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# RRB NTPC Expected Maths Questions PDF

Download RRB NTPC Expected Maths Questions and Answers PDF. Top 15 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: A train travels at a speed of 150 kmph without stoppages and at 100 kmph with stoppages. How many minutes per hour does the train stop?

a) 20 minutes

b) 15 minutes

c) 25 minutes

d) 10 minutes

Question 2: A, B and C are employed to do a piece of work for Rs.1400. A and B are supposed to finish $\dfrac{11}{14}th$ of the given work. The amount paid to C is?

a) Rs.600

b) Rs.300

c) Rs.250

d) Rs.700

Question 3: A car travels x km at 60 kmph and then 2x km at 40 kmph. Find its average speed for entire distance ?

a) 45 kmph

b) 48 kmph

c) 56 kmph

d) 60 kmph

Question 4: How many distinct roots does $x^2 – 26x + 169=0$ have?

a) 0

b) 1

c) 2

d) Can’t say

Instructions

Question 5: if $\csc\theta-\cot\theta$=3 then what is the value of $\csc\theta+\cot\theta$ ?

a) 3

b) 1/3

c) 2

d) 1/2

Question 6: If cosec θ – cot θ = 3, then find cosec θ + cot θ = ?

a) $\Large\frac{1}{9}$

b) $\Large\frac{1}{27}$

c) $\Large\frac{1}{4}$

d) $\Large\frac{1}{3}$

Question 7: If $a^3+b^3+c^3 = 3abc$, then find the value of $(a+b)^2-c^2$.

a) -1

b) 1

c) 0

d) 2

Question 8: A person can buy 20 balls with the money he has and if the cost of the ball is decreased by 1 rupee then he can buy 22 balls and he is left with 14 rupees. What is the total money he has ?

a) 60

b) 80

c) 100

d) 120

Question 9: 2[3x – 4/5] + 6x/7 = 8, then find x ?

a) 4/5

b) 5/6

c) 6/7

d) 7/5

Instructions

Question 10: Value of $221+45\div 4-7\times 12$

a) 144.50

b) 142.50

c) 146,25

d) 148.25

Question 11: Find $[((16)^{\frac{1}{4}})^{\frac{2}{3}}]^{\frac{6}{2}}$ = ?

a) 256

b) 64

c) 16

d) 4

Question 12: Sum of two numbers is 99 and difference is 63, then find the smaller number ?

a) 18

b) 36

c) 81

d) 54

Question 13: 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 14: What is the value of 8.33% of 72.72% of 28.57% of 462 ?

a) 2

b) 4

c) 6

d) 8

Question 15: Steph purchased 150 apples and 343 pineapples and after coming home he observed that 14.28% of pineapples were rotten and 16.66% of apples were rotten. Find the approximate percentage of good fruits available of all the fruits purchased ?

a) 87%

b) 85%

c) 83%

d) 81%

Due to stoppages, the train travelled 50 km less in an hour.
Time required to travel 50 km = $\dfrac{50}{150} \times 60 = 20$ minutes

Let the total work be 14 units.
11 units out of 14 units will be completed by A and B together.
14 units → Rs.1400
1 unit → Rs.100
Remaining 3 units will be completed by C.
Then, C will be paid 3*100 = Rs.300.

Time in case 1 = x/60 hrs

Time in case 2 = 2x/40 = x/20 hrs

Total time = x/60 + x/20 = x/15  hrs

Total distance = x+2x = 3x km

Average speed = total distance/ total time = (3x)/(x/15) = 45 kmph

So the answer is option A.

$x^2 – 26x + 169 =0$
=> $x^2 – 13x – 13x + 169 = 0$
$(x-13)(x-13)=0$
Hence, x=13

Thus, the roots of the equation are equal. Hence, it has only one distinct root.

we know that $\csc^{2}\theta-\cot^{2}\theta$=1
$(\csc\theta+\cot\theta)(\csc\theta-\cot\theta)$=1
$(\csc\theta+\cot\theta) \times$3=1
$\csc\theta+\cot\theta$=1/3

We know that $cosec^2 θ – cot^2 θ = 1$
$(cosec θ – cot θ) ( cosec θ + cot θ) = 1$
Given cosec θ – cot θ = 3
$\therefore 3 ( cosec θ + cot θ) = 1$
cosec θ + cot θ = $\Large\frac{1}{3}$

If $a^3+b^3+c^3 = 3abc$, then $a+b+c = 0$.
$(a+b)^2-c^2 = (a+b+c)(a+b-c) = 0$

let the cost of ball be x
20x=((x-1)*22)+14
20x=22x-22+14
2x=8
x=4
If x=4 total money with him is 20*4=Rs 80
Hence, option B is the correct answer.

2[3x – 4/5] + 6x/7 = 8

6x – 8/5 +6x/7 = 8

48x/7 = 8+8/5

48x/7 = 48/5

x = 7/5

So the answer is option D.

By simplification we get
=221+11.25-84
= 148.25

$[((16)^{\frac{1}{4}})^{\frac{2}{3}}]^{\frac{6}{2}}$ = $[16]^{\frac{1}{4}\times\frac{2}{3}\times\frac{6}{2}}$ = $[16]^{\frac{1}{2}}$ = $4$

So the answer is option D.

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.

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 $7^3 + 8^3 + 3 \times 7 \times 8 \times 15 – 15^3 = 0$

8.33%=1/12
72.72%=8/11
28.57%=2/7
Therefore we have (1/12)*(8/11)*(2/7)*462= (1/12)(8/11)(2/7)*11*7*6
=8