RRB JE ME (CBT I) Mock Test- 1 - Mechanical Engineering MCQ

Distance covered in crossing each other = (165 + 175) = 340 m.

∴ Required time = (340 x 9/255) = 12 s.

(2x - 3)(x - 4) = 0

x = 1.5 and 4

Since both the equations have one root in common-

Case 1: When x = 1.5 is common-

2(1.5)² - A(1.5) - 20 = 0

So let us assume the number of 20p coins = a

Then the number of 10p coins is 2a and the number of 5p coins is 3a.

Total worth of 20p coins = number of coins X worth of each coin = a X (1/5) = Rs (a/5)

Total worth of 10p coins = 2a X (1/10) = Rs (a/5)

Similarly total worth of 5p coins = 3a X (1/20) = Rs (3a/20)

Since total money with the Divya is Rs 44 hence

a/5 + a/5 + 3a/20 = 44 => 2a/5 + 3a/20 = 44

=> 8a/20 + 3a/20 = 44 => 11a/20 = 44 => a = 80

So the number of 5p coins = 3a = 80 x 3 = 240

∴ 42x - 23x = 304 => 19x = 304 => x = 16

∴ The number is 100 x 16 = 1600.

=> 72% of the number = 0.72 x 1600 = 1152

= 60 ÷ [{3 x √(25 + 11)} − 6] x 5 - 14 x 2

= 60 ÷ [{3 x √(36)} − 6] x 5 - 28

= 60 ÷ [{3 x 6} − 6] x 5 - 28

= 60 ÷ [18 − 6] x 5 - 28

= 60 ÷ 12 x 5 - 28

= 5 x 5 - 28

= -3

In one minute, pipe x can fill 1/20 part.

In t minutes, pipe x can then fill the t/20 part.

Therefore, t/20 + t - 2/25 + t/30 = 1

On solving, we get t = 324/37 minutes.

=> 4(x - 1/48x) = 8

=> x - 1/48x = 2 ------ (i)

Squaring both sides, we get

(x - 1/48x)² = 4

=> x² + 1/(2304x²) - 2x/48x = 4 ------ [(a - b)²]

= a² + b² - 2ab] x² + 1/2304 x²

= 4 + 1/24 = 97/24

Mass of ball = m = 4 kg, elevation from ground = h = 20 m; v = velocity at touch point

So, mgh = ½ (mv²)

Gives v = 20 m/s

mg = 1/2 (mv²)

2gh = v²

v² = 2 x 10 x 20 = 400

v = 20 m/sec

=> 100x = 78.414141.. (I)

=> 10000x = 7841.414141.. (II)

(II) - (I) gives, 9900x = 7763

=> x = 7763/9900

Interest obtained by Nidhi from the bank = Rs. (3500 x 2 x 12.5)/100 = Rs. 875

Interest obtained by Nidhi from Chit fund = 6500(1 + 0.1)2 - 6500 = Rs. 1365

So total interest obtained by Nidhi = Rs. (875 + 1365) = Rs. 2240

Profit = Rs. (2240 - 2000) = Rs. 240

(A) Initial marked price of book = Rs. 500 and purchased price for A = 92% of 500 = Rs. 460

(D) Marked price of book by A = 115% of 460 = Rs. 529

(B) Purchased price of book by B = 529 - 20 = Rs. 509

(E) Marked price of book by B = 120% of 509 = Rs. 610.8 and purchased price of book by C = 610.8 - 50.9 = Rs. 559.9

(C) Per cent profit = [(559.9 - 509)/509] x 100 = 10%

SP = 9600 x 95/100 = Rs.9,120

Second S.P. = 9120 x 105/100 = Rs.9,576

Loss = 9600 - 9576 = 24

So these two hands are apart by 2.5 blocks.

We know that the difference in angle of one block is 30°.

Hence the angle between the hands = 2.5 X 30° = 75°

Since 979 is frequently occurring, and the number of occurrence of 979 is 3.

Hence the mode of sex ratio in India in the last 10 years is 979.

=> x = -1 must satisfy the given equation

=> 7(-1)⁵ - 2(-1)⁴ + 3(-1)³ - 4(-1)² + 6(-1) + 2k = 0

=> -7 - 2 - 3 - 4 - 6 + 2k = 0

=> 2k = 22

=> k = 11

∴ Their H.C.F. = x = 6 (given).

=> The numbers are 42 and 30.

L.C.M. of 42 and 30 = 210.

Let the tangent be T

OT₂ = OA.OB

=> OA = 4

=> OB = OA + AB

=> 4 + 60

=> 64

OT₂ = 4 x 64

=> 256

=> OT = 16 cm

Length of wire = circumference of circle.

44 = 2 πr or r = (44 x 7)/(2 x 22) = 7 cm

Hence, area of circle = πr² = 22/7 x 7 x 7 = 154 cm²

Sum of ages of all employees in grade B = 40 x 24 = 960 years

Sum of ages of all employees in grade C = 15 x 32 = 480 years

Sum of ages of all employees in grade D = 20 x 22 = 440 years

Sum of ages of all the employees of the department P = 700 + 960 + 480 + 440 = 2580 years

Total employees in department P = 25 + 40 + 15 + 20 = 100

Average age of the employees of department P = 2580/100 = 25.8 years

Selling price = Rs. x/2

There is a loss of 12.5% when the selling price is Rs. x/2.

Cost Price = x/2 x 100/(100 - 12.5) = (100 x)/175 = 4x/7

Now, when selling price is Rs. x, % profit = [(x - (4x/7))/(4x/7)] x 100 = (3x/4x) x 100 = 75%

=> 3α + 6 - α = 71α - 3α - 6 = common difference of the arithmetic progression.

=> 2α = 68α - 6 - 6

=> 66α = 12

=> α = 12/66 = 2/11

∴ Her savings in Kisan Vikas Patra = ₹ (180,000 - x)

∴ (180,000 - x)/4 = x/5

=> 900,000 - 5x = 4x

=> x = ₹ 100,000

Second number = (40% of m ) - 21

According to the question:

20% of 40% of (m/2) + (3/4) x 20% of m = (40% of m) - 21

m = 100

Second number = (40% of m) - 21 = 40% of 100 - 21 = 19

Difference = 100 - 19 = 81

125% of [{a + 200 ÷ (16 ÷ 2 x 4 x (1/8))} ÷ 7.5 x 8 ÷ (18 ÷ 3 - 50% of 4)] = a

=> 125% of [{a + 200 ÷ (8 x 4 x (1/8))} ÷ 7.5 x 8 ÷ (6 - 2)] = a

=> 125% of [{a + 200 ÷ 4} ÷ 7.5 x 8 ÷ 4] = a

=> 125% of [{a + 50} ÷ 7.5 x 2] = a => 125% of 2(a + 50)/7.5 = a

=> (5/4) x 2(a + 50)/7.5 = a

=> a + 50 = 3a

=> a = 25

sin θ = cos (300 + θ)

sin θ = sin (900 - 300 - θ)

θ = 900 - 300 - θ

θ = 300

= (sin θ + cos 2θ)

= sin 300 + cos 600

= 1/2 + 1/2

= 1

According to the given information, if we square the fraction and perform the specified operations, we get:

((a/b)^2) = (a^2)/(b^2)
Numerator after division by 2: (a^2)/2
Denominator after increasing by 25%: (5/4)b

The new fraction obtained is 4 times the original fraction, so we can set up the following equation:

4(a/b) = (a^2)/2 / ((5/4)b)

To simplify, let's multiply both sides of the equation by 2b and 4 to eliminate the fractions:

8ab = a^2 / (5/4)
8ab = (4/5) * a^2

Now, let's solve for a/b:

8ab = (4/5) * a^2
40ab = 4a^2
10b = a

Substituting the value of a into the original fraction, we get:

New fraction = (10b)/b = 10

Therefore, the new fraction is 10.

17 + (2 x 3) = 23

23 + (3 x 4) = 35

35 + (4 x 5) = 55

55 + (5 x 6) = 85

Hence loss in 65 minutes = 65(5/11) - 65 = (5/11) minutes

Hence loss in 24 hours = (5 x 24 x 60) / (11 x 65) = 10(10/143) minutes

Let the present age of the son be 'x'.

(45 + 15)/(x + 15) = 2/1

2x + 30 = 60 x = 15

So, the present age of son is 15 years

Ramesh can do 6/5 work in 30 hours.

So Ramesh can complete the work in 30 x (5/6) = 25 hours.

Together they can finish 1/30 + 1/25 of work in 1 hour, i.e., 11/150.

So in 150/11 hours, they will finish the work together.