Solving Equations

Formulas are just equations with specific formats. Formula questions on the GMAT will supply values for some variables and expressions and ask you to find the other pieces in the formula.

Example

The arithmetic mean of 4, 6, 8, 12, and x is 9. What is the value of x?

Solution

4 + 6 + 8 + 12 + x/5 = 9

x + 30 = 9 × 5 = 45

x = 15

Substituting Given Values

You need to be able to substitute values into a formula, making sure to put the values into the right places.

Example

The set S consists of all non-negative integers x < y. If the probability of x2/4 < 4 is 1/2, what is the value of y?

Solution

This question gives the probability rather than asking for the probability.

Simplify the inequality.

x2/4 < 4

x2 < 16

Since x is positive, take the square root of both sides.

x < 4

Remember that the formula for probability is

P(A) = number of favorable outcomes A/number of possible outcomes

The non-negative integers x < 4 are 0, 1, 2, and 3, for 4 integers.

Since P = 1/2, 1/2 = one of the 4 integers/How many possible integers?, 2 × 4 = how many integers, so 8 integers.

Since there needs to be 8 integers less than y, the possibilities are 0, 1, 2, 3, 4, 5, 6, or 7. So y = 8.

Rewriting Formulas

You can solve an equation for any of the variables. Rewriting a formula by solving for one of the variables uses the same steps as solving an equation.

Example

The formula for calculating Celsius temperatures (°C) from Fahrenheit (°F) is
C = 5/9(F – 32)
Rewrite the equation to get the formula for calculating Fahrenheit from Celsius.

Solution

C = 5/9(F – 32)

Multiply both sides by the reciprocal.

9/5C = F – 32

Add 32 to both sides.

F = 9/5C + 32

Multiple Solutions

Solving a formula for one of the variables is a good approach when you are asked to find multiple values for that variable.

You can also substitute numbers to find values and ranges of values for a variable.

Example

The formula for the volume of a rectangular prism is V = lwh.

  1. If V = 126, l = 7 and h = 3, find w.
  2. If l = 7 and h = 3, find w when V = 84 and V = 63.

Solution

  1. To find just one value, substitution is often the easiest.
    V = lwh, 126 = (7)w(3), 126 = 21w, w = 6
  2. To find multiple values, solving the formula for the variable is faster.
    V = lwh, w = V</lh –> w = V</(7)(3) = V</21

When V = 84, w = 84/21 = 4.

When V = 63, w = 63/21 = 3.

Example

For which values of x and y are there no solutions?

xy = x + y

Solution

There are two methods.

Method 1

Solve the equation for x and for y.

xy = x + y –> xy – y = x –> y(x – 1) = x –> y = x/x – 1

If x = 1, it is division by zero so it’s undefined. There is no solution when x = 1.

Similarly, solving for x gives x = y/y – 1. There is no solution when y = 1.

Method 2

Try numbers and create a table of values.

Given xxy = x + yFind yGiven yxy = x + yFind x
-1-y = -1 + y1/2-1-x = x + (-1)1/2
00 = 0 + y000 = x + 00
1y = 1 + yno solution1x = x + 1no solution

Using More Than One Formula

Using two formulas that have the same variables is similar to solving systems of equations.

When making comparisons, don’t immediately solve all the way. Keeping the variables or constants that are in both formulas can make for easier comparisons.

Example

In a rectangle, the area A = 24 and the perimeter P = 22. Find the length and width of the rectangle.

Solution

This is combining two formulas to find two unknowns.
Solve both formulas for the same variable, either l or w.

A = lw

A/w = l

P = 2(l + w)

P/2 – w = l

A/w = P/2 – w

Write an equation that combines the formulas.

24/w = 22/2 – w

Substitute given values.

24/w = 11 – w

24 = 11w – w2

w2 – 11w + 24 = 0

Factor.

(w – 3)(w – 8) = 0

So, w = 3 or w = 8.

The side lengths are 3 and 8. Either value can be called the width or length.

Example

The diameter of the small circle is half of the diameter of the large circle. What fraction of the area of the large circle is not covered by the area of the small circle?

Solution

When comparing values using just one formula, don’t solve completely. Keep the variables and constants that are in both.
A = πr2

Let r be the radius of the small circle. Then 2r is the radius of the large circle.

Asmall = πr2, Abig = π(2r)2 = 4πr2

The ratio of the area that isn’t covered to the area of the big circle is:

3πr2/4πr2 = 3/4

Common Formulas

These are some of the formulas that commonly appear on the GMAT. They are reviewed in greater details in other chapters.

temperature

C = 5/9 (F – 32)

F = 9/5 C + 32

distance

distance = rate × time

d = rt

work

work = rate × time

w = rt

interest

P = principal, A = principal + interest paid, I = interest paid
r = annual rate, n = times per year, t = number of years
simple interest (interest paid only on the principal): I = Prt
compound interest (interest paid on principal and interest):
A = P (1 + r/n)nt

P = principal, A = principal + interest paid, I = interest paid
r = annual rate, n = times per year, t = number of years
simple interest (interest paid only on the principal): I = Prt
compound interest (interest paid on principal and interest):
A = P (1 + r/n)nt

sequences

linear
an = a1 + (n – 1)d

geometric (exponential)
an = a1rn

arithmetic mean

m = sum of all the terms/number of terms

probability

P(A) = number of favorable outcomes A/number of possible outcomes
independent (with replacement): P(A and B) = P(A) × P(B)
dependent (without replacement): P(A and B) = P(A) × P(B after A)

geometry

l = length, w = width, h = height, P = perimeter, A = area
V = volume, SA = surface area, B = area of the base
C = circumference, r = radius, d = diameter

triangle

A = 1/2 bh
sum of interior angles = 180°
Pythagorean theorem a2 + b2 = c2 or (side)2 + (side)2 = (hypotenuse)2

square

P = 4s

A = s2

all angles = 90°

rectangle

P = 2(l + w)

A = lw

all angles = 90°

parallelogram

P = 2(l + w)

A = lh

sum of interior angles = 360°

trapezoid

A = 1/2 (base 1 + base 2)(height) = 1/2 (b1 + b2)h

circle

2r = d, π ≈ 3.14 ≈ 22/7
C = 2πr = πd
A = πr2

rectangular prism

SA = 2lw + 2wh + 2lh = 2B + 2Ph
V = lwh

pyramid

V = 1/3 Bh

cylinder

SA = 2B + Ch

V = πr2h = Bh

cone

V = 1/3 πr2h = 1/3 Bh

sphere

SA = 4πr2

V = 4/3 πr3

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