QUESTION 1 - St Stithians Collegemaths.stithian.com/Maths Lit/Prelims/Brescia/12 Paper 2... · Web viewGRADE 12: MATHEMATICAL LITERACY: JULY PAPER 2 2016Page 19 of 19 GRADE 12 EXAMINATION

GRADE 12 EXAMINATION

JULY 2016

MEMOMATHEMATICAL LITERACY: PAPER 2

Time: 3 hours 150 marks

EXAMINER: Mrs J Hultzer MODERATOR(S): Mrs E Buytenhuys

QUESTION 1

1.1. Rodney is a public servant who owns two vehicles with engine capacities of 1,5 l and 2,3 l

respectively. The government has two vehicle subsidy schemes for distances travelled while

on official duty:

Scheme A:

The vehicles are subsidised* and

maintained by the government

(employer). Employees are reimbursed

(paid back) per kilometre travelled for

petrol cost only.

Scheme B:

The vehicles are owned and paid for by

the employee who also has to maintain

the vehicle. Employees are reimbursed

(paid back) per kilometre travelled at a

higher rate than that of Scheme A to

cover petrol and maintenance costs.

* Subsidised vehicles are proportionally paid for by both the employee and the employer.

1.1.1. Rodney has to submit a travel claim each month. The following table shows the

claim tariffs for 2015:

Engine capacity (in litres)Claim tariff (in cents per km)

Scheme A Scheme B

Up to 1,250 77,9 236,2

1,251 to 1,550 88,8 299,4

1,551 to 1,750 96,7 328,6

1,751 to 1,950 108,3 384,4

1,951 to 2,150 111,9 397,5

2,151 to 2,500 130,3 467,0

2,501 to 3,500 137,1 578,6

Greater than 3,500 160,6 660,8

GRADE 12: MATHEMATICAL LITERACY: JULY PAPER 2 2016 Page 2 of 19

Write down a .formula that can be used to calculate the amount that can be claimed

for a 2,3 l vehicle using Scheme B in the form:

Amount claimed (in Rand) = …..

Amount claimed (in Rand) = R4,67 a × km travelled a

1.1.2. Rodney, using Scheme B, claimed an amount of R9 430,00 for travelling 1 960 km

using his 2,3 l vehicle while performing official duties for the month of November

2015. Verify, showing ALL calculations, whether Rodney claimed the correct

amount.

R4,67 × 1 960 km = R9 153,20 a Subst

No, Rodney’s calculations were incorrect. a .Concl

OR R9 340 ÷ 1 960 = 4,8112

1.2. Rodney needs to determine whether it is better for him to use his 1,5 l or 2,3 l vehicle. He

travels approximately 1 960 km per month while performing his official duties.

The comparison of the monthly maintenance and petrol cost per kilometre is summarised in

the table below:

Engine capacity

Maintenance (in rand) Petrol(in rand / km)Service Tyres Insurance Warranty

1,5 l 450 125 500 200 1,0132,3 l 700 210 800 450 1,317

1.2.1. Calculate Rodney's total monthly cost for petrol and maintenance if he uses his 2,3 l vehicle.

R700,00 + R210,00 + R800,00 + R450,00 + (R1,317 × 1 960 km m)

= R2 160,00 a + R2 581,32 ca

= R4 741,32 ca


1.2.2. Would Rodney need to do calculations in order to determine which vehicle is the

cheapest to run when performing his official duties? Justify your answer.

No, all the costs in the table for the 2.3 l vehicle are higher. aJustify

1.2.3. Rodney, using Scheme B, used his 2,3 vehicle for his official duties. He travelled 1 960 km in total for the month.

a) Determine the percentage of Rodney’s travel claim that was not for

maintenance and petrol.

Travel claim = R9 153,20 a (from 1.1.2)

Total monthly cost for petrol and maintenance = R4 741,32 a (from 1..12)

R9 153,20 - R4 741,32 = R4 411,88

R 4411 , 88R 9153 ,20 × 100% = 48,2% ca

R 9430−R 4741 ,32R 9430 × 100% = 49,7%

b) What, do you think, would this money be allocated to?

Paying the car off / part of his travel allowance Any ONE

(NOT maintenance or petrol)

1.3. Rodney's wife is 66 years old. Her taxable income for 2015 / 2016 was R315 054,00. Use the

tables below to calculate how much tax she paid monthly:

TAX THRESHOLDS TAX YEAR: 2014/2015 TAX YEAR: 2015/2016Below age 65 R70 700 R73 650 Age 65 to 74 R110 200 R114 800 Age over 75 R123 350 R128 500

The new tax thresholds are due to increases in tax *rebates for individual taxpayers:TAX REBATES TAX YEAR: 2014/2015 TAX YEAR: 2015/2016Primary (for all taxpayers) R12 726 R13 257 Secondary (aged 65 and over) R7 110 R7 407 Tertiary (aged 75 and over) R2 367 R2 466

*A rebate is an amount by which an individual's calculated tax is reduced

THE NEW TAX RATESTax payable for the tax year ending 29 February 2016TAXABLE INCOME OF INDIVIDUALS (R) TAX PAYABLE (R)0 to 181 900 18% of taxable income181 901 to 284 100 32 742 + 26% of taxable income above 181 900 284 101 to 393 200 59 314 + 31% of taxable income above 284 100393 201 to 550 100 93 135 + 36% of taxable income above 393 200


550 101 to 701 300 149 619 + 39% of taxable income above 550 100701 301 and above 208 587 + 41% of taxable income above 701 300Trusts other than special trusts Rate of tax 41%

SARS A people’s guide to the budget 2015

R315 054,00 59 314 + 31% of taxable income above 284 100 Correct tax bracket

Annual tax

R59 314,00 + [31% × (R315 054,00 - R284 100)] = R59 314,00 + [31% × R30 954,00 a]

= R59 314,00 + R9 595,74 ca

= R68 909,74 ca

Rebates R13 257,00 a + R7 407 a = R20 664,00

Monthly tax

(R68 909,74 - R20 664,00) ÷ 12 = R48 245,74 ca ÷ 12 m

= 4 020,47833…

R4 020,48 ca

1.4. The cost of electricity has increased dramatically over the past two years. Individual households are charged according to the number of kilowatt-hours (kWh) of electricity used. Households using more electricity are charged at a higher rate per kWh than those using less electricity.

Use the table below to answer the questions that follow:

The average monthly increase in the cost of electricity (excluding VAT) between 2014 and 2015

Average monthly usage in kWh50 150 600 1 000 1 500

Amount payable in 2014 R27,35 R85,83 R393,67 R728,63 R1 147,33Amount payable in 2015 R28,83 R94,99 R467,43 R888,83 CIncrease between 2014 and 2015 R1,48 R9,16 APercentage increase between 2014 and 2015 5,39% 10,67% 18,74% B 23.38%

1.4.1. During 2015 the Vilakazi family used an average of 600 kWh per month while the

Nyati family used an average of 150 kWh per month. Use the table to calculate the

difference in cost per kWh that the Vilakazi and Nyati families had to pay.

(R467,43 a ÷ 600 kWh) – (R94,99 a ÷ 150 kWh)

=R0,77905 ca – R0,633267 ca

= R0,145783

R0,15 Round


1.4.2. The difference in the cost of electricity can be viewed by some consumers as fair and

by others as unfair. Give a suitable reason for each of these views.

Fair – 15c more / kWh is not much more OR the more you use the more you should

pay Reason

Unfair- Why should the rich always pay more? If a bigger family uses more why

should they be charged more? Reason

1.4.3. Calculate the missing values, A – C in the table showing all working out.

A

R467,43 – R393,67 = R73,76 a

B

R 888 ,83−R 728 ,63R 728 ,63 × 100%

=

R 160 , 20R 728 , 63 aa× 100%

= 21,9864… %

21,99% (22%) ca

C

R1 147,33 + (23 , 38100 × R1 147,33) m

= R1 147,33 + R268,2457…

= R1 415,577…

R1 415,58 a

1.4.4. Mrs Cronje used an average of 1 000 kWh of electricity per month. Calculate the

total annual increase, including VAT, of her electricity bill between 2014 and 2015.

Increase: (R888,83 – R728,63) × 12 months = R160,20 increase only × 12 ×12

= R1 922,40 a

Price with VAT R1 922,40 + (

14100 × R1 922,40) = R1 922,40 + R269,136

= R2 191,536

R2 191,54 ca

OR R1 922,40 x 114% = R2 191,56 ca


1.5. Nandi has saved up enough money to buy herself a popular PC Game. She looks on the internet to research where she can download the game and how much it will cost to download. She finds four sites and summarises the information she needs as follows:

Country Currency Cost Exchange rate (ZAR)America $ 49,99 15,8656England £ 35,99 23,1397Germany € 59,95 17,8627

South Korea W (Won) 49 950,00 0,0134

1.5.1. Which currency is the strongest? Justify your answer.

The British pound a as you will get more Rands for £1 than any of the other

currencies Reason

1.5.2. By looking at the exchange rate, where does the PC game appear to be the cheapest?

Justify your answer.

South Korea a the currency is the weakest. Reason

1.5.3. How much will she pay in Rand if she buys and downloads the game from America?

$1 : R7,74 = $49,99 : Rx

115 ,8656 =

49 ,99x m

x = R793,121….

x R793,12 a

1.5.4. The average inflation rate has been 5,8% for the last three years. What can Nandi

expect to pay for this type of game in two years’ time if she buys the game form

America?

Year 1: R793,12 + (

5,8100 × R793,12) = R793,12 + R46,00096a

= R839,12096 ca (=$52,89)

Year 2: R839,12096 + (

5,8100 × R839,12096 ) = R839,12096 + R48,66901…CI

= R887,7899…

R887,79 ca (=$55,96)


QUESTION 2

2.1. Lolly is the owner of the Tasty Sandwich Company. She supplies the following information

to her packers:

Take two slices of buttered bread and spread two tablespoons of filling

between them

Cut the sandwiches

diagonally and pack them

next to each other in the

triangular container

Fit two triangular containers

together as seen in the

diagram. Pack this pair of

containers into the carton


2.1.1. The dimensions of the sandwich container are 5% greater than the dimensions of the sandwich. A rectangular sticker with the Tasty Sandwich Company logo is pasted onto the diagonal side of the sandwich container. Lolly wants to change her logo on the sticker to include the type of filling in the sandwich.a) Calculate the length of the diagonal side of the sandwich container using the

Theorem of Pythagoras. Round to the nearest centimetre.

11 cm a Con + (5

100 × 11 cm) = 11,55 cm

Hyp2 = (11,55 cm)2 + (11,55 cm)2

Hyp2 = 133,4025 cm2 + 133,4025 cm2

√Hyp2=√266 ,805 cm2 ca

Hyp = 16,334.. cm ca

Hyp 16 cm Round

H2 = (11 cm)2 + (11 cm)2

H2 = 121 cm2 + 121 cm2

√ H2=√242 cm2 ca

H = 15,556.. cm ca

H 16 cm Round (4 only)

b) The ratio of the length of the sticker to the length of the diagonal side of the

container is 2 : 3. Calculate the length of the sticker, using the rounded value.

2 : 3

? cm : 16 cm m

?= 10,666666

?= 10,7 cm a from (a)

Hypotenuse2=(side 1 )2+(side 2 )2


c) The dimensions of the sandwich container are 5% greater than the dimensions of the sandwich. Lolly is having a heated debate with her cousin about the method for calculating the volume of the sandwich container. Lolly calculates the volume of the container to 420 cm3, to the nearest cm3. Her cousin says that the volume can also be determined by calculating the volume of the sandwich and then increasing this answer by 5% to get the volume of the container.

By calculation, determine whether her cousin arrives at the same answer and explain why this is so.

Volume = ( ½ base × height) Form × dist between the two parallel faces

= ( ½ × 11 cm × 11 cm) × 6 cm a Subst

= 363 cm3 ca

363 + (5

100 × 363 ) = 363 + 18,15

= 381,15 cm3 5% inc

The answers are not the same a . Her cousin is not correct as each dimension

must be increased by 5% and not just the final answer. Explanation

2.1.2. The sandwiches in containers need to be packed into a rectangular carton which is 94,6 cm by 58 cm by 36 cm. The sandwiches will be packed upright, as indicated in Lolly’s instructions. Determine the maximum number of sandwich containers that can be packed into one carton. Clearly show all working out.Sides of the containers = 11,55 cm

Depth of the container = 6 cm + (

5100 × 6 cm) = 6,3 cm

94,6 cm ÷ 11,55 cm a = 8.1 rectangles ca

58 cm ÷ 6,3 cm a = 9,206 rectangles ca

36 cm ÷ 11,55 cm a = 3,11 rectangles ca

Bottom layer = (8 × 2 a ) × 9 × 3 m

= 432 containers ca

Volume = area of the face × distance between the two parallel faces

58 ÷ 11,55 = 5,02 5

94,6 ÷ 6,3 = 15,0158 15

15 × 5 × 3 × 2 = 450 containers


2.1.3. a) The manufacturer of the cartons wants to increase his price because the price of

cardboard has gone up twice in one year. He now pays R0,40 / m2. How much

cardboard, to the nearest cm2, is needed to make one carton?

SA = 2(94,6 cm × 58 cm) + 2(94,6 cm × 36 cm) + 2(58 cm × 36 cm) m a

=10 973,6 cm2 + 6 811,2 cm2 + 4 176 cm2

= 21 960,8 cm2

21 961 cm2 ca (P1U)

b) What does it cost to manufacture one carton?

21 961 cm2 ÷ m (100)2 = 2,1961 m2 a

2,1961 m2 × R0,40 = R0,87844 ca

R0,88

2.2. The Den Coffee Shop orders cartons of sandwiches from Lolly as follows:

Type of bread: 70% low GI; 5% white bread; 25% brown bread

Filling: 40% chicken mayonnaise; 50% cheese and tomato; 10% ham and tomato

2.2.1. Draw a fully labelled tree diagram to show all the types of sandwiches available for

The Den Coffee Shop.

0,07


Use your tree diagram to answer the following questions:

When The Den opens the carton of sandwiches to unpack them, what is the probability that the

first the first container they pick up is a sandwich

2.2.2. made from brown bread? Write your answer as a fraction.

P(Brown) = 25100 a

=

14 Simp

2.2.3. made from White bread or Low GI bread with tomato in the filling? Write your

answer as a percentage.

P(Low GI or White with tomato) = 0,35 + 0,07 + 0,025 + 0,005 m

= 0,45 a

= 45% Correct unit

QUESTION 3

Study the elevation ad layout plans for this house.

Cupboard

14’

0”

KEY‘ = foot =ft

“ = inch(es) = in


3.1. What size family, do you think, lives in this house? Justify your answer.

One or two people a – there is only 1 bedroom / 1 shower Reason

2 bedrooms ONLY

3.2. How thick, in feet, are the walls of this house? Explain or show by calculation how you

arrived at the thickness.

Bedroom length = 14’ but interior dimensions =13’ m

2walls = 1 foot m and 1 wall = ½ ft a

3.3.

3.3.1 Show that the length of the house (in feet), including the porch is 28 ft 11 in

7’0” +13’11” + 8’0” aa (-1 for each X) = 28 feet 11 inches a

3.3.2 The breadth of the house is 14’ (feet) which is equivalent to 4,54 m. Use Table 1.1 A

factor table to convert the length of the house to metres. Round your answer off to 2

decimal places.

28 feet × 0,304 a = 8,512 m ca

11 inches × 2,540 a = 27,94 cm = 0,2794 m ca

Total length: 8,512 m + 0,2794 m = 8,7914 m

8,79 m ca Rounded


3.4. The wooden floor boards in the entire house as well as the porch are to be replaced with new ones. Each floor board is 30 cm wide and 2 m long. The owner uses the total length and breadth of the house as indicated by the shaded rectangle in the diagram below to do his calculations.

3.4.1 He increases the total number of floor boards needed by 5% to allow for the overlap when

joining them. Calculate how many floor boards are needed to complete the job.

Floor area: 4,54 m × 8,79 m a = 39,9066 m2 ca

Board area 0,30 m a Con × 2 m = 0,6 m2 a

Number of boards 39,9066 m2 ÷ 0,6 m2 m = 66,511 ca

66,511 + ( 5

100 × 66,511) = 66,511 +3,3255

= 69,8365 ca

70 boards Round up

4,54 m

Board area 0,28 m a × 2m = 0,56 m2 a

39,9066 m2 ÷ 0,56 = 71,26= 72 boards72 × 1,05 =75,6 75 or 76 boards


The floor boards are interlocking – as seen below:

3.4.2 Show how to calculate the number of floor boards needed when

the overlap is taken into consideration. Calculate the number of

interlocking floor boards needed to cover a portion of the floor

having a width of 170 cm as seen in the diagram on the right.

170 cm – 30 cm = 140 cm a

140 cm ÷ 28 cm m = 5 boards ca

Total boards: 5 + 1 boards = 6 boards ca

[24]

170 cm

Enlarged end of the floor boards showing how they fit together.

Floor boards are interlocking along the longest side of the piece of wood

This is what the finished floor looks like

The overlap is 2 cm of the 30 cm width


QUESTION 4

4.1. A fitness trainer introduces a new fitness programme at a school. Part of the program involved

doing as many push-ups as possible in 30 seconds. The data collected is organised in the

following frequency table:

Number of push-ups Frequency10 – 12 113 – 15 616 – 18 319 – 21 722 – 24 325 – 27 228 – 30 1Total 23

4.1.1. Represent the grouped data in a suitable graph on the Insert provided.

4.1.2. Explain why it is not possible to calculate the mean, mode or median.

You are not given the original raw data values and cannot do central tendency

calculations with grouped data. aa

a Histograma

a

a

a

a


4.1.3. What is the modal class for this set of data and what does it mean?

19 – 21 a means that the most number of learners were able to do between 19 and

21 push-ups in 30s a

4.1.4. What percentage of the group of learners managed more than 21 push-ups in 30

seconds?

3+2+123 × 100% =

623 aa × 100% m

= 26,0869…

26,09% ca (Accept 26,1%)

4.2. The group of learners from Question 4.1 above, followed the fitness programme for 4 weeks.

The scatter plot graph below shows the progress made by each learner.

4.2.1. Discuss the trend shown by this set of data.

Most the learners a increased the number of push-ups a by the end of the 4

week program. a

U

Z

Y

XW

V

T

S

R

Q

P

N

ML

KH

G

F

E

DC

BA


4.2.2. Identify and discuss the learners who made the most and the least progress during the

four week program.

Most progress = T a increased number of push-ups from 21 to 35 Explain

Least progress = P a no increase in number of push-ups 20 only Explain

(Accept - Least progress = B a increased in number of push-ups from 13 to 15

Explain)

4.3. There are 600 learners in a school. The Academic Head summarises their average, in

percentages, for the June Examination results in the graph below. The graph shows the results

with the percentiles on the vertical axis.

0 10 20 30 40 50 60 70 80 90 1000

10

20

30

40

50

60

70

80

90

100

Average percentage achieved by learners

Perc

entil

es


4.3.1. The achievements of these learners is summarised in the table below. Use this

information to fill in the information in the paragraph that follows. Only write the

letter and the missing information in your Answer Booklet.

Q1 = 38% Q2 =53% Q3 = 68% IQR = 30%

90th percentile = 80% 40th percentile = 47%

The A median / average for the June Examination is 53% meaning that B 300 /

50%/half the pupils scored less than 53% and C 300 / 50%/half the pupils scored

more than 53%. The bottom D 25% of the learners scored less than 38% i.e. 150

learners. The bottom 75% of learners scored E 68% or less while the top F 25% of

learners scored between 68% and 100%. The middle 50% of the learners’ marks

ranged between G 37%/38% and H 68%. Ninety percent of the learners achieved

marks I below 80% i.e. there were J 10% / 60 learners who achieved marks ranging

from 80% to 100%. K 240 learners i.e. 40% of the learners scored marks lower than

L 47% while M 60% of the learners scored marks higher than 47%.(6 - ½ for each

incorrect answer)

A = median / average/ Q2

B = 300 / 50% / half the

C = 300 / 50% / half the

D = 25%

E = 68%

F = 25%

G = 37% / 38%

H = 68%.

I = below 80%

J = 10% / 60

K = 240

L = 47%

M = 60% / 360 6 - ½ for each incorrect answer)


4.3.2. Use the graph to determine the following:

a) How many learners scored 60% or less?

From the graph we see 60% is on the 63rd percentile

63% a × 600 learners m = 378 learners scored 60% or less ca

b) If 0% - 29% is classified as a fail, how many learners failed the June

Examination?

From the graph we see 29% is on the 15th percentile

15% a × 600 learners m = 90 learners failed ca

Documents

QUESTION 1 - St Stithians Collegemaths.stithian.com/Maths Lit/Prelims/Brescia/12 Paper 2... · Web viewGRADE 12: MATHEMATICAL LITERACY: JULY PAPER 2 2016Page 19 of 19 GRADE 12 EXAMINATION