Solutions to CSEC Maths P2 June 2017

Solutions to CSEC Maths P2 June 2017

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Question 1a part (i)

[41

3− 1

2

5] ÷

4

15= [

13

3−

7

5] ÷

4

15

= [(5×13)−(7×3)

15] ÷

4

15

= [(65)−(21)

15] ÷

4

15

= [44

15] ×

15

4

= 11

Question 1a part (ii)

(3.1−1.15)2

0.005 =

(1.95)2

0.005 = 760.5

Question 1b part (i)

Total cost of phone under Plan A =Deposit +(Monthly Installments × # of months) + Tax

= 400 + (65 × 12) + 0

= $1180

Question 1b

Total cost of phone under Plan B =Deposit +(Monthly Installments × # of months) + Tax

= 600 + (80×6) + 0.05(600+(80×6))

= 600+480+ 54

= $1134

Plan B is the better deal as is has an overall cost (1180-1134 = $46) cheaper than that of

Plan A.

Question 1c part (i)



No. of kWh used = 0 3 3 0 3 0 0 7 1 1 -

0 0 2 9 6

= 296 kWh

1 kWh = $5.10

296 kWh = 5.10 × 296

= $1509.60

Question 1c part (ii)

At $5.10 per kWh, $2351.10 = 2351.10

5.10

= 461 kWh

At the end of April, the meter should read: = 0 3 3 0 0 4 0 7 6 1 +

0 3 7 6 8


6y2 – 18xy = 6y×y – 6y × 3x

= 6y (y-3x)


4m2 – 1 = (2m-1) (2m+1) (Difference of 2 squares)

Question 2a part (iii)

2t2 -3t – 2 = (2t + 1) (t – 2)

Question 2b

5𝑝+2

3−

3𝑝−1

4=

4(5𝑝+2)−3(3𝑝−1)

12



= 20𝑝+8−9𝑝+3

12

=11𝑝+11

12

= 11(𝑝+1)

12


𝑑 = √4ℎ

5 h=29

𝑑 = √4(29)

5= √23.2= 4.82


𝑑 = √4ℎ

5

d2 = 4ℎ

5

5d2 = 4h

h = 5𝑑2

4


U = {3, 4, 5, 6, 7, 8, 9, 10, 11}

M = {3, 5, 7, 9, 11}


R = {4, 9}







Question 3b part (ii)

Length of BQ = 6.9cm




f(x) = 1

3𝑥 − 2

f(3) + f(-3) = [1

33 − 2] + [

1

3(−3) − 2 ]

= (1 – 2) + (-1-2)

= -1 + -3

= -4


f(x) = 5

1

3𝑥 − 2 = 5

1

3𝑥 = 7

x = 21


f(x) = 1

3𝑥 − 2

Let y = f(x)

y = 1

3𝑥 − 2

Making x the subject of the formula

x = 3y + 6

f-1(x) = 3x + 6 = 3( x+2)


Line l1:

Points on l1: (0,1) and (2,5)



Gradient = 5−1

2−0=

4

2= 2

Line l2:

Points on l2: (12,0) and (0,6)

Gradient = 6−0

0−12=

6

−12= −

1

2


y intercept of l1 = 1 (c)

Gradient of l1 = 2 (m)

y = mx + c

Equation: y = 2x + 1

Question 4b part (iii)

Gradient of l1 = 2, Gradient of l2 = −1

2

The lines are the negative inverses of each other. This means they are perpendicular.


<RQT

Angles QRT and QTR are equal, due to it being an isosceles triangle.

So, Angle RQT = 180 – (76×2)



Angle RQT = 28o


<PRT

First, we must find angle PQR.

Angle PQR = 180 – 28

= 152o

Now, we have another isosceles triangle.

Angle PRQ = (180 – 152) ÷2

= 14o

Angle PRT = Angle PRQ +Angle QRT

= 14 + 76

= 90o


Given: Angle SRT = 145o, Angle PSR = 100o

Angle PRS = SRT - PRT

= 145 – 90

= 55o

Angle SPR = 180 – (100 + 55)

= 25o

Now, Angle SPT = 25 + 14

= 39o


A’B’C’ is a 90o clockwise rotation of ABC about the origin, O.




Translating A’B’C through the vector (4

−5)


Radius = 28

2 = 14 𝑚

Area = 90

360×

22

7× 142

Area = 154 m2




Perimeter = (2×radius) + ( 90

360𝜋𝑑)

= 28 + (90

360×

22

7× 28)

= 50 m


Using Pythagoras’ Theorem:

BC2 = AC2 + AB2

AC2 = 102 – 62

AC2 = 100 – 36 = 64

AC = √64 = 8𝑐𝑚

Area = 𝑏ℎ

2=

6×8

2= 24𝑐𝑚2


Volume of Prism = Cross Sectional Area × Length

540 = 24 x Length

Length = 540

24= 22.5𝑐𝑚


Surface Area = (2×Cross Sectional Area) + Areas of Rectangles (ABED + ADFE + BEFC)

2 × Cross Sectional Area = 48cm2

Area of ABED = 6 × 22.5 = 135 cm2

Area of ADFE = 8 × 22.5 = 180 cm2

Area of BEFC = 10 × 22.5 = 225 cm2

Surface Area = 48 + 135 + 180 + 225

Surface Area = 588cm2




The upper-class limit is 39


The class width is (39.5-19.5) = 20


Sixteen vehicles passed a checkpoint at no more than 39.5 kmh-1

Question 7b

Speed (in km-1) Frequency Cumulative Frequency

0-19 5 5

20-39 11 16

40-59 26 42

60-79 37 79

80-99 9 88



100-119 2 90

Question 7c

Question 7d part (i)

50% of vehicles = 90/2 = 45 vehicles



Question 7d part (ii)

The estimated speed is 62 km/h

Question 8a



Question 8b

1 + 2 + 3 + 4 + 5 + 6 = 21

Question 8c

Figure, n Number of Dots, d, in terms of n Number of dots used,

d

1 1

2× 1 × (1 + 1)

1

2 1

2× 2 × (2 + 1)

3

3 1

2× 3 × (3 + 1)

6

11 1

2× 11 × (11 + 1)

66

n

Question 8d



Figure, n Number of Dots, d, in terms of n Number of dots used,

d

1 1

2× 1 × (1 + 1)

1

2 1

2× 2 × (2 + 1)

3

3 1

2× 3 × (3 + 1)

6

11 1

2× 11 × (11 + 1)

66

n 1

2× 𝑛 × (𝑛 + 1)

1

2𝑛(𝑛 + 1)

Question 8e

Let 1

2𝑛(𝑛 + 1) = 1000

n(n+1) = 2000

This solution will not return an integer. Hence, 1000 is not a valid number of dots

in any diagram of this sequence.

Question 9a part (i) part a

O (0,0) A (25,10)



Gradient of OA = 10−0

25−10=

2

5

Question 9a part (i) part b

A (25,10) B (40,10)

Gradient of AB = 10−10

40−25= 0


The cyclist started from rest, where his velocity was 0 m/s, and steadily increased his velocity

by 2

5 m/s each second during the first 25 seconds. During the next 15 seconds, his velocity

remained constant, that is his acceleration was 0 ms-2


Average speed =𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒

Total distance = Area under graph

= 1

2[(40 − 25) + (40 − 0)] × 10

= 1

2[15 + 40] × 10

= 275 m

Average speed = 275

40=

6.875𝑚

𝑠


x2 + 2xy = 5 --- 1

x + y = 3 --------2

Substituting (1,2) in Eqn 1:

(1)2 + 2(1)(2) = 5



1 + 4 = 5

True for equation 1

Substituting (1,2) in Eqn 2:

1 + 2 = 3

True for both equations, hence proven.


x2 + 2xy = 5 --- 1

x + y = 3 --------2

From Eqn 2:

y = 3 – x ---------3

Substituting 3 in 1:

x2 + 2x(3-x) = 5

x2 + 6x - 2x2 = 5

- x2 + 6x = 5

x2 - 6x + 5 = 0

(x-1) (x-5) = 0


<SPQ = 180 – 58 = 122o (Opposite angles of a cyclic quadrilateral are supplementary)


x = 1:

y = 3-1

y = 2

x = 5:

y = 3-5

y = -2

When x = 1, y = 2

When x = 5, y = -2



<SOQ = 2 (SRQ) = 2(58) = 116o

<OQS = 180−116

2= 32o



<ABS = 44o (Alternate angles are equal)

<CBS = 180 – 105 = 75o (2 angles making a straight line are supplementary.)

<ABC = 11+75 = 119o




AC2 = 522 + 722 – 2(52)(72) cos119

= 2704 + 5184 + 3630.25

= 11518.25

AC = √11518.25 = 107.3 𝑘𝑚

Question 10b part (iv)

<BCN = 75o (Co-interior angles are supplementary)

Using the sine rule:

52

𝑠𝑖𝑛𝐴𝐶𝐵=

107.3

𝑠𝑖𝑛119

sinACB = 52𝑠𝑖𝑛119

107.3 = 0.4238

<ACB = sin-1(0.4238) = 25.07o

Bearing of A from C = 360 – (75+25.07) = 259.9o = 260o (To the nearest degree) – Hence

Proven


A×B = (3 2 5 4 )(4 0 3 − 1 )

= (𝑒11 𝑒12 𝑒21 𝑒22 )

e11 = (3×4) +(2×3) = 18

B×A = (4 0 3 − 1 )(3 2 5 4 )

= (𝑒11 𝑒12 𝑒21 𝑒22 )



e11 = (4×3) +(0×5) = 12

The first number in both matrix products do not match, hence A×B ≠ B×A


det A = (3×4) – (2×5) = 12-10 = 2

A-1 = 1

2(4 − 2 − 5 3 ) = (2 − 1

−5

2 3

2 )


A×A-1 = I

I = (1 0 0 1 )


(3 2 5 4 ) (𝑥

𝑦) = (

1

5)


(3 2 5 4 ) (𝑥

𝑦) = (

1

5)

A-1× 𝐴 (𝑥

𝑦) = (

1

5) × A-1

(𝑥

𝑦) = (

1

5) × A-1

(𝑥

𝑦) = (

1

5) (2 − 1

−5

2 3

2 )




OS:

OQ = (5

0)

QS = 3 (5

0) = (

15

0)

OS = OQ +QS

= (5

0) + (

15

0)

= (20

0) x = 20, y = 0

PQ:

PQ = PO + OQ

=- (4

3) + (

5

0)

PQ = (1

−3) x = 1, y = -3

RS:

PR = 3OP

= 3 (4

3)

= (12

9)

OR = OP + PR

= (4

3) + (

12

9)

= (16

12)

RS = RO + OS

= -(16

12)+(

20

0)

= (4

−12) x= 4, y = -12




PQ = (1

−3)

RS = (4

−12)

= 4 (1

−3) = 4PQ

RS is a scalar multiple of PQ, hence PQ and RS are parallel.

RS = 4(1

−3)

|RS| = 4|PQ| - This means that RS is quadruple the length of PQ.

Solutions to CSEC Maths P2 June 2017

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