CSEC MATHEMATICS JANUARY 2019 PAPER 2

[Pages:34]CSEC MATHEMATICS JANUARY 2019 PAPER 2

SECTION I

1. (a) Evaluate

(i) 3.8?102 +1.7?103, giving your answer in standard form

SOLUTION:

Required to evaluate: 3.8?102 +1.7?103 and write the answer in

standard form

Solution: 3.8 ?102 +1.7 ? 103 = 0.38 ?10 ?102 +1.7 ? 103

= 0.38 ?103 +1.7 ? 103 = (0.38 + 1.7) ? 103

= 2.08?103 expressed in standard form or in scientific notation, as required.

1?3 (ii) 2 5 , giving your answer as a fraction in its lowest terms.

31 2

SOLUTION:

Required to evaluate:

1 2

?

3 5

3

1 2

and give the answer as a fraction in its lowest terms

Solution:

Numerator:

13 3 2 ? 5 = 10

Denominator:

1 (3 ? 2) + 1 7

32 =

2

=2

So,

1 2

?

3 5

3

1 2

=

3 10 7 2

=

3 ?

10

2 7

=

3 35

as a fraction in its lowest terms.

(b) Express the number 6 as a binary number.

SOLUTION: Required to express: 6 as a binary number Solution: Binary means that the base is 2.

2 6 2 3 R0 2 1 R1

0 R1

Reading from the bottom upwards, 610 = 1102 . Re-checking,

1102 = (1? 22 ) + (1? 21 ) + (0? 20 )

= (1? 4) + (1? 2) + (0?1)

= 4+2+0 = 610

(c) John bought a car for $65 000. If the value of the car depreciates by 8% each year, how much will the car be worth at the end of 2 years?

SOLUTION: Data: John bought a car for $65 000 and its value depreciates by 8% each year. Required to calculate: The value of the car at the end of 2 years Calculation: Initial cost of the car is $65 000. Depreciation after 1 year is 8% = 8 ?$65000

100 = $5200

Value of the car after 1 year = $65000 - $5200 = $59800

Depreciation after second year = 8% of $59800 = 8 ? $59800 100 = $4 784

Value of the car after the second year = $59800 - $4784 = $55016

Alternative Method:

A

=

P

???1

+

R 100

n

? ??

,

where

P = the original cost of the car,

R = rate of depreciation,

n = the number of years, A = the value of the car.

= 65000(1 - / )2

011

= 65000(0.92)2

= $55016

(d) The table below shows the results obtained by a student in her CSEC Mathematic examination. The maximum mark for each paper is given in the third column of the table.

Paper

01 02 03 Total

Percentage Obtained

55 60 80

Maximum Mark for Paper 30 50 20 100

Determine, as a percentage, the student's final mark for the Mathematics examination.

SOLUTION:

Data: Table showing the results obtained, as percentages, by a student in each of

the three papers in CSEC Mathematics examinations.

Required to determine: The student's final mark as a percentage

Solution:

Actual mark obtained in Paper 01 is

55 ?30 = 16.5

100

Actual mark obtained in Paper 02 is 60 ?50 = 30 100

Actual mark obtained in Paper 03 is

80 ? 20 = 16

100

Total mark obtained in Paper 01, Paper 02, and Paper 03 = 16.5 + 30 +16 = 62.5

Since the mark is out of 100, the percentage mark is the same as the mark since 62.5 ?100 = 62.5% 100

2. (a) (i) Make x the subject of the formula

y = x +3p 5

SOLUTION: Data: y = x + 3p

5 Required to make: x the subject Solution:

y = x +3p 5

x +3p = y 5

x = y-3p 5

(?5) x = 5(y -3p)

(ii) Solve the following equation by factorisation.

2x2 -9x = 0

SOLUTION:

Data: 2x2 - 9x = 0

Required to find: x Solution:

2x2 -9x = 0

2x? x -9? x = 0

x(2x -9) = 0

So x = 0 or

2x -9 = 0 2x = 9 9 x= 2

(b) A farmer wishes to enclose a rectangular plot with a wire fence. The width of the plot is 3 metres less than its length, l.

Given that the area enclosed by the fence is 378 square metres, show that

l2 - 3l - 378 = 0.

SOLUTION: Data: The width of a rectangular plot of land is 3 metres less than its length, l. The area of the plot is 378 square metres.

Required to show: l2 - 3l - 378 = 0

Proof:

Since the width is 3 metres less than the length, then the width is (l - 3)m.

Area = length ? width

= l ?(l - 3) m2

( ) = l2 - 3l m2

The area is 378 m2.

So

l2 - 3l = 378

and l2 - 3l - 378 = 0

Q.E.D

(c) The force, F, applied to an object is directly proportional to the extension, e, produced by that object.

(i) Represent this information as an equation in terms of F, e and an appropriate constant, k.

SOLUTION: Data: Force, F, of an object is directly proportional to the extension, e. Required to write: An equation to represent this information

Solution:

F! orce F

is proportional to "###$###%

?

e"x#te$ns#io%n.

e

F ?e

F = k ? e (where k is the constant of proportionality)

(ii) The incomplete table below shows the corresponding values of F and e.

F

8

25

60

y

e

0.2

x

1.5

3.2

Using the equation obtained in (c)(i), or otherwise, determine the value of x and y.

SOLUTION: Data: Table showing corresponding values of F and e. Required to find: The value of x and of y Solution:

F = 8 when e = 0.2 F = ke So 8 = k ? 0.2 k= 8

0.2 k = 40 Hence, F = 40e (The same result for k would have been obtained if we had substituted F = 60 when k = 1.5 .)

When F = 25, e = x

So 25 = 40x x = 25 40 = 0.625

When F = y, e = 3.2 So y = 40?3.2

y = 128

\ x = 0.625 and y = 128

3. (a) Using a ruler, a pencil and a pair of compasses, construct the right-angled triangle ABC such that AB = 5 cm, ?ABC = 90? and ?BAC = 60? .

SOLUTION: Required to construct: The right-angled triangle ABC with AB = 5 cm, ?ABC = 90? and ?BAC = 60? . Solution:

(b) The diagram below shows a right-angled triangle with sides a units, b units and c units.

(i) Using the diagram, a) Express c in terms of a and b SOLUTION: Data: Diagram showing right-angled triangle with sides a units, b units and c units. Required to express: c in terms of a and b. Solution:

c2 = a2 + b2 (Pythagoras' Theorem) c = a2 + b2 (positive root taken)

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