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RRB JE ME (CBT I) Mock Test- 6 - Mechanical Engineering MCQ


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30 Questions MCQ Test - RRB JE ME (CBT I) Mock Test- 6

RRB JE ME (CBT I) Mock Test- 6 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The RRB JE ME (CBT I) Mock Test- 6 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The RRB JE ME (CBT I) Mock Test- 6 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- 6 below.
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RRB JE ME (CBT I) Mock Test- 6 - Question 1
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The area of a square park is 25 sq.km. The time taken to complete around of the field once, at a speed of 3 km/hour is

Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 1
Area = 25 km2

Side = √25 = 5 km

Perimeter = 4 × 5 = 20 km

S = D/T

Time taken in completing one round

= (Distance)/(Speed)

RRB JE ME (CBT I) Mock Test- 6 - Question 2
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1% of 1% of 25% of 1000 is?

Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 2
1% of 1% of 25% of 1000

(1/100)×(1/100)×(25/100) × 1000 = 0.025.

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RRB JE ME (CBT I) Mock Test- 6 - Question 3
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The average of all the perfect squares up to 100 is:

Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 3

RRB JE ME (CBT I) Mock Test- 6 - Question 4
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A is thrice as fast as B and is therefore able to finish a work in 60 days less than B. The time in which they can do it working together is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 4
Let B finish the work in 3x days

∴ A will finish the work in x days

Also 3x - x = 60

x = 30 days.

∴ A's one day work = 1/30

B"s one day work = 1/90

(A+B)'s one day work = 1/30+1/90 = 4/90

∴ Time taken to Finish the work

while (A+B) working together = 90/4

= 45/2 days.

RRB JE ME (CBT I) Mock Test- 6 - Question 5
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A rectangle with one side of length 12 cm is inscribed in a circle of diameter 15cm.Find the area of rectangle.
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 5

Now:OA = OC = 7.5 cm[Radius of circle]

⇒ AC = OA + OC = 7.5 + 7.5 = 15 cm

In right triangle ABC, by pythagoras theorem (AC)2 = (AB)2 + (BC)2

⇒ 152 = 122 + (BC)2

⇒ (BC)2 = 225 - 144 = 81

= BC = 9 cm

∴ Area of rectangle= AB x BC = 12 x 9 = 108 cm2

RRB JE ME (CBT I) Mock Test- 6 - Question 6
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The difference between the circumference and diameter of a circle is 150 m. The radius of that circle is (Take π = 22/7)
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 6

(Circumference of a circle - Diameter) = 150

RRB JE ME (CBT I) Mock Test- 6 - Question 7
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The total area of a circle and a square is equal to 5450 sq.cm. The diameter of the circle is 70 cms. What is the sum of the circumference of the circle and the perimeter of the square?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 7
If the side of the square be x cm then,

= 380 cm.

RRB JE ME (CBT I) Mock Test- 6 - Question 8
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The selling price of 12 objects is equal to the cost price of 9 objects. Find the loss or profit percentage?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 8
S.P. of 12 objects = C.P. of 9 objects

Clearly there is a loss as 12 articles are sold for what 9 articles are bought.

L% = L/C.P.×100 = 3/12 × 100 = 25%

RRB JE ME (CBT I) Mock Test- 6 - Question 9
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If a,b,c, are the distinct positive numbers then (a+b+c) (ab+bc+ca) is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 9
This is the standard inequality formula.
RRB JE ME (CBT I) Mock Test- 6 - Question 10
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The simple interest on a sum of money is 1/25 of the principal and the number of years is equal to the rate percent per annum. Find the rate percent per annum.
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 10

RRB JE ME (CBT I) Mock Test- 6 - Question 11
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If O is the orthocentre of ∆ABC, then ∠BOC + ∠BAC is equal to:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 11

∠BOC + ∠BAC = 1800

Orthocentre is the intersecting point of altitudes.

RRB JE ME (CBT I) Mock Test- 6 - Question 12
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The average monthly income of A and B is ₹7760. The average monthly income of B and C is ₹10990 and that of C and A is ₹9070. What is the annual income of B?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 12

Total income of A and B = 2 x 7760

⇒ A+B = 15520 ...(i)

Similarly,

B +C = 10990 × 2 = 21980...(ii)

and A+C = 9070 x 2 = 18140 ...(iii)

Adding Eqs. (i), (ii) and (iii),

2(A + B + C) = 55640

⇒A+B + C =27820 ,..(iv)

Subtracting Eq. (iii) from Eq. (iv),

B = 27820 – 18140 = 9680

Hence, annual income of

B = 9680 × 12 = ₹116160

RRB JE ME (CBT I) Mock Test- 6 - Question 13
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Two poles of the height 15 m and 20 m stand vertically upright on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 13

Applying Pythagoras theorem in triangle BDE

BD2 = BE2 + DE2

= 52 + 122

= 25 + 144 = 169

BD = 13m

RRB JE ME (CBT I) Mock Test- 6 - Question 14
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A mixture of 70 litres contains milk and water in ratio of 3 :4. How many litres of milk must be added to mixture so as to make the ratio 5 : 4?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 14

Milk : Water

New ratio 5 : 4

Old ratio 3 : 4

Difference 2 : 0

Sum of old ratios = 3 + 4 = 7

∴ Milk added = 2/7 × 70 = 20 litres

RRB JE ME (CBT I) Mock Test- 6 - Question 15
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Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 15

RRB JE ME (CBT I) Mock Test- 6 - Question 16
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In the given figure, BO and CO are the bisector of ∠CBD and ∠BCE respectively and ∠A = 40°, then ∠BOC is equal to:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 16
∠BOC

= 90° - 1/2 ∠A = 90° - 1/2 (40) = 70°

RRB JE ME (CBT I) Mock Test- 6 - Question 17
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For which of the following value of m the equation x2 - 6x + m = 0 has equal roots?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 17
When equation has equal roots then

b2 – 4ac = 0

∴(–6) 2 – 4 (1) (m) = 0

36 – 4m = 0

– 4m = – 36

4m = 36

m = 9

RRB JE ME (CBT I) Mock Test- 6 - Question 18
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Area of circle is equal to the area of a rectangle having perimeter of 100 cm and length is more than the breadth by 6 cm. What is the diameter of the circle?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 18
Let breadth = x, then length

= (x + 6)

∴ 2(x + x + 6) = 100

⇒ 2x + 6 = 50

⇒ x = 22 cm

∴ breadth = x = 22 cm & length

= 22 + 6 = 28 cm

∴ Area of circle = Area of rectangle

⇒ π r2 = 22 x 28

⇒ r2 = 22×28 x 22 / 7 =7×4×7

⇒ 7 × 2 = 14 cm

∴ Diameter = 2r = 28 cm

RRB JE ME (CBT I) Mock Test- 6 - Question 19
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In the given figure, ABCD is a trapezium such that AD II BC and P, Q are the points on AB and CD respectively such that PQ II AD and AP: PB = 5 : 3. Then PQ is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 19

Now In △APR and △ABC

∠APR = ∠ABC (∵ PQ I I AD I I BC)

And ∵ APR = ∠AC8 (∵ PQ I I BC)

RRB JE ME (CBT I) Mock Test- 6 - Question 20
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What is the difference between the simple interest on a principal of ₹600 being calculated at 4% per annum for 3 years and 3 % per annum for 3 years?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 20
4% for 3 years (S.I.) = 12% of the amount

3% for 3 years (S.I.) = 9% of the amount

The difference between the two is 3% of amount.

3% of 600 = 3/100× 600

= ₹18

RRB JE ME (CBT I) Mock Test- 6 - Question 21
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If 100 men can do 100 work in 100 days, then 1 man will do 1 work in how many days?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 21

RRB JE ME (CBT I) Mock Test- 6 - Question 22
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If tan θ = 2 - √3 , then tan (90 - θ)is equal to:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 22

RRB JE ME (CBT I) Mock Test- 6 - Question 23
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A tank is filled by pipe A in 32 minutes and by pipe B in 36 minutes. When it is full, it can be emptied by a pipe C in 20 minutes. If all the three pipes are opened simultaneously, half of the tank will be filled in how many minutes?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 23

A + B + C fill the tank in

A + B + C fil half the tank in

Alternate method

Half of the tank will be filled in :

= 720/13

or

RRB JE ME (CBT I) Mock Test- 6 - Question 24
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A shopkeeper earns a profit of 12% on selling a book at a 10% discount on the printed price. The ratio of the cost price and the published price of the book is
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 24
Let the CP be Rs. 100.

∴ SP = Es. 112

If the marked price is Rs. x, then

90% of x = 112

= 900 : 1120 = 45 : 56

RRB JE ME (CBT I) Mock Test- 6 - Question 25
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The sum of two numbers is 40, and their difference is 4. The ratio of the numbers is :
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 25
Let the two numbers be x and y. Then according to question

x + y = 40

x - y = 4

so x = 22 and y = 18

therefore required ratio = 22/18 = 11 : 9

RRB JE ME (CBT I) Mock Test- 6 - Question 26
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The average age of 30 students is 9 years. If the age of their teacher is included, it becomes 10 years. What is the age of the teacher?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 26
Total number of students = 30*9 = 270

When teacher's age is included then total age = 31*10 = 310

therefore, Required answer = 310- 270 = 40 years

RRB JE ME (CBT I) Mock Test- 6 - Question 27
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Two pipes can separately fill a tank in 20 hrs and 30 hrs respectively. Both the pipes are opened to fill the tank but when the tank is 1/3 full a leak develops in the tank through which 1/3 of the water supplied by both the pipes per hour leak out. What is the total time taken to fill the tank?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 27
Time taken by the two pipes to fill the tank

∴ 1/3 of the tank if filled in 12/3 = 4 hrs.

Now, 1/3 of the supplied water leaks out

⇒ the filler pipes are only as efficient as earlier.

⇒ the work of (12 - 4 =) 8 hrs will be completed now in

∴ total time = 4 + 12 = 16 hrs.

RRB JE ME (CBT I) Mock Test- 6 - Question 28
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If 3 tan θ - 4 = 0 and 180° < θ < 270° then cosec θ = _______
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 28

Given the equation 3 tan θ - 4 = 0, we can solve for tan θ: tan θ = 4/3 Since θ is between 180° and 270°, it lies in the third quadrant where both sine and cosine are negative, making tan θ positive. To find cosec θ, which is equal to 1 divided by sin θ, we first determine sin θ. From tan θ = opposite/adjacent = 4/3, we can find the hypotenuse using the Pythagorean theorem: hypotenuse = √(3² + 4²) = 5 Thus, sin θ = opposite/hypotenuse = 4/5 However, in the third quadrant, sine is negative, so sin θ = -4/5. Therefore, cosec θ = 1/sin θ = 1/(-4/5) = -5/4 Hence, the correct answer is option c) -5/4.

RRB JE ME (CBT I) Mock Test- 6 - Question 29
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DIRECTIONS: Study the following table carefully and answer the questions that follow.

The table represents the total number of students studying courses P, Q, K, S and T, across eight Institutes i.e., A, B, C, D, E, F, G and H.

What is the total number of students who are studying course T across all institutes?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 29
Total number of required students

= 370 + 480 + 380 + 250 + 180 + 370 + 590 + 660

= 3280

RRB JE ME (CBT I) Mock Test- 6 - Question 30
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DIRECTIONS: Study the following table carefully and answer the questions that follow.

The table represents the total number of students studying courses P, Q, K, S and T, across eight Institutes i.e., A, B, C, D, E, F, G and H.

What is the respective ratio between the total number of students studying in institute A and the total number of students studying in institute H?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 6 - Question 30
Total number of students studying in institute A

= 520 + 410 + 430 + 350 + 370

= 2080

Total number of student studying in institute H

= 450 + 310 + 480 + 460 + 660

= 2360

Now, required ratio = 2080/2360 = 52/59 = 52 : 59

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