# Elementary Algebra Questions for RRB NTPC

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## Elementary Algebra Questions for RRB NTPC

Download RRB NTPC Top-15 Elementary Algebra Questions PDF. Questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: Find the value of x for which the expression $2 – 3x- 4x^{2}$ has the greatest value.

a) $-\frac{41}{16}$

b) $\frac{3}{8}$

c) $-\frac{3}{8}$

d) $\frac{41}{16}$

Question 2: If 3x – 4 > 2 – x/3 and 3x + 5 > 4x – 5; then ﬁnd the value of x?

a) 7

b) 10

c) -11

d) 1

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

a) 2

b) 0

c) 10

d) 12

Question 4: If $3x^{2} = 10^{2} – 5^{2}$, find the value of x?

a) 7

b) 5

c) 9

d) 11

Question 5: If 2x+ 4(x – 3) < -2 – x < 2x – 1, then find the value of x.

a) -1

b) 1

c) 2

d) -2

Question 6: If $13x^2 = 17^2 – 9^2$, find the value of x?

a) 16

b) 12

c) 8

d) 4

Question 7: If x +$\frac{1}{x}$=c+$\frac{1}{c}$ then find the value of x?

a) C,$\frac{1}{c}$

b) C,$C^{2}$

c) C.2C

d) 0,1

Question 8: If x = 3 + 2√2 , then find the value of $x^2 + \frac{1}{x^2}$

a) 36

b) 30

c) 32

d) 34

Question 9: If $a + b +c= 6$, $a^{2}+ b^{2}+ c^{2} = 14$, find the value of $bc + ca + ab$.

a) 22

b) 25

c) 20

d) 11

Question 10: 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 11: 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

Question 12: If a:b = 5:8, find the value of (6a ­- 5b)/(a + 2b).

a) 10/21

b) ­21/10

c) -21/10

d) -10/21

Question 13: If $xy = -30$ and $x^{2} + y^{2} = 61$, then find the value of (x + y).

a) 2

b) 3

c) 1

d) 4

Question 14: If $7^{m+1}=2401$, then find the value of $2^{2m+2}$

a) 224

b) 256

c) 264

d) 286

Question 15: $If a + \frac{1}{a} = 1,$ find the value of $a^{3} + \frac{1}{a^{3}}$

a) 2

b) -2

c) 0

d) 1.5

NOTE :- To find the min/max value of an expression, we need to differentiate it, and put the first derivative equal to ‘0’ to find value of $x$

Then, we need to differentiate it again and put value of $x$, if second derivative is less than zero, then $x$ is maxima otherwise $x$ is minima.

Expression : $y$ = $2 – 3x- 4x^{2}$

=> $\frac{dy}{dx} = -3 – 8x$

Now, putting it equal to 0, we get :

=> $y’ = -3 -8x = 0$

=> $x = \frac{-3}{8}$

Differentiating it again :

=> $\frac{d^2y}{dx^2} = -8$

Since, it is less than ‘0’ ,=> $x = \frac{-3}{8}$ is maximum value.

Expression 1 : 3x – 4 > 2 – x/3

=> $3x + \frac{x}{3}$ > $2 + 4$

=> $\frac{10x}{3}$ > $6$

=> $x$ > $\frac{18}{10} = 1.8$ ——–(i)

Expression 2 : 3x + 5 > 4x – 5

=> $4x – 3x$ < $5 + 5$

=> $x$ < $10$ ———(ii)

Combining inequalities (i) and (ii), we get : $1.8$ < $x$ < $10$

Thus, the only value that $x$ can take among the options = 7

=> 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)

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)

Expression 1 : $-2 – x < 2x – 1$

=> $2x + x$ > $1 – 2$

=> $x$ > $\frac{-1}{3}$ ———-(i)

Expression 2 : $2x + 4(x – 3) < -2 – x$

=> $6x – 12$ < $-2 – x$

=> $6x + x$ < $12 – 2$

=> $x$ < $\frac{10}{7}$ ——(ii)

Combining inequalities (i) and (ii), we get : $\frac{-1}{3}$ < $x$ < $\frac{10}{7}$

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

=> Ans – (B)

Expression : $13x^2 = 17^2 – 9^2$

=> $13 x^2 = (17 – 9) (17 + 9)$

=> $13 x^2 = 8 \times 26$

=> $x^2 = 8 \times 2 = 16$

=> $x = \sqrt{16} = 4$

Solving the equation , you will get ,

x = $frac{(c+1)^2 }{2} , frac{(c-1)^2 }{2}$

Expression : $x = 3 + 2\sqrt{2}$

=> $\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}}$

=> $\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 – 2\sqrt{2}}{3 – 2\sqrt{2}}$

=> $\frac{1}{x} = 3 – 2\sqrt{2}$

$\therefore$ $x + \frac{1}{x} = 3 + 2\sqrt{2} + 3 – 2\sqrt{2} = 6$

Squaring both sides, we get :

=> $(x + \frac{1}{x})^2 = 6^2$

=> $x^2 + \frac{1}{x^2} + 2 = 36$

$\therefore x^2 + \frac{1}{x^2} = 34$

It is given that : $a + b + c = 6$ and $a^{2}+ b^{2}+ c^{2} = 14$

Also, $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

=> $6^2 = 14 + 2(ab + bc + ca)$

=> $ab + bc + ca = 22/2 = 11$

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)

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)

It is given that a : b = 5 : 8

Let $a = 5$ and $b = 8$

To find : $\frac{6a – 5b}{a + 2b}$

= $\frac{(6 \times 5) – (5 \times 8)}{(5) + (2 \times 8)}$

= $\frac{30 – 40}{5 + 16} = \frac{-10}{21}$

=> Ans – (D)

Given : $(x^2 + y^2) = 61$ and $xy = -30$

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

=> $(x + y)^2 = 61 + (2 \times -30)$

=> $(x + y) = \sqrt{61 – 60} = 1$

=> Ans – (C)

Given : $7^{m+1}=2401$

=> $7^{m+1}=7^4$

=> $m+1=4$

=> $m=4-1=3$

$\therefore$ $2^{2m+2}=2^{2\times3+2}$

= $2^8=256$

=> Ans – (B)

Given, $a + \frac{1}{a} = 1$
$a^{3} + \frac{1}{a^{3}} + 3(a + \frac{1}{a}) = 1$
$a^{3} + \frac{1}{a^{3}} = -2$ ( as we know $a + \frac{1}{a} = 1$)