# RRB Group-D Expected Maths Questions 2019 PDF

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## 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.

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) $6x^2 – x + 8 = 0$

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

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

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

Speed of the car = 270/3 = 90 km/hr
Speed of the car in m/sec = $90 \times \dfrac{5}{18} = 25$ m/sec.

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

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.

2, 5, 8, 11 ….. are in AP

a = 2, d = 3

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

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

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

$S_{10} = 5$

$S_{10} = 155$

So the answer is option C.

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.

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

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

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
$A = P(1+\frac{R/n}{100})^{nt}$
$A = 72,000(1+\frac{3/2}{100})^{2\times2}$
$A = 72,000(1+\frac{1.5}{100})^4$
$A = 72,000(1.015)^4$
$A = 76418.17$
$A$ ~ $Rs. 76418/-$

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

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

‘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

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

So the answer is option C.

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.

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$\sqrt{3}$

we have 4*s1=$\sqrt{2}$*s2
s2=32/$\sqrt{2}$