# Important Mathematics Questions for RRB Group-D Set – 2 PDF

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### Important Mathematics Questions for RRB Group-D Set – 2 PDF

Download Top-15 RRB Group-D Important Mathematics Questions set-2  PDF. RRB GROUP-D Important Mathematics questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

Question 1: In a moderately skewed distribution, it is known that the median is 5 and the mode is 7. What is the mean?

a) 10

b) 8

c) 6

d) 4

Question 2: If the ratio between sin a and cos a is x : y, then what is the value of $sin^2 a – cos^2 a$?

a) $\dfrac{x^2 – y^2}{x^2 + y^2}$

b) $\dfrac{xy}{x^2 – y^2}$

c) $\dfrac{x^2 + y^2}{x^2 – y^2}$

d) $\dfrac{xy}{x^2 + y^2}$

Instructions

Question 3: if $\sec\theta-\tan\theta$=4 then what is the value of $\sec\theta+\tan\theta$ ?

a) 4

b) 2

c) 1/4

d) 1/2

Instructions

Question 4: $\sin\theta$=5/13 and $\theta$ belongs to first quadrant then what is the value of $\sec\theta+\tan\theta$

a) 3/2

b) 13/25

c) 2/3

d) 25/12

Question 5: Arithmetic mean of two natural numbers is 15 and what is the minimum value of the product of two numbers ?

a) 225

b) 220

c) 210

d) 200

Question 6: Sum of 5 consecutive odd numbers is 125. What is the product of largest and smallest of the 5 numbers ?

a) 567

b) 609

c) 575

d) 675

Question 7: A person can buy 40 cakes with the money he has and if the cost of the cake is decreased by half rupee then he can buy 42 cakes and he is left with 9 rupees. What is the total money he has ?

a) 235

b) 220

c) 210

d) 240

Question 8: Given a = 3, b =4 and c = 6, which of the following statements is true?
3a+ 4b -c = 20
ab + bc – ca = 18
abc – $c^2$ + ca – b = 52
$a^2c > b^2a$

a) Only 2

b) Only 4

c) Only 1 and 3

d) Only 2 and 4

Question 9: If the radius of a cone is decreased to 2/5 th of the original radius and the volume of the cone is same then what is the ratio of the heights of original to the new one ?

a) 2:5

b) 5:2

c) 4:25

d) 25:4

Question 10: $\triangle$ ABC is a right angled triangle right angled at B. If AB = BC, then, find $\angle$ ACB

a) $60^\circ$

b) $35^\circ$

c) $45^\circ$

d) $30^\circ$

Question 11: A cake is made up of chocolate and bread. It is in the shape of cuboid whose length is 20cm, breadth is 14cm and height is 8 cm, Find the volume of chocolate if volume of bread is 880 cubic cm?

a) 1440

b) 1260

c) 1360

d) 1460

Question 12: Mean temperature of a city for a week is 28°C. If the mean temperature for Monday, Tuesday, Wednesday and Thursday is 27.5°C and the mean temperature for Thursday, Friday, Saturday and Sunday is 29°C, the temperature recorded for Thursday is

a) $30^{0}$

b) $29^{0}$

c) $31^{0}$

d) $28^{0}$

Question 13: A mother said to his son, “I was as old as you are at present at the time of your birth”. If the mother’s present age is 42 years, find the present age of her son.

a) 14 years

b) 18 years

c) 21 years

d) 24 years

Question 14: If log 2 = 0.3010 and log 3 = 0.4771, what is the value of log 432?

a) 1.6353

b) 3.6353

c) 5.6353

d) 2.6353

Question 15: Amar has 10 friends. In how many ways can he invite them to a party such that at least one friend is invited?

a) ${10}^2$ – 1

b) $2^{10}$ – 1

c) $2^{10}$

d) None of the above

Mode = 3(Median)-2(Mean)

7 = 3(5)-2(mode)

7-15 = -2(mode)

-8 = -2(mode)

mode = 4

So the answer is option D.

Given, $\dfrac{sin a}{cos a} = \dfrac{x}{y}$

⇒ $\dfrac{sin a}{x} = \dfrac{cos a}{y}$

Let $\dfrac{sin a}{x} = \dfrac{cos a}{y}$ = k

$\dfrac{sin a}{x} = k ⇒ sin a = kx$

$\dfrac{cos a}{y} = k ⇒ cos a = ky$
We know that $sin^2 a + cos^2 a = 1$
$sin^2 a = (kx)^2 = k^2 x^2$
$cos^2 a = (ky)^2 = k^2 y^2$

⇒ $k^2 x^2 + k^2 y^2 = 1$
⇒ $k^2 ( x^2 + y^2 ) = 1$
⇒ $k^2 = \dfrac{1}{x^2 + y^2}$

$sin^2 a – cos^2 a = k^2 x^2 – k^2 y^2 = k^2 (x^2 – y^2)$
Substituting $k^2 = \dfrac{1}{x^2 + y^2}$ in above equation
Therefore, $sin^2 a – cos^2 a = \dfrac{x^2 – y^2}{x^2 + y^2}$

we know that $\sec^{2}\theta-\tan^{2}\theta$=1
$(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)$=1
$(\sec\theta+\tan\theta) \times 4=1$
$\csc\theta+\cot\theta$=1/4

Given $\sin\theta$=5/13
From $\sin^{2}\theta+\cos^{2}\theta$=1
$\cos\theta$=12/13
$\sec\theta$=13/12
$\tan\theta$=$\sin\theta$/$\cos\theta$
=5/12
sum=(13/12)+(5/12)
=18/12
=3/2

We know that AM$\geq$GM
15$\geq$GM
Let the numbers be a and b
$\sqrt{ab}\leq15$
ab$\geq$225

Let the least odd number be 2n+1 then
2n+1+2n+3+2n+5+2n+7+2n+9=10n+25
10n+25=125
10n=100
n=10
Least odd number=21
Largest number=29
Product of 21*29=607
Hence, option B is the correct answer.

let the cost of cake be x
40x=((2x-1)/2)42+9
40x=42x-21+9
2x=12
x=6
If x=6 total money with him is 40*6=Rs 240
Hence, option D is the correct answer.

Substituting the values in (1), we get 3 * 3 + 4 * 4 – 6 = 19
So, (1) is false.

Substituting the values in (2), we get 3 * 4 + 4 * 6 – 6 *3 = 18
So, (2) is true

Substituting the values in (3), we get 3 * 4 * 6 – 4 * 4 + 6 * 3 – 4 = 50
So, (3) is false.

Substituting the values in (4), we get 3 * 3 * 6 > 4 * 4 * 3
So, (4) is true
Hence the correct option is (d)

Volume of a cone=$(1/3)\pi r^{2}h$
r2=2*r1/5
$(r1)^{2}h1$=$(r2)^{2}h2$
h1=(4/25)h2
h1:h2=4:25

Given $\angle ABC = 90^\circ$

In a triangle, if two sides are equal, then, opposite angles are equal.
Here, AB = BC ⇒ $\angle A = \angle C$
Let $\angle A = \angle C = x^\circ$
Sum of angles in a triangle is equal to $180^\circ$
⇒ $90^\circ + 2x = 180^\circ$
⇒ $2x = 90^\circ$
⇒ $x = 45^\circ$
Hence, $\angle ACB = 45^\circ$

Volume of cake = vol.of bread + vol.of chocolate

(20)(14)(8) = vol.of bread + 880

2240 = vol.of bread + 880

So the answer is option C.

Given,
Mean temperature of a week = $28^{0}$ i.e Total temperature 28 x 7 = 196
Mean temperature for Monday, Tuesday, Wednesday and Thursday is $27.5^{0}$ i.e total temperature = 27.5 x 4 = 110 and
The mean temperature for Thursday, Friday, Saturday and Sunday is $29^{0}$ i.e total temperature = 29 x 4 = 116
Now, the temperature recorded for Thursday is,
(110 + 116) – 196 = 226 – 196 = 30
Hence, option A is the correct answer.

Let the present age of the son be x years
Mother’s age at the time of his birth will be 42 – x years
Given, 42 – x = x ⇒ 2x = 42
⇒ x = 21 years
Hence, The present age of the son = 21 years.

$432 = 2^4 * 3^3$
So, log 432 = log $(2^4 * 3^3)$ = log $2^4$ + log $3^3$ = 4 log 2 + 3 log 3
So, the number of ways of inviting the friends is 2*2*2*…*2 (10 times) = $2^{10}$
From this, we have to subtract the case when none of the friends are invited. So, the total number of ways of inviting the friends such that at least one friend is invited is $2^{10}$ – 1