**RRB Group-D Expected Maths Questions 2019 PDF**

Download Top-15 RRB Group-D Expected Maths Questions PDF. RRB GROUP-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

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**Question 1: **A car travels 270 km in 3 hours. What is its speed?

a) 30 m/sec

b) 20 m/sec

c) 25 m/sec

d) 15 m/sec

**Instructions**

**Question 2: **Three people start to run around a circle at the same time.Time taken for them to complete a round is 15 sec,20 sec and 30 sec then after how many seconds will they meet for the 1st time ?

a) 0.8 min

b) 1 min

c) 1.5 min

d) 1.2 min

**Question 3: **If the roots of the quadratic equation $x^2 – 6x + 8 = 0$ are the reciprocals of the roots of another equation, which of the following is a correct representation of that equation?

a) $x^2 – 8x + 6 = 0$

b) $8x^2 – x + 6 = 0$

c) $8x^2 – 6x + 1 = 0$

d) $6x^2 – x + 8 = 0$

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**Question 4: **What is the sum of the first 10 terms of the series 2, 5, 8, 11, ……?

a) 135

b) 145

c) 155

d) 165

**Question 5: **What is the sum of $5^2 + 6^2 + 7^2 + . . .14^2$?

a) 1015

b) 985

c) 1065

d) 2010

**Question 6: **A sum of money yields an interest of Rs.3920 at a compound interest of 20% per annum for first year and 30% per annum for second year. Find the sum.

a) Rs.7500

b) Rs.70000

c) Rs.4500

d) Rs.7000

RRB Group D previous year papers

**Question 7: **A lent a sum of Rs.50000 to B at 15% Simple Interest for 2 years. B immediately lent it to C at 15% Compound Interest compounded annually for 2 years. Find the profit of B.

a) Rs.1025

b) Rs.1500

c) Rs.1125

d) Rs.1250

**Question 8: **Calculate the amount(approx in Rs.) on Rs. 72000 @ 3% p.a. compounded half yearly for 2 years.

a) 74692

b) 76492

c) 76418

d) None of these

**Question 9: **Find the probability of forming a 8 letter word with letters A,C,G,E,I,F,D,K such that each letter is used exactly once and all the vowels come together?

a) 1/7

b) 5/56

c) 1/14

d) 3/28

**Question 10: **Find the probability of forming a 4 letter words out of the letter A, C, E, I, G, H, Z, W such that the first letter is W.

a) 3/8

b) 1/8

c) 3/4

d) 4/5

RRB Group-D Important Questions (download PDF)

**Question 11: **Positive numbers ‘a’ and ‘b’ when divided by 7 leaves a remainder of 5 and 6 respectively then what is the remainder obtained when a-b is divided by 7 if a>b ?

a) 0

b) 2

c) 4

d) 6

**Question 12: **Convert 379 into binary system ?

a) 100011011

b) 100111011

c) 101111011

d) 101011011

**Question 13: **What is the last digit of the sum of first 32 whole numbers?

a) 8

b) 6

c) 5

d) 2

**Instructions**

**Question 14: **A square having an area of 81 square meters is made into an equilateral triangle then what is the area of the triangle ?

a) 6$\sqrt{3}$

b) 12$\sqrt{3}$

c) 36$\sqrt{3}$

d) 24$\sqrt{3}$

**Instructions**

**Question 15: **Perimeter of a square is 32 and is equal to the length of the diagonal of the other square.What is the area of the other square ?

a) 128

b) 512

c) 1024

d) 256

General Science Notes for RRB Exams (PDF)

**Answers & Solutions:**

**1) Answer (C)**

Speed of the car = 270/3 = 90 km/hr

Speed of the car in m/sec = $90 \times \dfrac{5}{18} = 25$ m/sec.

**2) Answer (B)**

As they all start at the same time an the time after which they three meet is simply the lcm of durations of time taken=LCM of 15,20 and 30

= 60 sec

**3) Answer (C)**

If the roots of the quadratic equation $ax^2 + bx + c = 0$ are the reciprocals of the roots of another quadratic equation, then the other equation is given by $cx^2 + bx + a = 0$

So, option c) is the correct answer.

**4) Answer (C)**

2, 5, 8, 11 ….. are in AP

a = 2, d = 3

$S_n = \frac{n}{2}[2a+(n-1)d]$

$S_{10} = \frac{10}{2}[2(2)+(10-1)(3)]$

$S_{10} = 5[4+(9)(3)]$

$S_{10} = 5[31]$

$S_{10} = 155$

So the answer is option C.

**5) Answer (B)**

Let S1 = $1^2 + 2^2 + 3^2 + . . .14^2$ and S2 = $1^2 + 2^2 + 3^2 + 4^2$. Hence, the given series is S1 – S2.

S1=14*15*29/6 = 7*29*5 = 1015

S2 = 4*5*9/6 = 2*5*3=30.

Hence, the sum of the given series = S2 – S1 = 1015 – 30 = 985.

**6) Answer (D)**

Let the principal be Rs.100P

Amount at 20% per annum for 1 year = 120% of Rs.100P = Rs.120P

Amount at 30% per annum for 2nd year = 130% of Rs.120 = Rs.156P

Compound Interest = Rs.156P – Rs.100P = Rs.56P

Given, 56P = 3920

⇒ P = 70

Therefore, Principal = 100P = 100*70 = Rs.7000

**7) Answer (C)**

A lent Rs.50000 to B at 15% Simple Interest for 2 years.

Then, Interest = 2*15% of 50000 = 30% of 50000 = Rs.15000

B lent it to C at 15% Compound Interest for 2 years.

Amount after 2 years = 115% of 115% of 50000 = Rs.66125

Compound Interest = Rs.66125 – Rs.50000 = Rs.16125

Profit of B = Rs.16125 – Rs.15000 = Rs.1125

**8) Answer (C)**

Given the Principal(P) = Rs. 72,000/-

Rate of Interest (R) = 3% (compounded half yearly, n=2)

Time period(t) = 2 years

Amount on compound Interest is calculated by

$A = P(1+\frac{R/n}{100})^{nt}$

$A = 72,000(1+\frac{3/2}{100})^{2\times2}$

$A = 72,000(1+\frac{1.5}{100})^4$

$A = 72,000(1.015)^4$

$A = 76418.17$

$A$ ~ $Rs. 76418/-$

**9) Answer (D)**

In these letters, no letter is used more than once. So the total no. of words formed with the letters A,C,G,E,I,F,D,K is 8! Ways.

A,C,G,E,I,F,D,K

In this group A, E, I are vowels. Let us consider them as one group. This group along with the other 5 letters can be arranged in (5+1)! = 6! Ways.

Now this group contains 3 letters which can be arranged in 3! Ways

Total number of ways of arranging all the letter such that vowels come together = 3!*6!

Probability = 3!*6!/8! = 6/56 = 3/28

**10) Answer (B)**

Number of ways of choosing 4 letters out of 8 letters and forming words with those 4 letters =$^8C_4*4!$

Now the first letter is W, we have to choose 3 letters out of 7 letters

Number of ways of choosing 3 letters out of 7 letters =$^7C_3*3!$

Probability =$\frac{^7C_3*3!}{^8C_4*4!}$ = 1/8

**11) Answer (D)**

‘a’ leaves a remainder of 5 when divided by 7 so a=7x+5(let x be the quotient)

‘b’ leaves a remainder of 6 when divided by 7 so b=7y+6(let y be the quotient)

a-b=7(x-y)-1

When divided by 7 it gives -1 as remainder i.e -1+7=6

**12) Answer (C)**

$(379)_{10} = (101111011)_2$

So the answer is option C.

**13) Answer (B)**

First 32 whole numbers = 0,1,2,…..31.

Sum of first 32 whole numbers = n(n+1)/2.

Here, n = 31 since the series starts with 0.

Sum = 31*32/2 = 31*16.

As we can clearly see, the last digit of the sum will be 6. Therefore, option B is the right answer.

**14) Answer (C)**

Area of the square $s^{2}$

$s^{2}$=81

s=9

4s=36

Triangle has 3 sides and each side has equal length as it is equilateral triangle and so a=36/3

=12

Area of an equilateral triangle=$\sqrt{3}\times a^{2}/4$

=36$\sqrt{3}$

**15) Answer (B)**

we have 4*s1=$\sqrt{2}$*s2

s2=32/$\sqrt{2}$

Area of a square=s2*s2

=32*32/2

=512

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