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.
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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
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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
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Question 15: On dividing $256a^2b^2c^2$ by $64a^2$, we get
a) $2c^2$
b) $2b^2$
c) 4
d) $4b^2c^2$
Answers & Solutions:
1) Answer (D)
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)
2) Answer (B)
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)
3) Answer (C)
Expression : $x^2 + 3x – 18$
= $x^2 + 6x – 3x – 18$
= $x(x + 6) – 3(x + 6)$
= $(x + 6)(x – 3)$
=> Ans – (C)
4) Answer (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)
5) Answer (D)
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)
6) Answer (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 : $(x + 8)(6 – 3x)$
= $6x – 3x^2 + 48 – 24x$
= $-3x^2 – 18x + 48$
$\therefore$ Coefficient of $x$ = -18
=> Ans – (C)
7) Answer (A)
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)
8) Answer (C)
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)
9) Answer (D)
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)
10) Answer (C)
$\frac{8+10x}{13x-2}=2$
$\Rightarrow 8+10x = 26x – 4$
$\Rightarrow 12 = 16x$
$\Rightarrow x = 3/4$
so the answer is option C.
11) Answer (B)
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)
12) Answer (A)
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)
13) Answer (A)
Expression : $3x – 8(2 – x) = -19$
=> $3x – 16 + 8x = -19$
=> $11x = 16 – 19 = -3$
=> $x = \frac{-3}{11}$
=> Ans – (A)
14) Answer (A)
Equation (i) : 3x + 2y = 7
Equation (ii) : 4x - y = 24
Multiplying by 2 on both sides, => $8x – 2y = 48$ ———–(iii)
Adding equation(i) and (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)
15) Answer (D)
Expression : $\frac{256a^2b^2c^2}{64a^2}$
= $\frac{256}{64} \times \frac{a^2b^2c^2}{a^2}$
= $4b^2c^2$
=> Ans – (D)
