Exercise 1


Directions (Qs. 1-5): In each question one/two equations are provided. On the basis of these you have to find out the relation between p and q.

Give answer (a) if p = q

Give answer (b) if p >q

Give answer (c) if q > p

Give answer (d) if p³ q, and

Give answer (e) if q ≥ p

Question.1
I. pq + 30 = 6p + 5q
Ans.c

I. pq + 30 = 6p + 5q

or,(6p – 30) + (5q – pq) = 0

or,6(p – 5) – q(p – 5) = 0

or,(p – 5)(6 – q) = 0

\ p = 5

or, q = 6

Hence, q > p

Question.2
I. 2p2 + 12p + 16 = 0
II. 2q2 + 14q + 24 = 0
Ans.b

I. 2p2 – 12p + 16 = 0

II. 2q2 + 14q + 24 = 0

or, p2 – 6p + 8 = 0

or,(p – 4)(p– 2) = 0

or,q2 + 7q + 12 = 0

or,(q + 4)(q + 3) = 0

p = +4 or, + 2

\ q = –3 or, – 4

When p = + 2, q = –3, then, p > q

When p = + 4, q = –4, then, p > q

When p = + 4, q = –3, then, p > q

Hence p > q

Question.3
I. 2p2 + 48 = 20p
II. 2q2 + 18 = 12q
Ans.b

I. 2p2 – 20p + 48 =0

p2 – 10p + 24 = 0

(p – 4)(p – 6) = 0

or p = 4 ; p = 6

II. 2q2 – 12q + 18 = 0

q2 – 6q + 9 = 0

(q – 3)(q – 3) = 0

or q = 3 ; q = 3

hence p > q

Question.4
I. q2 + q = 2
II. p2 + 7p + 10 = 0
Ans.e

I. q2 + q = 2

or,q2 + q – 2 = 0

II. p2 + 7p + 10 = 0

or, p2 + 5p + 2p + 10 = 0

or,(q + 2)(q – 1) = 0

or,(q + 5)(p+ 2) = 0

\ q = –2 or 1

\ p = –5 or –2

Hence, q ≥ p

Question.5
I. p2 + 36 = 12p
II. 4q2 + 144 = 48q
Ans.a

I. p2 + 36 = 12p

or, p2 – 12p + 36 = 0

II. 4q2 + 144 = 48q

or,(p – 6)2 = 0

or,q2 – 12q + 36 = 0

\p = 6

or,(q – 6)2 = 0

\q = 6

Hence, p= q