Em 2 Solutions
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COMMONWEALTH SECONDARY SCHOOL PRELIMINARY EXAMINATION 2009 SECONDARY FOUR EXPRESS/FIVE NORMAL MATHEMATICS 4016/02 1
(i)
Amt. of money paid = $1.870 × 42 = $78.54 50 1.870 ≈ 26.73797 litres Dist. = 26.73797 × 12.5 ≈ 334 km
(ii)
Amt. of petrol=
Let the max. volume be V litres. 1.687V − 5 < 0.85 × 1.767V 1.687V − 5 < 1.50195V 0.18505V < 5 V < 27.0197 Max. volume is 27.0 litres.
(iii)
(i)
Elegant handbags : Convenient handbags = 7500 : 7260 = 125 : 121 0.15 × 7500 + 50 × 100% 7500 + 7260 ≈ 7.96%
% discount =
(ii)
2
(a) ●
[B1]
[M1] [A1] [M1]
[A1] [A1]
[B1] [M1] [A1] [B1]
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(b)
●
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p = 33 q = 46
[B1] [B1]
2 (c)
2 = n + 2 ( n − 1) = n 2 + 2n − 2
no. of dots A =1 B=2 C = −2
3
[B1] [B1] [B1]
2 = 10 + 2 ( 10 ) − 2 = 118
(d)
no. of dots
(e)
n 2 + 2n − 2 = 253 n 2 + 2n − 255 = 0 ( n − 15 ) ( n + 17 ) = 0 n = 15 or n = −17 (N.A) Diagram 15 has 253 dots. (other methods are acceptable)
(a)
[B1]
a 2 − 6ab + 9b2 5a2 − 45b2 ÷ 6ac − 3ad 2ac − ad + 6bc − 3bd 2 ( a − 3b ) × a ( 2c − d ) + 3b ( 2c − d ) = 3a ( 2c − d ) 5 ( a + 3 b ) ( a − 3b )
(i)
[M1]
[A1]
[M1]
( a − 3b ) × ( 2c − d ) ( a + 3b ) = 3a ( 2c − d ) 5 ( a + 3b ) ( a − 3b ) 2
=
[M1]
a − 3b 15a
[A1]
3 5 −4= 2 ( x − 1) 3 ( x − 1)
(ii)
9 − 24 ( x − 1) = 10
[M1]
24 ( x − 1) = −1 x −1= − x=
(b)
4
1 24
23 24
[A1]
(i)
P(choosing a green disc) = 1 − p
(ii)
(a)
P(red disc chosen each time)
(b)
P(at least one green disc chosen) = 1 – P(red disc chosen each time) = 1 − p8
[B1] = p8
[B1]
[B1]
By Sine Rule, [M1]
CSS/Prelim2009/MATH/SEC4E5N/P2/AGS/Page 2 of 6
3 sin ∠CBE sin35° = 7.3 4.6 ∠CBE = 180° − 65.538659° ≈ 114.46134° ≈ 114.5°
[A1]
CA 4.6 CA ≈ 5.6805269 ≈ 5.68 cm
tan51° =
[B1]
By Cosine Rule, AD 2 = 5.68052692 + 42 − 2 ( 5.6805269 ) (4)cos103°
[M2]
AD ≈ 7.647948 ≈ 7.65 cm
[A1]
∠BCE = 180° − 35° − 114.46134° (∠s sum of ∆ ) = 30.53866° 1 Area of ∆BCE = ( 4.6 ) ( 7.3 ) sin30.53866° 2 ≈ 8.5313285 ≈ 8.53 cm2
[A1]
1 × 4.6 × shortest dist. = 8.5313285 2 Shortest dist. ≈ 3.70927 ≈ 3.71 cm 5
(a)
(b) (c)
[M1] [A1]
90 × 600 100 = 540 km/h 3500 Time taken (with wind) = h 540 + x
Speed of plane
[M1]
=
3500 Time taken (against wind) = h 540 − x 3500 3500 10 − =1 540 − x 540 + x 60 3000 3000 − =1 540 − x 540 + x 3000 ( 540 + x ) − 3000 ( 540 − x ) = ( 540 + x ) ( 540 − x ) 1620000 + 3000 x − 1620000 + 3000 x = 291600 − x 2 x 2 + 6000 x − 291600 = 0 (Shown)
[B1]
[B1] [M1]
[M1] [A1]
CSS/Prelim2009/MATH/SEC4E5N/P2/AGS/Page 3 of 6
4
(d)
x 2 + 6000 x − 291600 = 0 x=
−6000 ± 60002 − 4 ( 1) ( −291600 )
2 ( 1) ≈ 48.21259 or −6048.212 (N.A.) ≈ 48.2
(e)
6
(a)
Time taken
(i)
(ii)
(iii)
(b)
7
(a)
[A1] [A1] [M1]
3500 600 + 48.21259 ≈ 5.399 h ≈ 5 h 24 min
=
[A1]
∠TOP = 36° × 2 (∠ at ctr. = 2 × ∠ at circumference) = 72° Reflex ∠POT = 360° − 72° (∠s at a pt.) = 288°
[M1]
∠TSP = 180° − 36° (opp. ∠s of cyclic quad.) = 144° 180° − 144° ∠PTS = (base ∠ of isos. ∆ ) 2 = 18°
[M1]
∠PTR = 90° (∠ in a semicircle) ∠QTS = 180° − 90° − 18° (∠s on a st. line) = 72° ∠SQT = 180° − 2 ( 72° ) (∠s sum of isos. ∆ ) = 36° ∠QPR = 90° (tangent ⊥ rad.) ∠PQR = 180° − 90° − 36° (∠s sum of ∆ ) = 54° ∠PQS = 54° − 36° = 18°
∠SPT = 18° (isos. ∆ ) 180° − 72° ∠TPO = (base ∠ of isos. ∆ ) 2 = 54° ∠QPS = 90° − 54° − 18° = 18° ∠ Since QPS = ∠SPT = 18° , PS bisects ∠QPT . (i)
[M1]
62 × 2π ( 1.8 ) 360 ≈ 1.947787
dist. moved =
[A1]
[A1]
[M1] [M1] [A1] [M1] [M1] [A1] [M1] [A1]
CSS/Prelim2009/MATH/SEC4E5N/P2/AGS/Page 4 of 6
5 ≈ 1.95 m (ii)
(b)
(i)
(ii) 8
(i)
No. of oscillations in 1 min = 2.5 Dist. travelled in 1 min = 2.5 × 1.95 × 2 ≈ 9.7389 ≈ 9.74 m Speed is 9.74 m/min. Let the angle of depression be a° . 1.8 tan a° = 4 a ≈ 24.2277 ≈ 24.2 Angle of depression is 24.2° . dist.2 = 42 + 1.82 (by Pythagoras' Thm.) dist. ≈ 4.39 m
Vol. of hemisphere =
1 4 3 × ( π ) ( 7) 2 3 ≈ 718.3775201 ≈ 718 cm3
(ii) Let the base radius of the cone be r cm . 1 2 ( π ) ( r ) ( 10 ) × 3 = 718.3775201 3 Base radius of cone is 4.78 cm. (iii)
(iv)
r ≈ 4.781910357 ≈ 4.78
(b)
[A1]
[M1] [A1] [M1] [A1] [M1] [A1]
[M1] [A1]
1 Mass = 0.9 × × 718.3775201 3 ≈ 215.5132 ≈ 216 g Let the slanted height of the cone be l cm . l 2 = 102 + 4.78192 (by Pythagoras' Thm.) l ≈ 11.0845 Total external surface area 2 2 = 3π ( 7 ) + π ( 4.7819 ) ( 11.0845 ) + π ( 4.7819 ) ≈ 700.171587 ≈ 700 cm2
1 2 ( 6 ) θ = 98 2 4 θ = 5 rad. 9
[M1]
[M1] [A1]
[M1] [M2] [A1]
[M1]
CSS/Prelim2009/MATH/SEC4E5N/P2/AGS/Page 5 of 6
6
9
4 Perimeter of major sector = 6 5 + 2 ( 6 ) 9 2 = 44 cm 3
[A1]
(a)
Median daily wage ≈ $43
[B1]
(b)
Interquartile range ≈ $49 – $36 = $13
[M1] [A1]
(c)
77.5% of the male employees
22.5 × 80 100 = 18 =
z ≈ 35
[M1] [A1]
(d)
Median daily wage ≈ $38 Interquartile range ≈ $46 – $18 = $28
[B1] [M1] [A1]
(e)
The median wage of males ($43) is higher than that of females ($38). The interquartile range for wage of males (13) is smaller than that of females (28) and thus, the wage of males is less widespread as compared to that of females.
[B1]
(f)
26 + 30 140 2 = 5
P(wage less than or equal to $38) =
[B1]
[M1] [A1]
CSS/Prelim2009/MATH/SEC4E5N/P2/AGS/Page 6 of 6
(i)
Amt. of money paid = $1.870 × 42 = $78.54 50 1.870 ≈ 26.73797 litres Dist. = 26.73797 × 12.5 ≈ 334 km
(ii)
Amt. of petrol=
Let the max. volume be V litres. 1.687V − 5 < 0.85 × 1.767V 1.687V − 5 < 1.50195V 0.18505V < 5 V < 27.0197 Max. volume is 27.0 litres.
(iii)
(i)
Elegant handbags : Convenient handbags = 7500 : 7260 = 125 : 121 0.15 × 7500 + 50 × 100% 7500 + 7260 ≈ 7.96%
% discount =
(ii)
2
(a) ●
[B1]
[M1] [A1] [M1]
[A1] [A1]
[B1] [M1] [A1] [B1]
●
●
●
●
●
●
●
●
●
●
●
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●
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● ● ● ●
● ●
●
(b)
●
●
p = 33 q = 46
[B1] [B1]
2 (c)
2 = n + 2 ( n − 1) = n 2 + 2n − 2
no. of dots A =1 B=2 C = −2
3
[B1] [B1] [B1]
2 = 10 + 2 ( 10 ) − 2 = 118
(d)
no. of dots
(e)
n 2 + 2n − 2 = 253 n 2 + 2n − 255 = 0 ( n − 15 ) ( n + 17 ) = 0 n = 15 or n = −17 (N.A) Diagram 15 has 253 dots. (other methods are acceptable)
(a)
[B1]
a 2 − 6ab + 9b2 5a2 − 45b2 ÷ 6ac − 3ad 2ac − ad + 6bc − 3bd 2 ( a − 3b ) × a ( 2c − d ) + 3b ( 2c − d ) = 3a ( 2c − d ) 5 ( a + 3 b ) ( a − 3b )
(i)
[M1]
[A1]
[M1]
( a − 3b ) × ( 2c − d ) ( a + 3b ) = 3a ( 2c − d ) 5 ( a + 3b ) ( a − 3b ) 2
=
[M1]
a − 3b 15a
[A1]
3 5 −4= 2 ( x − 1) 3 ( x − 1)
(ii)
9 − 24 ( x − 1) = 10
[M1]
24 ( x − 1) = −1 x −1= − x=
(b)
4
1 24
23 24
[A1]
(i)
P(choosing a green disc) = 1 − p
(ii)
(a)
P(red disc chosen each time)
(b)
P(at least one green disc chosen) = 1 – P(red disc chosen each time) = 1 − p8
[B1] = p8
[B1]
[B1]
By Sine Rule, [M1]
CSS/Prelim2009/MATH/SEC4E5N/P2/AGS/Page 2 of 6
3 sin ∠CBE sin35° = 7.3 4.6 ∠CBE = 180° − 65.538659° ≈ 114.46134° ≈ 114.5°
[A1]
CA 4.6 CA ≈ 5.6805269 ≈ 5.68 cm
tan51° =
[B1]
By Cosine Rule, AD 2 = 5.68052692 + 42 − 2 ( 5.6805269 ) (4)cos103°
[M2]
AD ≈ 7.647948 ≈ 7.65 cm
[A1]
∠BCE = 180° − 35° − 114.46134° (∠s sum of ∆ ) = 30.53866° 1 Area of ∆BCE = ( 4.6 ) ( 7.3 ) sin30.53866° 2 ≈ 8.5313285 ≈ 8.53 cm2
[A1]
1 × 4.6 × shortest dist. = 8.5313285 2 Shortest dist. ≈ 3.70927 ≈ 3.71 cm 5
(a)
(b) (c)
[M1] [A1]
90 × 600 100 = 540 km/h 3500 Time taken (with wind) = h 540 + x
Speed of plane
[M1]
=
3500 Time taken (against wind) = h 540 − x 3500 3500 10 − =1 540 − x 540 + x 60 3000 3000 − =1 540 − x 540 + x 3000 ( 540 + x ) − 3000 ( 540 − x ) = ( 540 + x ) ( 540 − x ) 1620000 + 3000 x − 1620000 + 3000 x = 291600 − x 2 x 2 + 6000 x − 291600 = 0 (Shown)
[B1]
[B1] [M1]
[M1] [A1]
CSS/Prelim2009/MATH/SEC4E5N/P2/AGS/Page 3 of 6
4
(d)
x 2 + 6000 x − 291600 = 0 x=
−6000 ± 60002 − 4 ( 1) ( −291600 )
2 ( 1) ≈ 48.21259 or −6048.212 (N.A.) ≈ 48.2
(e)
6
(a)
Time taken
(i)
(ii)
(iii)
(b)
7
(a)
[A1] [A1] [M1]
3500 600 + 48.21259 ≈ 5.399 h ≈ 5 h 24 min
=
[A1]
∠TOP = 36° × 2 (∠ at ctr. = 2 × ∠ at circumference) = 72° Reflex ∠POT = 360° − 72° (∠s at a pt.) = 288°
[M1]
∠TSP = 180° − 36° (opp. ∠s of cyclic quad.) = 144° 180° − 144° ∠PTS = (base ∠ of isos. ∆ ) 2 = 18°
[M1]
∠PTR = 90° (∠ in a semicircle) ∠QTS = 180° − 90° − 18° (∠s on a st. line) = 72° ∠SQT = 180° − 2 ( 72° ) (∠s sum of isos. ∆ ) = 36° ∠QPR = 90° (tangent ⊥ rad.) ∠PQR = 180° − 90° − 36° (∠s sum of ∆ ) = 54° ∠PQS = 54° − 36° = 18°
∠SPT = 18° (isos. ∆ ) 180° − 72° ∠TPO = (base ∠ of isos. ∆ ) 2 = 54° ∠QPS = 90° − 54° − 18° = 18° ∠ Since QPS = ∠SPT = 18° , PS bisects ∠QPT . (i)
[M1]
62 × 2π ( 1.8 ) 360 ≈ 1.947787
dist. moved =
[A1]
[A1]
[M1] [M1] [A1] [M1] [M1] [A1] [M1] [A1]
CSS/Prelim2009/MATH/SEC4E5N/P2/AGS/Page 4 of 6
5 ≈ 1.95 m (ii)
(b)
(i)
(ii) 8
(i)
No. of oscillations in 1 min = 2.5 Dist. travelled in 1 min = 2.5 × 1.95 × 2 ≈ 9.7389 ≈ 9.74 m Speed is 9.74 m/min. Let the angle of depression be a° . 1.8 tan a° = 4 a ≈ 24.2277 ≈ 24.2 Angle of depression is 24.2° . dist.2 = 42 + 1.82 (by Pythagoras' Thm.) dist. ≈ 4.39 m
Vol. of hemisphere =
1 4 3 × ( π ) ( 7) 2 3 ≈ 718.3775201 ≈ 718 cm3
(ii) Let the base radius of the cone be r cm . 1 2 ( π ) ( r ) ( 10 ) × 3 = 718.3775201 3 Base radius of cone is 4.78 cm. (iii)
(iv)
r ≈ 4.781910357 ≈ 4.78
(b)
[A1]
[M1] [A1] [M1] [A1] [M1] [A1]
[M1] [A1]
1 Mass = 0.9 × × 718.3775201 3 ≈ 215.5132 ≈ 216 g Let the slanted height of the cone be l cm . l 2 = 102 + 4.78192 (by Pythagoras' Thm.) l ≈ 11.0845 Total external surface area 2 2 = 3π ( 7 ) + π ( 4.7819 ) ( 11.0845 ) + π ( 4.7819 ) ≈ 700.171587 ≈ 700 cm2
1 2 ( 6 ) θ = 98 2 4 θ = 5 rad. 9
[M1]
[M1] [A1]
[M1] [M2] [A1]
[M1]
CSS/Prelim2009/MATH/SEC4E5N/P2/AGS/Page 5 of 6
6
9
4 Perimeter of major sector = 6 5 + 2 ( 6 ) 9 2 = 44 cm 3
[A1]
(a)
Median daily wage ≈ $43
[B1]
(b)
Interquartile range ≈ $49 – $36 = $13
[M1] [A1]
(c)
77.5% of the male employees
22.5 × 80 100 = 18 =
z ≈ 35
[M1] [A1]
(d)
Median daily wage ≈ $38 Interquartile range ≈ $46 – $18 = $28
[B1] [M1] [A1]
(e)
The median wage of males ($43) is higher than that of females ($38). The interquartile range for wage of males (13) is smaller than that of females (28) and thus, the wage of males is less widespread as compared to that of females.
[B1]
(f)
26 + 30 140 2 = 5
P(wage less than or equal to $38) =
[B1]
[M1] [A1]
CSS/Prelim2009/MATH/SEC4E5N/P2/AGS/Page 6 of 6
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