F.5 CNY assignment 03 Solution

1. (a) minimum lower quartile median upper quartile maximum

City X 14 21 27 33 39

City Y 22 24 27 30 30

(b) Range of X = (39 - 14) °C = 25 °C

Range of Y = (30 - 22) °C = 8 °C

Since range of X > range of Y

The temperatures in city X are more dispersed.

(c) (i) Median = 80.6°F

inter-quartile range = 1.8(30-24)°F = 10.8°F

(ii) mean =[1.8(20) + 32]]°F = 68°F

variance = (1.8)(1.8)(2)°F = 6.48°F

(d) Minimum temperature in city Y = 22°C

Lower-quartile temperature in city X = 21°C

Since 22 > 21

Leo is correct.

2. (a) mean = 56.4%

(b) (i) least = 53% , greatest = 59%

(ii) x = 51% , y = 61.8%

(c) Note that the data are collected from only one magazine stall.

Also note that the week may not be randomly selected.

Thus, the claim is not agreed.

3. (a) In △ ABC

By cosine formula,

BC² = AC² + AB² - 2(AC)(AB)cos∠BAD

(41)² = (50)² + (39)² - 2(50)(39)cos∠BAD

∠BAD = 53.13010235°

AD = 39cos∠BAD = 39cos53.13010235°

AD = 23.4

Thus, the required distance is 23.4 cm.

(b) (i) (1) CD = AC – AD = 26.6

CD

In △ ACD

By cosine formula,

CD² = AD² + AC² - 2(AD)(AC)cos∠DAC

(26.6)² = (23.4)² + AC² - 2(23.4)ACcos50°

AC² - (30.08246013)AC -160= 0

AC = 34.694184

Thus, the required distance is 34.7 cm.

(2) s = 0.5(39 + 41 + 34.694184)

s = 57.347092

Area of △ABC

=√s(s - AB)(s - BC)(s - AC)

=√(57.347092)(57.347092 - 39)(57.347092 - 41)(57.347092 - 34.694184)

= 624.19682241 = 624.197

Thus, the required area is 624.197cm².

(3) Area of △ACD = 0.5(AD)(AC)sin∠DAC = 0.5(23.4)(34.694184)sin50°= 310.95425628

BD = ABsin∠BAD = (39)sin53.13010235° = 31.2

Let H be the projection of D on the horizontal plane.

Then the height of the tetrahedron ABCD is DH.

So, we have

DH(area of △ABC)

= BD(area of △ACD)

3 3

DH(624.19682241) = (31.2)(310.95425628)

DH = 15.54281029

Thus, the required distance is 15.543 cm.

The volume of the tetrahedron ABCD

= (BD)(AD)(CD)sin∠ADC

6

(ii)

So the volume of the tetrahedron varies directly as sin∠ADC.

When ∠ADC increases from 40° to 90°, the volume of the tetrahedron ABCD increases.

The volume of the tetrahedron ABCD attains maximum when ∠ADC = 90∘

When ∠ADC increases from 90° to 130°, the volume of the tetrahedron ABCD decreases.

4. (a) AB² = OA² + OB²

808 = 324 + OB²

OB = 22 or OB = - 22(rej.)

AC² = OA² + OC²

724 = 324 + OC²

OC = 20 or OC = - 20(rej.)

BC² = OB² + OC²

BC² = 484 + 400

BC = √884

s = √808 + √884 + √724

≈ 42.5324

2

area of ΔABC

= √(42.5324)(42.5324 - √808)(42.5324 - √884)(42.5324 - √724)

= 346.415935 cm²

= 346.42 cm²

(b) Volume of OABC

= 18 × 22 × 20

6

= 1320 cm³

Let d cm be the shortest distance from O to the plane ABC.

346.415935 × d

= 1320

3

d = 11.43

Thus, the shortest distance from O to ABC is 11.43 cm.

(c) Note that

sinα = d

sinβ = d

sinγ = d

OA

OB

OC

Since OA < OC < OB

Thus, sinα > sinγ > sinβ

i.e. α > γ > β (since α , β , γ less than 90°)

5. (a) In △OBQ,

OB = (52tan41°)m= 45.2m

(b) ∠QOP = 180° - 53° = 127°

In △POQ,

sin∠OPQ

= sin127°

52 91

∠OPQ = 27.15267596°

∠RQP = 127° + 27.15267596° = 154.2°

The true bearing of P from Q is 154.2°.

(c) ∠OQP = 180° - 154.15267596° = 25.84732404°

In △POQ,

91

= OP

sin127° sin25.84732404°

OP = 49.67683773

In △BOP,

tan∠BPO = 45.20291037

49.67683773

∠BPO = 42.30029403°

The angle of elevation of B from P is 42.3°.

6. (a) In △ABC

AC² = 18² + 18²

AC = 18√2

By symmetry , AH = CH = AC = 18√2

(b) (i) ∵ AC = AH = CH

∴ HAC = 60° (prop. of equil. △)

Area of △ACH = 0.5(18√2)(18√2)(sin60°) = 162√3

(c) Let O be the projection of D on △ACH, then DO is the shortest distance between D and △ACH.

Volume of tetrahedron HADC = 0.5(18)(18)(18) cm³

3

= 972 cm³

0.5(162√3)(DO)

= 972

3

DO = 10.39230485

The required distance is 10.392 cm.

7. (a) Prob. = (0.4)(0.25) = 0.1

(b) Prob. = (0.4)(0.75) + (0.6)(0.5) = 0.6

(c) Prob. = (0.1)(0.3) + (0.6)(0.2) + (0.3)(0.1) = 0.19

(d) Prob. = (0.4)(0.25)(0.3) + (0.6)(0.5)(0.7) = 0.24

(e) Prob. = (0.6)(0.6) + (1 - 0.1 - 0.6)(0.7)

= 57

1 - 0.19 - 0.1 101

8. Area of ΔAFD : Area of ΔAEF = 11 : 6

Area of ΔAFD = 198

Area of ΔDCF : Area of ΔAEF = 11² : 6²

Area of CDF = 363

Area of ΔABC = area of ΔCDF + area of ΔAEF = 363 + 108 = 561

Area of BCEF = area of ΔABC - area of ΔAEF =561 - 108 = 453

9. (a) Capacity of vessel

= 1

π(8)²(15)cm³ = 320πcm³

3

Let h cm be the depth of water.

h³

= 335

15³ 320π

h = 10.3993511

The depth of water is 10.4 cm.

(b) Surface area of vessel

= π(8)(17) cm² = 136π cm²

Let A cm² be the curve surface area covered by water.

(10.399+3)²

= A

15² 136π

A = 340.936738

The surface area covered by water is 341 cm²