RRB JE Previous Year Maths Questions PDF

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RRB JE Previous Maths Questions PDF
RRB JE Previous Maths Questions PDF

RRB JE Previous Year Maths Questions PDF

Download Top 15 RRB JE Maths Questions and Answers PDF. RRB JE Maths questions based on asked questions in previous exam papers very important for the Railway JE exam.

Download RRB JE Previous Year Maths Questions PDF

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Question 1: What is the minimum value of 3 sin θ + 4 cos θ ?

a) 25

b) -25

c) 5

d) -5

Question 2: Minimum value of $2\sin^{2}x+1\cos^{2}x$

a) 1.5

b) 2

c) 1

d) 0

Question 3: If a-b = 6, ab = 16, then find $a^2+b^2$.

a) 48

b) 68

c) 52

d) 72

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Question 4: Find the value of $1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{2}{7}}}}$.

a) $\frac{37}{25}$

b) $\frac{41}{22}$

c) $\frac{41}{25}$

d) $\frac{35}{19}$

Question 5: A number is increased by 12 and divided by 19 to get the result as 9 and if the number is decreased by 40 and divided by 17 then the remainder obtained is ?

a) 3

b) 0

c) 1

d) 2

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Question 6: Find the value of $x$ (from the given options), if $5 = 40-12x+x^2$ ?

a) 3

b) 4

c) 5

d) 6

Question 7: Find $ab+bc+ca$, if $a^2+b^2+c^2 = 24$ & $a+b+c=4\sqrt3$ ?

a) 12

b) 13

c) 14

d) 15

Question 8: What is the value of x if $2^{2x} * 2^{2^{x}} * 2^{x}$ = 32?

a) 0

b) 1

c) 2

d) 4

Question 9: The average score of a cricketer for his first 10 innings is 132 runs. He scored 0 runs in his next innings. What will be his average score after 6 innings?

a) 120 runs

b) 96 runs

c) 100 runs

d) 84 runs

Question 10: What is the third proportional of 12 and 36 ?

a) 24

b) 15

c) 72

d) 108

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Question 11: A theatre has 1st class tickets and 2nd class tickets. Number of people in 1st class to second class is 3:5 and the fares for 1st class to second class ticket is 5:2. If the total amount collected is 1250. What is the amount collected from second class tickets ?

a) 750

b) 500

c) 250

d) 800

Question 12: Find the equation of the line passing through (2, 3) and (-4, 15) ?

a) x-2y = 7

b) 2x-y = 7

c) x+2y = 7

d) 2x+y = 7

Question 13: Find the value of x, if 2x+3y = 4 and 4x+6y = 8 ?

a) 2

b) 3

c) 4

d) can not be determined

Question 14: A number when divided by 11 leaves a remainder of 10 and when divided by 10 leaves a remainder of 8. Which of the following is a possible value of the number?

a) 123

b) 194

c) 62

d) 98

Question 15: Find the value of x + y+ z if x+2y+5z = 14 and 3x+5y+11z = 46.

a) 18

b) 20

c) 35

d) 16

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Answers & Solutions:

1) Answer (D)

For the expression asin θ + bcos θ + c,

The maximum value = $c+\sqrt{a^{2}+b^{2}}$ &

The minimum value = $c-\sqrt{a^{2}+b^{2}}$

for 3 sin θ + 4 cos θ , a = 3, b = 4, c = 0

minimum value = $c-\sqrt{a^{2}+b^{2}}$
$= 0 – \sqrt{3^{2} + 4^{2}} = -\sqrt{25} = -5$

so the answer is option d.

2) Answer (C)

$(\sin^{2}x+\cos^{2}x)+\sin^{2}x$=1+$\sin^{2}x$
Minimum value of $\sin x$ is when $x$=0 i.e $\sin 0$=0
Therefore 1+0=1
Hence, option C is the correct answer.

3) Answer (B)

Given, a-b = 6, ab = 16
We know that $(a-b)^2 = a^2+b^2-2ab$
$6^2 = a^2+b^2-2\times16$
$a^2+b^2 = 36+32 = 68$

4) Answer (C)

$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{2}{7}}}} = 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{\dfrac{9}{7}}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{7}{9}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{\dfrac{16}{9}}}$

$= 1+\dfrac{1}{1+\dfrac{9}{16}}$

$= 1+\dfrac{1}{\dfrac{25}{16}}$

$= 1+\dfrac{16}{25}$

$= \dfrac{41}{25}$

5) Answer (B)

let the number be x
(x+12)/19=9
x=171-12
x=159
And then (159-40)=119
119/17 we get 0 as the remainder.

6) Answer (C)

$5 = 40-12x+x^2$

$x^2-12x+35 = 0$

$x^2-7x-5x+35 = 0$

$x(x-7)-5(x-7) = 0$

$(x-7)(x-5) = 0$

$x = 7$ or $x = 5$

So the answer is option C.

7) Answer (A)

$(a+b+c)^2$= $a^2+b^2+c^2+2(ab+bc+ca)$

$(4\sqrt3)^2 = 24 + 2(ab+bc+ca)$

$48 = 24 + 2(ab+bc+ca)$

$24 = 2(ab+bc+ca)$

$(ab+bc+ca) = \frac{24}{2} = 12$

So the answer is option A.

8) Answer (B)

Given $2^{2x} * 2^{2^{x}} * 2^{x}$ = $2^5$
So, $2x + 2^x + x$ = 5
We can observe that for x = 1, we can get required value.
So, option (b) is the correct answer.

9) Answer (A)

Total score made in his first 10 innings = 10*132 = 1320 runs
Score in his 11th innings = 0
Then, His new average = 1320/11 = 120 runs.

10) Answer (D)

For 3rd proportional we have 12:36::36:x
Therefore 12x=36*36
x=108

11) Answer (B)

Let the people in 1st class and 2nd class be 3x and 5x respectively
and the cost of ticket for 1st class and 2nd class be 5y and 2y respectively
Total amount from 1st class is 3x*5y=15xy
Total amount from 2nd class is 5x*2y=10xy
Total amount is 25xy=1250
xy=50
Amount from 2nd class is 10*50=Rs 500

12) Answer (D)

Slope = $\frac{y_2-y_1}{x_2-x_1} = \frac{15-3}{-4-2} = \frac{12}{-6} = -2$

Equation of a line passing through $(x_1, y_1)$ having slope m is $(y-y_1) = m(x-x_1)$

Equation of a line passing through $(2,3)$ having slope -2 is $(y-3) = (-2)(x-2)$ ==> $y-3 = -2x+4$ ==> $2x+y = 7$

So the answer is option D.

13) Answer (D)

(4x+6y=8) is 2*(2x+3y=4)

So both represents the same equation, hence we cant find x or y

So the answer is option D.

14) Answer (D)

By checking each of the options given above, we can notice that only 98 satisfies the conditions given in the question. Hence the answer is (d)

15) Answer (A)

Generally, when two equations are given with three variables, we cannot determine the value of any of the variables or their sum. However, in certain cases we can manipulate the equations to reduce them to a two variable form where one of the variables is the sum of the original variable. Let’s try that approach here:

x+2y+5z = 14 can be written as (x+y+z) + (y+4z) = 14 —(a)

3x+5y+11z = 46 can be written as 3(x+y+z) + 2(y+4z) = 46 —-(b)

Let x+y+z=A and y+4z=B. Replacing these in (a) and (b) we get

A + B = 14 — (c)

3A + 2B = 46 — (d)

We subtract 2*(c) from (d)

(3A+2B) – (2A+2B) = 46-2*14

A = 18

Hence, x+y+z=18

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We hope this Previous Year Maths Questions for RRB JE Exam will be highly useful for your preparation.

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