ACSI 2022 Paper 2 — Worked Solutions

Put both in the same unit: 35 km = 35 × 100 000 = 3 500 000 cm. So 5 cm : 3 500 000 cm, and dividing both sides by 5 gives 1 : 700 000.

The map scale is 1 cm to 7 km, so 1 cm² on the map is 7 × 7 = 49 km² of ground. 637 ÷ 49 = 13 cm². (The area scale factor is the square of the length scale factor, 700 000².)

A stem of 3 with leaves 5 7 7 7 8 gives 35, 37, 37, 37, 38 — 37 appears three times, more than any other value. Mode = 37 min

The 22 values in order start 21, 23, 23, 35, 37, 37, 37, 38, 40, 42, 43, 44 … The median is the mean of the 11th and 12th: (43 + 44) ÷ 2 = 43.5 min

Lower half (the first 11 values) has median 37, upper half (the last 11) has median 53, so the interquartile range is 53 − 37 = 16 min. If your class takes the quartiles at the (n + 1)/4 and 3(n + 1)/4 positions instead, they are 37 and 53.25, giving 16.25 min — check which method your teacher marks.

Use the midpoints 10, 30, 50, 75: (37×10) + (62×30) + (53×50) + (48×75) = 370 + 1860 + 2650 + 3600 = 8480. There are 37 + 62 + 53 + 48 = 200 students, so the mean is 8480 ÷ 200 = 42.4 min. It is an estimate because every value in a class is replaced by the midpoint of that class.

Work in the small right-angled triangle DBE first (the right angle is at B): DB² + BE² = DE², so 7² + BE² = 11² → 49 + BE² = 121 → BE = √72 = 8.485 cm.
Then CB = CE + EB = 9 + 8.485 = 17.485 cm.
In the big triangle ABC: AB² = AC² − CB² = 21² − 17.485² = 441 − 305.74 = 135.26 → AB = 11.630 cm.
Finally AD = AB − DB = 11.630 − 7 = 4.63 cm (3 s.f.). Estimated marks: M1 for BE, M1 for CB, M1 for AB, A1 for 4.63.

The two transversals cross at C, so angle ACB = angle ECD (vertically opposite). With AC = CE and BC = CD (both given), triangle ABC is congruent to triangle EDC, and the reason is SAS (two sides and the included angle).

Because AB ∥ FG, the transversal AG makes equal corresponding angles at C, and BF does the same. So triangle ABC is similar to triangle FGC.

The scale factor from ABC to FGC is FC ÷ BC = 9 ÷ 6 = 1.5, so CG = 1.5 × AC. But CG = CE + EG = AC + 4 (using AC = CE). So 1.5AC = AC + 4 → 0.5AC = 4 → AC = 8 cm (and CG = 12 cm).

Both triangles at C share the same angle, so their areas are in the ratio of the products of the sides around that angle. Write the areas as multiples of sin C:
Area ABC = ½ × 8 × 6 × sin C = 24 sin C
Area FGC = ½ × 12 × 9 × sin C = 54 sin C
Area CDE = ½ × 8 × 6 × sin C = 24 sin C
DEGF is FGC with CDE cut out of the corner: 54 sin C − 24 sin C = 30 sin C.
Ratio = area ABCarea DEGF = 24 sin C30 sin C = 4/5 (0.8)

Multiply both sides by (5x − 3): (3x − 2)(5x − 3) = 4 − x. Expand the left: 15x² − 9x − 10x + 6 = 15x² − 19x + 6. So 15x² − 19x + 6 = 4 − x → 15x² − 18x + 2 = 0.
Use the formula: x = [18 ± √(18² − 4×15×2)]/(2×15) = (18 ± √204)/30.
√204 = 14.2829, so x = 32.2829/30 = 1.08 or x = 3.7171/30 = 0.124 (both 3 s.f.). Both are valid — neither makes the original denominator 5x − 3 zero.

For similar solids the volumes are in the ratio (length ratio)³: 0.81 : 1.92 = 810 : 1920 = 27 : 64. So the length ratio is 3 : 4 and the area ratio is 3² : 4² = 9 : 16.
Smaller area = 300 × 9/16 = 169 cm² (168.75, 3 s.f.).

The total number of values is 5 + p + 10 + 9 = 24 + p.
Try p = 14: there are 38 values, so the median is the mean of the 19th and 20th. The five 1s take positions 1–5 and the fourteen 2s take 6–19, so the 19th is a 2 and the 20th is the first 3 — median = (2 + 3)/2 = 2.5. ✗
Try p = 15: there are 39 values and the median is the 20th value. The 1s take 1–5 and the fifteen 2s take 6–20, so the 20th value is a 2 — median = 2. ✓
So the smallest possible value is p = 15.

Sum of the eight numbers = 12 + 8 + 9 + 14 + 8 + 15 + 16 + 6 = 88, so their mean is 88 ÷ 8 = 11.
The mean of the nine is therefore 12, so the nine numbers add to 12 × 9 = 108.
Ninth number = 108 − 88 = 20.

It is a downward parabola (the x² coefficient is negative). Plot the y-intercept (0, 3), the two x-intercepts from (b) and the maximum from (c), then join them with a smooth curve.

Put y = 0: 3x² − 5x − 3 = 0, so x = [5 ± √(25 + 36)]/6 = (5 ± √61)/6. √61 = 7.8102 → x = 2.14 or x = −0.468 (3 s.f.).

The maximum is halfway between the roots, at x = 5/6 = 0.8333. y = 3 + 5(5/6) − 3(5/6)² = 3 + 25/6 − 25/12 = 61/12 = 5.0833. (0.833, 5.08) (or (5/6, 61/12) exactly).

The line of symmetry passes through the maximum: x = 5/6 (0.833).

A straight line of gradient 2 through (0, 2) — it also passes through (−1, 0) and (1, 4).

Solve 3 + 5x − 3x² = 2x + 2 → 3x² − 3x − 1 = 0 → x = [3 ± √(9 + 12)]/6 = (3 ± √21)/6, and √21 = 4.5826. So x = 1.2638 → y = 4.528, and x = −0.2638 → y = 1.472. (1.26, 4.53) and (−0.264, 1.47) (3 s.f.).

One solution means the horizontal line y = k just touches the curve at its maximum, so k = 61/12 = 5.08.

The fraction becomes:x − 2y + 2 = 13Cross-multiplying: 3(x − 2) = y + 2 → 3x − 6 = y + 2 → 3x − y = 8 ✓

From 3x − y = 8, y = 3x − 8. Substitute into 2x − 3y = −11: 2x − 3(3x − 8) = −11 → 2x − 9x + 24 = −11 → −7x = −35 → x = 5. Then y = 3(5) − 8 = 7. x = 5, y = 7 (so the fraction is 5/7 — check: 3/9 = 1/3 ✓).

A full cylinder holds π × 5² × 8 = 628.32 cm³. Removing a 30° slice removes 30/360 = 1/12 of it, so the solid is 11/12 of that: 11/12 × 628.32 = 575.96 = 576 cm³ (3 s.f.) ✓

Three surfaces are left:
• the curved face: 11/12 × 2π(5)(8) = 230.4 cm²
• the top and bottom sectors: 2 × 11/12 × π × 5² = 144.0 cm²
• the two flat cut faces: each is a rectangle 8 cm tall and as wide as the chord of the 30° gap, 2 × 5 × sin 15° = 2.588 cm, so 2 × 8 × 2.588 = 41.4 cm²
Total = 230.4 + 144.0 + 41.4 = 416 cm² (415.8, 3 s.f.)

The volume does not change: 576 cm³. The base is a square of area 121 cm² (side 11 cm). From V = ⅓ × base area × height: 576 = ⅓ × 121 × h → h = 3 × 576 ÷ 121 = 14.3 cm (14.28).

The slant height of a face runs from the apex to the midpoint of a base edge, so it is the hypotenuse of a triangle with sides 14.28 (the height) and 5.5 (half of 11): √(14.28² + 5.5²) = √234.2 = 15.30 cm.
Four triangular faces: 4 × ½ × 11 × 15.30 = 336.7 cm². Base: 121 cm².
Total = 336.7 + 121 = 458 cm² (457.7, 3 s.f.)

Time = capacity ÷ rate, so Tap A takes 600/x minutes.

Tap B's rate is (x − 2) litres per minute, so it takes 600/(x − 2) minutes, which is 10 minutes more than Tap A:
600/(x − 2) = 600/x + 10.
Multiply every term by x(x − 2): 600x = 600(x − 2) + 10x(x − 2) → 600x = 600x − 1200 + 10x² − 20x → 0 = 10x² − 20x − 1200. Dividing by 10 gives x² − 2x − 120 = 0 ✓

Factorise: (x − 12)(x + 10) = 0, so x = 12 or x = −10. A flow rate cannot be negative, so x = 12.

With both taps the rate is x + (x − 2) = 12 + 10 = 22 litres per minute, so the time is 600 ÷ 22 = 27.2727… minutes = 27 minutes and 0.2727 × 60 = 16.36 seconds. 27 minutes 16 seconds.