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

Answer: (d)

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

Answer: (c)

I. $4x^2$ = 25

or x =±$5/2$

II. $2y^2$ – 13y + 21 = 0

or, $2y^2$ – 6y – 7y + 21 = 0

or, (y – 3) (2y –7) = 0

or, y = 3, $7/2$

Hence, y > x

Answer: (c)

I. $p^2$ + 36 = 12p

or, $p^2$ – 12p + 36 = 0

II. $4q^2$ + 144 = 48q

or, ${(p – 6)}^2$ = 0

or, $q^2$ – 12q + 36 = 0

∴p = 6

or, ${(q – 6)}^2$ = 0

∴q = 6

Hence, p = q

Answer: (c)

I $3x^2$ – 7x + 2 = 0

or, $3x^2$ – 6x – x + 2 = 0

or, (x – 2) (3x – 1) = 0

or, x = 2, 1/3

II. $2y^2$ – 11y + 15 = 0

or, $2y^2$ – 6y – 5y+ 15 = 0

or, (2y – 5) (y – 3) = 0

or, y = 5/2, 3

Hence, y > x

Answer: (e)

I.$a^2$ + 9a + 20 = 0

Break 9 as $F_1 and F_2$ , so that $F_1 × F_2$ = 20

and $F_1 + F_2$ = 9.

Therefore, $F_1 = 5, F_2$ = 4

Now one value of a =${-5}/1$=-5

other value of ${-20}/5$ =-4

II. $2b^2$ + 10b + 12 = 0

The two parts of 10, ie $F_1$ = 6 and $F_2$ = 4

∴Value of b =${-6}/2=-3$ and ${-12}/6$=-2

Obviously b > a.

If general form of quadratic equation is

$ax^2$ + bx + c = 0,

then split b into two parts so that $b_1 + b_2$ = b and

$b_1 × b_{2e}$= a × c

Now say $b_1$ as $F_1$ and $b_2$ as $F_2$ . Then the values

of 'x' will be ${{-F_1}/a} and {{–C}/F_1}$ or ${{-F_2}/a} and {{–C}/F_2}$