Hard GCSE Ordering and comparing numbers Questions

Challenging, exam-style GCSE Ordering and comparing numbers questions with worked solutions. Stretch yourself on the hardest estimating surds, inequality chains, comparing to integers, comparing surds and decimals problems.

estimating surdsinequality chainscomparing to integerscomparing surds and decimalsorderingerror intervals
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
For positive a,b,c,da,b,c,d with ab<cd\frac{a}{b}<\frac{c}{d}, prove that ab<a+cb+d<cd\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}, and verify it by ordering the three fractions when ab=13\frac{a}{b}=\frac{1}{3} and cd=12\frac{c}{d}=\frac{1}{2}.
Show worked solution

Worked solution

  1. Write the starting fact

    ab<cd, a,b,c,d>0\frac{a}{b}<\frac{c}{d},\ a,b,c,d>0

    All four numbers are positive and the first fraction is smaller than the second.

  2. Turn it into a product statement

    ad<bcad < bc

    Cross-multiplying (safe because b,d>0b,d>0) turns ab<cd\frac{a}{b}<\frac{c}{d} into ad<bcad<bc.

  3. Introduce the middle fraction

    a+cb+d\frac{a+c}{b+d}

    This is the 'mediant' — add the tops and add the bottoms. We must show it lies between the two.

  4. Aim for the left inequality

    ab<a+cb+d?\frac{a}{b}<\frac{a+c}{b+d}?

    First prove the mediant is bigger than ab\frac{a}{b}.

  5. Cross-multiply the left

    a(b+d) □ b(a+c)a(b+d)\ \square\ b(a+c)

    Compare the two by cross-multiplying (denominators positive).

  6. Expand both sides

    ab+ad □ ab+bcab+ad \ \square\ ab+bc

    Multiply out the brackets.

  7. Cancel the common term

    ad □ bcad \ \square\ bc

    Take abab off both sides, leaving adad against bcbc.

  8. Use the known fact

    ad<bcad < bc

    We already know ad<bcad<bc, so the left side is smaller.

  9. Conclude the left inequality

    ab<a+cb+d\frac{a}{b}<\frac{a+c}{b+d}

    Therefore the mediant is greater than ab\frac{a}{b}.

  10. Aim for the right inequality

    a+cb+d<cd?\frac{a+c}{b+d}<\frac{c}{d}?

    Now prove the mediant is smaller than cd\frac{c}{d}.

  11. Cross-multiply the right

    (a+c)d □ c(b+d)(a+c)d\ \square\ c(b+d)

    Again compare by cross-multiplying.

  12. Expand and cancel

    ad+cd □ bc+cd⇒ad □ bcad+cd \ \square\ bc+cd \Rightarrow ad \ \square\ bc

    Multiplying out and cancelling cdcd again leaves adad against bcbc.

  13. Conclude the right inequality

    ad<bc⇒a+cb+d<cdad<bc \Rightarrow \frac{a+c}{b+d}<\frac{c}{d}

    Since ad<bcad<bc, the mediant is less than cd\frac{c}{d}.

  14. Combine the two halves

    ab<a+cb+d<cd\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}

    Both parts proved, so the mediant lies strictly between the two fractions.

  15. Verify with numbers

    1+13+2=25: 13<25<12\frac{1+1}{3+2}=\frac{2}{5}:\ \frac{1}{3}<\frac{2}{5}<\frac{1}{2}

    With 13\frac{1}{3} and 12\frac{1}{2} the mediant is 25=0.4\frac{2}{5}=0.4, and 0.333<0.4<0.50.333<0.4<0.5, confirming the result.

Answer
ab<a+cb+d<cd; e.g. 13<25<12\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d};\ \text{e.g. } \frac{1}{3}<\frac{2}{5}<\frac{1}{2}
Question 2
5 markschallenging
Six athletes finished a race in distinct times. Clues: nobody tied (≠\ne throughout); Runner 3 beat Runner 5; Runner 1 was slower than Runner 6 but faster than Runner 2; Runner 4 was fastest; Runner 5 was slowest; Runner 6 beat Runner 3; Runner 2 beat Runner 3. Reconstruct the unique finishing order.
Show worked solution

Worked solution

  1. Meaning of 'beat'

    beat=faster=finishes earlier\text{beat} = \text{faster} = \text{finishes earlier}

    In a race a faster time means finishing ahead, so 'beat' means 'ahead of'.

  2. Place the fastest

    R4=1stR4 = 1^{st}

    We are told Runner 4 was fastest, so R4 takes first place.

  3. Place the slowest

    R5=6thR5 = 6^{th}

    Runner 5 was slowest, so R5 is last.

  4. Read the Runner 1 clue

    R6 faster than R1R6 \text{ faster than } R1

    'R1 was slower than R6' means R6 finished ahead of R1.

  5. Continue the Runner 1 clue

    R1 faster than R2R1 \text{ faster than } R2

    'R1 was faster than R2' means R1 finished ahead of R2.

  6. Chain those two

    R6→R1→R2R6 \to R1 \to R2

    So in finishing order R6 comes before R1, which comes before R2.

  7. Read the Runner 6 clue

    R6 beat R3R6 \text{ beat } R3

    R6 also finished ahead of R3.

  8. Compare R6 with the middle group

    R6<R1, R6<R3R6 < R1,\ R6 < R3

    R6 is ahead of R1, R2 and R3, and only R4 is faster overall.

  9. Fix R6 in second

    R6=2ndR6 = 2^{nd}

    After the fastest R4, the next fastest must be R6.

  10. Read the Runner 3 clue

    R3 beat R5R3 \text{ beat } R5

    R3 finished ahead of the last-placed R5, so R3 is not last.

  11. List the remaining places

    3rd,4th,5th={R1,R2,R3}3^{rd},4^{th},5^{th}=\{R1,R2,R3\}

    R1, R2 and R3 must fill positions 3, 4 and 5.

  12. Keep R1 before R2

    R1→R2R1 \to R2

    From the earlier chain, R1 still finishes ahead of R2.

  13. Position Runner 3

    R2<R3⇒R3 after R2R2 < R3 \Rightarrow R3 \text{ after } R2

    The clue 'Runner 2 beat Runner 3' places R3 after R2. With R6 fastest of the middle group and R1 before R2, R3 must take 5th, just ahead of last-placed R5.

  14. Fill positions 3-5

    3rd=R1, 4th=R2, 5th=R33^{rd}=R1,\ 4^{th}=R2,\ 5^{th}=R3

    This keeps R1 before R2 and R3 ahead of R5.

  15. State the finishing order

    R4, R6, R1, R2, R3, R5R4,\ R6,\ R1,\ R2,\ R3,\ R5

    So the order from first to last is R4, R6, R1, R2, R3, R5.

Answer
R4, R6, R1, R2, R3, R5 (fastest to slowest)R4,\ R6,\ R1,\ R2,\ R3,\ R5\ (\text{fastest to slowest})
Question 3
5 markschallenging
Write these famous approximations to π\pi in ascending order and state which is the best: 227, 355113, π, 3.1416\frac{22}{7},\ \frac{355}{113},\ \pi,\ 3.1416.
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Worked solution

  1. Write pi accurately

    π=3.1415926…\pi = 3.1415926\ldots

    Use several decimal places of π\pi so we can compare closely.

  2. Convert 227\frac{22}{7}

    227=3.1428571…\frac{22}{7}=3.1428571\ldots

    Divide 22 by 7.

  3. Convert 355113\frac{355}{113}

    355113=3.1415929…\frac{355}{113}=3.1415929\ldots

    Divide 355 by 113 — a famously accurate approximation.

  4. Note the last value

    3.14163.1416

    The fourth number is already a decimal.

  5. Write all to 7 places

    3.1415927, 3.1415929, 3.1416000, 3.14285713.1415927,\ 3.1415929,\ 3.1416000,\ 3.1428571

    Line everything up to 7 decimal places for a fair comparison.

  6. Compare π\pi with 355113\frac{355}{113}

    3.1415927<3.14159293.1415927 < 3.1415929

    π\pi is a whisker smaller than 355113\frac{355}{113} (differ in the 7th place).

  7. Compare 355113\frac{355}{113} with 3.14163.1416

    3.1415929<3.14160003.1415929 < 3.1416000

    355113\frac{355}{113} is smaller than 3.14163.1416.

  8. Place 227\frac{22}{7}

    3.1428571 is the largest3.1428571 \text{ is the largest}

    227\frac{22}{7} is clearly the biggest of the four.

  9. Write the ascending order

    π, 355113, 3.1416, 227\pi,\ \frac{355}{113},\ 3.1416,\ \frac{22}{7}

    Smallest to largest.

  10. Set up closeness

    distance=∣value−π∣\text{distance}=|\text{value}-\pi|

    The best approximation is the one nearest to π\pi.

  11. Distance of 355113\frac{355}{113}

    ≈0.0000003\approx0.0000003

    355113\frac{355}{113} is only about 3 ten-millionths away from π\pi.

  12. Distance of 3.1416

    ≈0.0000073\approx0.0000073

    3.14163.1416 is about 7 millionths away, further than 355113\frac{355}{113}.

  13. Distance of 227\frac{22}{7}

    ≈0.00126\approx0.00126

    227\frac{22}{7} is much further, about a thousandth away.

  14. Compare the distances

    0.0000003<0.0000073<0.001260.0000003 < 0.0000073 < 0.00126

    355113\frac{355}{113} has by far the smallest distance.

  15. State the best

    355113 is best\frac{355}{113}\ \text{is best}

    So 355113\frac{355}{113} is the closest, most accurate approximation to π\pi.

Answer
π, 355113, 3.1416, 227; best is 355113\pi,\ \frac{355}{113},\ 3.1416,\ \frac{22}{7};\ \text{best is } \frac{355}{113}
Question 4
5 markschallenging
Order the numbers −3, 0.5, 4, 1.5, 3-3,\ 0.5,\ 4,\ 1.5,\ 3 by their distance from 22 (that is, by ∣x−2∣|x-2|), listing the closest first, and explain how this ordering differs from ordering the numbers themselves.
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Worked solution

  1. Define the measure

    distance=∣x−2∣\text{distance}=|x-2|

    The distance from 2 is how far a number is from 2, ignoring direction (the modulus).

  2. Distance of -3

    ∣−3−2∣=∣−5∣=5|-3-2|=|-5|=5

    Subtract 2 then drop the sign.

  3. Distance of 0.5

    ∣0.5−2∣=∣−1.5∣=1.5|0.5-2|=|-1.5|=1.5

    The gap from 0.5 to 2 is 1.51.5.

  4. Distance of 4

    ∣4−2∣=2|4-2|=2

    4 is 2 units from 2.

  5. Distance of 1.5

    ∣1.5−2∣=∣−0.5∣=0.5|1.5-2|=|-0.5|=0.5

    1.5 is only 0.50.5 from 2.

  6. Distance of 3

    ∣3−2∣=1|3-2|=1

    3 is 1 unit from 2.

  7. Collect the distances

    5, 1.5, 2, 0.5, 15,\ 1.5,\ 2,\ 0.5,\ 1

    These belong to −3,0.5,4,1.5,3-3, 0.5, 4, 1.5, 3 in that order.

  8. Order the distances

    0.5<1<1.5<2<50.5 < 1 < 1.5 < 2 < 5

    Sort the gaps from smallest to largest.

  9. Match smallest gap

    0.5→1.50.5 \to 1.5

    The closest number to 2 is 1.51.5.

  10. Match next gaps

    1→3, 1.5→0.5, 2→41\to3,\ 1.5\to0.5,\ 2\to4

    Then 3, then 0.5, then 4.

  11. Match largest gap

    5→−35 \to -3

    The furthest from 2 is −3-3.

  12. Write the closeness order

    1.5, 3, 0.5, 4, −31.5,\ 3,\ 0.5,\ 4,\ -3

    This lists the numbers closest to 2 first.

  13. Now order the numbers themselves

    −3, 0.5, 1.5, 3, 4-3,\ 0.5,\ 1.5,\ 3,\ 4

    By size alone, this is the normal ascending order.

  14. Compare the two orders

    different!\text{different!}

    The two lists are not the same because distance ignores whether a number is above or below 2.

  15. Explain the difference

    1.5 and 3 are both near 21.5 \text{ and } 3 \text{ are both near } 2

    Numbers on opposite sides of 2 (like 1.5 and 3) can be close in distance, while −3-3 is far below and comes last by distance but first by value.

Answer
1.5, 3, 0.5, 4, −3 (closest to 2 first)1.5,\ 3,\ 0.5,\ 4,\ -3\ (\text{closest to } 2 \text{ first})
Question 5
5 markschallenging
Determine all values of xx for which x<x2x<x^{2} holds. Present your solution using inequality notation and illustrate it on a number line.
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Worked solution

  1. Write the inequality

    x<x2x < x^2

    We want every xx whose square is bigger than itself.

  2. Move everything to one side

    0<x2−x0 < x^2 - x

    Subtract xx from both sides so we can factorise.

  3. Factorise

    x2−x=x(x−1)x^2 - x = x(x-1)

    Take out the common factor xx.

  4. Rewrite the inequality

    x(x−1)>0x(x-1) > 0

    We need the product of xx and (x−1)(x-1) to be positive.

  5. When is a product positive

    both+ or both−\text{both} + \text{ or both} -

    A product of two factors is positive when both factors have the same sign.

  6. Case A: both positive

    x>0 and x−1>0x>0 \text{ and } x-1>0

    Both factors positive means x>0x>0 and x>1x>1.

  7. Simplify Case A

    x>1x>1

    The stricter condition is x>1x>1.

  8. Case B: both negative

    x<0 and x−1<0x<0 \text{ and } x-1<0

    Both factors negative means x<0x<0 and x<1x<1.

  9. Simplify Case B

    x<0x<0

    The stricter condition is x<0x<0.

  10. Combine the cases

    x<0 or x>1x<0 \text{ or } x>1

    The inequality holds in either of these two regions.

  11. Test x=2x = 2

    2<4 ✓2 < 4\ \checkmark

    x=2x=2 is in x>1x>1 and works.

  12. Test x=−1x = -1

    −1<1 ✓-1 < 1\ \checkmark

    x=−1x=-1 is in x<0x<0 and works.

  13. Test a value in between

    x=0.5: 0.5<0.25? Nox=0.5:\ 0.5 < 0.25?\ \text{No}

    Between 0 and 1 it fails, confirming that region is excluded.

  14. Check the endpoints

    x=0: 0<0 false; x=1: 1<1 falsex=0:\ 0<0\ \text{false};\ x=1:\ 1<1\ \text{false}

    At x=0x=0 and x=1x=1 the two sides are equal, so these are not included (open circles).

  15. State the solution

    x<0 or x>1x<0 \ \text{or}\ x>1

    On a number line: open circles at 0 and 1, shading to the left of 0 and to the right of 1.

Answer
x<0 or x>1x < 0 \ \text{or}\ x > 1

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