Revision Libraryv1.6.8
Choose a subject →
Search
WorksheetsBooksFreePracticeHow it worksPricingFreeAboutFAQ
Choose a subject →

Admissions tests / ESAT / Maths 1 / Number, units and measures

Demanding. 15 questions, 15 marks, about 28 minutes.

ESAT Mathematics 1: Number, units and measures, set 3

Standard and compound units, ordering and operating on integers, decimals and fractions, primes and factors, powers and roots, standard form, surds, rounding, bounds and estimation.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  • This is a demanding set: expect multi-step working and less signposting than the real exam.
  1. 11 mark

    Water flows from a tap at a constant rate of 150 ml every 4 seconds. Work out this flow rate in litres per minute.

    1. A 2250 L/min
    2. B 0.0375 L/min
    3. C 2.25 L/min
    4. D 36 L/min
  2. 21 mark

    Four values are defined as P = 3 - 2 x 0.6, Q = 3/5 - 1/10, R = (0.5)^2 + 0.3 and S = 7/20. Which of the following correctly orders all four values from smallest to largest, using < throughout?

    1. A S < Q < R < P
    2. B S < P < Q < R
    3. C Q < S < R < P
    4. D R < S < Q < P
  3. 31 mark

    Work out (-3/4) x (2/9 - 5/6) + 1/3, giving your answer as a fraction in its simplest form.

    1. A -1/8
    2. B 5/24
    3. C 13/12
    4. D 19/24
  4. 41 mark

    A market trader has 84 kg of apples and 126 kg of pears. She wants to pack all of the fruit into boxes so that every box contains the same weight of apples and the same weight of pears, with no fruit left over, using the greatest possible number of boxes. How many boxes can she make?

    1. A 252
    2. B 42
    3. C 14
    4. D 84
  5. 51 mark

    Work out 5/6 divided by 5/9, then multiply the result by 3/10. Give your answer as a fraction in its simplest form.

    1. A 5/36
    2. B 5
    3. C 9/20
    4. D 9/5
  6. 61 mark

    Four candidates, Amy, Ben, Cai and Dan, stand for a two-person committee. Any two of the four may be chosen together. How many of the possible committees include Amy but not Ben?

    1. A 3
    2. B 6
    3. C 1
    4. D 2
  7. 71 mark

    Work out the positive square root of the cube of 25, minus the cube root of the square of 27.

    1. A 116
    2. B 98
    3. C 134
    4. D -2
  8. 81 mark

    Simplify (9^3 x 3^-4) / 3^5, giving your answer as a fraction in its simplest form.

    1. A 27
    2. B 1/27
    3. C 1/729
    4. D 243
  9. 91 mark

    The mass of a single hydrogen atom is 1.67 x 10^-27 kg. A sample contains 2 x 10^26 hydrogen atoms. Work out the total mass of the sample, in kg, giving your answer in standard form correct to 2 significant figures.

    1. A 3.34 x 10^-1
    2. B 3.3 x 10^-53
    3. C 3 x 10^-1
    4. D 3.3 x 10^-1
  10. 101 mark

    Which of the following fractions is equal to a terminating decimal?

    1. A 5/12
    2. B 11/15
    3. C 9/20
    4. D 7/24
  11. 111 mark

    A jar contains 80 sweets. Of these, 5/8 are red. Of the red sweets, 40% are strawberry flavoured. How many sweets are both red and strawberry flavoured?

    1. A 20
    2. B 32
    3. C 25
    4. D 12
  12. 121 mark

    Simplify sqrt(50) + sqrt(18) - sqrt(8), giving your answer in the form k sqrt(2).

    1. A 12 sqrt(2)
    2. B 10 sqrt(2)
    3. C 26 sqrt(2)
    4. D 6 sqrt(2)
  13. 131 mark

    A pack of 5 identical erasers costs 4.35 pounds in total, correct to the nearest 5p. Find the upper bound for the cost of one eraser, giving your answer in pounds.

    1. A 0.865
    2. B 0.875
    3. C 0.88
    4. D 0.87
  14. 141 mark

    A length is recorded as 350 cm, correct to 2 significant figures. Write down the error interval for the true length, x, using inequality notation.

    1. A 349.5 <= x < 350.5
    2. B 340 <= x < 360
    3. C 345 <= x < 355
    4. D 345 < x <= 355
  15. 151 mark

    By rounding each number to 1 significant figure, estimate the value of (0.48 x 612) / 240.

    1. A 1.5
    2. B 1.2
    3. C 1.44
    4. D 1

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. First find the rate per second: 150 ml every 4 seconds is 150 / 4 = 37.5 ml per second.
    2. Scale up to a rate per minute by multiplying by 60 (seconds in a minute): 37.5 x 60 = 2250 ml per minute.
    3. Convert millilitres to litres by dividing by 1000: 2250 / 1000 = 2.25.
    4. So the flow rate is 2.25 L/min, which is answer C.
    • Why not A: Correctly scales the rate up to millilitres per minute (150/4 x 60 = 2250) but then forgets to convert millilitres to litres by dividing by 1000, leaving the answer 1000 times too large.
    • Why not B: Correctly converts to a rate per second in litres (150/4 = 37.5 ml/s = 0.0375 L/s) but then forgets to scale this up to a per-minute rate by multiplying by 60.
    • Why not D: Inverts the rate calculation, multiplying 150 by 4 instead of dividing by it (as if a larger time interval should scale the volume up rather than down), giving 600 ml/s, then correctly scales to 36000 ml/min = 36 L/min.

  2. Question 2Answer: A

    1. Evaluate each value. P = 3 - 2 x 0.6 = 3 - 1.2 = 1.8, applying the multiplication before the subtraction.
    2. Q = 3/5 - 1/10 = 6/10 - 1/10 = 5/10 = 0.5, using a common denominator of 10.
    3. R = (0.5)^2 + 0.3 = 0.25 + 0.3 = 0.55.
    4. S = 7/20 = 0.35. Ordering the four decimals 0.35, 0.5, 0.55, 1.8 from smallest to largest gives S < Q < R < P, so the answer is A.
    • Why not B: Evaluates P by subtracting both numbers from 3 in turn instead of applying the multiplication first, computing 3 - 2 - 0.6 = 0.4 rather than 3 - (2 x 0.6) = 1.8, which places P between S and Q instead of after R.
    • Why not C: Evaluates Q by subtracting the numerators and denominators of the two fractions directly instead of using a common denominator, computing (3-1)/(5-10) = -0.4 rather than 3/5 - 1/10 = 0.5, which makes Q appear smaller than every other value.
    • Why not D: Evaluates R by adding 0.3 to 0.25 with the decimal point misaligned, computing 0.25 + 0.3 as 0.28 instead of 0.55, which places R below S instead of between Q and P.

  3. Question 3Answer: D

    1. Work out the bracket first: 2/9 - 5/6, using a common denominator of 18, gives 4/18 - 15/18 = -11/18.
    2. Multiply: (-3/4) x (-11/18) = 33/72, which simplifies (dividing by 3) to 11/24, since a negative times a negative is positive.
    3. Add 1/3, converting to twenty-fourths: 11/24 + 8/24 = 19/24.
    4. So the answer is 19/24, which is D.
    • Why not A: Correctly finds 2/9 - 5/6 = -11/18, but then treats a negative times a negative as negative, computing (-3/4) x (-11/18) as -11/24 instead of +11/24, giving -11/24 + 1/3 = -11/24 + 8/24 = -3/24 = -1/8.
    • Why not B: Includes the final + 1/3 inside the bracket before multiplying, computing (-3/4) x (2/9 - 5/6 + 1/3) instead of doing the multiplication first and adding 1/3 afterwards: 2/9 - 5/6 + 1/3 = -5/18, and (-3/4) x (-5/18) = 5/24.
    • Why not C: Evaluates 2/9 - 5/6 by subtracting the numerators and denominators separately instead of using a common denominator, getting (2-5)/(9-6) = -1 instead of -11/18, then computes (-3/4) x (-1) + 1/3 = 3/4 + 1/3 = 13/12.

  4. Question 4Answer: B

    1. The greatest number of equal boxes with nothing left over is the highest common factor (HCF) of 84 and 126.
    2. Write each as a product of primes: 84 = 2^2 x 3 x 7 and 126 = 2 x 3^2 x 7.
    3. The HCF takes each shared prime at its LOWER power: 2^1 x 3^1 x 7^1 = 42.
    4. So she can make 42 boxes, each with 84/42 = 2 kg of apples and 126/42 = 3 kg of pears, and the answer is B.
    • Why not A: Finds the lowest common multiple of 84 and 126 instead of the highest common factor, using the HIGHER power of each shared prime where the LOWER power is needed: 2^2 x 3^2 x 7 = 252.
    • Why not C: Identifies 2 and 7 as shared prime factors but overlooks the shared factor of 3 (84 = 2^2 x 3 x 7 and 126 = 2 x 3^2 x 7 both contain a factor of 3), giving 2 x 7 = 14 instead of 2 x 3 x 7.
    • Why not D: Assumes the smaller of the two weights is automatically the greatest possible number of boxes, without checking that it actually divides the larger weight exactly (126 / 84 = 1.5, which is not a whole number, so 84 boxes is not achievable).

  5. Question 5Answer: C

    1. To divide by a fraction, multiply by its reciprocal: 5/6 / (5/9) = 5/6 x 9/5.
    2. Cancel common factors before multiplying: the 5s cancel, leaving 9/6, which simplifies to 3/2.
    3. Multiply by 3/10, working left to right: 3/2 x 3/10 = 9/20.
    4. So the answer is 9/20, which is C.
    • Why not A: Multiplies the two fractions directly instead of using the reciprocal for the division: 5/6 x 5/9 = 25/54 (rather than 5/6 x 9/5), then multiplies by 3/10 to get 75/540, which simplifies to 5/36.
    • Why not B: Performs the operations in the wrong order, multiplying 5/9 by 3/10 first (getting 1/6) and then dividing 5/6 by that result: 5/6 divided by 1/6 = 5, instead of working left to right.
    • Why not D: Correctly divides 5/6 by 5/9 to get 3/2, but then adds 3/10 instead of multiplying by it: 3/2 + 3/10 = 15/10 + 3/10 = 18/10 = 9/5.

  6. Question 6Answer: D

    1. List all six possible committees systematically: Amy-Ben, Amy-Cai, Amy-Dan, Ben-Cai, Ben-Dan, Cai-Dan.
    2. A committee 'includes Amy but not Ben' if it contains Amy and does not also contain Ben.
    3. Checking the list, this rules out Amy-Ben (contains Ben) and leaves Amy-Cai and Amy-Dan.
    4. That is 2 committees, so the answer is D.
    • Why not A: Lists every committee that includes Amy (Amy-Ben, Amy-Cai, Amy-Dan) without applying the 'but not Ben' condition, counting all three instead of excluding the one that also contains Ben.
    • Why not B: Lists all six possible committees from the four candidates (Amy-Ben, Amy-Cai, Amy-Dan, Ben-Cai, Ben-Dan, Cai-Dan) without applying either condition at all.
    • Why not C: Misreads the condition as excluding BOTH Amy and Ben, counting only the committees with neither of them (just Cai-Dan), instead of committees that include Amy specifically.

  7. Question 7Answer: A

    1. The square root of the cube of 25 is 25^(3/2), which can be found as (sqrt(25))^3 = 5^3 = 125.
    2. The cube root of the square of 27 is 27^(2/3), which can be found as (cbrt(27))^2 = 3^2 = 9.
    3. Subtract: 125 - 9 = 116.
    4. So the answer is 116, which is A.
    • Why not B: Correctly finds the square root of the cube of 25 as 125, but on the second term takes the square root of the square of 27 instead of the cube root, treating sqrt(27^2) = 27 as the value to subtract: 125 - 27 = 98.
    • Why not C: Adds the two values instead of subtracting: 125 + 9 = 134.
    • Why not D: Swaps which root belongs with which term, taking the cube root of the cube of 25 (which is simply 25) and the square root of the square of 27 (which is simply 27), giving 25 - 27 = -2.

  8. Question 8Answer: B

    1. Rewrite 9 as 3^2, so 9^3 = (3^2)^3 = 3^6.
    2. Multiply in the numerator: 3^6 x 3^-4 = 3^(6-4) = 3^2.
    3. Divide by 3^5: 3^2 / 3^5 = 3^(2-5) = 3^-3.
    4. 3^-3 = 1/(3^3) = 1/27, so the answer is B.
    • Why not A: Subtracts the exponents the wrong way round in the final division, computing 3^(5-2) = 3^3 = 27 instead of 3^(2-5) = 3^-3.
    • Why not C: Converts 9^3 to base 3 by replacing the base without scaling the exponent, writing 9^3 as 3^3 instead of 3^6 (forgetting that 9 = 3^2 means the exponent must be multiplied by 2), giving 3^3 x 3^-4 = 3^-1, then dividing by 3^5 to get 3^-6 = 1/729.
    • Why not D: Drops the negative sign on 3^-4 and treats it as 3^4, giving 3^6 x 3^4 = 3^10, then dividing by 3^5 to get 3^5 = 243.

  9. Question 9Answer: D

    1. Multiply the coefficients: 1.67 x 2 = 3.34.
    2. Add the powers of ten, since the values are being multiplied: 10^-27 x 10^26 = 10^(-27+26) = 10^-1.
    3. This gives 3.34 x 10^-1. Rounding the coefficient to 2 significant figures gives 3.3.
    4. So the answer, in standard form to 2 significant figures, is 3.3 x 10^-1, which is D.
    • Why not A: Correctly multiplies to get 3.34 x 10^-1 but does not apply the instruction to round to 2 significant figures, leaving the unrounded coefficient.
    • Why not B: Subtracts the powers of ten instead of adding them when multiplying, computing the exponent as -27 - 26 = -53 rather than adding -27 + 26 = -1.
    • Why not C: Rounds the coefficient 3.34 to 1 significant figure instead of 2, giving 3 rather than 3.3.

  10. Question 10Answer: C

    1. A fraction in its simplest form gives a terminating decimal exactly when the only prime factors of its denominator are 2 and/or 5.
    2. 12 = 2^2 x 3, 15 = 3 x 5 and 24 = 2^3 x 3 each have a prime factor of 3, so 5/12, 11/15 and 7/24 all give recurring decimals.
    3. 20 = 2^2 x 5 has only 2 and 5 as prime factors, so 9/20 terminates: 9/20 = 0.45.
    4. So the answer is 9/20, which is C.
    • Why not A: 12 = 2^2 x 3. A candidate who notices 12 has a factor of 4 (a power of 2) may assume this is enough for the decimal to terminate, overlooking the leftover factor of 3, which forces the decimal to recur: 5/12 = 0.41666...
    • Why not B: 15 = 3 x 5. A candidate who notices 15 is a multiple of 5 may assume this guarantees a terminating decimal, overlooking the co-factor of 3: 11/15 = 0.7333...
    • Why not D: 24 = 2^3 x 3. A candidate who notices 24 is built mostly from factors of 2 may assume the single leftover factor of 3 does not matter, but any prime factor other than 2 or 5 forces the decimal to recur: 7/24 = 0.291666...

  11. Question 11Answer: A

    1. Find the number of red sweets first: 5/8 of 80 = 50.
    2. Find 40% of the red sweets, since the strawberry proportion applies only within that subset: 40% of 50 = (40/100) x 50 = 20.
    3. So 20 sweets are both red and strawberry flavoured.
    4. The answer is 20, which is A.
    • Why not B: Applies the 40% to the whole jar of 80 sweets instead of just to the red sweets, ignoring that the strawberry proportion only applies within the red subset: 40% of 80 = 32.
    • Why not C: Correctly finds the 50 red sweets, but then takes half of them instead of 40%, treating 40% loosely as roughly a half: 50% of 50 = 25.
    • Why not D: Uses the fraction of sweets that are NOT red (3/8) instead of the fraction that are red (5/8) before applying the percentage: 3/8 of 80 = 30, then 40% of 30 = 12.

  12. Question 12Answer: D

    1. Write each surd in terms of sqrt(2): sqrt(50) = sqrt(25 x 2) = 5 sqrt(2), sqrt(18) = sqrt(9 x 2) = 3 sqrt(2), and sqrt(8) = sqrt(4 x 2) = 2 sqrt(2).
    2. Substitute these into the expression: 5 sqrt(2) + 3 sqrt(2) - 2 sqrt(2).
    3. Collect the like surds by adding and subtracting their coefficients: 5 + 3 - 2 = 6.
    4. So the answer is 6 sqrt(2), which is D.
    • Why not A: Simplifies sqrt(18) as 9 sqrt(2) instead of 3 sqrt(2), pulling the whole factor of 9 out from under the root instead of its square root: 5 sqrt(2) + 9 sqrt(2) - 2 sqrt(2) = 12 sqrt(2).
    • Why not B: Treats the final subtraction as an addition, combining all three terms with plus signs instead of subtracting the last one: 5 sqrt(2) + 3 sqrt(2) + 2 sqrt(2) = 10 sqrt(2).
    • Why not C: Simplifies sqrt(50) as 25 sqrt(2) instead of 5 sqrt(2), pulling the whole factor of 25 out from under the root instead of its square root: 25 sqrt(2) + 3 sqrt(2) - 2 sqrt(2) = 26 sqrt(2).

  13. Question 13Answer: B

    1. Rounding to the nearest 5p means the true total could lie up to half of 5p, that is 2.5p or 0.025 pounds, either side of 4.35.
    2. The upper bound of the total cost is therefore 4.35 + 0.025 = 4.375 pounds.
    3. To maximise the cost of one eraser, divide this upper bound by the 5 erasers: 4.375 / 5 = 0.875.
    4. So the upper bound for the cost of one eraser is 0.875 pounds, which is B.
    • Why not A: Uses the lower bound of the total cost instead of the upper bound: (4.35 - 0.025) / 5 = 4.325 / 5 = 0.865.
    • Why not C: Treats 'correct to the nearest 5p' as if it meant 'correct to the nearest 10p', using a half-unit of 5p instead of 2.5p: upper bound total = 4.35 + 0.05 = 4.40, then 4.40 / 5 = 0.88.
    • Why not D: Does not find any bound at all, and simply divides the given total directly by 5: 4.35 / 5 = 0.87.

  14. Question 14Answer: C

    1. 350 to 2 significant figures has its last significant digit in the tens place (the 5), so the rounding unit is 10.
    2. The true value can lie up to half of that rounding unit, 5, either side of 350.
    3. The lower bound, 345, is itself included (it would round up to 350), but the upper bound, 355, is excluded (it would round to 360, not 350).
    4. So the error interval is 345 <= x < 355, which is C.
    • Why not A: Treats the precision as though 350 were given to 3 significant figures (or to the nearest whole unit) instead of 2, using a rounding unit of 1 and so a half-unit of 0.5, rather than the rounding unit of 10 that 2 significant figures on 350 actually implies.
    • Why not B: Uses double the correct half-unit, treating the rounding unit as 20 rather than 10, so bounds 10 either side of 350 instead of 5.
    • Why not D: Keeps the correct bounds of 345 and 355 but swaps which end of the interval is included, excluding the lower bound and including the upper bound instead of the other way round.

  15. Question 15Answer: A

    1. Round each number to 1 significant figure: 612 rounds to 600, 0.48 rounds to 0.5 (since the next digit, 8, rounds it up), and 240 rounds to 200.
    2. Substitute these rounded values: (0.5 x 600) / 200.
    3. 0.5 x 600 = 300, and 300 / 200 = 1.5.
    4. So the estimate is 1.5, which is A.
    • Why not B: Rounds 0.48 down to 0.4 to 1 significant figure instead of up to 0.5, truncating rather than rounding to the nearest value: (600 x 0.4) / 200 = 1.2.
    • Why not C: Forgets to round 0.48 at all, using it exactly while correctly rounding the other two numbers: (600 x 0.48) / 200 = 1.44.
    • Why not D: Rounds 240 up to 300 instead of down to 200, misjudging that 240 is closer to 200 than to 300 to 1 significant figure: (600 x 0.5) / 300 = 1.

More on this strand

More free ESAT practice

Every strand of the published ESAT specification, with worked solutions throughout.

Choose a subject →