Algebra Questions for IBPS Clerk set-2 PDF

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Algebra Questions for IBPS Clerk set-2 PDF
Algebra Questions for IBPS Clerk set-2 PDF

Algebra Questions for IBPS Clerk set-2 PDF

Download important  Algebra Questions set-2 PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Scheduling Questions and Answers for IBPS Clerk Exam.

Download Algebra Questions for IBPS Clerk set-2 PDF

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Question 1: The value of x for which the expressions 19 -­ 5x and 19x + 5 become equal is …………..

a) 7/12

b) -7/12

c) -12/7

d) ­12/7

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

a) 2

b) 3

c) 1

d) -5

Question 3: If $3x^{2} = 10^{2} – 5^{2}$, find the value of x?

a) 7

b) 5

c) 9

d) 11

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Question 4: If (7x – 13) – (12x + 3) = 14, then the value of x is ______ .

a) -6

b) 6

c) 2/5

d) -2/5

Question 5: If 2 + 2x < 5 – x/2 and 5x + 3 > 5 – 5x; then x can take which of the following values?

a) 2

b) 0

c) -2

d) 1

Question 6: If (x + y):(x – y) = 5:2, find value of (4x + 5y) / (x – 4y)

a) 43/5

b) -5/43

c) -43/5

d) 5/43

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Question 7: The value of x for which the expressions 11x + 7 and 17x – 1 become equal is ______.

a) -4/3

b) 3/4

c) 4/3

d) -3/4

Question 8: If 2x + 3y = 0 and 3x -­ 4y = 34, then x -­ y =

a) 10

b) -10

c) 2

d) -2

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Question 9: If 5 ­- 3x < 4 ­- x and 5(2 -­ x) > 2 -­ 2x, then x can take which of the following values?

a) 0

b) -1

c) 1

d) 3

Question 10: If a:b = 3:8, find the value of (5a – ­3b)/(2a + b).

a) 9/14

b) 14/9

c) -9/14

d) -14/9

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Answers & Solutions:

1) Answer (A)

Expressions : 19 -­ 5x and 19x + 5

=> $19 – 5x = 19x + 5$

=> $19x + 5x = 19 – 5$

=> $24x = 14$

=> $x = \frac{14}{24} = \frac{7}{12}$

=> Ans – (A)

2) Answer (C)

Expression 1 : 7 + 4x > 3 + 3x

=> $4x – 3x$ > $3 – 7$

=> $x$ > $-4$ ————–(i)

Expression 2 : 3x – 2 < 5 – x

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

=> $4x$ < $7$

=> $x$ < $\frac{7}{4}$ —————–(ii)

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

Thus, $x$ can take values = -3 , -2 , -1 , 0 , 1

=> Ans – (C)

3) Answer (B)

Expression : $3x^{2} = 10^{2} – 5^{2}$

=> $3x^2 = 100 – 25$

=> $3x^2 = 75$

=> $x^2 = \frac{75}{3} = 25$

=> $x = \sqrt{25} = 5$

=> Ans – (B)

4) Answer (A)

 

Expression : $(7x – 13) – (12x + 3) = 14$

=> $7x – 13 – 12x – 3 = 14$

=> $-5x – 16 = 14$

=> $-5x = 16 + 14 = 30$

=> $x = \frac{30}{-5} = -6$

=> Ans – (A)

5) Answer (D)

Expression 1 : 2 + 2x < 5 – x/2

=> $2x + \frac{x}{2}$ < $5 – 2$

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

=> $x$ < $\frac{6}{5}$ ———-(i)

Expression 2 : 5x + 3 > 5 – 5x

=> $5x + 5x$ > $5 – 3$

=> $10x$ > $2$

=> $x$ > $\frac{1}{5}$ ———-(ii)

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

Thus, the only value that $x$ can take = 1

=> Ans – (D)

6) Answer (C)

Given : $\frac{x + y}{x – y} = \frac{5}{2}$

=> $2x + 2y = 5x – 5y$

=> $2y + 5y = 5x – 2x$ => $7y = 3x$

=> $y = \frac{3x}{7}$

To find : $\frac{4x + 5y}{x – 4y}$

= $[4x + 5(\frac{3x}{7})] \div [x – 4(\frac{3x}{7})]$

= $(4x + \frac{15x}{7}) \div (x – \frac{12x}{7})$

= $(\frac{43x}{7}) \div (\frac{-5x}{7})$

= $\frac{43x}{7} \times \frac{-7}{5x} = \frac{-43}{5}$

=> Ans – (C)

7) Answer (C)

Expressions : 11x + 7 and 17x – 1

=> $11x + 7 = 17x – 1$

=> $17x – 11x = 7 + 1$

=> $6x = 8$

=> $x = \frac{8}{6} = \frac{4}{3}$

=> Ans – (C)

8) Answer (A)

Equation 1 : 2x + 3y = 0

Multiplying by 3 on both sides, we get : $6x + 9y = 0$ ———–(iii)

Equation 2 : 3x -­ 4y = 34

Multiplying by 2 on both sides, => $6x – 8y = 68$ ———–(iv)

Subtracting equation(iv) from (iii),

=> $(6x – 6x) + (9y + 8y) = (0 – 68)$

=> $17y = -68$

=> $y = \frac{-68}{17} = -4$

Substituting it in equation (i), we get : $2x + 3(-4) = 0$

=> $2x = 12$

=> $x = \frac{12}{2} = 6$

$\therefore (x – y) = 6 – (-4) = 6 + 4 = 10$

=> Ans – (A)

9) Answer (C)

Expression 1 : 5 ­- 3x < 4 ­- x

=> $3x-x$ > $5-4$

=> $2x$ > $1$

=> $x$ > $\frac{1}{2}$ ————(i)

Expression 2 : 5(2 -­ x) > 2 -­ 2x

=> $10-5x$ > $2-2x$

=> $5x-2x$ < $10-2$

=> $3x$ < $8$

=> $x$ < $\frac{8}{3}$ ————–(ii)

Combining inequalities (i) and (ii), we get : $\frac{1}{2}$ < $x$ < $\frac{8}{3}$

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

=> Ans – (C)

10) Answer (C)

It is given that $a$ : $b$ = 3 : 8

Let $a = 3$ and $b = 8$

To find : $\frac{5a – 3b}{2a + b}$

= $\frac{(5 \times 3) – (3 \times 8)}{(2 \times 3) + (8)}$

= $\frac{(15 – 24)}{(6 + 8)} = \frac{-9}{14}$

=> Ans – (C)

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