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Data Relationships practice questions — SAT Math

7 free multiple-choice problems on data relationships, ordered to match SAT Math difficulty, each with a full worked solution. No account needed — and there's a fresh problem set every day.

Problems & worked solutions

Problem #0941 US SAT

Problem 1 — Data Relationships

A class has 2020 students; 1212 are girls. The fraction of boys in the class is:

Show answer & worked solution
  1. A. 35\dfrac{3}{5}
  2. B. 14\dfrac{1}{4}
  3. C. 25\dfrac{2}{5}✓ correct
  4. D. 12\dfrac{1}{2}

Boys =20−12=8= 20 - 12 = 8.

Fraction of boys =820=25= \dfrac{8}{20} = \dfrac{2}{5}.

Problem #3557 International

Problem 2 — Data Relationships

A sixth form of 120120 students is surveyed. The two-way table below records whether a student takes Further Maths and whether a student plays a musical instrument; one entry is shown as xx.

InstrumentNo instrumentTotalFurther Mathsx3048No Further Maths244872Total4278120\begin{array}{c|cc|c} & \text{Instrument} & \text{No instrument} & \text{Total} \\ \hline \text{Further Maths} & x & 30 & 48 \\ \text{No Further Maths} & 24 & 48 & 72 \\ \hline \text{Total} & 42 & 78 & 120 \end{array}

One of the students who take Further Maths is chosen at random. The probability that this student plays a musical instrument is:

Show answer & worked solution
  1. A. 38\dfrac{3}{8}✓ correct
  2. B. 320\dfrac{3}{20}
  3. C. 37\dfrac{3}{7}
  4. D. 58\dfrac{5}{8}
  5. E. 720\dfrac{7}{20}

The Further Maths row must add up to its own total, so x=48−30=18x = 48 - 30 = 18. The instrument column confirms this reading: 18+24=4218 + 24 = 42.

Choosing at random from the students who take Further Maths restricts attention to that row alone, so the denominator is the row total 4848, not the grand total 120120. Of those 4848 students, 1818 play an instrument.

1848=38\dfrac{18}{48} = \dfrac{3}{8}

Problem #3556 International

Problem 3 — Data Relationships

A scatter diagram plots the number of cold drinks yy sold at a kiosk against the daily maximum temperature x ∘Cx\,^{\circ}\mathrm{C}, for days whose maximum temperature lay between 3 ∘C3\,^{\circ}\mathrm{C} and 18 ∘C18\,^{\circ}\mathrm{C}. A line of best fit is drawn on the diagram and passes through the plotted points (4, 17)(4,\,17) and (16, 41)(16,\,41). The number of drinks this line predicts for a day with maximum temperature 13 ∘C13\,^{\circ}\mathrm{C} is:

Show answer & worked solution
  1. A. 3535✓ correct
  2. B. 2626
  3. C. 2929
  4. D. 3333
  5. E. 4343

Both given points lie on the line, so they fix its gradient: m=41−1716−4=2412=2.m = \frac{41-17}{16-4} = \frac{24}{12} = 2. Use the point (4, 17)(4,\,17) to write the equation of the line: y−17=2(x−4),y - 17 = 2(x-4), y=2x−8+17=2x+9.y = 2x - 8 + 17 = 2x + 9. The temperature 13 ∘C13\,^{\circ}\mathrm{C} lies inside the range 3 ∘C3\,^{\circ}\mathrm{C} to 18 ∘C18\,^{\circ}\mathrm{C} covered by the data, so reading the line at x=13x=13 is interpolation and the estimate is a legitimate one. Substituting x=13x=13: y=2(13)+9=26+9=35.y = 2(13) + 9 = 26 + 9 = 35.

Problem #0944 US SAT

Problem 4 — Data Relationships

A two-way table: ABTotalX121830Y81220Total203050\begin{array}{c|cc|c} & \text{A} & \text{B} & \text{Total} \\ \hline \text{X} & 12 & 18 & 30 \\ \text{Y} & 8 & 12 & 20 \\ \hline \text{Total} & 20 & 30 & 50 \end{array} What fraction of all respondents are in row Y?

Show answer & worked solution
  1. A. 15\dfrac{1}{5}
  2. B. 35\dfrac{3}{5}
  3. C. 25\dfrac{2}{5}✓ correct
  4. D. 12\dfrac{1}{2}

Row Y total =20= 20. Grand total =50= 50.

Fraction =2050=25= \dfrac{20}{50} = \dfrac{2}{5}.

Problem #3555 International

Problem 5 — Data Relationships

A scatter diagram is drawn from seven readings, where xx is the number of weeks a seedling has been fed and yy is its height in centimetres: (1, 4), (2, 7), (3, 10), (4, 5), (5, 16), (6, 19), (7, 22)(1,\,4),\ (2,\,7),\ (3,\,10),\ (4,\,5),\ (5,\,16),\ (6,\,19),\ (7,\,22) The point that does not follow the linear trend of the others is:

Show answer & worked solution
  1. A. (4, 5)(4,\,5)✓ correct
  2. B. (5, 16)(5,\,16)
  3. C. (3, 10)(3,\,10)
  4. D. (7, 22)(7,\,22)
  5. E. (1, 4)(1,\,4)

Read off the pattern in the readings that agree with one another: yy rises by 33 for each step of 11 in xx, and at x=1x=1 the height is 44, so the trend line has gradient 33 and passes through (1, 4)(1,\,4): y=3x+1.y = 3x + 1. Compare the height predicted by this line with the height actually recorded: x12345673x+1471013161922y47105161922\begin{array}{c|ccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 3x+1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\ y & 4 & 7 & 10 & 5 & 16 & 19 & 22 \end{array} Six of the seven readings sit exactly on the line, so their vertical gaps are 00. At x=4x=4 the line predicts 3(4)+1=133(4)+1=13 while the recorded height is 55, a vertical gap of 13−5=8.13 - 5 = 8. That gap is far larger than any other, so the reading that breaks the trend is (4, 5).(4,\,5).

Problem #0943 US SAT

Problem 6 — Data Relationships

A scatter plot shows hours studied vs. exam score, with a clear upward trend. The correlation is:

Show answer & worked solution
  1. A. positive✓ correct
  2. B. negative
  3. C. zero
  4. D. impossible to tell without the data

An upward (rising-left-to-right) trend means higher xx is associated with higher yy — a positive correlation.

Problem #0018 International

Problem 7 — Data Relationships

A linear regression of test score yy on hours studied xx gives y^=50+8x\hat{y} = 50 + 8x. By how much does the predicted score increase when xx increases by 0.50.5?

Show answer & worked solution
  1. A. 0.50.5
  2. B. 44✓ correct
  3. C. 88
  4. D. 5454

The slope is 88 score points per additional hour. For a 0.50.5-hour increase, the predicted change is 0.5×8=40.5 \times 8 = 4.

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