SAT Math Problem Solving and Data Analysis

Problem Solving & Data Analysis — SAT Topic Page

Ultra exam-oriented. Master proportional reasoning and data inference. Expect multi-step setups and interpretation under time pressure.

Quick Reference (Minimal Theory)

Idea	Formula / Cue	Tooltip
Percent change	\(\dfrac{N – O}{O}\times100\%\)	Chain by factors
Rate	\(r=\dfrac{\text{amount}}{\text{time}}\)	Track units
Weighted mean	\(\bar{x}=\dfrac{\sum w_ix_i}{\sum w_i}\)	Weight pulls
Two-way table	Conditional: \(P(A\mid B)=\dfrac{n(A\cap B)}{n(B)}\)	Condition then count
Line of best fit	\(y=mx+b\)	Interpret\, don’t assume

Worked Examples (10) — Statistical Visuals Included

Example 1 (Ivy)

A quantity increases by 30% and then decreases by 20%. What single percent change is equivalent?

Show solution

Factor chain: \(\times1.30\) then \(\times0.80\) → net factor \(1.04\). Net change = \(+4\%\).

Answer: \(+4\%\).

Example 2 (MIT)

A solution is \(25\%\) acid. How many liters of pure acid must be added to 20 L of the solution to make it \(40\%\) acid?

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Acid now: \(0.25\times20=5\) L. Add \(x\) L pure → total acid \(5+x\), total volume \(20+x\). Solve \(\dfrac{5+x}{20+x}=0.40\Rightarrow 5+x=8+0.4x\Rightarrow 0.6x=3\Rightarrow x=5\) L.

Answer: 5 L.

Example 3 (Ivy)

A car uses 8.2 L per 100 km. What is the fuel economy in miles per gallon? (Use 1 mile = 1.609 km, 1 gallon = 3.785 L.)

Show solution

\(8.2\,\text{L}/100\,\text{km} = 0.082\,\text{L}/\text{km}\). Convert: \(\dfrac{1}{0.082}\,\text{km/L} \approx 12.195\,\text{km/L}\). \(12.195\,\text{km/L} \times \dfrac{1\,\text{mile}}{1.609\,\text{km}} \times \dfrac{3.785\,\text{L}}{1\,\text{gal}} \approx 28.7\,\text{mpg}\).

Answer: \(\approx 28.7\) mpg.

Example 4 (Ivy)

A factory makes widgets at rate \(r\) per hour. After a 15% efficiency gain, it produces 1{,}150 in 10 hours. Find \(r\).

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New rate \(=1.15r\). So \(10(1.15r)=1150\Rightarrow r=100\) per hour.

Answer: 100/hour.

Example 5 (MIT)

Given data pairs \((x,y)\): (1,2), (2,5), (3,7), (4,10). Estimate the line of best fit and predict \(y\) at \(x=6\).

Show solution

Rough slope from endpoints: \(m\approx\dfrac{10-2}{4-1}=\dfrac{8}{3}\approx2.67\). Using point (1,2): \(y\approx2.67(x-1)+2\Rightarrow y\approx2.67x-0.67\). Predict at 6: \(\approx 15.35\).

Answer: \(\hat y\approx15.3\text{–}15.4\).

Example 6 (Ivy)

Dataset of 10 values has mean 50. A new value 80 is added. What is the new mean?

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Total originally \(=10\cdot50=500\). New total \(=580\) with 11 values → mean \(=52.73\).

Answer: \(\approx 52.73\).

Example 7 (Ivy)

In a class, 60% take biology and 45% take chemistry; 25% take both. What percent take neither?

Show solution

\(P(B\cup C)=P(B)+P(C)-P(B\cap C)=0.60+0.45-0.25=0.80\). Neither \(=1-0.80=0.20=20\%\).

Answer: 20%.

Example 8 (MIT)

A two-way table shows 200 people: 120 prefer tea, 80 prefer coffee. Of tea-preferrers, 54 are morning persons; of coffee-preferrers, 48 are morning persons. Find \(P(\text{morning}\mid\text{tea})\) and \(P(\text{morning}\mid\text{coffee})\).

Show solution

\(P(M\mid T)=54/120=0.45\). \(P(M\mid C)=48/80=0.60\).

Answer: 0.45 and 0.60.

Example 9 (MIT)

A sample of 8 measurements: 12, 14, 17, 18, 22, 25, 29, 35. Find median and IQR.

Show solution

Median is mean of 4th and 5th: \((18+22)/2=20\). Q1 median of first half: (14+17)/2=15.5. Q3 median of last half: (25+29)/2=27. IQR=11.5.

Answer: median 20, IQR 11.5.

Example 10 (Ivy)

Correlation is \(r=0.85\) between study time and score. Which is a valid interpretation?

Show solution

Strong positive linear association; does not imply causation. Outliers or lurking variables may exist.

Answer: Strong positive linear trend; no causal claim.

Practice Questions (20) — Exam-Oriented, with Solutions

1. A price is discounted 15% and then increased 10%. What is net percent change?
   Solution

   \(0.85\times1.10=0.935\Rightarrow -6.5\%\).

2. Item A is 30% more than B. B is 25% less than C. A is what percent of C?
   Solution

   \(A=1.3B=1.3(0.75C)=0.975C\Rightarrow 97.5\%\).

3. Mixture: 12% saline with 5% saline to make 20 L at 8%. How much 12%?
   Solution

   Let \(x\). \(0.12x+0.05(20-x)=0.08\cdot20\Rightarrow x=\tfrac{60}{7}\approx8.57\,\text{L}.\)

4. Population grows 8% yearly for 3 years; equivalent single percent?
   Solution

   \(1.08^3\approx1.2597\Rightarrow +25.97\%\).

5. A recipe ratio flour:sugar = 7:3. If sugar is 240 g, find flour.
   Solution

   Flour \(=\tfrac{7}{3}\cdot240=560\,\text{g}.\)

6. Pump fills pool in 6 h; leak alone drains in 9 h. Together time?
   Solution

   Rate \(=1/6-1/9=1/18\Rightarrow 18\,\text{h}.\)

7. Runner averages 4.5 min/km. What is pace in min/mile? (1 mile = 1.609 km)
   Solution

   \(4.5\times1.609\approx7.24\,\text{min/mile}.\)

8. Machine produces 240 items in 36 min. At 12% faster, how many in an hour?
   Solution

   Base rate \(=240/36=6.667\,\text{/min}\). New rate \(=7.467\Rightarrow 448\) per hour.

9. Car travels 150 km at 60 km/h and returns at 75 km/h. Average speed?
   Solution

   Time \(=2+2=4\,\text{h}\Rightarrow 300/4=75\,\text{km/h}.\)

10. Convert 5.2 L/100 km to mpg.
    Solution

    \(\approx45.2\,\text{mpg}.\)

11. Line of best fit is \(y=3x+7\). Predict at \(x=12\).
    Solution

    \(y=43\).

12. Mean of 12 numbers is 40. If one number 28 is replaced by 44, new mean?
    Solution

    Sum +16 → mean increases by \(16/12\approx1.33\Rightarrow 41.33\).

13. Boxplot has Q1=18, median=24, Q3=31. What is IQR?
    Solution

    \(31-18=13\).

14. Two categories A/B: \(P(A)=0.55, P(B)=0.45, P(A\cap B)=0.20\). Find \(P(A\mid B)\).
    Solution

    \(0.20/0.45\approx0.444\).

15. Dataset mean 70, SD 8. A value 94 is how many SD above mean?
    Solution

    \((94-70)/8=3.0\).

16. Of 300 voters, 180 are urban, 120 rural. 60 urban support policy, 72 rural support. \(P(\text{support}\mid\text{urban})\)?
    Solution

    \(60/180=1/3\).

17. Deck: probability of drawing two hearts without replacement?
    Solution

    \(\tfrac{13}{52}\cdot\tfrac{12}{51}=\tfrac{1}{17}\cdot\tfrac{4}{17}\approx0.0588\).

18. Bag with 5 red, 7 blue. Draw two with replacement: \(P(\text{both red})\).
    Solution

    \((5/12)^2=25/144\).

19. Given \(P(A)=0.6\), \(P(B)=0.5\), independent. \(P(A\cup B)\)?
    Solution

    \(0.6+0.5-0.3=0.8\).

20. Conditional: In a group, 40% are left-handed, and 25% of left-handed play piano. If 18% overall play piano, what percent of right-handed play piano?
    Solution

    Let RH proportion 0.6. Piano overall: \(0.4\cdot0.25 + 0.6\cdot x = 0.18\Rightarrow x=0.20\Rightarrow 20\%\).

Next Steps

Practice with strict timing to mirror real SAT conditions. When a problem includes graphs or charts, always read the axes, units, and scales first, then compute. Interpreting visuals correctly before solving ensures accuracy and confidence under test pressure.. Aim for \(\ge 80\%\) accuracy at high difficulty.