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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  RRB JE ME (CBT I) Mock Test- 9 - Mechanical Engineering MCQ

RRB JE ME (CBT I) Mock Test- 9 - Mechanical Engineering MCQ


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30 Questions MCQ Test - RRB JE ME (CBT I) Mock Test- 9

RRB JE ME (CBT I) Mock Test- 9 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The RRB JE ME (CBT I) Mock Test- 9 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The RRB JE ME (CBT I) Mock Test- 9 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE ME (CBT I) Mock Test- 9 below.
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RRB JE ME (CBT I) Mock Test- 9 - Question 1
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If the rate of interest is 8% per annum and Rs. 10000 lent at the compound interest half yearly then calculate the equivalent simple rate of interest for the first year?

Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 1
Let the principal be Rs.100

the rate of interest is 8% per annum lent at the compound interest half yearly.

Compound interest for 6 months = 4%

Compound interest next 6 months = 4% + 4of 4% = (4 +.16)% = 4.16%

the equivalent simple rate of interest for the first year = 4 + 4.16 = 8.16%

RRB JE ME (CBT I) Mock Test- 9 - Question 2
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Three taps A, B, and C can fill a tank in 20, 24, and 30 hours, respectively. How long (in hours) would the three taps take to fill the tank if all of them are opened together?

Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 2
A can fill 1 unit of water in the tank in 1/20 hours

B can fill 1 unit of water in tank in 1/24 hours

C can fill 1 unit of water in tank in 1/30 hours

(A+B+C) can fill 1 unit of water in tank in 1/20+1/24+1/30 = 6+5+4/120 = 15/120 = 18 days

Therefore,

(A+B+C) can fill full tank in 8 days.

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RRB JE ME (CBT I) Mock Test- 9 - Question 3
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PAB and PCD are two secants to a circle. if PA =10cm , AB =12cm and PC =11cm, then what is the value of PD?

Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 3

PA x PB = PC x PD

Let PD = X cm

10 x (10 + 12) = 11 x (11 + X)

11 + X = 20

X = 9

PD = 9 = 11 = 20 cm

RRB JE ME (CBT I) Mock Test- 9 - Question 4
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The speed of a boat in still water is 10 km/hr. It covers a distance of 50 km upstream in 6 hours. What is the speed (in km/hr) of the stream?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 4
Speed of boat = 10 km/hr

Upstream speed 50/6 = 25/3 km/hr

So, Downstream speed 35/3 km/hr.

RRB JE ME (CBT I) Mock Test- 9 - Question 5
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500 litres of milk solution has 40% milk in it. How much more milk should be added to make it 60% in the solution?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 5
Milk = 40% of 500 = 200 litres

Water is 60% of 500 = 300 litres

ATQ,

40%of mixture = water = 300 litres

New mixture quantity = 300/40% = 750 liters

The quantity of milk in the mixture = 60 % of 750 = 450 liters

Required milk to be mixed = 450 - 200 = 250 liters

Therefore,

250 liter of milk should be added to make it 60% in the solution.

RRB JE ME (CBT I) Mock Test- 9 - Question 6
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The shadow of a tower on the ground level is increased by 20 m, when the angle of depression by the sun changes from 60∘ to 30∘. The height of the tower is?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 6

Let the height of tower DC = h meter

tan 600 = h/BC

Putting the value of BC from equation(l) in equation (2), we get

RRB JE ME (CBT I) Mock Test- 9 - Question 7
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What will be the sum of all odd whole numbers between 10 and 38?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 7
To calculate sum, we need to find the number of odd whole numbers between 10 and 38 using the Arithmetic Progression formula.

That is,

Last term = First term + (number of terms-1) × Common difference

ATQ,

37 = 11 + (N-1) × 2

So, N or Number of terms = 14

Now,

Sum of terms = N/2 [2×first term(N−1)×d]

ATQ,

Sum = 14/2 [2×11+(14−1)×2]

Sum = 336

Therefore,

The Sum of all odd whole numbers between 10 and 38 is 336

RRB JE ME (CBT I) Mock Test- 9 - Question 8
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If the volume of two cubes are in the ratio 125 : 1, then what is the ratio of their edges?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 8

RRB JE ME (CBT I) Mock Test- 9 - Question 9
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A , B, and C invest in a business in the ratio 3:6:5. A and C are working partners. Only B is a sleeping partner; hence his share will be 3/4th of what it would have been if he were a working partner. If they make Rs. 50000 profit, half of which is reinvested in the business and the other half is distributed between partners, then how much does C get (in Rs)?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 9

Given the investment ratio old ratio

A : B: C

3 : 6 : 5

ATQ,

total profit = 50000

after reinvestment the profit

= 50000/2 = 25000

RRB JE ME (CBT I) Mock Test- 9 - Question 10
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The HCF and LCM of the two numbers are 45 and 270, respectively. If the ratio of the two numbers is 2:3 and the smaller of the two numbers is X, then find X2.
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 10
We know that

Product of numbers = product of lcm and hcf

2y × 3y = 45 × 270

y2 = 45 × 45

y = 45

then smaller no.

X = 2 × 45 = 90

X2 = (90)2 = 8100

RRB JE ME (CBT I) Mock Test- 9 - Question 11
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What is the value of cos40∘ − cos140∘/sin80∘ + sin120∘?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 11

RRB JE ME (CBT I) Mock Test- 9 - Question 12
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What is the value of tan (π/4 + A) × tan (3π/4 + A)?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 12

RRB JE ME (CBT I) Mock Test- 9 - Question 13
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The distance between the centers of the two circles is 61cm and their radius is 35cm and 24cm. What is the length(in cm) of the direct common tangent to the circles?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 13

RRB JE ME (CBT I) Mock Test- 9 - Question 14
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What is the value of 2cot P−A/2/1+tan2 2P−A/2 ?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 14

RRB JE ME (CBT I) Mock Test- 9 - Question 15
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Find the standard deviation of {11, 7, 10, 13, 9}
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 15
First we have to find the mean

= (11 - 10)2 + (7 - 10)2 + (10 - 10)2 + (13 - 10)2 + (9 - 10)2

= 1 + 9 + 0 + 9 + 1 = 20

Standard deviation =

RRB JE ME (CBT I) Mock Test- 9 - Question 16
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If a dairy mixes cow’s milk which contains 10%fat with buffalo’s milk which contains 20% fat,then the resulting mixture has the fat of 120/7%. In what ratio was the cow’s milk mixed with buffalo’s milk?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 16

Use alligation rule

RRB JE ME (CBT I) Mock Test- 9 - Question 17
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If two complementary angles are in the ratio of 4:5, find the greater angle.
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 17
Let the angles be 4x , 5x

4x + 5x = 90∘

9x = 90∘

x = 10∘

Greater angle = 5x = 50∘

RRB JE ME (CBT I) Mock Test- 9 - Question 18
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If cosθ = 4/5, then (secθ+cosecθ)=
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 18

RRB JE ME (CBT I) Mock Test- 9 - Question 19
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In the given figure, in a right angle triangle ABC, AB = 12 cm and AC = 15cm. A square is inscribed in the triangle. One of the vertices of the square coincides with the vertex of the triangle. What is the maximum possible area of the square (cm2)?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 19

AB = 12 cm, AC = 15 cm

then, BC = 152 - 122 = 9cm

side of square

where, P = perpendicular, B = base

side of the square

area of the largest square that can be formed

inside the triangle = (36/7)2 = 1296/49

RRB JE ME (CBT I) Mock Test- 9 - Question 20
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What is the measure of the two equal angles of a right isosceles triangle?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 20

A.T.Q

∠B = 90°

and AB = BC

So, ∠A = ∠C

we know that

∠A + ∠B + ∠C = 180°

x + 90° + x = 180°

2x = 90°

x = 45°

RRB JE ME (CBT I) Mock Test- 9 - Question 21
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If 1082 = 11664, then what is the value of √1.1664 + √116.64 ?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 21

RRB JE ME (CBT I) Mock Test- 9 - Question 22
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The length and breadth of a rectangular field are in the ratio of 5 : 2. If the perimeter of the field is 140 metres, then what is the breadth (in metres)?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 22
Let Length and Breadth of rectangle are 5x and 2x respectively.

Perimeter = 2(Length + Breadth)

140 = 2(5x+2x)

70 = 7x

So, x = 10 m

Therefore, Breadth of rectangle = 2×10=20 metre

RRB JE ME (CBT I) Mock Test- 9 - Question 23
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Calculate the amount on Rs. 37,500 @ 8% p.a compounded half yearly for 1 1/2 years.
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 23

As the interest compounded half yearly

So time will be =3 half year

Rate = 8/2 = 4%

= 42182.4

RRB JE ME (CBT I) Mock Test- 9 - Question 24
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The mean of 17 numbers is 12. If two numbers 9 and 15 are removed, then what will be the mean of remaining number?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 24
Sum of 17 numbers = 17 ×12 = 204

After removal of 9 and 15, sum becomes 180.

Therefore,

Mean of remaining 15 numbers = 180/15=12

RRB JE ME (CBT I) Mock Test- 9 - Question 25
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Ram Naresh got two successive discounts, a bag with a list price of Rs. 400 is available at Rs. 160. If the second discount is 20%, find the first discount.
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 25

Let the first discount be x%

First a discount of 20% means that you multiply the price by. 80(1 - 0.20)

The initial price is therefore first multiplied by x, then by 0.80, and the result is 160.

Multiply by 0.5 means reduce by 50%.

RRB JE ME (CBT I) Mock Test- 9 - Question 26
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If the diagonal of a square is 12 cm, then what is the area (in cm2) of the square?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 26

Diagonal of square = side x √2

Area of square = side2

ATQ,

Diagonal = 12cm

Side x √2 = 12

So, side = 12/√2

Therefore,

Area = (12/√2)2 ⇒ 144/2 ⇒ 72

RRB JE ME (CBT I) Mock Test- 9 - Question 27
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A line passing through (3, 5) makes the angle of 135∘ with x-axis. What will be its equation?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 27

Slope = tan 1350 = −1 = m

⇒ A (3,5)

(y - 5) = -(x - 3)

Equation of line

x + y - 8 = 0

RRB JE ME (CBT I) Mock Test- 9 - Question 28
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Three toys are in the shape of the cylinder, hemisphere, and cone . The three toys have the same base. The height of each toy is 2 √2. What is the ratio of the total surface areas of the cylinder, hemisphere and cone respectively ?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 28

Height of each toy = 2√2

since height of hemisphere = radius of hemisphere

= radius of each toy = 2√2

slant height of cone(l)

ratio = T.S.A of cylinder: T.S.A of the hemisphere : T.S.A of cone

RRB JE ME (CBT I) Mock Test- 9 - Question 29
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A sum of Rs 4000 becomes Rs 5800 in 3 years, when invested in a scheme of simple interest. If the same sum is invested in a scheme of compound interest with same yearly interest rate (compounding of interest is done yearly), then what will be the amount (in Rs) after 2 years?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 29

RRB JE ME (CBT I) Mock Test- 9 - Question 30
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Average of the runs of 133 players of a team is 38. If the average of the runs of the male players is 43 and the average of the runs of the female players is 24, then what will be the ratio of the total runs of male players and the total runs of female players respectively?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 9 - Question 30
The ratio of male females can be calculated by the formula of the allegation.

M :F = 1 4 :1 0 = 14 : 5

No. of males

Total runs of males

= 98 x 43

No. of females

= 133 - 9 8 = 35

Total runs of females

= 35 x 24

Required ratio

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