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

Answer: (e)

I. $y^2$ + 2y – 3 = 0⇒(y – 1)(y + 3) = 0

or, y = 1, – 3

II. $2x^2$ – 7x + 6 = 0

or, $2x^2$ – 4x – 3x + 6 = 0

or, (2x – 3)(x – 2) = 0

or, x = 2, $3/2$

Hence, x > y

Answer: (e)

I. $12a^2$ – 7a + 1 = 0

II. $15b^2$ – 16b + 4 = 0

Sum of the two values of a, i.e., $(a_1 , + a_2)$

${-(-7)}/ 12=7/12$

Similarly, Sum of the two values of b,

i.e.,$(b_1+b_2)={-(-16)}/15=16/15$

Since $7/12$ < $16/15$

Therefore, a < b,

Now check the equality of root

${(12×4-15× 1)}^2$= {12×( 16)-15×(- 7)} {(-7) ×4-(-16)×1}

⇒ $33^2$={-87}{-12}

⇒ 1089 = 1044, which is not true.

Therefore, our answer should be a < b.

Answer: (e)

(i) p – 7 = 0 or, p = 7

(ii) $3q^2$ – 10q + 7 = 0

or,$3q^2$ – 3q – 7q + 7=0

or,3q(q – 1) – 7(q – l) = 0

or, (3q –7) (q –1) or, q = 1 or, $7/3$

∴p > q

Answer: (e)

$5/{28}×9/8p=15/14×13/16a$

or, ${45p}/224={195q}/224$

or, 3p = 13q

∴ p > q

Answer: (d)

I. $2p^2$ + 20p + 50 = 0

or, $p^2$ + 10p + 25 = 0

or, ${(p + 5)}^2$ = 0

or, p = – 5

II. $q^2$ = 25

or, q = ±5

Hence, p ≤ q