hapter 1 Prime Numbers, Factors and Multiples Preface Secondary 1 (N) Mathematics Tutorial 1 and 1 are designed to prepare Secondary 1 students in their understanding and application of mathematical concepts, skills and processes. What s covered in this book? Written in accordance with the latest syllabus, each chapter includes Objectives, Key oncepts and Formulae and Worked Examples to supplement and complement the lessons taught in school. Practice questions are structured as core, consolidation and challenging to ensure steady improvement and quick mastery of concepts. ooks 1 and 1 cover all the topics for the entire school year. dditional Feature that provides important notes, tips, examples and common student errors. Important concepts are highlighted to enhance understanding and retention. This encourages students to self-study, regardless of the level of competency they are at. nswer Key Fully worked solutions are provided for students to understand better how each problem is solved. These also serve as a tool for self study and assessment. Enhanced resources available Learning www.onlineresources.sapgrp.com There are 4 assessment papers: Mid Year Examination Paper 1 and 2 (found in ook 1) and End of Year Examination Paper 1 and 2 (found in ook 1). These assessment papers are available online, FREE. This book will help students gain mastery and confidence in learning Mathematics systematically. The questions are also closely aligned to examinations in Singapore and hence students will be better equipped to face and excel in the examinations. The Editorial Team
ontents hapter 1 Prime Numbers, Factors and Multiples ontinuation from Secondary 1 (N) Mathematics Tutorial 1 (hapters 1 6) hapter Objectives Page 7 8 9 10 11 12 Ratio, Rate and Speed Find ratios involving two or more quantities alculate average speed onvert speed from one unit to another Solve problems involving ratio, rate and speed Lines, ngles and Triangles pply properties of vertically opposite angles, angles on a straight line and angles at a point pply properties of interior angles, corresponding angles and alternate angles formed by parallel lines Solve problems involving angles in triangles Percentage Express one quantity as a percentage of another quantity ompare two quantities in terms of percentage alculate percentage increase and decrease in quantities Solve problems involving reverse percentages Solve problems involving discounts and GST rea and Perimeter of Plane Figures alculate the area of squares, rectangles, parallelograms, trapeziums and circles alculate the area and perimeter of composite figures Volume and Surface rea of Solids alculate volume and surface area of cubes, cuboids, prisms and cylinders Solve problems involving volume and surface area of composite solids Statistics onstruct and interpret tables, bar graphs, pictograms, line graphs and pie charts Explain why certain statistical diagrams can lead to misinterpretation of data 1 25 49 71 91 109 Fully worked solutions S1 S27 omplete your learning with End of Year Examination Paper 1 and 2. ownload from www.onlineresources.sapgrp.com
hapter 7 Objectives Find ratios involving two or more quantities alculate average speed onvert speed from one unit to another Solve problems involving ratio, rate and speed Ratio, Rate and Speed Key oncepts and Formulae 1 The ratio of a is to b is written as a : b, where a and b represent similar quantities. ratio has no units. 2 Rate is a comparison of two different quantities, e.g. 35 litres/km. 3 Speed, Time and istance are related to each other by the following triangle: S T istance = Speed Time S T Time = istance Speed S T Speed = istance Time Total istance travelled 4 verage Speed = Total Time taken Note: Rest time is also taken into consideration when Total Time taken is calculated! 5 onversion of speed to other units: (1000 x) m x km/h = 3600 s x km/h = x 3.6 m/s y m/s = y 1000 km 1 = 3600 y 3600 h 1000 km/h y m/s = 3.6y km/h km/h 3.6 3.6 m/s hapter 7 Ratio, Rate and Speed
Worked Example 1 (a) Simplify 1 1 2 : 2 : 5 1 4. (b) The ratio of aron s height to etty s height is 7 : 6. The ratio of etty s height to harlie s height is 4 : 5. Find the ratio of aron s height to that of harlie s height. Solution: LM of 4 and 6 is 12. Hence, we convert etty s quantity in both ratios to 12. We can compare aron s quantity to harlie s quantity directly only if etty s quantity is the same. (a) 1 1 1 2 : 2 : 5 4 = 3 2 : 2 : 21 4 4 4 = 6 : 8 : 21 (b) aron : etty = 7 : 6 = 14 : 12 etty : harlie = 4 : 5 = 12 : 15 Required ratio of aron : harlie = 14 : 15 (LM of 2 and 4 is 4) (Multiply each term by 4) (b) lternative method Students 25 units Teachers 2 units Male teachers = 2 5 of teachers = 2 5 2 units = 4 5 units Students : Male teachers 25 : 4 5 = 125 : 4 Worked Example 2 In a school of 1500 students, the ratio of students to teachers is 25 : 2. The ratio of male teachers to female teachers is 2 : 3. Find (a) the number of teachers in the school, (b) the ratio of students to male teachers. Solution: (a) Number of teachers = 1500 25 2 = 120 (b) Number of male teachers = 2 2 + 3 120 = 48 Ratio of students to male teachers = 1500 : 48 = 125 : 4 hapter 7 Ratio, Rate and Speed
Worked Example 3 (a) James car travelled 1160 km on 80 l of petrol. Find the rate of petrol consumption in km/l. (b) typist typed 180 words in the first 3 minutes, 70 words in the next 2 min and 388 words in another 6 minutes. Find the average rate of typing. Solution: 1160 km (a) Rate of petrol consumption = 80l = 14.5 km/l (b) verage rate of typing = 180 + 70 + 388 3 + 2 + 6 = 638 11 = 58 words/min If the rate is not constant, then average rate is used to find the rate. Worked Example 4 (a) onvert 100 km/h to m/s. (b) car travels 320 km from Singapore to Kuala Lumpur at an average speed of 100 km/h. The car travels the first 220 km in 2 hours. Find the average speed for the second part of the journey. Solution: 100 km (a) 100 km/h = 1h 100 000 m = 3600 s = 27.8 m/s (correct to 3 s.f.) (b) Time taken for entire journey = 320 100 h = 3.2 h Time taken for second part = 3.2 h 2 h = 1.2 h istance travelled in second part = 320 km 220 km = 100 km verage Speed = Total istance travelled Total Time taken = 100 km 1.2 h = 83 1 3 km/h hapter 7 Ratio, Rate and Speed 1 h = 60 min = 60 60 s = 3600 s S T s the car might not travel at constant speed for the second part, hence the term average speed is used.
Ratio 1 Simplify the following ratios. ore Practice Remember to: Express a ratio in its simplest term. onvert quantities to the same unit. You may use the fraction sign calculator to help you simplify the ratio. in the (a) 100 : 250 (b) 0.18 : 0.32 (c) 70 g : 1 kg (d) 150 min : 2 h (e) 1 1 4 1 : 3 : 6 2 (f) 45 mm : 6 cm : 1.35 m (g) 3500 ml : 1.2 l (h) $0.50 : 200 : $3 hapter 7 Ratio, Rate and Speed
Speed 9 Solve the following problems. S T Ensure that the units of speed, time and distance are consistent when applying the formula. E.g. km/h; km; h (a) car travels at a constant speed of 90 km/h from 10.30 am to 1.00 pm. alculate the distance travelled in km. (b) plane travelled 8494 km from Singapore to New Zealand in 10 h 20 min. alculate the average speed of the plane in km/h. (c) Sam Whittingham was the fastest cyclist on record (in 2009) with a speed of 37.5 m/s. alculate the time taken in hours to cycle 53.2 km. (d) cheetah sprints at a speed of 112 km/h. alculate the time taken in seconds it takes to sprint 1.4 km. 13 hapter 7 Ratio, Rate and Speed
hapter 8 Objectives Lines, ngles and Triangles pply properties of vertically opposite angles, angles on a straight line and angles at a point pply properties of interior angles, corresponding angles and alternate angles formed by parallel lines Solve problems involving angles in triangles and quadrilaterals Key oncepts and Formulae 1 Types of triangles Right triangle Isosceles triangle Equilateral triangle Scalene triangle 60º a a 60º 60º 2 sides are perpendicular 2 sides are equal; base s are equal ll 3 sides are equal ll 3 sides are unequal 2 Types of angles cute angle Right angle Obtuse angle Straight angle Reflex angle 90º 180º angle < 90º angle = 90º 90º < angle < 180º angle = 180º 180º < angle < 360º 3 ngles on a straight line aº bº aº + bº = 180º ( s on a st. line) 25 hapter 8 Lines, ngles and Triangles
4 ngles at a point aº dº bº cº aº + bº + cº + d º = 360º ( s at a point ) 5 Vertically opposite angles b a c d a = c b = d (vert. opp. s) (vert. opp. s) 6 ngles formed by parallel lines orresponding angles Interior angles lternate angles aº cº eº bº dº f º aº = bº (corr. s, // ) cº + dº = 180º (int. s, // ) eº = f º (alt. s, // ) 7 ngles in a Triangle aº aº bº cº aº + bº + cº = 180º ( sum in ) bº cº = aº + bº (ext. of ) cº 26 hapter 8 Lines, ngles and Triangles
Worked Example 1 In the figure, and E are straight lines. 2xº 5xº xº F Only assume lines are straight if the question states so. E Find (a) the value of x, (b) E, (c) obtuse E. Solution: (a) 2xº + 5xº + xº = 180º ( s on a st. line) 8xº = 180º x = 22.5 (b) E = (vert. opp. s) = 2xº = 2(22.5º) = 45º When an unknown angle is denoted with its degree, e.g. xº, then x represents a number, i.e. x itself has no unit. (c) E = 180º E ( s on a st. line) = 180º 45º = 135º When stating reasons, you may use abbreviations. 27 hapter 8 Lines, ngles and Triangles
Types of ngles ore Practice 1 In each of the following, name the type of angle. cute angle angle < 90º Right angle angle = 90º Obtuse angle 90º < angle < 180º Straight angle angle = 180º Reflex angle 180º < angle < 360º 90º 180º (a) (b) (c) (d) (e) (f) (g) (h) 30 hapter 8 Lines, ngles and Triangles
ngles on a Straight Line 2 Given that O is a straight line, find the value of x in each of the following figures. Sum of s on a straight line = 180º. ( s on a st. line) Observe that x is a number with no units. (a) (b) xº xº 140º O xº 70º O (c) (d) 2xº 3xº xº O xº O xº + 16º 2xº 28º 31 hapter 8 Lines, ngles and Triangles
hapter 7 ore Practice 1. (a) 100 : 250 = 100 50 : 250 = 2 : 5 50 (b) 0.18 : 0.32 = 18 : 32 = 18 2 : 32 2 = 9 : 16 (c) 70 g : 1 kg = 70 g : 1000 g = 70 10 : 1000 10 = 7 : 100 (d) 150 min : 2 h = 150 min : 120 min = 150 30 : 120 30 = 5 : 4 (e) 1 1_ 4 : 3 : 6 1_ 2 = 5_ 4 : 3_ 1 : 13 2 13 ): 4(3) : 4( 2 ) = 4( 5_ 4 = 5 : 12 : 26 (f) 45 mm : 6 cm : 1.35 m = 45 mm : 60 mm : 1350 mm = 45 15 : 60 15 : 1350 15 = 3 : 4 : 90 (g) 3500 ml : 1.2 l = 3500 ml : 1200 ml = 3500 100 : 1200 100 = 35 : 12 (h) $0.50 : 200 : $3 = 50 : 200 : 300 = 50 50 : 200 50 : 300 50 = 1 : 4 : 6 2. (a) x = 2y x_ y = 2 x : y = 2 : 1 (b) x = 1_ 2 y x_ y = 1_ 2 x : y = 1 : 2 divide both sides by y divide both sides by y (c) 2x = 3y 2x y = 3 divide both sides by y x_ y = 3_ 2 x : y = 3 : 2 divide both sides by 2 (d) 7x = 8y 7x y = 8 divide both sides by y x_ y = 8_ 7 divide both sides by 7 x : y = 8 : 7 S hapter 7 (e) x y = 3y 5x x + 5x = 3y + y 6x = 4y 6x y = 4 divide both sides by y x_ y = 4_ 6 divide both sides by 6 (f) = 2_ 3 x : y = 2 : 3 7x y 2 = x 7x y = 2x multiply both sides by 2 7x 2x = y 5x = y 5x y = 1 divide both sides by y x_ y = 1_ 5 divide both sides by 5 x : y = 1 : 5 3. (a) a : b and b : c = 2 : 3 = 3 : 4 Since b is the same value i.e. 3, a : b : c = 2 : 3 : 4 (b) a : b = 1 : 3 b : c = 6 : 7 = 2 : 6 Since b is the same value i.e. 6, a : b : c = 2 : 6 : 7 (c) a : b = 3 : 8 and b : c = 4 : 7 = 8 : 14 Since b is the same value i.e. 8, a : b : c = 3 : 8 : 14 (d) a : b = 2 : 5 and b : c = 4 : 3 = 8 : 20 = 20 : 15 Since b is the same value i.e. 20, a : b : c = 8 : 20 : 15 (e) a : b = 1 : 13 and c : b = 23 : 26 = 2 : 26 Since b is the same value i.e. 26, a : b : c = 2 : 26 : 23 (f) b : a = 7 : 5 and c : b = 7 : 2 = 14 : 10 = 49 : 14 Since b is the same value i.e. 14, a : b : c = 10 : 14 : 49 4. (a) 2 x : 2 2 = 6 : 4 2x : 4 = 6 : 4 \ 2x = 6 x = 3 (b) x : 3 = 3 7 : 1 3 x : 3 = 21 : 3 (c) \ x = 21 3 11 : x 3 = 33 : 15 33 : 3x = 33 : 15 \ 3x = 15 x = 5