Engineering and Science Admissions Test

ESAT Maths 1 Units questions

2 original ESAT Maths 1 units 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: Average speed = total distance ÷ total time

Question 1ESAT Maths 1Standard

A training session repeats this 3-part routine several times:
Walk 400 m400\ \text{m} at an average speed of 24 km h−124\ \text{km h}^{-1}
Jog 100 m100\ \text{m} at an average speed of 4 km h−14\ \text{km h}^{-1}
Run 100 m100\ \text{m} at an average speed of 12 km h−112\ \text{km h}^{-1}
What is the average speed for the whole session, in km h−1\text{km h}^{-1}?
  1. A1010
  2. B1111
  3. C1212
  4. D1313
  5. E403\dfrac{40}{3}
  6. F1414
  7. G1515
  8. H563\dfrac{56}{3}
Show the answer and worked solution

Answer: C · Method: Average speed = total distance ÷ total time

  1. Total distance: 600 m=35 km600\ \text{m} = \dfrac{3}{5}\ \text{km}.
  2. Total time: 0.424+0.14+0.112=120\tfrac{0.4}{24} + \tfrac{0.1}{4} + \tfrac{0.1}{12} = \dfrac{1}{20} hours.
  3. Average speed =35÷120=12 km h−1= \dfrac{3}{5} \div \dfrac{1}{20} = 12\ \text{km h}^{-1}.

Why the tempting wrong answers are wrong

  • E: You can't average the speeds: each part takes a different amount of time.
  • H: Weighting the speeds by distance still isn't right. Find the total time.

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Question 2ESAT Maths 1Standard

A training session repeats this 2-part routine several times:
Walk 300 m300\ \text{m} at an average speed of 12 km h−112\ \text{km h}^{-1}
Jog 600 m600\ \text{m} at an average speed of 6 km h−16\ \text{km h}^{-1}
What is the average speed for the whole session, in km h−1\text{km h}^{-1}?
  1. A6.26.2
  2. B7.27.2
  3. C88
  4. D415\dfrac{41}{5}
  5. E99
  6. F465\dfrac{46}{5}
Show the answer and worked solution

Answer: B · Method: Average speed = total distance ÷ total time

  1. Total distance: 900 m=910 km900\ \text{m} = \dfrac{9}{10}\ \text{km}.
  2. Total time: 0.312+0.66=18\tfrac{0.3}{12} + \tfrac{0.6}{6} = \dfrac{1}{8} hours.
  3. Average speed =910÷18=7.2 km h−1= \dfrac{9}{10} \div \dfrac{1}{8} = 7.2\ \text{km h}^{-1}.

Why the tempting wrong answers are wrong

  • C: Weighting the speeds by distance still isn't right. Find the total time.
  • E: You can't average the speeds: each part takes a different amount of time.

Try it interactively →

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