Reference Sheet · SAT Math · Problem Solving & Data
Numbers, Rates & Statistics
Percent change, interest formulas, conversion factors, distance/rate/time, and the basic statistics measures. The SAT's "Problem Solving and Data Analysis" content domain in one reference.
Fractions, decimals & percents
Part-whole relationship
$$\frac{\text{part}}{\text{whole}} = \frac{\text{percent}}{100}$$
Set up a proportion when you know two of the three (part, whole, percent). Cross-multiply to solve.
EXAMPLE
$30$ is what percent of $80$?
$\dfrac{30}{80} = \dfrac{p}{100}$
$p = \dfrac{3000}{80} = $ $37.5\%$
Percent change
$$\% \text{ change} = \frac{\text{new} - \text{old}}{\text{old}} \cdot 100\%$$
Always divide by the OLD value, not the new one. Positive answer means increase; negative means decrease.
EXAMPLE
Price goes from $\$80$ to $\$100$. Find the percent increase.
$\dfrac{100 - 80}{80} \cdot 100\% = \dfrac{20}{80} \cdot 100\% = $ $25\%$
Increase by a percent
$$\text{new} = \text{old} \cdot (1 + r)$$
r = the percent as a decimal (e.g., $20\% = 0.20$)
Faster than computing the dollar amount of the increase and adding it. One multiplication does the whole job.
EXAMPLE
Increase $50$ by $30\%$.
$50 \cdot (1 + 0.30) = 50 \cdot 1.3 = $ $65$
Decrease by a percent
$$\text{new} = \text{old} \cdot (1 - r)$$
r = the percent as a decimal
Same trick as percent increase, just with subtraction instead of addition.
EXAMPLE
Decrease $200$ by $15\%$.
$200 \cdot (1 - 0.15) = 200 \cdot 0.85 = $ $170$
Interest formulas
Simple interest
$$A = P(1 + rt)$$
A = total amount after interest
P = principal (starting amount)
r = annual interest rate (as decimal)
t = time in years
P = principal (starting amount)
r = annual interest rate (as decimal)
t = time in years
Interest is calculated only on the original principal, not on accumulated interest. Linear growth.
EXAMPLE
$\$1000$ at $5\%$ simple interest for $3$ years.
$A = 1000(1 + 0.05 \cdot 3) = 1000(1.15) = $ $\$1150$
Compound interest (annual)
$$A = P(1 + r)^t$$
A, P, r, t = same as simple interest
Interest is calculated on principal AND accumulated interest. Exponential growth.
EXAMPLE
$\$1000$ at $5\%$ compound interest for $3$ years.
$A = 1000(1.05)^3 = 1000(1.157625) \approx $ $\$1157.63$
Compare to simple interest above — about $\$7.63$ more.
Compound interest ($n$ times per year)
$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$
n = number of compounding periods per year (12 for monthly, 4 for quarterly, etc.)
More frequent compounding $\Rightarrow$ slightly more growth. The annual formula above is just the case $n = 1$.
EXAMPLE
$\$1000$ at $5\%$ compounded monthly for $3$ years.
$A = 1000\left(1 + \dfrac{0.05}{12}\right)^{36} \approx $ $\$1161.47$
Rates & conversions
Distance, rate, time
$$d = rt$$
d = distance, r = rate (speed), t = time
Most flexible formula on the SAT. Solve for any of the three. Make sure units agree — if rate is mph, time must be hours.
EXAMPLE
A car travels at $60$ mph for $2.5$ hours. How far?
$d = 60 \cdot 2.5 = $ $150$ miles
Conversion factor
$$\text{starting} \cdot \frac{\text{ending units}}{\text{starting units}}$$
Set up the fraction so that the units you DON'T want cancel diagonally and the units you DO want remain.
EXAMPLE
Convert $5$ feet to inches.
$5 \text{ ft} \cdot \dfrac{12 \text{ in}}{1 \text{ ft}} = $ $60$ inches
The "ft" units cancel; "in" remains.
Mixture problems (concentration)
$$C_A V_A + C_B V_B = C_{\text{final}}(V_A + V_B)$$
C = concentration, V = volume
When two solutions are combined, the total amount of solute equals the sum of the parts.
EXAMPLE
Mix $4$ L of $30\%$ solution with $6$ L of $50\%$ solution. Final concentration?
$0.30(4) + 0.50(6) = C_{\text{final}}(10)$
$1.2 + 3.0 = 10 C_{\text{final}}$
$C_{\text{final}} = $ $0.42 = 42\%$
Statistics
Mean (average)
$$\text{mean} = \frac{\text{sum of values}}{\text{count of values}}$$
Add all values, divide by how many. Affected by outliers.
EXAMPLE
Find the mean of $3, 5, 7, 9, 11$.
$\dfrac{3 + 5 + 7 + 9 + 11}{5} = \dfrac{35}{5} = $ $7$
Median
$$\text{middle value when ordered}$$
First sort the values. If there's an odd count, take the middle one. If even, average the two middle ones. Resistant to outliers.
EXAMPLE
Find the median of $4, 8, 2, 10, 6$.
Sort: $2, 4, 6, 8, 10$.
Middle is $6$.
Mode
$$\text{most frequent value}$$
The value that appears most often. A data set can have no mode, one mode, or multiple modes.
EXAMPLE
Find the mode of $3, 5, 5, 7, 8, 5, 9$.
$5$ appears three times — more than any other value.
Mode $= $ $5$
Range
$$\text{range} = \text{max} - \text{min}$$
Difference between the largest and smallest values. A measure of spread.
EXAMPLE
Find the range of $12, 4, 18, 7, 22, 9$.
Max is $22$, min is $4$.
Range $= 22 - 4 = $ $18$
Probability
Basic probability
$$P(\text{event}) = \frac{\text{desired outcomes}}{\text{total outcomes}}$$
Always between $0$ (impossible) and $1$ (certain). Make sure outcomes are equally likely before applying this.
EXAMPLE
A bag has $3$ red marbles and $5$ blue marbles. Probability of drawing red?
$P(\text{red}) = \dfrac{3}{3 + 5} = \dfrac{3}{8} = $ $0.375$
Complement rule
$$P(\text{not A}) = 1 - P(A)$$
Probability of "not A" is one minus the probability of A. Often much easier than computing "not A" directly.
EXAMPLE
If $P(\text{rain}) = 0.3$, find $P(\text{no rain})$.
$1 - 0.3 = $ $0.7$