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# Algebra Previous Questions For SSC CGL PDF

Download SSC CGL Algebra Previous Year questions with answers PDF based on previous papers very useful for SSC CGL exams. 20 Very important Algebra objective questions (MCQ’s) for SSC exams.

Question 1: What is the value of $\frac{3}{4}+\frac{8}{9}$ ?

a) $\frac{57}{27}$

b) $\frac{11}{13}$

c) $\frac{59}{36}$

d) $\frac{11}{9}$

Question 2: The sum of 2xy(3x + 4y – 5z) and 5yz(2x – 3y) is

a) $6x^{2}y-8xy^{2}+15y^{2}z$

b) $6x^{2}y+8xy^{2}-15y^{2}z$

c) $6x^{2}y+8xy^{2}-15y^{2}z-20xyz$

d) $6x^{2}y-8xy^{2}+15y^{2}z+20xyz$

Question 3: If 4x-7<x-2 and $5x+\frac{2}{3}\geq3x+1$; then x can take which of the following values?

a) 2

b) -1

c) -2

d) 1

Question 4: The solution set of 4x – 3y = 47 and 3x + y = 32 is

a) {(15, 3)}

b) {(4, 12)}

c) {(11, -1)}

d) {(12, 3)}

Question 5: What is the value of $\frac{(4a^{2} + 8b + 14c + 2)}{2}$ ?

a) $2a^{2} + 4b + 7c + 1$

b) $a^{2} + 4b + 7c + 1$

c) $2a^{2} + 4b + 7c + 2$

d) $a^{2} + 4b + 7c + 2$

Question 6: Coefficient of $x^{2}$ in (x + 9)(6 – 4x) is

a) 54

b) -4

c) -30

d) 4

Question 7: If 7x + 6y = 5xy and 10x – 4y = 4xy, then value of x and y is

a) 3, 2

b) 2, 3

c) 4, 2

d) 5, 6

Question 8: Coefficient of $x^{2}$ in $6x^{3}+4x^{2}+2x+3$ is

a) 4

b) 6

c) 3

d) 2

Question 9: On dividing $8a^{2}b^{2} c^{2}$ by $4a^{2}$, we get

a) $2b^{2}$

b) $2c^{2}$

c) $2b^{2}c^{2}$

d) $2$

Question 10: If 2x + 6y = 3xy and 10x -­ 3y = 4xy, find x, y.

a) 3, 2

b) 2, 3

c) 4, 6

d) 6, 4

Question 11: If $2x-3(4-2x)<4x-5<4x+\frac{2x}{3}$, then x can take which of the following values?

a) 2

b) 8

c) 0

d) -8

Question 12: If a – b = 11 and ab = 24, then value of $a^{2}+b^{2}$  is

a) 169

b) 37

c) 73

d) 48

Question 13: The simpliﬁed form of  $(x+3)^{2}+(x-1)^{2}$  is

a) $(x^{2}+2x+5)$

b) $2(x^{2}+2x+5)$

c) $(x^{2}-2x+5)$

d) $2(x^{2}-2x+5)$

Question 14: What should be added to 5(2x-y) to obtain 4(2x – 3y) + 5(x + 4y)?

a) 3x – 13y

b) 3x + 13y

c) 13x – 3y

d) 13x + 3y

Question 15: If 3(2 – 3x) < 2 – 3x ≥ 4x -6; then x can take which of the following values?

a) 2

b) -1

c) -2

d) 1

Question 16: If (4x – 3) – (2x + 1) = 4, then the value of x is

a) 0

b) 1

c) 4

d) 3

Question 17: Which of the following equations has real and distinct roots?

a) $3x^{2} – 6x + 2 = 0$

b) $3x^{2} – 6x + 3 = 0$

c) $x^{2} – 8x + 16 = 0$

d) $4x^{2} – 8x + 4 = 0$

Question 18: Value of $(4a^{2}+12ab+9b^{2}/(2a+3b)$ is

a) 2a – 3b

b) 2a + 3b

c) 2a

d) 3b

Question 19: Coefﬁcient of $x^{2}$  in (x + 9)(6 – 4x)(4x – 7) is

a) 216

b) -4

c) -92

d) 108

Question 20: Given: 5x -3(2x-7) > 3x – 1 < 7 + 4x; then x can take which of the following values?

a) 6

b) 9

c) -6

d) -9

Expression : $\frac{3}{4}+\frac{8}{9}$

= $\frac{3(9)+8(4)}{36}$

= $\frac{27+32}{36} = \frac{59}{36}$

=> Ans – (C)

Sum of 2xy(3x + 4y – 5z) and 5yz(2x – 3y)

= $(6x^2y+8xy^2-10xyz)+(10xyz-15y^2z)$

= $6x^2y+8xy^2-15y^2z$

=> Ans – (B)

Expression 1 : 4x – 7 < x – 2

=> $4x-x$ < $7-2$

=> $3x$ < $5$

=> $x$ < $\frac{5}{3}$ ————(i)

Expression 2 : $5x+\frac{2}{3}\geq3x+1$

=> $5x-3x \geq 1-\frac{2}{3}$

=> $2x \geq \frac{1}{3}$

=> $x \geq \frac{1}{6}$ ———–(ii)

Combining inequalities (i) and (ii), we get : $\frac{1}{6} \leq x$ < $\frac{5}{3}$

The only value that $x$ can take among the options = 1

=> Ans – (D)

Equation 1 : 4x – 3y = 47

Equation 2 : 3x + y = 32

Multiplying equation (ii) by 3 and adding it to equation (i)

=> $(4x+9x)+(-3y+3y)=(47+96)$

=> $13x=143$

=> $x = \frac{143}{13}=11$

Substituting it in equation (ii), => $y=32-3(11) = 32 – 33 = -1$

$\therefore (x,y)=(11,-1)$

=> Ans – (C)

Expression : $\frac{(4a^{2} + 8b + 14c + 2)}{2}$

= $\frac{2(2a^2+4b+7c+1)}{2}$

= $2a^2+4b+7c+1$

=> Ans – (A)

A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Eg : In $ax^2$, coefficient is $a$

Expression : $(x + 9)(6 – 4x)$

= $6x – 4x^2 + 54 – 36x$

= $-4x^2 – 30x + 54$

$\therefore$ Coefficient of $x^2$ = -4

=> Ans – (B)

Equation 1 : 7x + 6y = 5xy

Equation 2 : 10x -­ 4y = 4xy

Dividing both equations by $(xy)$

=> $\frac{7}{y} + \frac{6}{x} = 5$

and $\frac{10}{y}-\frac{4}{x} = 4$

Let $\frac{1}{y} = u$ and $\frac{1}{x} = v$

=> $7u+6v=5$ ————(iii)

and $10u-4v=4$ ————(iv)

Multiplying equation (iv) by 3 and equation (iii) by 2 and adding them, we get :

=> $(14u+30u) = (10+12)$

=> $u = \frac{22}{44} = \frac{1}{2}$

Substituting it in equation (iv), => $4v = 5 – 4 = 1$

=> $v = \frac{1}{4}$

$\therefore (x,y) = (4,2)$

=> Ans – (C)

A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Eg : In $ax^2$, coefficient is $a$

Expression : $6x^{3}+4x^{2}+2x+3$

=> Coefficient of $x^2 = 4$

=> Ans – (A)

On dividing $8a^{2}b^{2} c^{2}$ by $4a^{2}$

= $\frac{8a^2b^2c^2}{4a^2}$

= $\frac{8}{4} \times \frac{a^2}{a^2} \times (b^2c^2)$

= $2b^2c^2$

=> Ans – (C)

Equation 1 : 2x + 6y = 3xy

Equation 2 : 10x -­ 3y = 4xy

Dividing both equations by $(xy)$

=> $\frac{2}{y} + \frac{6}{x} = 3$

and $\frac{10}{y}-\frac{3}{x} = 4$

Let $\frac{1}{y} = u$ and $\frac{1}{x} = v$

=> $2u+6v=3$ ————(iii)

and $10u-3v=4$ ————(iv)

Multiplying equation (iv) by 2 and adding it to equation (iii), we get :

=> $22u = 11$

=> $u = \frac{11}{22} = \frac{1}{2}$

Substituting it in equation (iii), => $6v = 3 – 1 = 2$

=> $v = \frac{2}{6} = \frac{1}{3}$

$\therefore (x,y) = (3,2)$

=> Ans – (A)

Expression 1 : 2x – 3(4 – 2x) < 4x – 5

=> $2x-12+6x$ < $4x-5$

=> $8x-4x$ < $-5+12$

=> $4x$ < $7$

=> $x$ < $\frac{7}{4}$ ———–(i)

Expression 2 : 4x – 5 < 4x + 2x/3

=> $\frac{2x}{3}$ > $-5$

=> $x$ > $\frac{-15}{2}$ ———-(ii)

Combining inequalities (i) and (ii), we get : $\frac{-15}{2}$ < $x$ < $\frac{7}{4}$

The only value that $x$ can take among the options = 0

=> Ans – (C)

Given : $(a – b) = 11$ and $ab = 24$

Using $(a – b)^2 = a^2 + b^2 – 2ab$

=> $(11)^2 = (a^2 + b^2) – (2 \times 24)$

=> $(a^2 + b^2) = 121 + 48 = 169$

=> Ans – (A)

Expression : $(x+3)^{2}+(x-1)^{2}$

= $(x^2+9+6x)+(x^2+1-2x)$

= $2x^2+4x+10$

= $2(x^2+2x+5)$

=> Ans – (B)

Let $m$ should be added to 5(2x-y) to obtain 4(2x – 3y) + 5(x + 4y)

=> $(m) + [5(2x-y)] = 4(2x-3y)+5(x+4y)$

=> $m + 10x-5y=8x-12y+5x+20y$

=> $m + 10x-5y=13x+8y$

=> $m = (13x-10x)+(8y+5y)$

=> $m=3x+13y$

=> Ans – (B)

Expression 1 : 3(2 – 3x) < 2 – 3x

=> $6-9x$ < $2-3x$

=> $9x-3x$ > $6-2$

=> $6x$ > $4$

=> $x$ > $\frac{2}{3}$ ————(i)

Expression 2 : 2 – 3x ≥ 4x -6

=> $4x+3x \leq 2+6$

=> $7x \leq 8$

=> $x \leq \frac{8}{7}$ ———–(ii)

Combining inequalities (i) and (ii), we get : $\frac{2}{3}$ < $x \leq \frac{8}{7}$

The only value that $x$ can take among the options = 1

=> Ans – (D)

Expression : (4x – 3) – (2x + 1) = 4

=> $4x-3-2x-1=4$

=> $2x-4=4$

=> $2x=4+4=8$

=> $x=\frac{8}{2}=4$

=> Ans – (C)

A quadratic equation : $ax^2 + bx + c = 0$ has real and distinct roots iff Discriminant, $D = b^2 – 4ac$ > $0$

(A) : $3x^{2} – 6x + 2 = 0$

=> D = $(-6)^2 – 4(3)(2) = 36 – 24 = 12$

(B) : $3x^{2} – 6x + 3 = 0$

=> D = $(-6)^2 – 4(3)(3) = 36 – 36 = 0$

(C) : $x^{2} – 8x + 16 = 0$

=> D = $(-8)^2 – 4(1)(16) = 64 – 64 = 0$

(D) : $4x^{2} – 8x + 4 = 0$

=> D = $(-8)^2 – 4(4)(4) = 64 – 64 = 0$

Thus, the equation : $3x^{2} – 6x + 2 = 0$ has real and distinct roots.

Expression : $(4a^{2}+12ab+9b^{2}/(2a+3b)$

= $\frac{(2a)^2+(3b)^2+(2.2a.3b)}{(2a+3b)}$

= $\frac{(2a+3b)^2}{(2a+3b)}$

= $2a+3b$

=> Ans – (B)

A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Eg : In $ax^2$, coefficient is $a$

Expression : $(x + 9)(6 – 4x)(4x – 7)$

= $(6x – 4x^2 + 54 – 36x)(4x – 7)$

= $(-4x^2 – 30x + 54)(4x – 7)$

= $4x(-4x^2 – 30x + 54) – 7(-4x^2 – 30x + 54)$

= $-16x^3 – 120x^2 + 216x + 28x^2 + 210x – 378$

= $-20x^3 – 92x^2 + 426x – 378$

$\therefore$ Coefficient of $x^2$ = -92

=> Ans – (C)

Expression 1 : 5x -3(2x-7) > 3x – 1

=> $5x-6x+21$ > $3x-1$

=> $3x+x$ < $21+1$

=> $4x$ < $22$

=> $x$ < $\frac{11}{2}$ ———-(i)

Expression 2 : 3x – 1 < 7 + 4x

=> $4x-3x$ > $-1-7$

=> $x$ > $-8$ ———-(ii)

Combining inequalities (i) and (ii), we get : $-8$ < $x$ < $\frac{11}{2}$

The only value that $x$ can take among the options = -6

=> Ans – (C)

We hope this Algebra Previous Year questions for SSC Exam will be highly useful for your preparation.