 16$ , so $p + q < 8$ .

Official answer: ASource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Two inequalities that point the same way can be added side by side: if $a < b$ and $c < d$ , then $a + c < b + d$ .
Add (i) and (ii): $(3 p - q) + (3 q - p) < 4 + 12$
Collect like terms: $2 p + 2 q < 16$
Divide both sides by the positive number 2. Dividing by a positive number keeps the direction of the sign: $p + q < 8$
So every pair $(p, q)$ that satisfies both inequalities has $p + q < 8$ . That rules out (B) $p + q = 8$ , (C) $8 \leq p + q < 16$ and (D) $p + q \geq 16$ , because each of them needs $p + q \geq 8$ .
Check with one pair: $p = 1, q = 1$ gives $3 p - q = 3 (1) - 1 = 2 < 4$ and $3 q - p = 3 (1) - 1 = 2 < 12$ . Both inequalities hold, and $p + q = 2 < 8$ , which agrees with (A).

KEY IDEA

Inequalities that point the same way can be added. Adding (i) and (ii) cancels part of each variable: $(3 p - p) + (3 q - q) = 2 (p + q)$ , which leaves $2 (p + q) < 16$ .

WATCH OUT FOR

Trying to solve for $p$ and $q$ separately, as if the inequalities were equations, or subtracting one inequality from the other. Subtracting two inequalities that point the same way is not valid. Another mistake is to pick (C) or (D) by adding the right-hand sides $4 + 12 = 16$ and forgetting to divide by 2.

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 58 Numerical questions, with notes

GATE 2022 IN GA Q.8: solution and answer

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