Montfort Math PRWC Worksheet QP+MS

1 Question 1 [TKSS 2020]The cash price of a new motorcycle is $25 120. Jack paid a down payment of 20% of the price of the motorcycle and the balance with a fixed simple interest rate of 4% per annum for a period of 3 years. (a) Calculate his monthly instalment. Answer $………………………………….. [3] MATHEMATICS DEPARTMENTREVISION (PROBLEMS IN REAL WORLD CONTEXT)NAME: ____________________CLASS: ____________________DATE: _____________________

2 The technical specification of his new motorbike, as stated by the manufacturer, is shown in Table 1.Fuel capacity :3.7 gallonsFuel efficiency:6 litres/100 kmDimension front tyre :120 / 70 R17Dimension rear tyre :180 / 55 R17Recommended replacement:Front tyre, every 6000kmRear tyre, every 2900kmConversions:1 gallon = 3.785 litres1 inch = 2.54 cm (b) (i) Find the fuel capacity of his motorbike, correct to the nearest litre.Answer ……………………………litres [1]Jack starts riding with a full tank of petrol. After riding for a total distance of 190 km, he filled up the petrol tank to the brim again, by pumping 13 litres of petrol. (ii) Determine if the fuel efficiency stated by the manufacturer in Table 1 is true. Answer ………………………………………………………………...…...…………………… …………………………………………………………………..……………………… …………………………………………………………………..……………………… .….……………………………………………………………..……………………… [2] An example of the dimensions for a motorbike tyre 195 / 55 R16 is explained below.

3 The cross-section of a wheel can be presented in Diagram II, where d is the diameter of inner rim, and h the height of the sidewall.dh 195 is the section width of the tyre in millimetres55% aspect ratio represents the height of the tyre sidewall as a percentage of the section widthDiameter of the tyre’s inner rim is 16 inches

4 (c)Jack replaces his tyres at the recommended distance travelled.What would be the number of complete revolutions made, when Jack has to first replace a front tyre? Use the information in Table 1 and show your calculations clearly. Answer ……………………… [5]

5 Question 2 [Bedok View 2020] 1st National TapWith an area of about 724 km2 and growing urban areas, Singapore lacks the space to collect and store all the rain that falls on it. However, through a network of rivers, canals and drains, rain that falls on two-thirds of Singapore's land area is channelled to its 17 reservoirs. This network forms the water catchment area in Singapore which has plans to extend the catchment area to 90 per cent by 2060.(a)What is the current area of water catchment in Singapore? Answer (a) ………………………….. km2[1]The diagram below shows Singapore’s domestic water consumption from the year 2000 onwards.

6 (b)Population wise, Bedok is home to an estimated 289 000 residents. Given that Bedok Reservoir has a storage capacity of 71.2810m3, and assuming that Bedok residents maintain the rate of water consumption as in 2018, for how long can the supply of water from Bedok Reservoir last its residents? Give your answer correct to the nearest day. Answer (b) ………………………… days[3]2nd National TapThe table below shows five NEWater plants and their capacities for producing clean recycled water from used water.NEWater Plant NameCapacity per day (m3)Bedok (PUB)85 000 Kranji (PUB)80 000Ulu Pandan (Keppel Segher)148 000Changi (Sembcorp)228 000Changi (BEWG-UESH)227 0003rd National TapThe table below shows four Desalination plants which can convert sea water to fresh water.Desalination Plant NameSingSpring Desalination PlantThese 4 desalination plants have a combined capacity of producing 160 mgd of water.Tuas South Desalination PlantTuas Desalination PlantMarine East Desalination PlantAdditional Information1) 1 mgd represents 1 million gallons of water per day. 1 mgd = 4546.09 m3/day2) Water demand in Singapore is currently about 430 million gallons a day (mgd)

7 with homes consuming 45% and the non-domestic sector taking up the rest.The following is a statement reported by Uma Sharkeri in Business Times 29 June 2010.“PUB’s near-term targets called for 40% of Singapore’s water demand to be met by NEWater and 25% to be met by desalinated water by 2020.”(c)Has PUB achieved its targets as stated in the Business Times reported 10 years ago? Justify your decision with calculations. Answer (c)…………………………………………………………………………………….…………………………………………………………………………………….…………………………………………………………………………………….…………………………………………………………………………………….[5](More writing space on the next page)

1(a)Principal = 0.825120 = $20096Monthly instalment =(__0.043)36TheirPTheirP= $625.21B1M1A1Correct PFind I = PRT(bi)3.7853.714lB1(bii)610013216.67lkmlkm It is not true because he travelled less than 216.67km. Fuel efficiency is lower than stated.190131006.84orkmlkmlIt is not true because it consumes more than 6 litres. Fuel efficiency is lower than stated.Or 100 km used 6 litres190 km used 11.4 litresIt is not true because he should have just used 11.4 litres, not 13 litres. Since he used up more than 11.4 litres, fuel efficiency is lower than stated.Or 13 litres travel 190km6 litres travel 87.7kmIt is not true as he only travelled 87.7km instead of 100km with 6 litres. Fuel efficiency is lower than stated.Or comparison for how much distance can be travelled per 1 litre:13 litres travel 190km (actual)1 litre travel 14.61km (actual)6 litres travel 100km (as stated)1 litre travel 16.67km (as stated)Since he can only travel 14.61km per litre of petrol instead of 16.67 km per litre of petrol as stated, fuel efficiency is actually lower. B1B1190km < 216.67km 6.84l > 6l

(c)Front tyre:84 mmh = 0.7 ×120 = 84 mm or 8.4 cmd=17(2.54)+2(their height) = 59.98 cm ORr = 17 (2.54)(0.5) + their height = 29.99 cmCircumference (front tyre) = 2π (their radius) = 188.43 cm = 1.884 m = 0.001884 kmNumber of complete revolutions= 6000 ÷ (their circumference)= 3184514 B1M1M1M1A1 Find height using % of widthFind diameter or radius in cm using their heightFind circumferenceFind no. of revolutions using circumference11 marks 1784 mm

2(a)Water catchment area = 27243 = 482.666 = 483 km2 (3 s.f.) [B1](b)1 litre ------------ 1000 cm3141 litres -------- 141000 cm3 [M1] = 3141000100 m3 = 0.141 m3Vol of water used by one person per day = 0.141 m3 Vol of water used by Bedok residents per day = 2890000.141 = 40749 m3 Number of days = 71.281040749 [M1] = 314.118 = 314 (to nearest day) [A1] To award M1 if correct conversions of all units are observed.(c)Total volume of NEWater produced per day[M1] = (85 + 80 + 148 + 228 + 227) 1000 = 768 000 m3 = 168.9363 mgd Percentage of demand = 168.9363100430% [M1] = 39.2875… = 39.3% (3 s.f.) Desalinated water as percentage of demand = 160100430% [M1] = 37.209… = 37.2% (3 s.f.)PUB falls short of its target of 40% of NEWater production by about 0.7% (accept 40% > 39.3%). [A1]PUB exceeds its target of 25% of desalinated water production by 12.2% (accept 25% < 37.2%). [A1] To award M1 if correct conversions of all units are observed. [Total : 9m]