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# Algebra Questions For SSC MTS PDF

SSC MTS Algebra Questions download PDF based on previous year question paper of SSC exams. 20 Very important Algebra Questions for SSC MTS Exam.

1: The value of x for which the expressions 15 -­ 7x and 15x + 7 become equal is .

a) -4/11

b) -11/4

c) ­11/4

d) 4/11

Question 2: If 5x ­- 3 ≥ 3 + x/2 and 4x -­ 2 ≤ 6 + x; then x can take which of the following values?

a) 1

b) 2

c) -1

d) -2

Question 3: Factorise $x^{2}$ + 3x ­-18

a) (x+18)(x­-1)

b) (x­+1)(x+18)

c) (x+6)(x­-3)

d) (x­+6)(x+3)

Question 4: If 5x + 4(1 – x) > 3x -4 > 5x/3 – x/3; then x can take which of the following values?

a) 2

b) 1

c) 3

d) -2

Question 5: If x + y = 10 and $x^2+ y^2$ = 68, then find xy

a) 21

b) 24

c) 25

d) 16

SSC MTS Study Material (FREE Tests)

Question 6: Coefficient of x in (x + 8)(6 – 3x) is

a) 18

b) 30

c) -18

d) -30

Question 7: What is the value of (91 + 92 + 93 + ……… +140)?

a) 5775

b) 11550

c) 17325

d) 23100

Question 8: If 2x – 3(x + 2) < 5 – 2x < – x + 2, then ﬁnd the value of x.

a) 2

b) 0

c) 10

d) 12

Question 9: If $xy = -18$ and $x^{2} + y^{2} = 85$, then ﬁnd the value of (x + y).

a) 8

b) 10

c) 9

d) 7

Question 10: If (8 + 10x) : (13x – 2) = 2, then the value of x is

a) -3/4

b) -4/3

c) 3/4

d) 4/3

Question 11: If 2x – 3(2x – 2) > x – 1 < 2 + 2x; then  x can  take which of the following values?

a) 2

b) -2

c) 4

d) -4

Question 12: If x – y = 6 and xy = 40, then ﬁnd $x^2 + y^2$

a) 116

b) 80

c) 89

d) 146

Question 13: If 3x – 8(2 – x) = -19, then the value of x is

a) -3/11

b) -33/11

c) -3/5

d) -33/5

Question 14: If 3x + 2y = 7 and 4x – y = 24, then x – y = .

a) 9

b) 1

c) -9

d) -1

Question 15: On dividing $256a^2b^2c^2$ by $64a^2$, we get

a) $2c^2$

b) $2b^2$

c) 4

d) $4b^2c^2$

Expressions : 15 -­ 7x and 15x + 7

=> $15 – 7x = 15x + 7$

=> $15x + 7x = 15 – 7$

=> $22x = 8$

=> $x = \frac{8}{22} = \frac{4}{11}$

=> Ans – (D)

Expression 1 : 5x ­- 3 ≥ 3 + x/2

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

=> $\frac{9x}{2} \geq 6$

=> $x \geq \frac{4}{3}$ ————(i)

Expression 2 : 4x -­ 2 ≤ 6 + x

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

=> $3x \leq 8$

=> $x \leq \frac{8}{3}$ ————(ii)

Combining inequalities (i) and (ii), we get : $\frac{4}{3} \leq x \leq \frac{8}{3}$

The only value that $x$ can take among the given options = 2

=> Ans – (B)

Expression : $x^2 + 3x – 18$

= $x^2 + 6x – 3x – 18$

= $x(x + 6) – 3(x + 6)$

= $(x + 6)(x – 3)$

=> Ans – (C)

Expression 1 : $3x – 4$ > $\frac{5x}{3} – \frac{x}{3}$

=> $9x – 12$ > $4x$

=> $9x – 4x$ > $12$

=> $x$ > $\frac{12}{5}$ ———-(i)

Expression 2 : $5x + 4(1 – x)$ > $3x – 4$

=> $x + 4$ > $3x – 4$

=> $3x – x$ < $4 + 4$

=> $x$ < $4$ ——(ii)

Combining inequalities (i) and (ii), we get : $\frac{12}{5}$ < $x$ < $4$

Thus, only value that $x$ can take among the options = 3

=> Ans – (C)

Given : $(x + y) = 10$ and $x^2 + y^2 = 68$

Using $(x + y)^2 = x^2 + y^2 + 2xy$

=> $(10)^2 = 68 + (2 \times xy)$

=> $2 xy = 100 – 68 = 32$

=> $xy = \frac{32}{2} = 16$

=> Ans – (D)

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 : $(x + 8)(6 – 3x)$

= $6x – 3x^2 + 48 – 24x$

= $-3x^2 – 18x + 48$

$\therefore$ Coefficient of $x$ = -18

=> Ans – (C)

Expression : (91 + 92 + 93 + ……… +140)

This is an arithmetic progression with first term, $a = 91$ , last term, $l = 140$ and common difference, $d = 1$

Let number of terms = $n$

Last term in an A.P. = $a + (n – 1)d = 140$

=> $91 + (n – 1)(1) = 140$

=> $n – 1 = 140 – 91 = 49$

=> $n = 49 + 1 = 50$

$\therefore$ Sum of A.P. = $\frac{n}{2} (a + l)$

= $\frac{50}{2} (91 + 140)$

= $25 \times 231 = 5775$

=> Ans – (A)

Expression 1 : $5 – 2x < -x + 2$

=> $2x – x$ > $5 – 2$

=> $x$ > $3$ ———-(i)

Expression 2 : $2x – 3(x + 2) < 5 – 2x$

=> $-x – 6$ < $5 – 2x$

=> $2x – x$ < $5 + 6$

=> $x$ < $11$ ——(ii)

Combining inequalities (i) and (ii), we get : $3$ < $x$ < $11$

Thus, only value that $x$ can take among the options = 10

=> Ans – (C)

Given : $(x^2 + y^2) = 85$ and $xy = -18$

Using $(x + y)^2 = x^2 + y^2 + 2xy$

=> $(x + y)^2 = 85 + (2 \times -18)$

=> $(x + y)^2 = 85 – 36 = 49$

=> $(x + y) = \sqrt{49} = 7$

=> Ans – (D)

$\frac{8+10x}{13x-2}=2$

$\Rightarrow 8+10x = 26x – 4$

$\Rightarrow 12 = 16x$

$\Rightarrow x = 3/4$

so the answer is option C.

Expression 1 : $x – 1 < 2 + 2x$

=> $2x – x$ > $-1 – 2$

=> $x$ > $-3$ ———-(i)

Expression 2 : $2x – 3(2x – 2)$ > $x – 1$

=> $-4x + 6$ > $x – 1$

=> $4x + x$ < $6 + 1$

=> $x$ < $\frac{7}{5}$ ——(ii)

Combining inequalities (i) and (ii), we get : $-3$ < $x$ < $\frac{7}{5}$

Thus, only value that $x$ can take among the options = -2

=> Ans – (B)

Given : $(x – y) = 6$ and $xy = 40$

Using $(x – y)^2 = x^2 + y^2 – 2xy$

=> $(6)^2 = (x^2 + y^2) – (2 \times 40)$

=> $(x^2 + y^2) = 36 + 80 = 116$

=> Ans – (A)

Expression : $3x – 8(2 – x) = -19$

=> $3x – 16 + 8x = -19$

=> $11x = 16 – 19 = -3$

=> $x = \frac{-3}{11}$

=> Ans – (A)

Equation (i) : 3x + 2y = 7

Equation (ii) : 4x -­ y = 24

Multiplying by 2 on both sides, => $8x – 2y = 48$ ———–(iii)

=> $(3x + 8x) + (2y – 2y) = (7 + 48)$

=> $11x = 55$

=> $x = \frac{55}{11} = 5$

Substituting it in equation (ii), we get : $4(5) – y = 24$

=> $y = 20 – 24 = -4$

$\therefore (x – y) = 5 – (-4) = 5 + 4 = 9$

=> Ans – (A)

Expression : $\frac{256a^2b^2c^2}{64a^2}$
= $\frac{256}{64} \times \frac{a^2b^2c^2}{a^2}$
= $4b^2c^2$