Paper 1: checked solutions

(a) 6a - 5(2a - 3) = 6a - 10a + 15 = 15 - 4a.

(b) cd - c + d - 1 = c(d - 1) + (d - 1) = (c + 1)(d - 1).

(a) 120 = 2³ × 3 × 5.

(b)(i) For A = pq²r⁴ and B = p²qr³, take the smaller power of each prime:
HCF = pqr³.

(ii) pqr³ = 120 = 5 × 3 × 2³. Since p > q > r, p = 5, q = 3 and r = 2.
A = 5 × 3² × 2⁴ = 720.

(iii) Take the larger prime powers: LCM = p²q²r⁴ = 2⁴ × 3² × 5².

Swipe across the diagram if needed. Its method is explained in the text below.

(a) OA = OC because both are radii. Thus ∠OCA = ∠OAC = 16° (base angles of an isosceles triangle).
∠AOC = 180° - 16° - 16° = 148° (angles in a triangle).
∠BOC = 148° - 44° = 104°.

(b) ∠BAC = 104° ÷ 2 = 52°, because the angle at the centre is twice the angle at the circumference subtended by the same arc BC.

(5x - 3y + 16x + 17y) / (5x - 3y + x + 7y)
= (21x + 14y) / (6x + 4y)
= 7(3x + 2y) / [2(3x + 2y)] = 7/2.

Since x and y are positive, 3x + 2y is non-zero. The simplified value contains neither variable, so it is the same for all allowed x and y.

There will be eight numbers, so the median is the mean of the 4th and 5th numbers. To make it 9, add another 9 and one number at or below 9. The maximum stays 12, so the required minimum is 12 - 7 = 5.
The integers are 5 and 9.

Check: 5, 8, 8, 9, 9, 10, 11, 12 has median (9 + 9)/2 = 9 and range 12 - 5 = 7.

(a)(i) Total time = x/18 + x/12 = 5x/36 hours.

(ii) Average speed = total distance / total time
= 2x / (5x/36) = 72/5 = 14.4 km/h.

(b) 12 km/h = (12 × 1000)/3600 = 10/3 m/s (= 3.33 m/s to 3 s.f.).

The acute reference angle is sin^-1(0.9867) = 80.644953…°.
x = 180° - 80.644953…° = 99.355047…° = 99.4°.

(a) Theatre only = x - 4; both = x; cinema only = x + 10; neither = 12.
(x - 4) + x + (x + 10) + 12 = 72
3x + 18 = 72, so x = 18.

(b) (T ∩ C) ∪ (T ∪ C)′ means “both or neither”.
Number = 18 + 12 = 30.

(c) Select only from the theatre visitors: n(T) = 14 + 18 = 32.
Required probability = 18/32 = 9/16.

(a) 0.0000123 = 1.23 × 10^-5.

(b) 5B/A = (5 × 4.8 × 10ⁿ⁺¹) / (1.5 × 10ⁿ)
= 16 × 10 = 160 = 1.6 × 10².

(a) The class midpoints are 15, 22.5, 27.5, 35 and 55.
Estimated mean = [1(15) + 2(22.5) + 8(27.5) + 7(35) + 7(55)] / 25
= 910/25 = 36.4 minutes.

Brian : Amal = 3 : 4 = 12 : 16.
Brian : Priya = 4 : 7 = 12 : 21.
Hence Brian : Amal : Priya = 12 : 16 : 21.
21 - 16 = 5 parts represent $150, so one part is $30.
Brian receives 12 × $30 = $360.

Swipe across the diagram if needed. Its method is explained in the text below.

Comment 1: The 20-year-olds generally react faster, as their median reaction time is about 384 ms, compared with about 412 ms for the 50-year-olds.

Comment 2: The 20-year-olds’ reaction times are more consistent, as their interquartile range is 412 - 340 = 72 ms, smaller than 444 - 360 = 84 ms for the 50-year-olds.

Use group comparisons, not claims about every individual. The figures are readings from the plotted scale; the original scan was enlarged to check them.

21 < 5 - 4x ≤ 47
16 < -4x ≤ 42
Dividing by -4 reverses both inequality signs: -10.5 ≤ x < -4.
Integers: -10, -9, -8, -7, -6, -5.

(a) H = (5 - 8, -14 + 20) = (-3, 6).

(b) Gradient of GH = (6 - (-14))/(-3 - 5) = 20/(-8) = -5/2.
Using G(5, -14): y + 14 = -(5/2)(x - 5).
Therefore y = -(5/2)x - 3/2.

(c) FG = √[(5 - (-3))² + (-14 - (-4))²]
= √(8² + (-10)²) = √164 = 2√41 = 12.8 (3 s.f.).

Check: both G(5, -14) and H(-3, 6) satisfy the line equation.

2x + 6y = -13 … (1)
3x - 8y = 6 … (2)
3 × (1): 6x + 18y = -39
2 × (2): 6x - 16y = 12
Subtract: 34y = -51, so y = -3/2.
Substitute into (1): 2x - 9 = -13, so x = -2.

Check: 2(-2) + 6(-3/2) = -13 and 3(-2) - 8(-3/2) = 6.

3c/(2c - 3)² + 4/(2c - 3)
= [3c + 4(2c - 3)] / (2c - 3)²
= (11c - 12) / (2c - 3)².

The original expression requires c ≠ 3/2.

The S$141.56 includes a 2.75% fee, so it is 102.75% of the converted amount.
Converted amount before the fee = 141.56/1.0275 = S$137.7712895…
Exchange rate in UK pounds per Singapore dollar
= 86/137.7712895… = 0.6242229443…
S$1 = £0.624 (3 s.f.).

Swipe across the diagram if needed. Its method is explained in the text below.

Let T be the point of tangency, with outer radius R and inner radius r. OT is perpendicular to AB, and the perpendicular from the centre bisects the chord, so AT = 19/2 = 9.5 cm.
By Pythagoras, R² = r² + 9.5².
Shaded area = π(R² - r²) = 90.25π = 284 cm² (3 s.f.).

(a) P(blue or yellow) = 1 - P(pink) = 1 - 0.37 = 0.63.

(b) There are 10 red and 10 non-red counters. Exactly two reds can occur as RRN, RNR or NRR, without replacement.
P(exactly two red) = 3 × (10/20) × (9/19) × (10/18) = 15/38.

Swipe across the diagram if needed. Its method is explained in the text below.

(a) Reverse bearing = 304° - 180° = 124°.

(b) ∠BLP = 90° - 56° = 34°.
tan 34° = 18/PL, so PL = 18/tan 34° = 26.686097… = 26.7 km.

(c) ∠BPC = 360° - 331° = 29°.
cos 29° = 18/PC, so PC = 18/cos 29° = 20.580373… = 20.6 km.

Use the difference of two squares in the numerator:
(3x - 2)² - (x - 3)²
= [(3x - 2) - (x - 3)][(3x - 2) + (x - 3)]
= (2x + 1)(4x - 5).

Factorise the denominator:
4x² - 8x - 5 = (2x + 1)(2x - 5).

Therefore [(3x - 2)² - (x - 3)²] / (4x² - 8x - 5)
= [(2x + 1)(4x - 5)] / [(2x + 1)(2x - 5)]
= (4x - 5) / (2x - 5).

The original expression is undefined at x = -1/2 and x = 5/2; these restrictions remain after cancellation.

(2x + 5)/3 - (2x² - 1)/(3x - 1) = 1, where x ≠ 1/3.
Multiply throughout by 3(3x - 1):
(2x + 5)(3x - 1) - 3(2x² - 1) = 3(3x - 1).

6x² + 13x - 5 - 6x² + 3 = 9x - 3
13x - 2 = 9x - 3
4x = -1, so x = -1/4.

(27x⁶y^-3/z⁹)^-4/3
= 27^-4/3 × x^6(-4/3) × y^(-3)(-4/3) × z^(-9)(-4/3)
= 3^-4 × x^-8 × y⁴ × z¹²
= (1/81)x^-8y⁴z¹².

Thus a = 1/81, p = -8, q = 4 and r = 12. The question asks for the form ax^py^qz^r, so leave the negative power of x in that form.

Swipe across the diagram if needed. Its method is explained in the text below.

(a) AB is parallel to DC because both posts are vertical.
∠BAE = ∠CDE (alternate angles, with AD as transversal).
∠AEB = ∠DEC (vertically opposite angles).
Therefore △ABE is similar to △DCE by AA.

(b) From this similarity, BE/CE = AB/DC = 21/15 = 7/5.
Hence BE/BC = 7/(7 + 5) = 7/12.
Since EF is parallel to CD, △BFE is similar to △BDC.
EF/CD = BE/BC = 7/12, so EF = 15 × 7/12 = 8.75 m.

Swipe across the diagram if needed. Its method is explained in the text below.

(a) Let the sector radius be R. Arc length = (216/360) × 2πR = 18π.
R = 18 × 360 / (216 × 2) = 15 cm.

(b) The sector radius becomes the cone’s slant height, l = 15 cm.
The major arc becomes the base circumference: 2πr = 18π, so r = 9 cm.
The perpendicular height is h = √(15² - 9²) = √144 = 12 cm.
Volume = (1/3)πr²h = (1/3)π(9²)(12) = 324π cm³
= 1017.876… cm³ = 1.02 × 10³ cm³ (3 s.f.).

Geometry check: sector area = (216/360)π(15²) = 135π cm², matching the cone’s curved surface area πrl = π(9)(15).

Multiply every number by 2, then add 15.

Multiplication by 2 changes the mean from 30 to 60 and the standard deviation from 10 to 20. Adding 15 changes the mean from 60 to 75 and leaves the standard deviation unchanged.
In symbols, replace each value x with 2x + 15.

f ∝ √T, so f = k√T for a constant k.
A 100% increase in frequency means the new frequency is 2f.
Since T is proportional to f², the new tension is 2²T = 4T.
Percentage increase = [(4T - T)/T] × 100% = 300%.

The coefficient of x² is 1 and the minimum point is (3, -11).
Therefore the vertex form is y = (x - 3)² - 11.
Expanding: y = x² - 6x + 9 - 11 = x² - 6x - 2.
Comparing with y = x² + bx + c gives b = -6 and c = -2.

Check: (x - 3)² is non-negative, so its least value is 0 at x = 3. The minimum value of y is therefore -11, as given.