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.
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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
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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
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Answers & Solutions:
1) Answer (C)
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.
2) Answer (A)
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)
3) 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)
4) 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)
5) Answer (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)
6) Answer (D)
Expression : $13x^2 = 17^2 – 9^2$
=> $13 x^2 = (17 – 9) (17 + 9)$
=> $13 x^2 = 8 \times 26$
=> $x^2 = 8 \times 2 = 16$
=> $x = \sqrt{16} = 4$
7) Answer (A)
Solving the equation , you will get ,
x = $frac{(c+1)^2 }{2} , frac{(c-1)^2 }{2}$
8) Answer (D)
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 :
=> $(x + \frac{1}{x})^2 = 6^2$
=> $x^2 + \frac{1}{x^2} + 2 = 36$
$\therefore x^2 + \frac{1}{x^2} = 34$
9) Answer (D)
It is given that : $a + b + c = 6$ and $a^{2}+ b^{2}+ c^{2} = 14$
Also, $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
=> $6^2 = 14 + 2(ab + bc + ca)$
=> $ab + bc + ca = 22/2 = 11$
10) 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)
11) 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)
12) Answer (D)
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)
13) Answer (C)
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)
14) Answer (B)
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)
15) Answer (B)
Given, $a + \frac{1}{a} = 1$
Cubing on both sides we get,
$a^{3} + \frac{1}{a^{3}} + 3(a + \frac{1}{a}) = 1$
$ a^{3} + \frac{1}{a^{3}} = -2$ ( as we know $a + \frac{1}{a} = 1$)
Hence, option B is the correct answer.
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