RRB JE Mathematics Questions Set-2 PDF

Download Top 15 RRB JE Mathematics Set-2 Questions and Answers PDF. RRB JE Mathematcis Questions based on asked questions in previous exam papers very important for the Railway JE exam.

Download RRB JE Mathematics Questions Set-2 PDF

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Question 1: If tan(θ/2) = 5/12, then find sinθ = ?

a) 5/13

b) 119/120

c) 120/169

d) 12/13

Question 2: Find the value of $x$ (from the given options), if $x^2-16x+64=4$ ?

a) 8

b) 10

c) 12

d) 14

Question 3: Sum of a natural number and 5 times its reciprocal is 14/3, then find that number ?

a) 6

b) 5

c) 4

d) 3

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Question 4: If p, q are the sum and product of the roots of $3x^3-4x^2+5x-6=0$, then find p/q ?

a) 3

b) 1/3

c) 2/3

d) 3/2

Question 5: 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
$a^{2}c>b^{2}a$

a) Only 2

b) Only 4

c) Only 1 and 3

d) Only 2 and 4

Question 6: Find the value of the expression:$\frac{{114}^3+7^3}{{114}^2+7^2-(114\ast 7)}$

a) 118

b) 124

c) 125

d) 121

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Question 7: Three numbers A, B and C are such that A is 50% more than B which is twice of C. Then find the ratio between A and C.

a) 2 : 3

b) 3 : 2

c) 3 : 1

d) 1 : 2

Question 8: The ratio of two numbers is 2:5. If the difference between the squares of the numbers is 1029, then find the sum of the two numbers.

a) 35

b) 56

c) 49

d) 28

Question 9: The ratio of boys to girls in a class is 10:7. After a few days, 60 boys were added to the class. Now, the ratio of boys to girls changed to 5:2. What was the total number of student in the class initially?

a) 80

b) 120

c) 136

d) 196

Question 10: A beaker contains acid solution of concentration 60%. What fraction of the acid solution should be replaced with water, so that the resultant solution has a concentration of 50%?

a) 1/3

b) 1/4

c) 1/6

d) 1/5

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Question 11: The difference of two positive numbers is 10. 2 and 48 are the HCF and LCM of those two numbers, then find the smaller number ?

a) 12

b) 16

c) 6

d) 8

Question 12: In an exam, Amit is supposed to calculate the product of 34.1 * 4.67. He instead calculates the product of 0.0341 * 46.7 as 1.59. What is the actual answer in the exam?

a) 1.59

b) 15.90

c) 0.159

d) 159.0

Question 13: Given that log2 = 0.3 approx, one billion would be approximately

a) $2^9$

b) $2^{10}$

c) $2^{20}$

d) $2^{30}$

Question 14: If the hypotenuse of an isosceles right angled triangle is $6\times \sqrt{2}$m, then find the area of that triangle ?

a) 36 sq.m

b) 18 sq.m

c) 24 sq.m

d) 48 sq.m

Question 15: A third angle orthographic projection of an object is given below. What is the object ?

a) Triangle

b) Trapezium

c) Cone

d) Frustrum of a cone

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Answers & Solutions:

1) Answer (C)

$Sin\theta$ = $\frac{2tan(\theta /2)}{1+ta{n}^2(\theta /2)}$ = $\frac{2(5/12)}{1+(5/12{)}^2}$ = $120/169$

So the answer is option C.

2) Answer (B)

$x^2-16x+64=4$

$x^2-16x+60=0$

$x^2-10x-6x+60=0$

$x(x-10)-6(x-10)=0$

$(x-10)(x-6)=0$

$x=10$ or $x=6$

So the answer is option B.

3) Answer (D)

Let the number be x

$x+5(1/x)=14/3$

$x^2+5=14x/3$

$3x^2+15=14x$

$3x^2-14x+15=0$

$3x^2-9x-5x+15=0$

$(3x-5)(x-3)=0$

$x=5/3$ or $x=3$

So the answer is option D.

4) Answer (C)

Sum of roots = p = -b/a = 4/3

Product of roots = -d/a = 6/3  = 2

p/q = (4/3)/2 = 2/3

So the answer is option C.

 

5) Answer (D)

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)

6) Answer (D)

$\frac{{114}^3+7^3}{{114}^2+7^2-(114\ast 7)}=\frac{(114+7)({114}^2+7^2-(114\ast 7))}{{114}^2+7^2-(114\ast 7)}$ = 121

7) Answer (C)

Let C be x.
Then, B = 2*x = 2x
A = 150% of 2x = 3x
Then, A : C = 3x : x = 3 : 1.

8) Answer (C)

Let the two numbers be 2x and 5x.
Given, $(5x{)}^2–(2x{)}^2=1029\Rightarrow$25x^2 – 4x^2 = 1029
⇒ $21x^2 = 1029
⇒ x^2 = 49
⇒ x = 7.
Then, the two numbers will be,
2x = 2*7 = 14
5x = 5*7 = 35
Then, Sum of two numbers = 14+35 = 49.

9) Answer (C)

Boys : Girls = 10 : 7
Let the number of boys and girls initially be 10x and 7x respectively.
Now, 60 boys were added and ratio changed to 5 : 2.
Total number of boys = 10x + 60
Total number of girls = 7x
=> $\frac{10x+60}{7x}$ = $\frac{5}{2}$
=> 20x + 120 = 35x
=> 15x = 120
=> x = 8
=> Total number of boys initially = 80 and Total number of girls initially = 56
=> Total number of students initially = 80 + 56 = 136

10) Answer (C)

Pure water => concentration is 0%
So, the situation can be represented as follows:

The required ratio of acid solution to water in the resultant solution is 50 : 10 = 5 : 1
So, the fraction of acid solution that has to be replaced with water is 1/(5+1) = 1/6

11) Answer (C)

Let one number is X, then another number is 10+X

Product of two numbers  = (LCM)*(HCF)

(X)(X+10) = (2)(48)

X^2+10X = 96

X^2+10X-96 = 0

(X+16)(X-6) = 0

X = -16 or 6

Since X is positive, X = 6 = smaller number

So the answer is option C.

12) Answer (D)

0.0341 * 46.7 = 1.59
So, 34.1 * 46.7 = 1.59 * 1000
34.1 * 4.67 = 1.59 * 100
Hence the answer is 159

13) Answer (D)

1 billion = ${10}^9$

log (1 billion) = 9

log 2 = 0.3

$lo{g}_2$1billion = 30

=> 1 billion = $2^{30}$

14) Answer (B)

Let a, a are the adjacent sides of the isosceles triangle,

$a^2+a^2=(6\sqrt{2}{)}^2$

$2a^2=72$

$a=6$

Area of triangle = $\frac{1}{2}\times 6\times 6$ = $18sq.m$

So the answer is option B.

15) Answer (D)

By cutting a portion of the cone parallel to its base at the top, a frustum is formed.

The first and second figures are top view and side view of a frustum.

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