Simplification Questions for Railway Exams PDF
Download Simplification Questions for Railway Exams PDF. Top 15 questions based on asked questions in previous exam papers very important for the Railway NTPC exam.
Download Simplification Questions for Railway Exams PDF
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Question 1: Simplify : $5 + 4 \times 5^2 + 4 \times 5^3 + 4 \times 5^4 + 4 \times 5^5$
a) $5^8$
b) $5^4$
c) $5^6$
d) $5^2$
Question 2: simplify $\sqrt{25+10\sqrt{6}}+ \sqrt{25-10\sqrt{6}}$?
a) $2\sqrt{15}$
b) $2\sqrt{5}$
c) $\sqrt{50}$
d) $\sqrt{55}$
Question 3: Simplify: $6 \div [2(1 + 2)]$
a) 3
b) 9
c) 1
d) 2
Question 4: $(8-(28-53))\div(-4\times5-(-9))=?$
a) 11
b) -3
c) 3
d) -11
Question 5: $45-[38-(80\div4-(8-12\div3)\div4)]=?$
a) 25
b) 27
c) 26
d) 28
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Question 6: The value of $ 14 \div $ { ( 5 of 2 – 3 ) } $ \times 4 ( 7 – 2 ) $ is :
a) 44
b) 40
c) $ \frac{1}{10} $
d) $ \frac{14}{19} $
Question 7: $ 4 + \frac{1}{6} \times $ [ { $ – 12 \times ( 24 – 13 – 3) $ } $ \div ( 20 – 4) $ ] = ?
a) 4
b) 6
c) 5
d) 3
Question 8: $384 \div 2^5 \times 3 + 8 = ?$
a) 12
b) 132
c) 3
d) 44
Question 9: $\frac{3}{4} + \left\{\frac{3}{4} + \frac{3}{4} \div (\frac{3}{4} + \frac{3}{4})\right\} = ?$
a) $\frac{3}{4}$
b) 1
c) 2
d) $2\frac{3}{4}$
Question 10: The simplest form of $\frac{391}{667}$ is
a) $\frac{23}{31}$
b) $\frac{19}{23}$
c) $\frac{15}{19}$
d) $\frac{17}{29}$
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Question 11: $\frac{9x^{2}-24xy+16y^{2}}{3x-4y}$ = ?
a) 3x+4y
b) 3x-4y
c) 4x+3y
d) 4x-3y
Question 12: Evaluate :- $\frac{16 \times2^{n+1 }-4\times2^{n}}{16\times 2^{n+2}-2\times2^{n+2}}$
a) $\frac{1}{2}$
b) $0$
c) $4$
d) $\frac{1}{3}$
Question 13: $ (1-\frac{1}{2}) (1-\frac{1}{3}) (1-\frac{1}{4}) …..(1-\frac{1}{40}) = ?$
a) $ \frac{1}{40} $
b) $ \frac{1}{20} $
c) Countless
d) Zero
Question 14: Evaluate: $\left[\frac{\sqrt{3} + 1}{\sqrt{3} – 1}\right]^2 + \left[\frac{\sqrt{3} – 1}{\sqrt{3} + 1}\right]^2$
a) 16
b) 12
c) 14
d) 24
Question 15: Simplify: $\frac{1\frac{1}{4} \div 1\frac{1}{2}}{\frac{1}{15} + 1- \frac{9}{10}}$
a) 2
b) 5
c) 3
d) 4
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Answers & Solutions:
1) Answer (C)
2) Answer (A)
3) Answer (C)
4) Answer (B)
$(8-(28-53)) = 8-28+53 = 33$
$(-4\times5-(-9)) = -20+9 = -11$
Therefore, $(8-(28-53))\div(-4\times5-(-9))=\dfrac{33}{-11} = -3$
5) Answer (C)
$45-[38-(80\div4-(8-12\div3)\div4)] = 45 – (38 – (20 – \dfrac{8-4}{4}))$
$= 45 – (38 -(20-\dfrac{4}{4})) = 45-(38-(20-1))$
$= 45-(38-19) = 45-19 = 26$
6) Answer (B)
$ 14 \div $ { ( 5 of 2 – 3 ) } $ \times 4 ( 7 – 2 ) $
=$ 14 \div $ 7 $ \times 4 ( 5) $
=2$ \times 4 ( 5) $
=2$ \times 20 $
=40
7) Answer (D)
$ 4 + \frac{1}{6} \times $ [ { $ – 12 \times ( 24 – 13 – 3) $ } $ \div ( 20 – 4) $ ]
=$ 4 + \frac{1}{6} \times $ [ { $ – 12 \times 8 $ } $ \div 16 $ ]
=$ 4 + \frac{1}{6} \times -6$
=4-1
=3
8) Answer (D)
9) Answer (C)
$\frac{3}{4} + \left\{\frac{3}{4} + \frac{3}{4} \div (\frac{3}{4} + \frac{3}{4})\right\}$
First solve small bracket as per BODMAS rule.
$\frac{3}{4} + \left\{\frac{3}{4} + \frac{3}{4} \div (\frac{6}{4} )\right\}$
$\frac{3}{4} + \left\{\frac{3}{4} + \frac{1}{2} \right\}$
Now solve curly bracket as per BODMAS rule.
$\frac{3}{4} + \frac{5}{4} $
$\frac{8}{4} $
2
10) Answer (D)
$\frac{391}{667}$ = $\frac{17 \times 23}{29 \times 23}$ = $\frac{17}{29}$
11) Answer (B)
$\frac{9x^{2}-24xy+16y^{2}}{3x-4y}$
=$\frac{(3x-4y)^{2}}{3x-4y}=3x-4y$
12) Answer (A)
$\frac{16 \times2^{n+1 }-4\times2^{n}}{16\times 2^{n+2}-2\times2^{n+2}}$
=$\frac{2^{n}(32-4)}{2^{n+2}(16-2)}$
=$\frac{28}{4(14)}$=$\frac{1}{2}$
13) Answer (A)
1 – 1/2 = 1/2
1 – 1/3 = 2/3
1 – 1/4 = 3/4
The denominator of the first term gets cancelled by the numerator of the second term and so on…
So, the final value = 1/40
14) Answer (C)
Using the identities
$ (a + b)^2 = a^2 + 2ab + b^2 $
$ (a – b)^2 = a^2 – 2ab + b^2 $
$ (a + b)(a – b) = a^2 – b^2 $
Rationalizing the denominator,
$\left[\frac{\sqrt{3} + 1}{\sqrt{3} – 1}\right] = \frac{\sqrt{3} + 1}{\sqrt{3} -1} \times \frac{\sqrt{3} + 1}{\sqrt{3} +1} $
Solving the equation using identities we get
$ \frac{\sqrt{3} + 1}{\sqrt{3} -1} \times \frac{\sqrt{3} + 1}{\sqrt{3} +1} = \frac{4 + 2\sqrt3}{2} $
= $ 2 + \sqrt 3 $
$\left[\frac{\sqrt{3} + 1}{\sqrt{3} – 1}\right]^2 = (2 + \sqrt 3)^2$
= $ 7 + 4\sqrt3 $
Rationalizing the denominator,
$\left[\frac{\sqrt{3} – 1}{\sqrt{3} + 1}\right] = \frac{\sqrt{3} – 1}{\sqrt{3} +1} \times \frac{\sqrt{3} -1}{\sqrt{3} -1} $
Solving the equation using identities we get
$ \frac{\sqrt{3} -1}{\sqrt{3} +1} \times \frac{\sqrt{3} – 1}{\sqrt{3} -1} = \frac{4 – 2\sqrt3}{2} $
= $ 2 – \sqrt 3 $
$\left[\frac{\sqrt{3} -1}{\sqrt{3} + 1}\right]^2 = (2 – \sqrt 3)^2$
= $ 7 – 4\sqrt3 $
Thus,
$\left[\frac{\sqrt{3} + 1}{\sqrt{3} – 1}\right]^2 + \left[\frac{\sqrt{3} – 1}{\sqrt{3} + 1}\right]^2 = 7 + 4\sqrt 3 + 7 – 4\sqrt 3 $
= 14
15) Answer (B)
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