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.
The arithmetic mean of 4, 6, 8, 12, and x is 9. What is the value of x?
Solution
The formula for the arithmetic mean is
\dfrac{4 + 6 + 8 + 12 + \textit{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.
The set S consists of all non-negative integers x < y. If the probability of \dfrac{\textit{x}^{\displaystyle{2}}}{4} < 4 is \dfrac{1}{2}, what is the value of y?
Solution
This question gives the probability rather than asking for the probability.
Simplify the inequality.
\dfrac{\textit{x}^{\displaystyle{2}}}{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) = \dfrac{number\, of\, favorable \,outcomes\, \textit{A}}{number \,of \,possible \,outcomes}
The non-negative integers x < 4 are 0, 1, 2, and 3, for 4 integers.
Since P = \dfrac{1}{2},
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.
The formula for calculating Celsius temperatures (°C) from Fahrenheit (°F) is
C = \dfrac{5}{9} (F – 32)
Rewrite the equation to get the formula for calculating Fahrenheit from Celsius.
Solution
C = \dfrac{5}{9}(F – 32) …Multiply both sides by the reciprocal.
\dfrac{9}{5}C = F – 32 …Add 32 to both sides.
F = \dfrac{9}{5}C + 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.
The formula for the volume of a rectangular prism is V = lwh.
If V = 126, l = 7 and h = 3, find w.
Solution
To find just one value, substitution is often the easiest.
V = lwh 126 = (7)w(3)
126 = 21w w = 6
The formula for the volume of a rectangular prism is V = lwh.
If l = 7 and h = 3, find w when V = 84 and V = 63.
Solution
To find multiple values, solving the formula for the variable is faster.
V = lwh
w = \dfrac{\textit{V}}{\textit{lh}} → w = \dfrac{\textit{V}}{(7)(3)} = \dfrac{\textit{V}}{21}
When V = 84, w = \dfrac{84}{21} = 4.
When V = 63, w = \dfrac{63}{21} = 3.
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 = \dfrac{\textit{x}}{\textit{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 = \dfrac{\textit{y}}{\textit{y} \,-\, 1}
There is no solution when y = 1.
Method 2: Try numbers and create a table of values.
given x | xy = x + y | find y |
-1 | –y = -1 + y | \dfrac{1}{2} |
0 | 0 = 0 + y | 0 |
1 | y = 1 + y | no solution |
given y | xy = x + y | find x |
-1 | –x = x + (-1) | \dfrac{1}{2} |
0 | 0 = x + 0 | 0 |
1 | x = x + 1 | no 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.
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 \dfrac{\textit{A}}{\textit{w}} = l
P = 2( l + w) \dfrac{\textit{P}}{2} – w = l
\dfrac{\textit{A}}{\textit{w}} = \dfrac{\textit{P}}{2} – w …Write an equation that combines the formulas.
\dfrac{24}{\textit{w}} = \dfrac{22}{2} – w …Substitute given values.
\dfrac{24}{\textit{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.
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 = πr 2
Let r be the radius of the small circle. Then 2r is the radius of the large circle.
Asmall = πr 2 Abig = π(2r) 2 = 4πr 2
The area of the big circle that is not covered by the small circle is:
4πr 2 – πr 2 = 3πr 2
The ratio of the area that isn’t covered to the area of the big circle is:
\dfrac{3π\textit{r}^{\displaystyle{2}}}{4π\textit{r}^{\displaystyle{2}}} = \dfrac{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 = \dfrac{5}{9} (F – 32)
F = \dfrac{9}{5} C + 32
distance
distance = rate × time
d = r t
work
work = rate × time
w = r t
simple interest (interest paid only on the principal) I = Prt
compound interest (interest paid on principal and interest)
A = P \Big(1 + \dfrac{\textit{r}}{\textit{n}}\Big)^{\displaystyle{\textit{nt}}}
sequences
linear
an = a1 + (n – 1)d
geometric (exponential)
an = a1r n
arithmetic mean
m = \dfrac{sum\, of\, all \,the\, terms}{number \,of \,terms}
probability
P(A) = \dfrac{number\, of \,favorable\, outcomes\, \textit{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
V = volume SA = surface area
C = circumference r = radius
P = perimeter A = area
B = area of the base
d = diameter
triangle
A = \dfrac{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 = \dfrac{1}{2} (base 1 + base 2)(height)
= \dfrac{1}{2} (b1 + b2)h
circle
2r = d C = 2πr = πd A = πr 2
π ≈ 3.14 ≈ \dfrac{22}{7}
rectangular prism
SA = 2lw + 2wh + 2lh = 2B + 2Ph V = lwh
pyramid
V = \dfrac{1}{3} Bh
cylinder
SA = 2B + Ch
V = πr 2h = Bh
cone
V = \dfrac{1}{3} πr 2h = \dfrac{1}{3} Bh
sphere
SA = 4πr 2
V = \dfrac{4}{3} πr 3