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ESAT Maths 1 Statistics questions

5 original ESAT Maths 1 statistics questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Histograms: frequency = frequency density × class widthUse the mean (a total) and the median (a position)Work with totals, not means

Question 1ESAT Maths 1Warm-up

The mean of 10 numbers is 28.528.5. When one of the numbers is removed, the mean of the remaining 9 numbers is 3030. What number was removed?
  1. A−15-15
  2. B−1.5-1.5
  3. C1414
  4. D1515
  5. E1616
  6. F2525
Show the answer and worked solution

Answer: D · Method: Work with totals, not means

  1. The original total is 10×28.5=28510 \times 28.5 = 285.
  2. The new total is 9×30=2709 \times 30 = 270.
  3. So the removed number is 285−270=15285 - 270 = 15.

Why the tempting wrong answers are wrong

  • A: Work with totals: the removed number is the old total minus the new total.
  • B: That's the change in the mean, not the number removed.

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Question 2ESAT Maths 1Standard

A histogram shows the times tt (in minutes) that some students spent on homework. The frequency densities are:
10≤t<2010 \le t < 20: 1.51.5
20≤t<3020 \le t < 30: 11
30≤t<5030 \le t < 50: 2.52.5
50≤t<8050 \le t < 80: 22
80≤t<10080 \le t < 100: 11
Estimate the number of students who spent between 6565 and 8080 minutes.
  1. A22
  2. B2929
  3. C3030
  4. D3131
  5. E3535
  6. F6060
Show the answer and worked solution

Answer: C · Method: Histograms: frequency = frequency density × class width

  1. Frequency = frequency density × class width.
  2. Add up the parts of each bar between 6565 and 8080 (assuming students are spread evenly within a class): 3030.

Why the tempting wrong answers are wrong

  • A: Frequency density isn't the frequency: multiply by the class width.
  • F: Only part of the 5050–8080 class is included, so take that fraction of its frequency.

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Question 3ESAT Maths 1Standard

Six numbers in increasing order are 3,5,8,x,12,y3, 5, 8, x, 12, y. Their mean is 99 and their median is 99. What is their range?
  1. A1111
  2. B1212
  3. C1313
  4. D1414
  5. E1515
  6. F1616
  7. G1717
  8. H5151
Show the answer and worked solution

Answer: C · Method: Use the mean (a total) and the median (a position)

  1. With six numbers the median is the mean of the 3rd and 4th: 8+x2=9\tfrac{8 + x}{2} = 9, so x=10x = 10.
  2. The total is 6×9=546 \times 9 = 54, so y=54−(3+5+8+10+12)=16y = 54 - (3 + 5 + 8 + 10 + 12) = 16.
  3. Range =16−3=13= 16 - 3 = 13.

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Question 4ESAT Maths 1Standard

Six numbers in increasing order are 4,6,x,14,18,y4, 6, x, 14, 18, y. Their mean is 1212 and their median is 1212. What is their range?
  1. A1414
  2. B1515
  3. C1616
  4. D1717
  5. E1818
  6. F6868
Show the answer and worked solution

Answer: C · Method: Use the mean (a total) and the median (a position)

  1. With six numbers the median is the mean of the 3rd and 4th: x+142=12\tfrac{x + 14}{2} = 12, so x=10x = 10.
  2. The total is 6×12=726 \times 12 = 72, so y=72−(4+6+10+14+18)=20y = 72 - (4 + 6 + 10 + 14 + 18) = 20.
  3. Range =20−4=16= 20 - 4 = 16.

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Question 5ESAT Maths 1Standard

Six numbers in increasing order are 3,4,x,7,9,y3, 4, x, 7, 9, y. Their mean is 77 and their median is 6.56.5. What is y−xy - x?
  1. A55
  2. B66
  3. C77
  4. D88
  5. E99
  6. F1919
Show the answer and worked solution

Answer: C · Method: Use the mean (a total) and the median (a position)

  1. With six numbers the median is the mean of the 3rd and 4th: x+72=6.5\tfrac{x + 7}{2} = 6.5, so x=6x = 6.
  2. The total is 6×7=426 \times 7 = 42, so y=42−(3+4+6+7+9)=13y = 42 - (3 + 4 + 6 + 7 + 9) = 13.
  3. y−x=13−6=7y - x = 13 - 6 = 7.

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