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Test: Cylinders - 2 - GMAT MCQ


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10 Questions MCQ Test - Test: Cylinders - 2

Test: Cylinders - 2 for GMAT 2024 is part of GMAT preparation. The Test: Cylinders - 2 questions and answers have been prepared according to the GMAT exam syllabus.The Test: Cylinders - 2 MCQs are made for GMAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Cylinders - 2 below.
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Test: Cylinders - 2 - Question 1

 Volume of a cylinder having radius r and height h is given by __________

Detailed Solution for Test: Cylinders - 2 - Question 1

Volume of a cylinder can be given by multiplying base area with its height.
Hence, volume of a cylinder having radius r and height h is equal to πr2h where πr2 is base area and h is height.

Test: Cylinders - 2 - Question 2

A cylinder has radius of 7cm and its volume is 1540cm3, what is the height of the cylinder? (Take π = 22/7)

Detailed Solution for Test: Cylinders - 2 - Question 2

Volume of a cylinder having radius r and height h is equal to πr2h.
πr2h = 1540
22/7 * 7 * 7 * h = 1540
h = 10cm.

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Test: Cylinders - 2 - Question 3

A cylinder has volume of 176cm3 and height equal to 14cm, what is the diameter of the cylinder? (Take π = 22/7)

Detailed Solution for Test: Cylinders - 2 - Question 3

Explanation: Volume of a cylinder having radius r and height h is equal to πr2h.
πr2h = 176
22/7 * 14 * r2 = 176
r2 = 4
Therefore, r = 2cm
We know that diameter = 2 * radius
= 2 * 2
= 4cm.

Test: Cylinders - 2 - Question 4

A steel pipe has outer radius equal to 3cm and inner radius equal to 2cm. Its length is equal to 5m. What is the mass of the pipe if one cm3 of steel is 8.05 gram? (Take π = 22/7)

Detailed Solution for Test: Cylinders - 2 - Question 4

Let the outer radius = ro = 3cm
Inner radius = ri = 2cm
Height (here length) = l = 5m = 500cm
Volume of pipe = π * (ro 2 – ri 2) * l
= 22/7 * (32 – 22) * 500
= 7857.14cm3
Mass of 1cm3 is 8.05 gram
Therefore, mass of 7857.14cm3 is equal to 8.05 * 7857.14
= 63249.97gm
≈ 63250 gm
= 63.25 kg.

Test: Cylinders - 2 - Question 5

A cylinder has a height of 9 and a radius of 4. What is the total surface area of the cylinder?

Detailed Solution for Test: Cylinders - 2 - Question 5

We are given the height and the radius of the cylinder, which is all we need to calculate its surface area. The total surface area will be the area of the two circles on the bottom and top of the cylinder, added to the surface area of the shaft.
If we imagine unfolding the shaft of the cylinder, we can see we will have a rectangle whose height is the same as that of the cylinder and whose width is the circumference of the cylinder.
This means our formula for the total surface area of the cylinder will be the following:
SA = 2(πr2) + 2πrh
SA  =2π(4)2 + 2π(4)(9)
SA = 32π + 72π
SA = 104π

Test: Cylinders - 2 - Question 6

Find the surface area of a cylinder whose height is 6 and radius is 3.

Detailed Solution for Test: Cylinders - 2 - Question 6

To find the surface area of a cylinder, you must use the following equation.
SA = 2πrh + 2πr2
Thus,
SA = 2π(3)(6) + 2π(32) = 54π

Test: Cylinders - 2 - Question 7

What is the volume of a cylinder that is 12 inches high and has a radius of 6 inches?

Detailed Solution for Test: Cylinders - 2 - Question 7

volume = πr2h = π∗62∗12 = 432π

Test: Cylinders - 2 - Question 8

A cylindrical gas tank is 30 meters high and has a radius of 10 meters. How much oil can the tank hold?

Detailed Solution for Test: Cylinders - 2 - Question 8

V = πr2h = π(102)30 = 3000π

Test: Cylinders - 2 - Question 9

A given cylinder has a radius of 10 and a height of 15. What is the volume of the cylinder?

Detailed Solution for Test: Cylinders - 2 - Question 9

The volume V of a cylinder with radius r and height h is defined as V = πr2h.
Plugging in our given values:
V = πr2h
V = π(10)2(15)
V = π(100)(15)
V = 1500π

Test: Cylinders - 2 - Question 10

Find the volume of a cylinder whose height is 8 and radius is 2.

Detailed Solution for Test: Cylinders - 2 - Question 10

To find the volume of a cylinder, you must use the following equation:
V = πr2h
Thus,
V = π(22)(8) = 32π

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