Volume - GMAT Math Study Guide

Table of Contents

  1. Rectangular Solids
    1. Surface Area
  2. Right Cylinders
    1. Surface Area
  3. Types of GMAT Problems

Rectangular Solids

A rectangular solid is a three-dimensional object composed of rectangular sides. Each of the six rectangular sides is a face. The solid and dotted lines are the edges. Opposing faces are parallel and have the same length and width. If all of the six sides are of equal length, the rectangular solid is called a cube.

Volume = (length)(width)(height)
Volume = lwh

Consider the following example:

What is the volume of the figure above if DJ = 2, GJ = 4, and DB = 3?
Volume = lwh
V = 2 * 4 * 3 = 24

Surface Area

Surface Area = 2(length)(width) + 2(height)(width) + 2(length)(height)
SA = 2lw + 2hw + 2lh

In other words, the surface area is equal to the summation of the area of each of the faces. For example, if a rectangular solid had a length of 2, height of 4, and width of 3, then SA = 2(2)(3) + 2(4)(3) + 2(2)(4)

Right Cylinders

A right circular cylinder has two circular bases of equal size whose centers are both perpendicularly intersected by the same line. In the diagram below, the surface with point P and the surface with point Q are both the same size. Surface P and surface Q are both perpendicularly intersected by line PQ, making these surfaces parallel.

Volume = (area of base) * (height)
V = πr2h where r is the radius, which is QR in the above diagram

In other words, the volume is equal to the area of the base multiplied by the height.

Consider the following example:

What is the area of the figure above if PQ = 5 and QR = 3?
V = (π32)(5) = 9π(5) = 45π

Surface Area

Surface Area = 2(pi)(radius)2 + 2(pi)(radius)(height)
SA = 2(πr2) + 2πrh

In other words, the surface area is equal to the summation of the area of each of the faces. For example, if a right circular cylinder had a radius of 3 and height of 5, then SA = 2(π*9) + 2π*3*5 = 18π + 30π = 48π

Types of GMAT Problems

  1. Filling Problems

    Often, GMAT questions will ask how long it takes to fill an object given the rate that a substance is being poured into the object. For time based problems, the volume of the object in question should be determined. Then the volume can be divided by the rate at which the substance is being poured to determine how long it will take to fill the object in the question. Another twist to the problem occurs when the time and rate are given, but one measurement of the object is left unknown. In order to calculate the length of the unknown side, calculate the volume with the given time. Then substitute the volume, V, for the corresponding volume in the corresponding function and solve for the unknown. (This probably sounds quite confusing, but an example will make it seem much simpler).

    If a gas tank in the shape of a right circular cylinder is 10 meters high and is being filled at a rate of 4π gallons a minute until it is filled after 10 minutes, what is the radius of the base of the cylinder?
    Correct Answer: A
    1. When the gas tank is full, the volume of gasoline in the tank will equal the volume of the cylinder.
    2. For each minute that elapses, another 4π gallons of gasoline occupy the tank. When the tank is full, the volume of the cylinder will be 10(min) * 4π(gal/min) = 40πgal since it took 10 minutes to fill the tank and every minute another 4π gallons is poured into the tank.
    3. V = πr2h (this is the formula for the volume of a cylinder)
      h = 10, given in the problem
      Substitute 40π in for V
      40π = hπr2 = 10πr2
    4. Solve for r.
      40π = 10πr2
      4 = r2
      2 = r
      r = 2