Mensuration - MCQ 1


20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations | Mensuration - MCQ 1


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This mock test of Mensuration - MCQ 1 for Quant helps you for every Quant entrance exam. This contains 20 Multiple Choice Questions for Quant Mensuration - MCQ 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Mensuration - MCQ 1 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Mensuration - MCQ 1 exercise for a better result in the exam. You can find other Mensuration - MCQ 1 extra questions, long questions & short questions for Quant on EduRev as well by searching above.
QUESTION: 1

A right circular cone is placed over a cylinder of the same radius. Now the combined structure is painted on all sides. Then they are separated now the ratio of area painted on Cylinder to Cone is 3:1. What is the height of Cylinder if the height of Cone is 4 m and radius is 3 m?

Solution:

Cylinder painted area = 2πrh+πr²
Cone painted area = πrl
2h+r/√ (r² +h1² ) = 3:1
h = 6

QUESTION: 2

The diameter of Road Roller is 84 cm and its length is 150 cm. It takes 600 revolutions to level once on a particular road. Then what is the area of that road in m²?

Solution:

600*2*22/7*42/100*150/100=2376

QUESTION: 3

A smaller triangle is having three sides. Another big triangle is having sides exactly double the sides of the smaller triangle. Then what is the ratio of Area of Smaller triangle to Area of the Bigger triangle?

Solution:

Smaller triangle sides = a, b, c
Area= √s(s-a) (s-b) (s-c);
s=a+b+c/2
= √(a+b+c)(b+c-a)(a+c- b)(a+b-c)/4
Bigger triangle =2a, 2b, 2c
Area = √(a+b+c)(b+c-a)(a+c- b)(a+b-c)
Ratio = 1:4

QUESTION: 4

ABCD is a square of 20 m. What is the area of the least-sized square that can be inscribed in it with its vertices on the sides of ABCD?

Solution:

It touches on midpoints on the sides of the square ABCD
Side= √ (10² +10²) = √200
Area= 200 m²

QUESTION: 5

A hemispherical bowl of diameter 16cm is full of ice cream. Each student in a class is served exactly 4 scoops of ice cream. If the hemispherical scoop is having a radius of 2cm, then ice cream is served to how many students?

Solution:

2/3*π*8³ = n*4*2/3*π*2³
n = 16

QUESTION: 6

A hollow cylindrical tube is made of plastic is 4 cm thick. If the external diameter is 18 cm and length of the tube is 59cm, then find the volume of the plastic?

Solution:

R = 9, r =5
V =22/7*59(9 2 -5 2) = 10384

QUESTION: 7

What is the radius of the circle whose area is equal to the sum of the areas of two circles whose radii are 20 cm and 21 cm?

Solution:

πR² = πr1² + πr2²
πR² = π(r1² + r2²)
R² = (400 + 441)
R = 29

QUESTION: 8

A well with 14 m diameter is dug up to 49 m deep. Now the soil taken out during dug is made into cubical blocks of 3.5m side each. Then how many such blocks were made?

Solution:

22/7*7²*49 = n*(7/2)³
n =176

QUESTION: 9

If the ratio of radius two Cylinders A and B are in the ratio of 2:1 and their heights are in the ratio of 2:1 respectively. The ratio of their total surface areas of Cylinder A to B is?

Solution:

Cylinder A: 2πr1 (r1 + h1)
Cylinder B: 2πr2 (r2 + h2)
r 1 /r 2 = 2:1; h 1 /h 2 = 2:1
T A /T B = 2πr1 (r1 + h1)/ 2πr2 (r2 + h2)

QUESTION: 10

The area of the Circular garden is 88704 m². Outside the garden a road of 7m width laid around it. What would be the cost of laying road at Rs. 2/m².

Solution:

88704 = 22/7*r
r =168
Outer radius = 168+7 = 175
Outer area = 22/7*175 2 = 96250
Road area = 96250 – 88704 = 7546
Cost = 7546*2 = 15092

QUESTION: 11

A cylindrical cistern whose diameter is 14 cm is partly filled with water. If a rectangular block of iron 22 cm in length, 18 cm in breadth and 7 cm in thickness is wholly immersed in water, by how many cm will the water level rise?

Solution:

Volume of the Rectangular Block = 22 * 18 * 7
Radius of the Cistern = 7 cm
Area of the Cylinder = π * r² * h = 22/7 * 7 * 7 * h
22/7 * 7 * 7 * h = 22 * 18 * 7; h = 18 cm

QUESTION: 12

If the radius of cylinder is doubled, but height is reduced by 50%. What is the percentage change in volume?

Solution:

R1/R2 = R/2R & H1/H2 = H/H/2
Original Volume = πr²h
New Volume = π(2r)²h/2
Change in Volume = (2 – 1)/1 * 100 = 100

QUESTION: 13

The perimeter of a rectangle and a square is 80 cm each. If the difference between their areas is 100 cm. Find the sides of the rectangle.

Solution:

2(l + b) = 4a = 80
l + b = 40; a = 20 ⇒ a² = 400
a² – lb = 100; 400 – lb =100; lb = 300
(l – b)² = (l + b)² – 4lb
(l – b)² = 1600 – 1200; l – b = 20;

QUESTION: 14

The length of a wall is 5/4 times of its height. If the area of wall will be 500m². What is the sum of the length and height of the wall?

Solution:

l = 5x; h = 4x ; l * h = 500
20x² = 500 ; x = 5
l + h = 25 + 20 = 45m

QUESTION: 15

A circular path runs round a circular garden. If the difference between the circumference of the outer circle and inner circle is 88m. Find the width of Path?

Solution:

Width of the Road = R – r
2πR – 2πr = 88
R – r = 14 m

QUESTION: 16

The radius of a circle is 4 m. What is the radius of another circle whose area is 16 times of that first?

Solution:

Ratio of areas = (ratio of radii)²
16/1 = (ratio of radii)²
ratio of radii = 4/1
The required radius = 16 m

QUESTION: 17

What is the radius of the circle whose area is equal to the sum of the areas of two circles whose radii are 20 cm and 21 cm?

Solution:

πR² = πr1² + πr2²
πR² = π(r1² + r2²)
R² = (400 + 441)
R = 29

QUESTION: 18

Find the area of a white sheet required to prepare a cone with a height of 21cm amd the radius of 20cm.

Solution:

r = 20 ; h = 21
l = √(400 + 441) = 29 cm
Total surface area = πrl + πr² = πr(l + r) = 22/7 * 20(49) = 3080 cm²

QUESTION: 19

A hollow cylindrical tube open at both ends is made of plastic 4 cm thick. If the external diameter be 54 cm and the length of the tube be 490 cm, find the volume of plastic.

Solution:

External Radius = 27; Internal Radius = 23
Volume of Plastic =πh(R² – r²) = 22/7 * 490(27² – 23²) = 308000 cm³

QUESTION: 20

A solid metallic cylinder of base radius 5 cm and height 7 cm is melted to form cones, each of height 1 cm and base radius 1 mm. Find the number of cones?

Solution:

Number of cones =Volume of Cylinder / Volume of one cone
= π*5*5*7 / (1/3π * 1/10 * 1/10 * 1)
= 52500

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