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TMUA 2020 · Paper 2 · Question 17 of 20

TMUA 2020 Paper 2 Question 17

Number and divisibility — Mean and median · minimising a range. Try it first; the answer and a full worked solution are below.

TMUA 2020 · Paper 2Number and divisibilityMean and median · minimising a range6 options
A set of six distinct integers is split into two sets of three.

The first set of three integers has a mean of 10 and a median of 8.

The second set of three integers has a mean of 12 and a median of 9.

What is the smallest possible range of the set of all six integers?

  1. A8
  2. B10
  3. C11
  4. D12
  5. E14
  6. F15
Show the answer and worked solution
answer · E
  1. A8
  2. B10
  3. C11
  4. D12
  5. E14
  6. F15
Write the first set as {p,  8,  q} with p< 8 <q and p+q= 22, and the second as {r,  9,  s} with r< 9 <s and r+s= 27. To make the range small you want the two small values as large as possible and the two large values as small as possible, so push p and r up. The obstruction is distinctness: 8 is already used, so r 8, forcing r 7 and s 20; and if r= 7 then p 6, so q 16. That gives {6,  8,  16} and {7,  9,  20}, all six distinct, with range 20  6 = 14. Anything smaller is impossible: s 19 always, and the minimum of the six is at most 7 with equality only if r 7 and p= 7, which distinctness forbids.