CHAPTER 3 Ratios, Rates and Proportions 86 E Conversion problems in rate EXAMPLE 21 (a) Mutton is sold at the rate of $15/kg. Convert the price rate to cents/g. (b) The speed of a car is 90 km/h. Change the speed to m/s. Solution: (a) 1 kg of mutton costs $15. \ Cost of 1000 g of mutton = (15 × 100) cents = 1500 cents Cost of 1 g of mutton = 1500 1000 = 1.5 cents/g \ Price rate = 1.5 cents/g (b) The car travels 90 km in 1 hour. \ Distance travelled in 3600 s = (90 × 1000) m = 90 000 m Distance travelled in 1 s = 90 000 3600 = 25 m/s \ Speed = 25 m/s EXAMPLE 22 (a) A reptile crawls 3 m across a road at the speed of 2 m/s. How much time does it take in seconds? (b) A car 40 m away from the reptile is travelling at a speed of 72 km/h. How long does it take to travel 40 m in seconds? Will the reptile cross the road safely? Speed = Distance Time Solution: (a) Time = Distance Speed = 3 m 2 m/s = 1.5 s (b) Speed = 72 km/h = 72 × 1000 m 3600 s = 20 m/s Time = 40 m 20 m/s = 2 s The time taken for the reptile to cross 3 m of road is less than the time taken for the car to travel 40 m. \ The reptile will cross the road safely. ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 87 EXAMPLE 23 In a fitness test, Nurul’s heartbeat rate was 630 times in 6 minutes. If Nurul repeated the test for 10 minutes, estimate the number of heartbeats that will be recorded. F Problem-solving Solution: Stage 1: Understand the problem List the facts and the question. Facts: Nurul's heartbeat rate = 640 times 6 minutes Question: Estimate the number of heartbeats recorded in 10 minutes. Stage 2: Think of a plan • Let the number of heartbeats be y times in 10 minutes. • Use the cross-multiplication method to find the value of y. Stage 3: Carry out the plan 630 6 = y 10 630 × 10 = 6 × y y = 630 × 10 6 = 1050 The number of Nurul's hearbeats in 10 minutes was 1050 times. Stage 4: Look back Work backwards to check. The number of heartbeats in 6 minutes = 1050 ÷ 10 × 6 = 630 Duration Rate ($/h) First 1 hour or part of it 2.00 Every additional 1 2 hour or part of it 0.50 The table shows the parking rates at a car park. Siti parked her car for 3 1 2 hours. How much she has to pay? ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 88 Discuss with your teammates. Explain how ratios and rates are different. Describe a situation where you might use a ratio. Describe a situation where you might use a rate. INTERACTIVE ZONE Practice 3.2 Basic Intermediate Advanced A Find the rate of each of the following. (a) Water temperature increases from 30°C to 54°C in 12 min. (b) A toy car moves 300 cm in 5 s. (c) 10 l of milk cost $45. (d) 8 loaves of white bread cost $12. B 12 kg of potatoes cost $66. (a) What is the price rate of potatoes? (b) Find the selling price of 3 kg of potatoes. C The speed of an aircraft is 756 km/h. Express the speed of the aircraft in m/s. 4 Determine the unit rate for each of the following situations. (a) The price for 12 oranges is $9. (b) Aman walks 800 metres in 10 minutes. (c) Rima pays $15 for 5 kg of rice. (d) John types 135 words in 3 minutes. 5 Convert (a) price rate of $18/m to $/cm, (b) speed of 72 km/h to m/s, (c) density of 4 g/cm3 to kg/m3 , (d) pressure of 12 N/m2 to N/cm2 , (e) acceleration of 0.2 m/s2 to km/h2 . 6 (a) The temperature of a cup of coffee dropped constantly from 85°C to 60°C in 20 minutes. Find the cooling rate of the coffee in °C/min. (b) Water flow out from a tank at a constant rate. If the height of water decreases from 50 cm to 41 cm in one and a half hour, what is the flow rate of the water in cm/min. 7 (a) A bakery used 350 kg of flour in 7 days. Find the consumption rate of the flour in kg/day. (b) A car travelled a distance of 364 km on 20 litres of petrol. What is its average rate of consumption in km per l? 8 A factory worker received a total wage of $600 per week. Find (a) the worker’s daily wage rate if he works 6 days a week, (b) the worker’s hourly wage rate if he works 45 hours a week. 9 The population of a town was 250 000 in 2011. In 2021, the population increased to 400 000. (a) Find the size of the population increased. (b) What is the average rate of increase per year? (a) Teddy won 42 out of 54 tennis matches. Find his average win rate. (b) Karan won 20 out of 30 tennis matches. Find his average win rate. (c) Who do you consider as the better player? Why? K Shop A B C Quantity (kg) 2 4.5 5 Price ($) 7.00 14.40 19.00 The table above shows the price paid for the same item X at three shops. Arrange the shops from the lowest to the highest price rates. L Car Distance travelled (km) Petrol consumption ( ) P 54 4.5 Q 42 3 R 63 6 The table above shows the quantity of petrol consumed and the corresponding distance travelled by three cars P, Q and R. Determine the car that has the best rate and the worst rate of petrol consumption. ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 89 3.3 Proportions A Understanding proportions From the findings of Activity 2, you have use proportions to find the number of words that you and your friend copied in 5 minutes. Proportion is a relationship between two ratios that are equal. Two pairs of quantities are in proportions if they are equivalent ratios. It can be written in a : b = c : d or a b = c d . B Direct proportions (I) Direct proportions When two quantities are in direct proportion, they increase or decrease in the same rate. In other words, the ratio of the quantities is equivalent. For example, the more hours you work, the more you earn. A quantity x is proportional to another quantity y if the ratios of two pairs of quantities are the same. For example, x1 : x2 = y1 : y2 or x1 x2 = y1 y2 where x1 and x2 are values of quantity x whereas y1 and y2 are values of quantity y. Objective : To determine the relationship between ratio and proportion. Instruction : Do this activity in pairs. 1. Choose one storybook. Select a page and copy the text. Time allocation for copying is 5 minutes. 2. Your partner will act as a timekeeper, telling you when to start copying and when to stop copying. 3. Copy pages comfortably to minimise spelling mistakes. 4. Stop copying when the time is up. Mark the place at where you stopped copying. 5. Count the total number of words copied in the given 5 minutes. 6. Exchange roles with your partner and repeat Steps 1 to 5. and obtain data for your friend. Make sure to use the same page when copying. 7. Record your findings in the following table. Name of student Number of words Time (minutes) 5 5 2 ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 90 EXAMPLE 24 Determine whether the following pairs of ratios are a proportion. (a) $15 : $75 and 2 kg : 10 kg (b) 1 l : 3 l and 16 km : 50 km Solution: (a) $15 : $75 = 15 75 1 5 = 1 5 2 kg : 10 kg = 2 10 1 5 = 1 5 ∴ The two pairs of ratios are in proportion. (b) 1 l : 3 l = 1 3 16 km : 50 km = 16 50 8 25 = 8 25 ∴ The two pairs of ratios are not in proportion. EXAMPLE 25 The table shows the distance travelled and the amount of petrol used by a car. Distance travelled (km) 51 255 Amount of petrol (l) 3 15 Determine whether the distance travelled is directly proportional to the amount of petrol used. Solution: 51 km : 255 km and 3 l : 15 l 51 km : 255 km = 51 255 1 5 = 1 5 3 l : 15 l = 3 15 = 1 5 ∴ The distance travelled is directly proportional to the amount of petrol used. EXAMPLE 26 The length of a rope is directly proportional to its mass. If a 5 m rope is 1.2 kg, what is the mass of 15 m rope? Solution: Let the mass of 15 m rope = x kg x kg : 1.2 kg = 15 m : 5 m x 1.2 = 15 5 5x = 1.2 × 15 x = 1.2 × 15 5 = 3.6 kg ∴ The mass of 15 m rope is 3.6 kg. ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 91 EXAMPLE 27 If 3 kg of avocados cost $111, (a) find the cost of 1 kg of avocados. (b) what is the cost of 10 kg of avocados? (c) how many kilograms of avocados can be bought for $185? Solution: Using unitary method (a) 3 kg = $111 ∴ 1 kg = $ 111 3 = $37 Using proportion method (c) Let the quantity of avocados that can be bought = x kg x kg : 1 kg = $185 : $37 x 1 = 185 37 37x = 185 x = 185 37 = 5 kg ∴ 5 kg of avocados can be bought for $185. (b) Cost of 10 kg of avocados = $37 × 10 = $370 (II) Determining whether a quantity is directly proportional to another quantity There are many instances in our daily lives which involve direct proportion. For example, the price of sugar sold in a supermarket depends on the mass of the sugar. The price of sugar and the mass of sugar are quantities which varies directly with each other. When one quantity increases, the other quantity also increases. One quantity y is directly proportional to another quantity x if and only if y x = k, where k is called the constant of proportion. y directly proportional to x is written as y ∝ x. When y is directly proportional to x, the graph of y against x is a straight line which passes through the origin. EXAMPLE 28 Number of apples, N 3 4 5 6 Total cost ($), P 12 16 20 24 The table shows the number of apples and the total cost of apples that are sold at a fruit stall. Determine whether P is directly proportional to N and draw a graph to show the relation between N and P. x 0 y ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 92 Solution: Number of apples, N 3 4 5 6 Total cost ($), P 12 16 20 24 P N 4 4 4 4 The value of P N is a constant. Therefore, P is directly proportional to N. N P 20 24 28 16 8 4 0 1 2 3 4 5 6 12 EXAMPLE 29 x (years) 5 6 7 8 y (kg) 18 20 21 24 The table shows the relationship between the mass, y kg, of a boy and his age, x years. Determine whether y is directly proportional to x. Solution: x (years) 5 6 7 8 y (kg) 18 20 21 24 y x 3.6 3.33 3 3 y x is not a constant. So, y is not directly proportional to x. When two quantities are in direct proportion, they increase or decrease at the same rate. The ratio of quantities is equivalent. For example, the more hours you work, the more you earn. When one quantity decrease at the same rate as the other quantity increase, the two quantities are in indirect proportion with each other. For example, time to complete a task and number of workers are in inverse proportion. (III) Expressing a direct proportion in the form of an equation involving two variables When a quantity y is directly proportional to another quantity x, we can express the proportion in the equation form y = kx where k is a constant which can be determined. EXAMPLE 30 The price, $ y, of a watermelon is directly proportional to the mass, m kg, of the watermelon. Given that y = 25 when m = 5, write an equation in terms of y and m. ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 93 Solution: y ∝ m y = km When m = 5, y = 25 25 = k(5) k = 25 5 = 5 Therefore, y = 5m. (IV) Finding the value of a variable in a direct proportion When y is directly proportional to x, the value of an unknown variable can be determined by writing the proportion in the equation form, then substituting the given value of a variable into the equation to find the value of the other variable. EXAMPLE 31 Given that p is directly proportional to q, and p = 18 when q = 6. Find the value of p when q = 5. Solution: p is directly proportional to q. So, the equation is p = kq. When q = 6, p = 18, 18 = k(6) k = 18 6 = 3 Hence, p = 3q. When q = 5, p = 3(5) = 15 When y ∝ x, the value of x or y can be found by using the relation (a) y = kx (b) y1 x1 = y2 x2 Using the proportion method, p1 q1 = p2 q2 Let p1 = 18, q1 = 6, q2 = 5 Then, 18 6 = p2 5 p2 = 3 × 5 = 15 (V) Solving problems involving direct proportion for cases y ∝ xn, where n = 1 2 , 2, 3 If y ∝ x n , where n = 1 2 , 2, 3, then the equation is y = kx n where k is a constant. The graph of y against x n is a straight line which passes through the origin. If y ∝ x n and sufficient information is given, the values of variable x or variable y can be determined. ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 94 EXAMPLE 32 The area, A, of a circle is directly proportional to the square of its radius, r. Given that the area of the circle is 154 cm2 when its radius is 7 cm. Find (a) the equation for A in terms of r, (b) the value of r when A = 28 2 7 cm2 . Solution: (a) A is directly proportional to r2 . So, the equation is A = kr2 . When r = 7, A = 154, 154 = k × 72 k = 154 49 = 22 7 Therefore, A = 22 7 r 2 . (b) When A = 28 2 7 , 28 2 7 = 22 7 r 2 r = 198 22 = 9 r = 3 cm EXAMPLE 33 Given that y is directly proportional to the cube root of x, and y = 6 when x = 8. Find the value of x when y = 12. Solution: y ∝ 3 x, so y = k 3 x. When y = 6, x = 8 6 = k 3 8 6 = k(2) k = 3 y = 33 x The value of x or y for the proportion y ∝ x n can be obtained by using the relation (a) y = kxn (b) y1 x1 n = y2 x2 n When y = 12, 12 = 33 x 3 x = 4 x = 43 = 64 ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 95 C Inverse proportions (I) Inverse proportion When one quantity decreases in the same rate as the other quantity increases, the two quantities are in indirect proportion with each other. For example, time to complete a task and number of workers are in inverse proportion. EXAMPLE 34 It takes 12 workers to plaster a building in 25 days. How long will 15 workers take to complete the same job if they work at the same rate? Solution: Using unitary method 12 workers = 25 days 1 worker = 12 × 25 = 300 days 15 workers = 300 15 = 20 days \ 15 workers will take 20 days to complete the same job. (II) Determining whether a quantity is inversely proportional to another quantity In a situation where one quantity increases, the other quantity decreases, such a situation involves inverse proportion. For example, the period of time taken to build a house would increase if the number of workers building the house is decreased. Conversely, if the number of workers increases, then the period of time to build the house decreases. If the variable y is inversely proportional to the variable x, then xy is a constant, that is, xy = k where k is called the constant of proportion. The symbol for ‘y is inversely proportional to x’ is y ∝ 1 x . When y is directly proportional to 1 x , the graph of y against 1 x is a straight line. EXAMPLE 35 Number of pupils, N 1 2 3 4 Time taken, T (minutes) 72 36 24 18 The table shows the number of pupils needed to clear some rubbish in a school and the time taken to clear the rubbish. Determine whether T is inversely proportional to N. x – 0 1 y ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 96 Solution: Number of pupils, N 1 2 3 4 Time taken, T (minutes) 72 36 24 18 TN 72 72 72 72 The value of TN is a constant. Therefore, T is inversely proportional to N. (III) Expressing an inverse proportion in the form of an equation involving two variables When a quantity y is inversely proportional to another quantity x, we can express the proportion in the equation form y = k x where k is a constant which can be determined. EXAMPLE 36 Given that y is inversely proportional to x, and y = 12 when x = 5. Write an equation which relates x and y. Solution: y is inversely proportional to x, So, y = k x . When x = 5, y = 12 12 = k 5 k = 60 Therefore, the equation is y = 60 x . (IV) Finding the value of a variable in an inverse proportion When y is inversely proportional to x, the value of an unknown variable can be determined by writing the proportion in the equation form, then substituting the given value of a variable into the equation to find the value of the other variable. EXAMPLE 37 Given that J is inversely proportional to V, and J = 6 when V = 8. Find the value of J when V = 12. Solution: J is inversely proportional to V. So, the equation is J = k V . When V = 8, J = 6 6 = k 8 k = 6 × 8 = 48 Hence, J = 48 V . When V = 12, J = 48 12 = 4 Using the proportion method, J1 V1 = J2 V2 Let J1 = 6, V1 = 8, V2 = 12 Then, 6 × 8 = J2 × 12 ∴ J2 = 6 × 8 12 = 4 When y ∝ 1 x , the value of x or y can be found using the relation (a) y = k x (b) x1 y1 = x2 y2 ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 97 (V) Solving problems involving inverse proportion for cases y ∝ 1 xn , where n = 1, 2, 3, 1 2 If y ∝ 1 xn , where n = 1, 2, 3, 1 2 , then the equation is y = k xn where k is a constant. The graph of y against 1 xn is a straight line which passes through the origin. If y ∝ 1 xn and sufficient information is given, the value of variable x or variable y can be determined. EXAMPLE 38 Given that V is inversely proportional to the square root of x, and V = 12 when x = 36. (a) Form an equation expressing V in terms of x. (b) Calculate the value of V when x = 9. (c) Calculate the value of x when V = 144. Solution: (a) V is inversely proportional to the square root of x. So, V = k x . When x = 36, V = 12, 12 = k 3 6 12 = k 6 k = 12 × 6 = 72 Therefore, V = 72 x . (b) When x = 9, (c) When V = 144, 144 = 72 x 144x = 72 x = 72 144 = 1 2 x = 1 1 2 2 2 = 1 4 V = 72 9 = 72 3 = 24 The value of x or y for the proportion y ∝ 1 xn can be found using the relation (a) y = k xn (b) x1 n y1 = x2 n y2 ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 98 D Solving problems related to proportions EXAMPLE 39 The ratio of the number of Mathematics books to the number of Science books in a bookshop is 9 : 4. If there are 24 Science books in the bookshop, how many Mathematics books are there in the bookshop? Solution: Using unitary method There are 9 parts Mathematics books and 4 parts Science books. Thus, 4 parts are equivalent to 24 books. 1 part is equivalent to 24 4 = 6 books. 9 parts are equivalent to 9 × 6 = 54 books. There are 54 Mathematics books. Using proportion method Let x be the number of Mathematics books. So, x : 24 = 9 : 4 x 24 = 9 4 x = 9 4 × 24 = 54 There are 54 Mathematics books. EXAMPLE 40 If the average speed of an airplane is 600 km/h, the flight time from Town P to Town Q is 8 hours. If the average speed of the airplane increases to 640 km/h, how much time will be saved? Solution: Distance travelled = Average speed × Time = 600 × 8 = 4800 km New flight time = Distance travelled Average speed = 4800 km 640 km/h = 7.5 h ∴ Time saved = 8 h – 7.5 h = 0.5 h ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 99 EXAMPLE 41 The ratio of the base to the height of a right-angled triangle is 3 : 4. If its perimeter is 6 cm and the length of the other side is 2.5 cm, find the length of the base and the height of the triangle. Solution: Let the base be b cm and the height be h cm. b : h = 3 : 4 b : b + h = 3 : 3 + 4 b : 3.5 = 3 : 7 b 3.5 = 3 7 b = 3 7 × 3.5 = 1.5 h = 3.5 – 1.5 = 2 Hence, base = 1.5 cm, height = 2 cm. Practice 3.3 Basic Intermediate Advanced h cm b cm 2.5 cm b + h + 2.5 = 6 b + h = 6 – 2.5 = 3.5 cm A Determine whether the following pairs of ratios are in proportion. (a) $70 : $168 and 5 kg : 12 kg (b) 2 l : 10 l and 36 km : 180 km (c) 160 km : 220 km and 2 h : 3 h (d) $200 : $75 and 8 l : 3 l B The table shows the number of eggs needed for baking a certain number of cakes. Number of cake baked 5 16 Number of eggs used 30 96 Determine whether the number of cakes baked is directly proportional to the number of eggs needed. C The length of a rod is directly proportional to its mass. If a 5 m rod is 8 kg, what is the mass of a 9 m rod? D A car used 1 litre of fuel to travel 18 km. How far could the car travel on 30 litres of fuel at the same rate? E A tray of 12 eggs costs $3. (a) Find the cost of an egg. (b) What is the cost of 30 eggs? (c) How many eggs can be bought for $270? F The time taken, t hours, to do a paint job is inversely proportional to the number of workers N. Given that 16 workers take 15 days to paint a factory. (a) How long will 24 workers take to paint the same factory? (b) How many workers are needed to paint the same factory in 16 days? G An airplane usually flies from City A to City B in 5 hours at an average speed of 680 km/h. Due to bad weather, the airplane decreased its average speed to 600 km/h. (a) How long would the same journey take at average speed of 600 km/h? (b) Find the time difference for the flight at two different average speeds. ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 100 H E 2 4 F 16 m The table shows some values of the variables E and F. Given that F ∝ E2 , find the value of m. I G 16 128 y H 2 x 6 The table shows some values of the variables G and H. Given that G is directly proportional to the cube of H, calculate (a) the value of x, (b) the value of y. J G 16 36 F 12 m The table shows some values of the variables F and G such that F is inversely proportional to the square root of G. Find the value of m. K Given that b ∝ 1 a3 , and b = 3 when a = 2. Find the value of b when a = 4. 3.4 Relationship between Ratios, Rates and Proportions with Percentages, Fractions and Decimals A Determining the relationship between percentages and ratios Percentage is a fraction with 100 as its denominator. (a) Convert fraction to percentage. For example: 1 2 = 1 × 50 2 × 50 = 50 100 = 50% or 1 2 × 100% = 50% (b) Convert percentage to fraction. For example: 20% = 20 100 = 20 100 = 1 5 Fractions Percentages ×100% ÷100% A decimal can be converted to a fraction with power of 10 as the denominator. 0.1 = 1 10 , 0.01 = 1 100 Hence, decimals and percentages are interchangeable like fractions. Decimals Percentages ×100% ÷100% Percentage is a ratio comparing a part to the whole with the value of the whole set as 100. Percentage can be expressed in the fraction or decimal form. 1 5 ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 101 EXAMPLE 42 The number of red radishes to the number of white radishes in a basket is in the ratio of 3 : 7. Find the percentage of white radishes in the basket. Solution: The ratio of the number of white radishes to the total number of radishes = 7 : 10 7 10 = 7 × 10 10 × 10 = 70 100 = 70% Hence, percentage of white radishes in the basket is 70%. EXAMPLE 43 Mr Zaid has a piece of land. He used 45% of the area of the land to plant durian trees. Find the ratio of the area of land used to plant durian trees to the total area of his land. Solution: Percentage of land used = 45% = 45 100 The ratio of the area of land used to the total area of the land = 45 : 100 = 9 : 20 B Determine the percentage of a quantity by applying the concept of proportions EXAMPLE 44 Determine the percentage of each of the following quantities by applying the concept of proportion. (a) A class has 25 students. 10 of the students are boys. What is the percentage of boys in the class? (b) Rahim has $400. He used $220 to buy pants and T-shirts. What is the percentage of money spent? Solution: (a) Let the percentage of boys in the class be m. Number of boys Total number of students = m 100 Proportion ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 102 ×4 10 25 = m 100 ×4 m = 40 The percentage of boys in the class is 40%. (b) Let the percentage of money spent be s. Money spent Sum of the money = s 100 220 400 = s 100 220 × 100 = 400 × s s = 220 × 100 400 = 55 The percentage of money spent by Rahim is 55%. EXAMPLE 45 Determine the percentage of changes in each of the following quantities. (a) The population of a village increases from 35 600 to 37 380 people in 6 months. Find the percentage of the population increase of the village. (b) The temperature of some rice decreases from 80°C to 48°C after several minutes. Find the percentage of the temperature decrease of the rice. Solution: (a) The increase of the number of the population = 37 380 – 35 600 = 1780 Let the percentage increase in the population be p. The increase in population Total initial population = p 100 1780 35 600 = p 100 1780 × 100 = 35 600 × p p = 1780 × 100 35 600 = 5 The percentage increase in the population is 5%. ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 103 (b) Decrease in temperature = 80 – 48 = 32°C Let the percentage decrease in temperature be s. Decrease in temperature Initial temperature = s 100 32 80 = s 100 32 × 100 = 80 × s s = 32 × 100 80 = 40 The percentage decrease in the temperature of the rice is 40%. C Problem- solving EXAMPLE 46 Rafi sold his bicycle for $180 and made a loss of 10%. Calculate the original price of the bicycle. Solution: Stage 1: Understand the problem List the facts and the question. Facts: Rafi sold his bicycle at $180. He made a loss of 10% Question: What is the original price of his bicycle? Stage 2: Think of a plan • He sold the bicycle at $180 but gain a loss of 10%. • Loss means not giving any profit to Rafi. • Calculate the percentage of the Rafi's bicycle selling price. Stage 3: Carry out the plan Percentage of the selling price = 100% – 10% = 90% 90% of the original price = $180 1% of the original price = 180 90 = $2 100% of the original price = 100 × $2 = $200 ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 104 Stage 4: Look back Work backwards to check. Selling percentage = 100% – 10% = 90% Original price = Selling price Selling percentage = $180 90 100 = $200 EXAMPLE 47 Nadia bought 90 pieces of biscuits. She gave 3 5 of the biscuits to her friends. Then, she gave 18 pieces of the biscuits to her brother. Find the percentage of the biscuits remained with Nadia. Solution: Stage 1: Understand the problem List the facts and the question. Facts: Nadia has 90 pieces of biscuits. She gave to her friends 3 5 of her biscuits. She gave 18 pieces of biscuits to her brother. Question: What is the percentage of Nadia's remaining biscuits? Stage 2: Think of a plan • How many pieces of biscuits is 3 5 out of 90 pieces? • Then, subtract 8 pieces when she gave them to her brother. • Calculate the percentage of Nadia's remaining biscuits. Stage 3: Carry out the plan Number of remaining biscuits = 90 – 1 3 5 × 902 – 18 = 90 – 54 – 18 = 18 Percentage of the remaining biscuits = 18 90 × 100% = 20% ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 105 Stage 4: Look back Work backwards to check. Percentage of remaining biscuits = 90 – 1 3 5 × 902 – 18 90 × 100% = 20% Practice 3.4 Basic Intermediate Advanced A The number of students who wear watches to the number of students who do not wear watches in a class is 9 : 16. Find the percentage of students who wear watches. B The number of female workers in a factory decreased from 72 to 54. Find the percentage decrease in the number of female workers. C The number of members of a badminton club has declined 16% to 42 people. Determine the initial number of members of the club. 4 Jun Hong answered 65% of the questions correctly in a Science quiz. He answered 7 questions wrongly. If Jun Hong answered all the questions in the quiz, how many questions are there in the Science quiz? 5 The initial price of a box of chocolate is $22.00. The selling price of the chocolate is 4 5 of the initial price. Maria used 40% of her pocket money to buy the chocolate. How much pocket money does Maria have? 6 The mass of John is 20% heavier than the mass of Marie. The mass of Wong is 12% lighter than Marie’s mass. If John’s mass is 66 kg, find the mass of Wong. 7 The money collected for an excursion is divided into two portions. 0.4 of the money collected is used to pay for transport. 3 5 of the second portion is used to pay for food and drinks. The balance of the second portion which is $480 is used to buy souvenirs. Determine the percentage and the total amount of money used to pay for food and drinks. 3.5 Scale A scale is a ratio. It can be expressed as the length on a drawing (map / plan / model) to the length of the actual object. A scale is used to show a real object with sizes reduced or enlarged by a certain ratio / number. Map scales are good examples of ratios in everyday life. The scale of a map is usually expressed in the form 1 : n. For example, a scale of 1 : 250 000 means that 1 cm on the map represents 250 000 cm on the ground. They must be in the same unit. Scale of a drawing = Size of the drawing Size of the object or Length of side of the drawing Length of corresponding side of the object ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 106 For example, A 3 cm 6 cm B 4 cm 8 cm Drawing Object If A is the scale drawing of B, then the scale is 3 : 4. The scale of a scale drawing is usually expressed in the form 1 : n which means 1 unit in the drawing represents n units in the actual object. For example, the scale of 3 : 4 is expressed as 1 : 4 3 . In the scale of 1 : n, if (a) n = 1, then the drawing and object have the same size. (b) n . 1, then the drawing has a smaller size than the object. (c) n , 1, then the drawing has a larger size than the object. If a drawing has a scale of 1 : n, then the drawing is 1 n times the size of the object. The picture above (not drawn to scale) is a 1 : 48 000 000 scale map of the sailing route from Kalimantan to Java. The sailing route shown on the map is about 1.5 cm. What is the distance of the boat trip in kilometres? Give the answer correct to 3 significant figures. 4 INDONESIA Java Kalimantan 480 km 1 cm ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 107 EXAMPLE 48 ? 4 cm Drawing Actual The diagram shows a drawing of an actual horse. If the scale is 1 : 50, what is the height of the actual horse? Solution: Scale = 1 : 50 Drawing length : Actual length = 1 cm : 50 cm = (4 × 1) cm : (4 × 50) cm = 4 cm : 200 cm = 4 cm : 2 m ∴ The height of the actual horse is 2 m. EXAMPLE 49 Write the following scales in the form 1 : n. (a) 1 cm represents 20 m (b) 2 cm represents 5 km Solution: (a) 1 cm : 20 m = 1 cm : (20 × 100) cm = 1 cm : 2000 cm ∴ The scale is 1 : 2000. (b) 2 cm : 5 km = 2 cm : (5 × 1000 × 100) cm = 2 cm : 500 000 cm = 1 cm : 250 000 cm ∴ The scale is 1 : 250 000. EXAMPLE 50 A map has a scale of 1 : 10 000. If the distance between two buildings on a map is 6 cm, what is the actual distance on the ground between two buildings? A scale is usually expressed in the form of 1 : n. For example, 1 cm to 1 km. Actual length = Drawing length × Scale Drawing length = Actual length ÷ Scale Convert: m to cm km to cm ©Praxis Publishing_Focus On Maths
CHAPTER 3 Ratios, Rates and Proportions 108 Solution: Scale = 1 : 10 000 1 cm : 10 000 cm = 1 cm : 100 m Map distance : Actual distance = (6 × 1) cm : (6 × 100) m = 6 cm : 600 m ∴ The actual distance between two buildings on the ground is 600 m. EXAMPLE 51 A length of 3 cm on the map represents a distance of 6 km. (a) Find the scale of the map in the form of 1 : n. (b) Find the length on the map that represents a distance of 900 m on the ground. Solution: (a) Given that 3 cm represents 6 km. 6 km = 6 × 1000 m = 6000 m = 6000 × 100 = 600 000 cm Scale = 3 cm : 600 000 cm = 3 3 : 600 000 3 = 1 : 200 000 ∴ The scale of the map is 1 : 200 000. (b) Let the map distance = x cm 200 000 cm = 200 000 ÷ 100 = 2000 m Scale = 1 cm : 2000 m x cm : 900 m = 1 cm : 2000 m x 1 = 900 2000 x = 0.45 cm ∴ 0.45 cm on the map represents 900 m. Scale enables comparison between the actual size of an object with the size of its drawing. A scale can be expressed as ratio (for example, 1 : 15 000) or fraction (for example, 1 15 000 ). Explain how would you know whether a drawing drawn to scale 5 : 2 is bigger or smaller than the actual object. ©Praxis Publishing_Focus On Maths
Ratios, Rates and Proportions CHAPTER 3 109 A map designer measures the main street from the airport to the bank and get 5 cm. Cycling along the main street from the airport to the bank is known to be 1 km 200 m. What ratio is the map using? Give your answer in the form 1 : x. Critical Thinking EXAMPLE 52 The homeowner wishes to do some renovations to his bedroom. He wants to add a bookshelf, which is approximately 150 cm. He will place it on the floor plan marked with a X. Given that the floor plan of the house is 1 : 200. What length should the bookshelf be drawn to in the floor plans? Solution: He wanted to add a bookshelf of about 150 cm, which he would place on the floor plan marked with a X. The ratio for the scale drawing is 1 : 200. This means 200 cm in the real world is represented by 1 cm on the floor plan. 00:00 MIN 00:00 MIN 00:00 MIN 00:00 MIN Main Street ©Praxis Publishing_Focus On Maths
110 CHAPTER 3 Ratios, Rates and Proportions Let l be the length of the bookshelf on floor plan. The ratio 1 : 200 is equivalent to l 150 = 1 200 200 l = 150 l = 150 200 = 0.75 cm The length of the bookshelf that should be drawn on the floor plan is 0.75 cm. Practice 3.5 Basic Intermediate Advanced A 3.5 cm The diagram shows a drawing of an elephant. If the scale is 1 : 100, what is the actual height of the elephant? B Write the following scales in the form 1 : n. (a) 1 cm represents 10 m (b) 5 cm represents 100 m (c) 2 cm represents 1 km (d) 5 cm represents 2 km C Simplify each of the following scales. (a) 1 mm to 5 m (b) 2 cm to 3 km D A map of Mauritius has a scale of 1 : 500 000. If the distance between Port Louis and Quatre Bornes on the map is 3.2 cm, what is the actual distance (on the ground) between these two towns? E A map has a scale of 1 : 50 000. If the actual distance between two bus stations is 3 km, what is the distance between these two bus stations on the map? 6 A particular map shows a scale of 1 : 500. A rectangular pond measures 9 cm by 6.5 cm on the map. (a) Find the actual length and breadth of the pond in metres. (b) What is the actual area of the pond in hectares? (1 ha = 10 000 m2 ) 7 D E C B A 4 cm 12 cm 5 cm The diagram shows a rose garden ABE in a farm ABCDE. Using a scale of 1 cm to represent 2 m, (a) draw an accurate scale drawing of the farm, (b) find the length of AE in the diagram, (c) find the perimeter of the rose garden. 8 The map of Mauritius shows that the distance between Port Louis and Curepipe is 5.2 cm. Estimate the actual distance (on the ground) between these two towns. Port Louis Rose Hill Quatre Bornes Curepipe Vacoas Goodlands Bon Accueil Bel Air Mahebourg Chemin Grenier 403020100 50 Kilometres ©Praxis Publishing_Focus On Maths
111 Ratios, Rates and Proportions CHAPTER 3 9 A map has a scale of 1 : 500 000. What length on the map represents a 54 km long railway line? On the floor plan of an office with a scale of 1 : 200, the dimensions of rectangular meeting room are 11 m by 8 m. (a) Draw an accurate scale drawing of the meeting room. (b) Find the area of the meeting room on the map in cm2 . Summary Summary Summary Ratio of two quantities • Comparison of two quantities in the same unit and written as a : b or a b . • a : b and c : d are equivalent ratios if a b = c d . • Given a ratio a : b, where a > b – Sum of two quantities can be found by adding the quantities, a + b. – Difference of two quantities can be found by subtracting the quantities, a – b. • When a number is increased in the ratio p : q, where p > q, the resulting value is obtained by multiplying the value by the multiplying factor p q . • When a number is decreased in the ratio p : q, where p < q, the resulting value is obtained by multiplying the value by the multiplying factor p q . • A rate is usually expressed as one quantity per unit of another quantity. Rates must include the units of quantities. Scale • A scale can be expressed as ‘length on drawing : actual length’. It is usually given as a ratio form of 1 : n. Map scale = map distance : actual distance Ratio of three quantities • Comparison of three quantities in the same unit and written as a : b : c. • Given that the ratios a : b and b : c, the ratio a : b : c can be found by using the common quantity b. – If the value of b is the same in both ratios, then the two ratios can be combined and written as a : b : c. – If the values of b are different, then the common value for b can be found by determining their LCM. Direct proportion • If y is directly proportional to x, then (a) y ∝ x (b) y = kx, k is a constant • If y ∝ x n, then (a) y = kxn , k is a constant (b) y1 x1 n = y2 x 2 n Inverse proportion • If y is inversely proportional to x, then (a) y ∝ 1 x (b) y = k x , k is a constant • If y ∝ 1 xn , then (a) y = k xn , k is a constant (b) x1 n y1 = x2 n y2 Integers Ratios, Rates and Proportions Proportion • If two pairs of ratios are equivalent, they are proportional. For example, if a : b is equivalent to c : d, then a : b = c : d or a b = c d . • When quantities are in direct proportion, they increase or decrease at the same rate. • When quantities are inversely proportional, one increases as the other decreases. ©Praxis Publishing_Focus On Maths
112 CHAPTER 3 Ratios, Rates and Proportions Section A 1. Given that x : y = 35 : 27, find x – y : x + y . A 4 : 31 C 7 : 3 B 5 : 3 D 31 : 4 2. Given that s : t = 18 : 5, find the ratio of t : s – t. A 2 : 5 C 4 : 11 B 3 : 8 D 5 : 13 3. Given that p : q = 9 : 13 and p = 135, what is the value of q? A 180 C 200 B 195 D 205 4. Given that m : n = 15 : 7 and m + n = 143, what is the value of m – n? A 52 C 64 B 54 D 68 5. 2 m for $25 7 m for $86 K 4 m for $48 10 m for $120 N 3 m for $39 10 m for $139 L 5 m for $68 7 m for $94 M The diagram shows the prices of cloth sold by four shops, K, L, M and N. In which shop was the cloth sold at a price that is in proportion to the length of cloth? A K C M B L D N 6. The ratio of the time Jalil spent on reading storybooks to the time he spent on watching television was 65 : 72. If the total time spent was 6.85 hours, calculate the time spent on reading storybooks. A 2 hours 50 minutes B 2 hours 55 minutes C 3 hours 10 minutes D 3 hours 15 minutes 7. A few months ago, the ratio of Suzie’s mass to Ming’s mass was 22 : 23 and their total mass was 90 kg. Now, Suzie is 0.5 kg lighter and Ming is 1 kg heavier. Find the new ratio of Suzie’s mass to Ming’s mass. 3 A 45 : 47 C 87 : 94 B 46 : 47 D 90 : 93 8. For a mathematics test in March, Alan and Sim scored a total mark of 190. The ratio of Alan’s marks to Sim’s marks was 49 : 46. If Alan scored 90 marks in another mathematics test in April, how many marks must Sim get so that his total mark for March and April is the same as Alan’s? A 94 C 96 B 95 D 97 9. 14 cm 4 cm 3 cm 3.2 cm 2.8 cm 1 cm Stall 3Stall 2Stall 1 Stall 6 Stall 5 Stall 4 The diagram shows the plan of a hawker centre. Each stall occupies a rectangular floor space. The plan is drawn to a scale of 1 : 200. Calculate the area of the floor space, in m2 , which stall 5 occupies. A 44.8 C 58.4 B 49.0 D 78.4 10. 5 cm 8 cm S T 9 cm The diagram shows two parallelograms, S and T. The length of the base and the height of S are in proportion to the length of the base and the height of T. Given that the height of S is 6 cm, calculate the height, in cm, of T. A 12 C 24 B 18 D 30 Section B 1. A van travels a distance of 175 km in 2 1 2 hours. Calculate the average speed of the van in (a) km/h, (b) m/s. ©Praxis Publishing_Focus On Maths
113 Ratios, Rates and Proportions CHAPTER 3 2. The table shows the mass of three pupils. Name of pupils Mass (kg) Jonah 54 Karan 60 Ryan 48 Express each of the following ratios in its simplest form. (a) Ryan’s mass : Karan’s mass (b) Jonah’s mass : Karan’s mass : Ryan’s mass 3. A small airplane flies a distance of 100 km on 25 litres of fuel. (a) Find the rate of fuel consumption in km/l. (b) If 1 litre of fuel costs 60 cents, what is the expected cost of fuel for a journey of 350 km? 4. Ritesh is paid $780 for 29 hours of work. (a) If he works for 24 hours, find his wage. (b) How many hours does he need to work in order to earn $1350? 5. (a) A 9.6 MB internet file takes 12 seconds to download. What is the rate at which file is downloaded? Give answer in kilobytes per second. (1 megabyte = 1024 kilobytes) (b) Anita's computer is downloading a 4.6 MB internet file at the rate of 340 KB/s. How long it take to download the file? Give your answer to the nearest second. 6. (a) The scale of a street map is 1 : 50 000. If the actual length of a street is 4 km, what is the length of the street on the map in cm? (b) A map has a scale of 1 : 500 000. If the distance between 2 towns is 6.4 cm on the map, what is the actual distance between the two towns in km? (c) The floor plan of a flat is drawn with a scale of 1 : 50. On the floor plan, the length of the flat is 30 cm. Find the actual length of the flat in m. 7. The price of a queen-size mattress is $950 and that of a king-size mattress is $1197. (a) The price of the queen - size mattress decreases to $800. In what ratio has the price decreased? (b) If the price of the king-size mattress decreases in the same ratio, what is its new price? 8. In an online sale, the prices of certain kitchenware were reduced in the ratio 4 : 5. (a) Find the multiplying factor. (b) What is the new price of an article which cost $450 initially? (c) Aurelie bought an article and paid $220, what was the original price? 9. The ratio of ingredients to make butter cookies is one part of sugar to two parts of butter and three parts of flour. (a) Find the ratio of sugar, butter and flour. (b) How much flour would Mrs Smith need if she used 25 kg of butter to bake butter cookies? 10. (a) Joy earned a total of $75 by selling 5 glasses of mixed fruit juice. After selling a total of 24 glasses of mixed fruit juice, how much money will she earn? Assume the relationship is directly proportional. (b) Anne made 14 finger puppets in 4 hours. (i) Find the number of finger puppets made by Anne in 18 hours. (ii) How many hours did she spend to make 56 finger puppets? Assume the relationship is directly proportional. 11. Given that F = k T , and T = 4 + S, where k is the constant of the proportion. (a) Find the value of k when F = 4 and S = 5. (b) State the equation that relates F and T. (c) Calculate the value of F when S = 32. 12. Given that R is inversely proportional to S and the cube of T. R = 1 4 when S = 5 and T = 2. Find the value of R when S = 10 and T = 3. ©Praxis Publishing_Focus On Maths
ALGEBRAIC 4EXPRESSIONS Applications of this chapter Some of the common examples and applications of algebra use in daily life includes: performing basic activities such as throwing a ball into a basketball net, cooking with the right amount of ingredients, making activity schedules and finding tax liabilities. Algebra is in action when children try to play spatial reasoning games. How is algebra used in sports? 114 ©Praxis Publishing_Focus On Maths
Concept Map Learning Outcomes • Distinguish between coefficients, variables, constants and terms in the algebraic expressions. • Distinguish between like and unlike terms of algebraic expressions. • Recognise the algebraic expression in variety of form, addition and subtraction of algebraic form, and to simplify the algebraic form. • Perform multiplication and division involving algebraic expressions. • Simplify the algebraic expressions. • Perform computations involving algebraic expressions. • Simplify algebraic fractions. • Add and subtract two algebraic fractions. • Multiply and divide two algebraic fractions. • Find the factors of algebraic terms. • Find the common factors of algebraic terms. • Factorise algebraic expressions. • Coefficient • Variable • Term • Constant • Like term • Unlike term • Algebraic expression • Commutative Law • Associative Law • Distributive Law • Expansion • Factorisation Key Terms Algebraic Fractions Expansion and Factorisation Algebraic Expressions Constant Subtraction Term Addition Coefficient Multiplication Variable Division Expressions Operations on Algebraic Expressions Maths History Al-Khwarizmi is best known in revolutionising algebra and arithmetic. The word ‘algorithm’ was derived from the word ‘al-jabr’ which is part of the title of his famous book, that he first introduced the fundamental algebraic methods and techniques for solving equations. 115 ©Praxis Publishing_Focus On Maths
CHAPTER 4 Algebraic Expressions 116 What do ants eat on to survive? Thinking :1 Flashback Fill in the blanks based on the properties of mathematics. (b) (–6) × 4 = 4 × 4. Associative property of multiplication. (a) (5 × 10) 2 = 5 (10 × ) (b) 2 [(–6) × 4] = [ × (–6)] × 4 5. Distributive property of multiplication. (a) 2 × (1 + 4) = 2 × + 2 × 4 (b) 3 × (5 + 10) = 3 × 5 + 3 × 1. Commutative property of addition. (a) + 12 = 12 + 3 (b) 46 + = 17 + 2. Associative property of addition. (a) (3 + 2) + 1 = 3 + ( + 1) (b) 6 + (4 + 10) = ( + 4) + 10 3. Commutative property of multiplication. (a) 5 × 10 = 10 × Suppose we use b to represent the number of ants in the nest. (a) If 25 ants die, write an expression for the number of ants in the nest. (b) If the original number of ants doubled, write an expression for the number of ants in the nest. (c) If the original population has increased by 50, write an expression for the number of ants in the nest. (d) If we said that a nearby nest contained b + 100 ants, what would it mean? (e) Another nest in very poor soil contains b 2 ants. How much smaller than the original is this nest? 1 ©Praxis Publishing_Focus On Maths
Algebraic Expressions CHAPTER 4 117 4.1 Algebraic Expressions A Representing and identifying variables in given situations The first four figures of a sequence are as shown. (a) Draw the next two figures of the sequence. (b) Copy and complete the table. Figure number Number of triangles Number of lines 1 1 1 + 1 × 2 = 3 2 2 1 + 2 × 2 = 5 3 3 1 + 3 × 2 = 7 4 4 1 + 4 × 2 = 9 5 6 n 2 Algebra is part of Mathematics language that describes various patterns around us. If there is any repetitive pattern, we can use algebra to simplify and then have a general expression to describe this pattern. For example, let’s make the triangle pattern using matchsticks as shown in the diagram below. Here, the triangles are connected with each other. In this matchstick pattern, the number of matchsticks is 3, 5, 7 and 9, which is one more than twice the number of triangles in the pattern. Therefore, the pattern is 2n + 1, where n is the number of triangles. ©Praxis Publishing_Focus On Maths
CHAPTER 4 Algebraic Expressions 118 2n + 1 A variable is a quantity whose value has not been determined. For example, Mandy bought some oranges for her children. The variable is the number of oranges Mandy bought. Generally, letters or pictorial symbols can be used to represent variable. For example, Mandy bought n oranges for her children. Fixed value Always constant at any time Varied value Changes over the time Variables The variable in a situation can be represented by an appropriate letter. For example, (a) the distance between Darren’s house and his school is d km. d is a variable and it has a fixed value because the distance between Darren’s house and his school is always constant. (b) the profit of a grocery store is $x per month. x is a variable and it has a varied value because the profit of the grocery store depends on the sales in every month. (c) the height of Ravi is y cm. y is a variable and it has a varied value because the height of Ravi changes according to his age. Coefficient Variable Constant EXAMPLE 1 Rewrite each of the following statements, replacing the variable with suitable letters. (a) Ms Rose gave some pencils to her students. (b) There are a number of students who scored A in the mathematics test. Solution: (a) Ms Rose gave n pencils to her students. (b) There are x students who scored A in the mathematics test. Variables can also be represented by illustration symbols such as +, ✯, … Sometimes, letters can be used to represent objects too. EXAMPLE 2 Determine whether each of the following letters represents a variable quantity or an object. (a) Mr y sold his car for k dollar. (b) There are m teachers in school z. ©Praxis Publishing_Focus On Maths
Algebraic Expressions CHAPTER 4 119 Solution: (a) y is an object. k is a variable. (b) m is a variable. z is an object. B Algebraic terms with one variable Term in an algebraic expression is a number or a variable or the product of a number and variable(s) in the expression which separated by plus (+) or minus (–) signs. For example, there are 3 terms in the expression 6m – xy + 2, i.e. 6m, xy and 2. • 6m is the product of a number and a variable. • xy is the product of two variables. • 2 is a number. An algebraic term with one is the product of a variable and a number. For example, 3 × y = y + y + y = 3y 3y is known an algebraic term. In the term 3y, the number 3 is known as the coefficient of y. 3y Coefficient Variable Examples of algebraic terms with one variable: k, –2m, x 7 , 1.6p An algebraic term can be written as the product of variable(s) and its factors. For example, Algebraic term Product Variable (s) Coefficient 2pq 2 × pq pq 2 2q × p p 2q 2p × q q 2p 5r 2 st 5r × rst rst 5r 5r 2 × st st 5r 2 5r 2 s × t t 5r 2 s y represents the name of a person; k represents the selling price. m represents the number of teachers; z represents the school. ©Praxis Publishing_Focus On Maths
CHAPTER 4 Algebraic Expressions 120 EXAMPLE 3 Write down the coefficient of each of the following algebraic terms. (a) – 6x (b) 2w 5 (c) q (d) n 8 Solution: (a) –6 –6x = –6 × x (b) 2 5 2w 5 = 2 5 w (c) 1 q = 1 × q (d) 1 8 n 8 = 1 8 n EXAMPLE 4 In the algebraic term – xy 5 , state the coefficient of (a) xy, (b) y, (c) x. Solution: (a) – xy 5 = – 1 5 × x × y = – 1 5 × xy Therefore, the coefficient of xy is – 1 5 . (b) – xy 5 = – 1 5 × x × y = – 1 5 x × y Therefore, the coefficient of y is – 1 5 x. (c) – xy 5 = – 1 5 × x × y = – 1 5 y × x Therefore, the coefficient of x is – 1 5 y. • 3 × y is written as 3y, not y 3. • A coefficient of a variable can be positive or negative. • A coefficient of a variable can be a whole number, a fraction or a decimal. • 1 × k = 1k = k When we write the coefficient of a variable for a term, the negative sign must be considered. For example, –x = –1 × x ©Praxis Publishing_Focus On Maths
Algebraic Expressions CHAPTER 4 121 C Like and unlike terms Like terms are terms with the same variable. For example, 3m and 7m are like terms. Unlike terms with one variable are terms with different variables. For example, 3m, 7n, 9x and 10p are unlike terms. 3m 7m The same variable 3m 7n 9x 10p Different variables EXAMPLE 5 Determine whether each of the following pairs are like terms or unlike terms. (a) 4a and 4b (b) 7f and – f 3 Solution: (a) 4a and 4b are unlike terms. (b) 7f and – f 3 are like terms. EXAMPLE 6 Write the like terms in each of the following lists of terms. (a) 4a, 4ab, 5a (b) hk, h ²k, 5kh Solution: (a) 4a and 5a are like terms. (b) hk and 5kh are like terms. EXAMPLE 7 Identify whether each of the following pairs of algebraic terms is like terms or unlike terms. (a) mn and –8mn (b) 6xy and x (c) 0.6p2 and 2p (d) xyz 3 and –yzx Solution: (a) Like terms because they have the same variable, mn, with the same power. (b) Unlike terms because they have different variables, xy and x. (c) Unlike terms because the power of variable p in each term are different. (d) Like terms because they have the same variable, xyz, with the same power. They have different variables, a and b. Both have the same variable, f. If two algebraic terms have the same variable with the same power but in different order, these algebraic terms are also like terms. For example, 2py2 and 3y2 p are like terms. ©Praxis Publishing_Focus On Maths
CHAPTER 4 Algebraic Expressions 122 D Algebraic expressions with one variable We can derive an algebraic expression from a situation as below. Situation Total number of balls A box contains y balls. y y 5 balls are added into the box. y y + 5 7 balls are taken out from the box. y y – 7 The balls in the box are divided equally between among 3 students. y y 3 ©Praxis Publishing_Focus On Maths
Algebraic Expressions CHAPTER 4 123 Two boxes each contains y balls respectively. y y 2 × y = 2y Two boxes each contains y balls respectively and another box contains x balls, and 3 extra balls are added. y y x 2y + x + 3 Two boxes each contains y balls respectively where the mass of each ball is w gram. y y Total mass of the balls in both boxes = 2 × y × w = 2yw From the findings above, we found that the total number of balls for each situation can be written as follows: y, y + 5, y – 7, y 3 , 2 × y = 2y, 2y + x + 3 The total mass of the balls is 2yw. The total number of balls or the total mass of the balls which are written in numbers and variables are called algebraic expressions. An algebraic expression is a combination of numbers, variables and operational signs such as – or +. For example, y + 5 is an expression with two terms. 2y + x + 3 is an expression with three terms. Try the following activities in groups of 4. (a) Think of a number. (b) Add 5 to the number. (c) Double the number. (d) Subtract 6 from the number. (e) Divide the number by 3. (f) Subtract the original number. What answers did each group in the class get? Present each group's answer in class. team work ©Praxis Publishing_Focus On Maths
CHAPTER 4 Algebraic Expressions 124 EXAMPLE 8 State the number of terms in each of the following expressions. (a) 7m – 3 + m (b) 9m – 5 + 6y + 1 – w Solution: (a) 3 terms (b) 5 terms A number in an expression is considered as a term. Practice 4.1 Basic Intermediate Advanced A Identify and represent the variable in each of the following statements with suitable letter. (a) There are a number of monkeys in the zoo. (b) Alex cut a reel of string into several pieces. (c) A bus took a few hours to travel from Tokyo to Osaka. (d) Simon bought more than 20 kg of durians. B Determine whether each of the following letters represents an unknown quantity or an object. (a) Bookshop x sold y science books in a day. (b) There are m students in school n. (c) Scooter H is made in the US. The price of the scooter is k dollar. (d) An amount of y dollar is shared among z students. C Write the coefficient of each of the following terms. (a) –7k (b) 4 5 m (c) 20h (d) p (e) y 12 (f) – 3n 10 D In the term 2ghk 3 , state the coefficient of (a) g, (b) h, (c) k, (d) gh, (e) gk, (f) ghk. E Given that the term – 5 6 px3 y, state the coefficient of (a) px, (b) x3 y, (c) 5py. F State the number of terms in each of the following expressions. (a) 3 + 6r – 2 (b) 5m – 1 2 m + 8 – 7m + 400 (c) 0.2r + 4 – 2r 3 + 10 + r (d) 7x + 6y – 3x + y 4 (e) 20h – 3m 4 + 0.75y + 4m – 1 G Determine whether each of the following pairs of algebraic terms are like terms. (a) 6s, –s (b) 7p, 7q (c) 2 3 x, x 5 (d) m 10 , – 0.2m (e) 19w, 99y (f) 23h, 6n H Determine if these pairs of terms are like terms or unlike terms. (a) 3a2 and 4a2 (b) m and m 2 (c) 3 2 z2 and 3 2 y2 (d) 7n2 and 8n3 I Identify whether each of the following pairs of terms is like terms or unlike terms. (a) 3a2 b and 6ab (b) f 8 and – 0.2f (c) –pqr and 15rpq (d) 20xyz and 10xy J Write an algebraic expression for each of the following situations. (a) Ashley bought x slices of vanillaflavoured cake and y slices of chocolate-flavoured cake. If the cost of one slice of vanilla-flavoured cake and one slice of chocolate-flavoured cake are $5 and $6 respectively, express the amount to be paid by Ashley in terms of x and y. (b) A classroom can accommodate m students and a hall can accommodate n students. If the hall can accommodate more students, state the difference between the number of students can be accommodated by 2 classrooms and 3 halls. ©Praxis Publishing_Focus On Maths
Algebraic Expressions CHAPTER 4 125 4.2 Addition and Subtraction of Algebraic Expressions A microbiologist places n bacteria onto an agar plate. She counts the number of bacteria at approximately 3 hours intervals. The results are shown in the table below. Time Number of bacteria 8.00 a.m. n 11.00 a.m. 2n 2.18 p.m. 4n 5.20 p.m. 8n 8.05 p.m. 16n 11.00 p.m. 32n – 1240 (a) Explain what happens to the number of bacteria in the first 5 intervals. (b) What might be causing the number of bacteria to increase in this way? (c) What is different about the last bacteria count? (d) What may have happened to cause this? 3 A Simplifying algebraic expressions Algebraic expressions with like terms can be simplified by adding or subtracting the coefficients of the variables in algebraic terms. An algebraic expression consisting of two unlike terms cannot be simplified. For example, 3h + 4k cannot be simplified. To simplify an algebraic expression, the like terms are gathered first and then addition or subtraction of the like terms are performed. EXAMPLE 9 Simplify each of the following. (a) 3h + 4h (b) y – 3y (c) 7x + (–3x) (d) –12y – (–9y) Solution: (a) 3h + 4h = (3 + 4)h (b) y – 3y = (1 – 3)y = 7h = –2y (c) 7x + (–3x) = 7x – 3x +(–3x) = –3x = (7 – 3)x = 4x (d) –12y – (–9y) = –12y + 9y –(–9y) = +9y = (–12 + 9)y = –3y • + (+) = + • + (–) = – • – (+) = – • – (–) = + The addition and / or subtraction of algebraic expressions can be solved by the following steps. Identify algebraic terms. Rearrange the algebraic terms to gather the like terms. Add and subtract the coefficients of like terms. ©Praxis Publishing_Focus On Maths
CHAPTER 4 Algebraic Expressions 126 We follow the steps below to simplify an expression with several like terms and unlike terms. 1 Group the like terms first. 2 Simplify them. EXAMPLE 10 Simplify each of the following. (a) x + 5y + 2x – y (b) 3p – 4 + 5p – 2q + 9 Solution: (a) Like terms (b) 3p – 4 + 5p – 2q + 9 x + 5y + 2x – y Like terms = x + 2x + 5y – y = (1 + 2)x + (5 – 1)y = 3x + 4y EXAMPLE 11 Simplify each of the following. (a) (8m – 6n) + (7m + 2n – 3) (b) (4x + 3xy) – 112xy – x 2 + 62 + 11 – 10 3 x2 (c) (0.8r + 6.7s) – 1 6st 5 – 9s2 + 2st Solution: (a) (8m – 6n) + (7m + 2n – 3) Like terms = 8m – 6n + 7m + 2n – 3 Like terms = 8m + 7m – 6n + 2n – 3 = 15m – 4n – 3 (b) (4x + 3xy) – 112xy – x 2 + 62 + 11 – 10 3 x2 = 4x + 3xy – 12xy + x 2 – 6 + 1 – 10 3 x x 2 = 1 2 x = 4x + 1 2 x – 10 3 x + 3xy – 12xy – 6 + 1 = 14 + 1 2 – 10 3 2x + 3xy – 12xy – 5 = 7 6 x – 9xy – 5 The operational sign lying before a term must be moved together with the term. = 3p + 5p – 2q – 4 + 9 = (3 + 5)p – 2q + 5 = 8p – 2q + 5 When rearrange the like terms, the operation sign in front of the term must transferred together with the term. ©Praxis Publishing_Focus On Maths
Algebraic Expressions CHAPTER 4 127 (c) (0.8r + 6.7s) – 1 6st 5 – 9s2 + 2st = 0.8r + 6.7s – 6st 5 + 9s + 2st = 6.7s + 9s – 6st 5 + 2st + 0.8r = 15.7s + 0.8st + 0.8r B Determine the values of algebraic expressions The value of an algebraic expression can be determined by substituting the variables with the given values. EXAMPLE 12 Find the value of algebraic expression 3x – y 2 + 1 when (a) x = 5 and y = 8, (b) x = 2 and y = –4. Solution: (a) 3x – y 2 + 1 = 3(5) – 8 2 + 1 = 15 – 4 + 1 = 12 (b) 3x – y 2 + 1 = 3(2) – –4 2 + 1 = 6 + 2 + 1 = 9 EXAMPLE 13 The manager of an optical shop analysed the sales of contact lens at his shop on a certain day. He found that 2 5 of the adult customers buy the contact lenses while 14 teenage customers did not buy the contact lenses. (a) Based on the above situation, write an algebraic expression to represent the total customers who buy the contact lenses. (b) If there are 25 adult customers and 27 teenage customers, how many contact lenses are sold if each customer buys two boxes of contact lenses? Scientific calculator can be used to determine the value of an algebraic term by using the following steps: 1 Enter the algebraic term by pressing 3 ALPHA X – ALPHA Y ab/c 2 + 1 CALC B When screen X? is displayed, press 5 = to enter the value of x. C When screen Y? is displayed, press 8 = to enter the value of y. Answer: 12 Substitute x = 2 and y = –4. Substitute x = 5 and y = 8. Tira has three wooden blocks, A, B and C. The length of Wooden block B is (6y + x) cm. The difference between the length of Wooden blocks B and A is 2x cm while the difference between the length of Wooden blocks C and B is (3y – 2x) cm. Given Wooden block B is longer than Wooden block A but shorter than Wooden block C. Express the total length of all the three wooden blocks in an algebraic expression. ©Praxis Publishing_Focus On Maths
CHAPTER 4 Algebraic Expressions 128 Solution: (a) Let x = number of adult customers and y = number of teenage customers Therefore, the total number of customers that buy the contact lenses = 2 5 x + y – 14 (b) When x = 25 and y = 27, The total number of customers = 2 5 (25) + 27 – 14 = 10 + 27 – 14 = 23 Total sales of contact lenses = 23 × 2 = 46 boxes Practice 4.2 Basic Intermediate Advanced A Simplify each of the following. (a) 6k + 3k (b) 5m + 7m (c) –3y + 8y (d) 9x + (–4x) (e) –10h + 2h (f) –4f + (–7f ) B Simplify each of the following. (a) 7x – 4x (b) 11y – 8y (c) 6m – 9m (d) –2p – 5p (e) 3z – (–z) (f) –4k – (–6k) C Simplify. (a) 2x + 4 + 6x – 1 (b) 8 – 3y – 2y + 4 (c) 10m + 9 – 6m – 10 (d) 7p + 1 – 6p – 3 + 2p (e) 5e – 8 + 2e – 7e + 3 (f) w 8 – 2 + w 4 – 9 + 1 2 w D Simplify. (a) x + 3y + 4x – y (b) 6m – 4n + 5n – 8m (c) 7p + 6q + 8 – 3p – 4q (d) 9w – 9f + 3w – 2 + 5f + 7 (e) 10k + 2k – 3e + 6k – 2e (f) 4h – 2g + 3h – 7h + 6g E Simplify. (a) 1 3 b – 1 6 b (b) f + 1 3 f – 1 4 g F Simplify each of the following. (a) (7a + 10b) + (3a – 2b + 7) (b) (xy – 8y + 11) – (7xy + 5y – 2) (c) (3p + q) + (9p – 5q + 4) – (p – 4q – 1) (d) (–4m – 8n + 7mn) – (– 3m + mn) (e) 1 2 5 pq – 3 4 pr2 – 1 3 4 pq + pr – 82 + PQ 10 (f) 1 2 3 kh + 6pq2 + 1 1 6 kh – pq2 + kh G Given that p = 3, q = –8 and r = 4 5 , find the value of each of the following algebraic expressions. (a) 2pq + p2 – 5q (b) – 1 2 q + 6pr – 7 H Given that (8p – 5pq + 2) – (p + 6pq – 9) + (10p – pq) = ap + bpq + c, where a, b and c are integers. Determine the values of a, b and c. ©Praxis Publishing_Focus On Maths
Algebraic Expressions CHAPTER 4 129 4.3 Multiplication of Algebraic Expressions A Multiplying algebraic expressions (I) Repeated multiplication of algebraic expressions Objective : To make generalisation on repeated multiplication of algebraic expressions. Instruction : Do this activity in a group of four. 1. Find the area for each of square below in the form of repeated multiplication. 4 cm 4 cm p p q q Area = cm × cm = 2 cm2 Area = × = Area = × = 2. Find the volume for each of cube below in the form of repeated multiplication. 4 cm 4 cm 4 cm p p p q q q Volume = cm × cm × cm = 3 cm3 Volume = × × = Volume = × × = 3. From the findings, what generalisations can be made about (a) p × p × p × p? (b) p × p × p × p × p ... × p, where the multiplication of p is repeated n times? (c) (i) (p + q) × (p + q)? (ii) (p + q) × (p + q) × (p + q)? (iii) (p + q) × (p + q) × (p + q) × (p + q)? (iv) (p + q) × (p + q) × (p + q) × (p + q) × ... × (p + q) Where the multiplication of (p + q) is repeated n times? 1 From the findings in Activity 1, it is found that: p × p = p2 p × p × p = p 3 p × p × p × ... × p = pn Repeated multiplication of p by 2 times. Repeated multiplication of p by 3 times. Repeated multiplication of p by n times. ©Praxis Publishing_Focus On Maths
CHAPTER 4 Algebraic Expressions 130 Diagram below shows a cube of length 2 m. 2 m 2 m 2 m Therefore, the area of each surface of the cube = 2 m × 2 m = 22 m2 The volume of the cube = 2 m × 2 m × 2 m = 23 m3 Hence, generalisations for repeated multiplication of algebraic expressions are as follows: (p + q ) × (p + q ) = (p + q ) 2 (p + q ) × (p + q ) × (p + q ) = (p + q ) 3 (p + q ) × (p + q ) × (p + q ) × (p + q ) = (p + q ) 4 In general, (p + q ) × (p + q ) × (p + q ) × ... × (p + q ) = (p + q ) n Power on n Formulae for repeated multiplication: ● a × a × a × ... × a = an (a repeated n times) ● (a + b) × (a + b) × (a + b) × … × (a + b) = (a + b) n [(a + b) repeated n times] When the side of a cube is a m, a m a m a m The area of each surface of the cube = a m × a m = a 2 m2 The volume of the cube = a m × a m × a m = a3 m3 Generally, the power of a is equal to the number of a in the repeated multiplication. For example, a × a = a2 a is repeated twice in the repeated multiplication. Therefore, the power of a in a2 is 2. a × a × a = a3 a is repeated thrice in the repeated multiplication. Therefore, the power of a in a3 is 3. Repeated multiplication of the algebraic expression (a + b) by n times. EXAMPLE 14 Simplify each of the following. (a) p × p × p × p × p × p (b) (2n – 5) × (2n – 5) × (2n – 5) × (2n – 5) (c) (xy + 7y) × (xy + 7y) × (xy + 7y) ©Praxis Publishing_Focus On Maths
Algebraic Expressions CHAPTER 4 131 Solution: (a) p × p × p × p × p × p = p6 (b) (2n – 5) × (2n – 5) × (2n – 5) × (2n – 5) = (2n – 5)4 (c) (xy + 7y) × (xy + 7y) × (xy + 7y) = (xy + 7y)3 Multiplication is repeated 6 times. Multiplication is repeated 4 times. Multiplication is repeated 3 times. (II) Product of like terms To find the product of like terms, multiply the coefficients and the variables. EXAMPLE 15 Find the product of each of the following. (a) 3p × 2p (b) 15a2 b × (–2a 2 b) Solution: (a) 3p × 2p = 3 × p × 2 × p = 3 × 2 × p × p = 6 × p2 = 6p2 • (+) × (+) = + • (+) × (–) = – • (–) × (+) = – • (–) × (–) = + (b) 15a2 b × (–2a2 b)= 15 × a × a × b × (–2) × a × a × b = 15 × (–2) × a × a × a × a × b × b = –30 × a 4 × b2 = – 30a4 b2 (III) Product of unlike terms We group together the same variables and then multiply the coefficients and the variables to find the product of unlike terms. EXAMPLE 16 Find the product of each of the following. (a) 4mn × 7mt (b) 6xyz × 1– 2 4 xy 2 2 Solution: (a) 4mn × 7mt = 4 × m × n × 7 × m × t = 4 × 7 × m × m × n × t = 28m2 nt (b) 6xyz × 1– 2 4 xy 2 2 = 6 × x × y × z × 1– 2 4 2 × x × y × y = 6 × 1– 2 4 2 × x × x × y × y × y × z 2 times 3 times = –3x2 y3 z ©Praxis Publishing_Focus On Maths
CHAPTER 4 Algebraic Expressions 132 (IV) Multiplying algebraic expressions by a number Objective : Multiplying algebraic expressions with a number. Material : Manila card, tracing paper, scissors, ruler and pencil. Instruction : Do this activity in pair. P 2 a 2 Q b 1. Trace both of the rectangles P and Q in the diagram above on the manila card. Cut out each of the rectangle. 2. Then write the area of rectangles P and Q on the manila card. 3. Arrange the cards so that they will exactly cover the rectangle R. R 4. Copy and complete: (a) The area of rectangle R = Length × Width = × (b) The total area of rectangles that cover rectangle R = Area of rectangle P + Area of rectangle Q = + 5. Based on the results in 4(a) and 4(b), discuss and present you findings in class. 2 From the findings of Activity 2, we found that the area of rectangle R is 2(a + b) = 2a + 2b. Note that 2a + 2b can be obtained when multiplying each term in expression a + b with term 2. That, 2(a + b) = 2 × a + 2 × b = 2a + 2b In multiplication of an algebraic expression by a number, every term in the expression is multiplied by the same number. When multiplying two or more algebraic expressions that consist of one term, the number is multiplied with number and the variable is multiplied with variable. Multiply each term in the bracket with the number. ©Praxis Publishing_Focus On Maths
Algebraic Expressions CHAPTER 4 133 EXAMPLE 17 (a) 3(5x + 2y) – 7(2x – 3y) (b) 1 2 (8uv + 4w – 3) + 1 3 (6uv – 9w – 2) Solution: (a) 3(5x + 2y) – 7(2x – 3y) = 15x + 6y – 14x + 21y = 15x – 14x + 6y + 21y = x + 27y a(b + c) = ab + ac (b) 1 2 (8uv + 4w – 3) + 1 3 (6uv – 9w – 2) = 4uv + 2w – 3 2 + 2uv – 3w – 2 3 = 4uv + 2uv + 2w – 3w – 3 2 – 2 3 = 6uv – w – 13 6 EXAMPLE 18 Find the product for each of the following. (a) 2mn2 × 7m3 n (b) 15pq × 1– 2 3 pq3 r2 Solution: (a) 2mn2 × 7m3 n = 2 × m × n × n × 7 × m × m × m × n = 2 × 7 × m × m × m × m × n × n × n = 14m4 n3 (b) 15pq × 1– 2 3 pq3 r2 = 15 × p × q × 1– 2 3 2 × p × q × q × q × r = 15 × 1– 2 3 2 × p × p × q × q × q × q × r = –10p2 q 4 r Repeated multiplication of algebraic expression can be simplified to the following form. a × a × a × b × b = a3 b2 Write each expression as multiplication of factors. Gather the numbers and the same variables. ©Praxis Publishing_Focus On Maths
CHAPTER 4 Algebraic Expressions 134 B Properties of multiplication of algebraic expressions (I) Commutative property of multiplication The commutative property of multiplication says that the order in which we multiply do not matter. For example, p × q = q × p pq = qp From the example above, we can say that the commutative property of multiplication indicates that changing the order of factors does not change the product. (II) Associative property of multiplication The associative property of multiplication states that the terms in a multiplication expression can be regrouped using parentheses. For example, (a × b) × c = a × (b × c) = ab × ac (III) Distributive property of multiplication The distributive property of multiplication indicates that multiplying a number by a group of numbers added together is the same as doing each multiplication separately. For example, 8 × (2 + 6) = 8 × 2 + 8 × 6 = 16 + 48 = 64 Similarly, in algebra, we can apply this property to remove the brackets. a(x + y) = a × x + a × y and a(x – y) = a × x – a × y This is known as expansion. For example, 3(x + y) = 3 × x + 3 × y = 3x + 3y EXAMPLE 19 Expand. (a) 4(x + y) (b) 6y(a – b) (c) 7a(2x + m) (d) 10(2y – 4z) Solution: (a) 4(x + y) = 4 × x + 4 × y = 4x + 4y (b) 6y(a – b) = 6y × a – 6y × b = 6ay – 6by ©Praxis Publishing_Focus On Maths
Algebraic Expressions CHAPTER 4 135 (c) 7a(2x + m) = 7a × 2x + 7a × m = 14ax + 7am EXAMPLE 20 Expand. (a) –2a(a + b) (b) –5ab(m – n) (c) –6(4x + 2y) (d) –11(–6m – 4n) Solution: (a) –2a(a + b) = –2a × a + (–2a) × b = –2a2 – 2ab (c) –6(4x + 2y) = (–6) × 4x + (–6) × 2y = –24x – 12y (d) 10(2y – 4z) = 10 × 2y – 10 × 4z = 20y – 40z (b) –5ab(m – n) = –5ab × m – (–5ab) × n = –5abm + 5abn (d) –11(–6m – 4n) = (–11) × (–6m) – (–11) × 4n = 66m + 44n Practice 4.3 Basic Intermediate Advanced A Simplify each of the following. (a) k × k × k × k × k (b) xy × xy × xy × xy × xy × xy × xy (c) (h + 5) × (h + 5) × (h + 5) × (h + 5) (d) (2a – 5b) × (2a – 5b) B Write each of the following in the form of repeated multiplication. (a) (7 – x)3 (b) (2pq)4 (c) (3ab + 2p) 2 C Find the product of each of the following. (a) 8x × (–9x) (b) 3ab × 4ab (c) (–2a2 c) × 7a2 c (d) 4xyz × 6xyz D Find the product of each of the following. (a) 5pr × 2qr (b) –4kmn × 3m (c) –6xy 2 × 7yz (d) –1 1 2 pq × 4 9 p2 r E Find the product for each of the following. (a) 2a × 5a7 (b) –6x3 y × (– 8x4 y) (c) 1 2 ab5 × 14bc2 (d) 18p2 q6 × 1– 1 3 p2 × 2 5 pq3 F Expand each of the following. (a) 13(x + y) (b) 6a(a + 4b) (c) 15(10 – c) (d) 3ab(3b – 2a) G Expand each of the following. (a) –3(m + 2) (b) –2a(7b + 2a) (c) – 6pq(c – 5) (d) –3b(–k – m) 8 The volume of a cube is (x – 2y) 3 cm3 . What is the total surface area of the cube? ©Praxis Publishing_Focus On Maths