# RRB NTPC Simplification Questions PDF

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## RRB NTPC Simplification Questions PDF

Download RRB NTPC Simplification Questions and Answers PDF. Top 20 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: 3245 – 123 – 456 – 347 + x = 3650 ?

a) 2134

b) 1331

c) 1567

d) 1467

Question 2: $600\div 12 \times \frac{1}{2} \div 5 = ?$

a) 4

b) 5

c) 6

d) 8

Question 3: The number of even factors of 28 is ?

a) 2

b) 3

c) 4

d) 5

Question 4: Sum of two numbers is 99 and difference is 63, then find the smaller number ?

a) 18

b) 36

c) 81

d) 54

Question 5: Find the value of $\frac{19}{121}$ of $209 – \frac{3}{187}$ of $51$.

a) $\frac{360}{11}$

b) 43

c) 32

d) 12

Question 6: What is the approximate value of $\frac{(\sqrt{334} + \sqrt{223}) * (\sqrt{334} – \sqrt{223})}{(\sqrt{111} + \sqrt{11}) * (\sqrt{111} – \sqrt{11})}$?

a) 1.01

b) 1.10

c) 1.11

d) 1.21

Question 7: Arrange the following fractions in descending order of value $\frac{3}{4}, \frac{11}{12},\frac{7}{9},\frac{13}{16}$

a) $\frac{11}{12},\frac{7}{9},\frac{13}{16},\frac{3}{4}$

b) $\frac{11}{12},\frac{13}{16},\frac{3}{4},\frac{7}{9}$

c) $\frac{11}{12},\frac{13}{16},\frac{7}{9},\frac{3}{4}$

d) $\frac{7}{9},\frac{13}{16},\frac{11}{12},\frac{3}{4}$

Question 8: What is the value of $7^3 + 8^3 + 3 \times 7 \times 8 \times 15 – 15^3$?

a) 900

b) 169

c) 225

d) 0

Question 9: If $x^{2}-5x-24=0$, then find the value of x

a) +8,-3

b) +8,+3

c) -8,-3

d) -8,+3

Question 10: The simplest form of $\frac{391}{667}$ is

a) $\frac{23}{31}$

b) $\frac{19}{23}$

c) $\frac{15}{19}$

d) $\frac{17}{29}$

Question 11: $0.5 \div 12 + 0.25 \times 0.05$ = ?

a) 0.0525

b) 0.7

c) 0.00196

d) None of these

Question 12: $\sqrt{6400}$ = ?

a) 40

b) 80

c) 60

d) None of these

Question 13: 4% of 400 – 2% of 800 = ?

a) 0

b) 1

c) 8

d) 16

Question 14: When polynomial $x^{4}-3x^{2}+2x+5$ is divided by (x-1), the remainder is

a) 2

b) 3

c) 4

d) 5

Question 15: 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 16: Evaluate :- $(-216 \times 1728)^{\frac{1}{3}}$

a) -72

b) 27

c) 72

d) -27

Question 17: 612/991 = ?

a) 0.65

b) 0.56

c) 0.6355

d) 0.6175

Question 18: 67 * 89 + 78 * 56= ?

a) 14911

b) 9991

c) 10201

d) 10331

Question 19: 50% of (37 * 11/ 25)

a) 8.14

b) 9.24

c) 7.6

d) 8.4

Question 20: $67^2 – 33^2$ = ?

a) 3700

b) 3400

c) 4532

d) 3467

3245 – 123 – 456 – 347 + x = 3650

2319 + x = 3650

x = 1331

So the answer is option B.

$600\div 12 \times \frac{1}{2} \div 5 = 50 \times 1/10 = 5$

So the answer is option B.

Factors of 28 are 1, 2, 4, 7, 14 & 28.

28 has 4 even numbers as it’s factors

So the answer is option C.

x+y = 99 —-(1)

x-y = 63 —–(2)

(1)-(2) ==> x+y-x+y = 99-63 ==> 2y = 36 ==> y = 18

So the answer is option A.

$\frac{19}{121}$ of $209 – \frac{3}{187}$ of $51$

$\frac{361}{11} – \frac{9}{11}$ = 32

The formula we use to simplify the expression is (a+b)(a-b) = $a^2- b^2$
So, $\frac{(\sqrt{334} + \sqrt{223}) * (\sqrt{334} – \sqrt{223})}{(\sqrt{111} + \sqrt{11}) * (\sqrt{111} – \sqrt{11})}$ = $\frac{334 – 223}{111-11}$ = 1.11

The four fractions given are $\frac{3}{4}, \frac{11}{12},\frac{7}{9},\frac{13}{16}$
The LCM of the denominators is $144$.
So, the fractions can be written as $\frac{108}{144},\frac{132}{144},\frac{112}{144},\frac{117}{144}$
Based on the numerators, we can arrange the four fractions in descending order as below

$\frac{11}{12},\frac{13}{16},\frac{7}{9},\frac{3}{4}$

We know that $A^3 + B^3 + C^3 – 3ABC = (A+B+C) \times (A^2 + B^2 + C^2 – 3ABC)$
In this case, $A+B+C = 7+8-15 = 0$
Therefore, the value of $7^3 + 8^3 + 3 \times 7 \times 8 \times 15 – 15^3 = 0$

$x^{2}-5x-24=0$

So, $x^{2}-8x + 3x -24=0$

ie x(x- 8) + 3(x-8) = 0

=> (x+3)(x-8)=0

or x = 8 or -3

$\frac{391}{667}$ = $\frac{17 \times 23}{29 \times 23}$ = $\frac{17}{29}$

.5 $\div$ 12 = .04

.25* .05 = .0125

.04 + .0125 = .0525

$\sqrt{6400} = \sqrt {80 \times 80}$ = 80

4% of 400 = $\frac {4}{100} \times 400$= 16

2% of 800 = $\frac {2}{100} \times 800$ = 16

16 – 16 = 0

$f(x)=x^{4}-3x^{3}+2x +5$

to find the remainder put x=1 in $f(x)$

$f(1)$=1-3+2+5=5

∴5 is the remainder

$\frac{16 \times2^{n+1 }-4\times2^{n}}{16\times 2^{n+2}-2\times2^{n+2}}$

=$\frac{2^{n}(32-4)}{2^{n+2}(16-2)}$

=$\frac{28}{4(14)}$=$\frac{1}{2}$

$(-216 \times 1728)^{\frac{1}{3}}$

=$(-6^{3} \times 12^{3})^{\frac{1}{3}}$

=-72

We see that 991 is quite close to 1000. The 1000 is 9 greater than 991 which is approximately 1%.
1% of 612 is approximately 6.
Hence we increase numerator and denominator to get
612/991 ~ 618/1000 = 0.618.
The option closest to the answer we have gotten is option D.
Hence, the answer is option D.

67 * 89 = (70-3)*(90-1) = 6300-270-70+3 = 5963.
78*56 = (80-2)*(60-4) = 4800-120-320+8 = 4368
67 * 89 + 78 * 56 = 5963+4368 = 10331

$a^2-b^2$= (a+b)*(a-b)
$67^2 – 33^2$ = $(67+33)(67-33)$ = 100*34 =3400.