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
$a^2c > b^2a$
a) Only 2
b) Only 4
c) Only 1 and 3
d) Only 2 and 4
Question 6: Find the value of the expression:$\frac{114^3+7^3}{114^2+7^2-(114*7)}$
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)
$Sinθ$ = $\frac{2tan(θ/2)}{1+tan^2(θ/2)}$ = $\frac{2(5/12)}{1+(5/12)^2}$ = $120/169$
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)
$\frac{114^3+7^3}{114^2+7^2-(114*7)} = \frac{(114+7)(114^2+7^2-(114*7))}{114^2+7^2-(114*7)}$ = 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
⇒ $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
$log_2$1billion = 30
=> 1 billion = $2^{30}$
14) Answer (B)
Let a, a are the adjacent sides of the isosceles triangle,
$a^2+a^2 = (6\sqrt2)^2$
$2a^2 = 72$
$a = 6$
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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