Algebra Questions for SSC CGL Set-3 PDF

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Algebra Questions for SSC CGL Set-3 PDF
Algebra Questions for SSC CGL Set-3 PDF

Algebra Questions for SSC CGL Set-3 PDF

Download SSC CGL Algebra Questions set-3 PDF. Top 10 SSC CGL questions based on asked questions in previous exam papers very important for the SSC exam.

Download Algebra Questions for SSC CGL Set-3 PDF

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Question 1: If xy = 56 and $x^{2} + y^{2} = 113$, then what will be the value of (x + y)?

a) 29

b) 21

c) 36

d) 15

Question 2: If a + b = 11 and $a^2 + b^2$ = 61, then value of ab is

a) 12

b) 96

c) 24

d) 30

Question 3: If 4(2x -4) – 2 > 3x – 1 ≥ 4x -7, then x can take which of the following values?

a) 7

b) 6

c) 2

d) 0

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Question 4: Factors of $48x^{3} – 8x^{2} – 93x – 45$ are

a) (4x + 3)(4x ­- 3)(3x ­- 5)

b) (4x – 3)(4x ­- 3)(3x -­ 5)

c) (4x + 3)(4x + 3)(3x -­ 5)

d) (4x -­ 3)(4x + 3)(3x + 5)

Question 5: Divide 32 into two parts such that the sum of the square of the parts is 674. What is the value of the parts?

a) 22, 10

b) 30, 2

c) 25, 7

d) 20, 12

Question 6: If  (4x ­-5) = (3x ­-1), then the numerical value of $(x + 4)^{2}$ is

a) 16

b) 64

c) 32

d) 8

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

a) -8

b) 6

c) ­8

d) ­10

Question 8: If 51.97 -(81.18 ­-x ) ­-59.39 = 5.268, then value of x will be

a) ­24.912

b) ­68.492

c) 93.868

d) 197.808

Question 9: What should be added to 3(x­-2y) to obtain 2(3x + y) ­- 5(2x + 3)?

a) 8y -7x -15

b) 8y-7x + 15

c) 8y + 7x + 15

d) 8y + 7x ­-15

Question 10: If 1/6 of x – 7/2 of 3/7 equals – 7/4, then the value of x is

a) -1.5

b) 3

c) -2.5

d) 6

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Answers & Solutions:

1) Answer (D)

Given : $(x^2 + y^2) = 113$ and $xy = 56$

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

=> $(x + y)^2 = 113 + (2 \times 56)$

=> $(x + y)^2 = 113 + 112 = 225$

=> $(x + y) = \sqrt{225} = 15$

=> Ans – (D)

2) Answer (D)

Given : $(a + b) = 11$ and $a^2 + b^2 = 61$

Using $(a + b)^2 = a^2 + b^2 + 2ab$

=> $(11)^2 = 61 + (2 \times ab)$

=> $2 ab = 121 – 61 = 60$

=> $ab = \frac{60}{2} = 30$

=> Ans – (D)

3) Answer (B)

Expression 1 : 4(2x -4) – 2 > 3x – 1

=> $8x-16-2$ > $3x-1$

=> $8x-3x$ > $-1+18$

=> $x$ > $\frac{17}{5}$ ————(i)

Expression 2 : 3x – 1 ≥ 4x -7

=> $4x-3x \leq -1+7$

=> $x \leq 6$ ———–(ii)

Combining inequalities (i) and (ii), we get : $\frac{17}{5}$ < $x \leq 6$

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

=> Ans – (B)

4) Answer (C)

(A) : (4x + 3)(4x ­- 3)(3x ­- 5)

= $(16x^2 – 12x + 12x – 9)(3x – 5)$

= $(16x^2 – 9)(3x – 5)$

= $48x^3 – 80x^2 – 27x + 45$

(B) : (4x – 3)(4x ­- 3)(3x -­ 5)

= $(16x^2 – 24x + 9)(3x – 5)$

= $48x^3 – 80x^2 – 72x^2 + 120x + 27x – 45$

= $48x^3 – 152x^2 + 147x – 45$

(C) : (4x + 3)(4x + 3)(3x -­ 5)

= $(16x^2 + 24x + 9)(3x – 5)$

= $48x^3 – 80x^2 + 72x^2 – 120x + 27x – 45$

= $48x^3 – 8x^2 – 93x – 45$

=> Ans – (C)

5) Answer (C)

Let the first part = $x$ and second part = $(32 – x)$

According to ques, => $(x)^2 + (32 – x)^2 = 674$

=> $x^2 + (x^2 + 1024 – 64x) = 674$

=> $2x^2 – 64x + 1024 – 674 = 0$

=> $x^2 – 32x + 175 = 0$

=> $x^2 – 25x – 7x + 175 = 0$

=> $x(x-25) – 7(x-25) = 0$

=> $(x-25)(x-7) = 0$

=> $x = 25,7$

=> Ans – (C)

6) Answer (B)

Given : (4x ­- 5) = (3x ­-1)

=> $4x – 3x = 5 – 1$

=> $x = 4$

To find : $(x + 4)^2$

= $(4 + 4)^2 = 8^2 = 64$

=> Ans – (B)

7) Answer (B)

Expression 1 : 2(3x + 5) > 4x -­ 5

=> $6x+10$ > $4x-5$

=> $6x-4x$ > $-5-10$

=> $2x$ > $-15$

=> $x$ > $\frac{-15}{2}$ ——–(i)

Expression 2 : 4x -­ 5 < 3x + 2

=> $4x-3x$ < $2+5$

=> $x$ < $7$ ———(ii)

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

The only value that $x$ can take = 6

=> Ans – (B)

8) Answer (C)

Expression : 51.97 -(81.18 ­-x ) ­-59.39 = 5.268

=> 51.97 – 81.18 + x = 5.268 + 59.39

=> -29.21 + x = 64.658

=> x = 64.658 + 29.21

=> x = 93.868

=> Ans – (C)

9) Answer (A)

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

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

=> $m + 3x-6y=6x+2y-10x-15$

=> $m = (2y+6y)+(-4x-3x)-15$

=> $m=8y-7x-15$

=> Ans – (A)

10) Answer (A)

According to ques,

=> $(\frac{1}{6} \times x) – (\frac{7}{2} \times \frac{3}{7}) = \frac{-7}{4}$

=> $\frac{x}{6} – \frac{3}{2} = \frac{-7}{4}$

=> $\frac{x}{6} = \frac{3}{2} – \frac{7}{4}$

=> $\frac{x}{6} = \frac{-1}{4}$

=> $x = \frac{-6}{4} = -1.5$

=> Ans – (A)

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