algebraic expressions Model Questions & Answers, Practice Test for ssc cgl tier 1 2023

Answer: (d)

q = 5, 6 & p = $3/2$ , 2

Answer: (e)

I. $3P^2$ – 10P + 7 = 0

$3P^2$ – 3P – 7P + 7 = 0

3P (P – 1) – 7 (P – 1) = 0

⇒ (3P – 7) ( P – 1) = 0

⇒ P = $7/3$, 1

II. $15Q^2$ – 22Q + 8 = 0

$15Q^2$ – 10Q – 12Q + 8 = 0

5Q (3Q – 2) – 4(3Q – 2) = 0

(5Q – 4) (3Q – 2) = 0

⇒ Q = $4/5 , 2/3$

∴P > Q

Answer: (b)

I. $3x^2$ + 7x – 6 = 0

or, $3x^2$ + 9 x – 2x – 6 = 0

or, (x + 3) (3x – 2) = 0

or, x = – 3, $2/3$

II. 6($2y^2$ + 1) = 17y

or, $12y^2$ + 6 – 17y = 0

or, $12y^2$ – 9y – 8y + 6 = 0

or, (4y – 3) (3y – 2) = 0

or, y=$3/4,2/3$

Hence, y ≥ x

Answer: (e)

I. $6p^2$ + 5p + 1 = 0

or, $6p^2$ + 3p + 2p + 1 = 0

or, 3p(2p + 1) + 1 (2p + 1) = 0

or, (3p + 1) (2p + 1) = 0

Hence, p = $-1/3,-1/2$

II. $20q^2$ + 9q + 1 = 0

or, $20q^2$ + 5q + 4q + 1 = 0

or, 5q(4q + l) + 1(4q + 1) = 0

or, (5q + 1)(4q + l) = 0

Hence, q = $-1/5,-1/4$ Thus, p < q.

Answer: (e)

I. $2p^2$ – 12p + 16 = 0

II. $2q^2$ + 14q + 24 = 0

or, $p^2$ – 6p + 8 = 0

or, (p – 4) (p – 2) = 0

or, $q^2$ + 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