Algebra Questions for RRB NTPC Set-3 PDF
Download RRB NTPC Algebra Questions Set-3 PDF. Top 10 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.
Download Algebra Questions for RRB NTPC Set-3 PDF
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Question 1: $\sqrt{8 + \sqrt{57 + \sqrt{38 + \sqrt{108 + \sqrt{169}}}}}$
a) 4
b) 6
c) 8
d) 10
Question 2: If a * b = 2a + 3b – ab, then the value of (3 * 5 + 5 * 3) is
a) 10
b) 6
c) 4
d) 2
Question 3: If a * b = $a^{b}$, then the value of 5 * 3 is
a) 125
b) 243
c) 53
d) 15
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Question 4: If $x = 1 + \sqrt{2} + \sqrt{3}$ , then the value of $(2x^4 – 8x^3 – 5x^2 + 26x- 28)$ is __?
a) $6\sqrt{6}$
b) $0$
c) $3\sqrt{6}$
d) $2\sqrt{6}$
Question 5: If $a^2+b^2+c^2=2(a-2b-c-3)$ then the value of a+b+c is
a) 3
b) 0
c) 2
d) 4
Question 6: Find the simplest value of $2\sqrt{50} + \sqrt{18} – \sqrt{72}$ is __? $(\sqrt{2} = 1.414)$.
a) 9.898
b) 10.312
c) 8.484
d) 4.242
Question 7: If $a^{3}-b^{3}-c^{3}=0$ then the value of $a^{9}-b^{9}-c^{9}-3a^{3} b^{3} c^{3}$ is
a) 1
b) 2
c) 0
d) -1
Question 8: If x + y + z = 6 and $x^{2}+y^{2}+z^{2}$=20 then the value of $x^{3}+y^{3}+z^{3}$-3xyz is
a) 64
b) 70
c) 72
d) 76
Question 9: If a and b are the roots of the equation $x^2+4x-2=0$, find the equation, which will have the roots, $\frac{a}{b}$ and $\frac{b}{a}$.
a) $x^2 +5x +6 = 0$
b) $x^2 -10x +5 = 0$
c) $x^2 +10x +1 = 0$
d) $2x^2 +5x +1 = 0$
Question 10: Find x in the equation $81^{0.75} * 6561^{0.25} = 3^x $
a) 6
b) 8
c) 2
d) 5
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Answers & Solutions:
1) Answer (A)
Start from the root of 169 then second root will reduce to 11, thrid root will reduce to 7, fourth root will reduce to 8, and finally it reduce to value 4
2) Answer (A)
For 3*5 put a=3 and b=5 in given equation
and for 5*3 put a=5 and b=3 in equation
now add both values
3) Answer (A)
Put a=5 and b=3 in given equation
hence it will be $5^{3}$ = 125
4) Answer (A)
x = 1+ $\sqrt {2} + \sqrt {3} $
$(x-1)^{2}$ = $(\sqrt {2} + \sqrt {3}) ^ {2} $
$x^{2} +1 – 2x = 5 + 2 \sqrt {6}$
$x^{2} – 2x = 4 + 2 \sqrt {6}$ ( eq. (1) )
$(x^{2} – 2x)^{2} = x^{4} + 4x^{2} – 4x^{3} = 40 + 16\sqrt{6} $ eq (2)
Now in $2x^{4} – 8x^{3} – 5x^{2} + 26x – 28 $
or $2(x^{4} – 4x^{3}) – 5x^{2} + 26x – 28 $ ( putting values from eq (1) and eq (2) )
After solving we will get it reduced to $6\sqrt{6}$
5) Answer (B)
Given $a^2+b^2+ c^2=2(a-2b-c-3)$,
So, $(a-1)^2+(b+2)^2+(c-1)^2=0$
Hence, a=1, b=-2 and c=1
So, the sum of the equation is
6) Answer (A)
Given equation can be reduced in the form of $10\sqrt2 + 3\sqrt2 – 6\sqrt2 = 7\sqrt2$
Hence $7\sqrt2$ will be around 9.898
7) Answer (C)
shortcut :
put c = 0 in $a^{3}-b^{3}-c^{3}=0$ $\Rightarrow$ $a^{3}=b^{3}$
$a^{9}-b^{9}-(0)^{9}-3a^{3} b^{3} (0)^{3}$ = $a^{9}-b^{9}$ = $(a^{3})^{3}-(b^{3})^{3}$ = $(a)^{3}-(a)^{3}$ = 0 ( $\because$ $a^{3}=b^{3}$ )
so the answer is option C.
normal method :
$a^{3}-b^{3}-c^{3}=0$
$a^{3}=b^{3}+c^{3}$
cubing on both sides,
$(a^{3})^{3}=(b^{3}+c^{3})^{3}$
$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(b^{3}+c^{3})$
$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(a^{3})$
$a^{9}-b^{9}-c^{9}-3a^{3}b^{3} c^{3}=0$
so the answer is option C.
8) Answer (C)
We know that $x^{3}+y^{3}+z^{3}-3xyz = (x + y + z)(x^2 +y^2 + z^2 -xy-yz-xz)$
$x^{3}+y^{3}+z^{3}-3xyz = (6)(20 -xy-yz-xz)$
Hence the solution must be a multiple of 6.
Out of the given options only Option C is a multiple of 6.
Hence Option C is the correct answer.
9) Answer (C)
Sum of the roots of new equation = $\frac{a}{b} + \frac{b}{a}$ = $\frac{a^2 + b^2}{ab}$ = $\frac{(a+b)^2 – 2ab}{ab}$ = $\frac{(-4)^2 – 2*(-2)}{(-2)}$ =-10
Product of the roots = $\frac{a}{b} * \frac{b}{a}$ = 1
Therefore, the required equation is $x^2 +10x +1 = 0$.
10) Answer (D)
$81^{0.75} = 27$
$6561^{0.25} = 9$
==> $81^{0.75} * 6561^{0.25} = 3^5 $
==> x = 5.
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