Math class 8

95 Oasis School Mathematics-8 95 (iii) R(–9, –3) Reflection y-axis R'(9, –3) (iv) M(4, 7) Reflection y-axis M'(–4, 7) Example: 4 Find the reflection image of ∆ABC with vertices A(3, 3), B(–3, –2) and C(5, 1) on (i) x–axis (i.e. y = 0) (ii) y–axis (i.e. x = 0). Also, show ∆ABC and ∆A'B'C' on the same graph paper. Solution: (i) x–axis (.i.e. y = 0) We have, P(x, y) Reflection x-axis P'(x, –y) ∴ A(3, 3) Reflection x-axis A'(3, –3) ∴ B(–3, –2) Reflection x-axis B'(–3, 2) ∴ C(5, 1) Reflection x-axis C'(5, –1) Hence, ∆ABC and its image ∆A'B'C' are shown on the graph. (ii) Y–axis (i.e. x = 0) We have, P(x, y) Reflection y-axis P'(–x, y) ∴ A(3, 3) Reflection y-axis A'(–3, 3) ∴ B(–3, –2) Reflection y-axis B'(3, –2) ∴ C(5, 1) Reflection y-axis C'(–5, 1) Hence, ∆ABC and its image ∆A'B'C' are shown on the graph. Exercise 6.1 1. Copy the following figures in your copy and find the images of the following figures under the reflection on the line 'M'. (a) (d) (b) (e) (c) Y Y' X' X B A C C' A' B' Y Y' X' X B A C B' C' A'

96 Oasis School Mathematics-8 2. Find the images of the following points under the reflection on x–axis (i.e. on y = 0). (a) A(5, 3) (b) B(7, –4) (c) C(–2, –3) (d) D(–5, 4) 3. Find the images of the following points under the reflection on y–axis i.e. on x = 0). (a) N(3, 4) (b) Q(–4, 5) (c) R(–4, –9) (d) S(6, –5) 4. Find the axis of reflection in each of the following cases: (a) A(4, 6) → A'(4, –6) (b) M(6, 5) → M'(–6, 5) 5. Find the reflection image of ∆ABC with vertices A(3, 2), B(5, 6) and C(8, 1) on (a) x–axis (i.e. y = 0) (b) y–axis (i.e. x = 0). Also show ∆ABC and ∆A'B'C' on the same graph paper. 6. Find the reflection image of ∆PQR with vertices P(4, –2), Q(8, –2) and R(8, 2) on (a) y–axis (i.e. x = 0) (b) x–axis (i.e y = 0) Also show ∆PQR and ∆P'Q'R' on the same graph paper. 7. Draw a parallelogram on graph paper where A(–2, 3), B(–1, 1), C(4, 1) and D(3,3). Find the image of the parallelogram and plot on graph paper after reflecting on (a) x–axis and (b) y–axis. 8. Find the image of the quadrilateral ABCD whose vertices are A(–5, 4), B(–3, 1), C(–1, 3) and D(–2, 6) under reflection on y–axis. Present the quadrilateral ABCD and its image on the graph. 1. Consult your teacher. 2. (a) (5, –3) (b) (7, 4) (c) (–2, 3) (d) (–5, –4) 3. (a) (–3, 4) (b) (4, 5) (c) (4, –9) (d) (–6, –5) 4. (a) X-axis (b) Y-axis 5. (a) A'(3, -2), B' (5, -6) C' (8, -1) (b) A'(-3, 2), B' (-5, 6) C' (-8, 1) 6. (a) P'(-4, -2), Q' (-8, -2) R' (-8, 2) (b) P'(4, 2), Q' (8, 2) R' (8, -2) 7. (a) A'(-2, -3), B' (-1, -1) C' (4, -1), D' (3, -3) (b) A'(2, 3), B' (1, 1) C' (-4, 1), D' (-3, 3) 8. A'(5, 4), B' (3, 1) C' (1, 3), D' (2, 6) Answer

97 Oasis School Mathematics-8 97 6.2 Rotation–Review Study the following illustrations and get the idea of rotation of a point. Now, we take a point P, that rotates about a point O as shown in figures. In figure (i), ∠POP'= 90° and OP = OP' i.e. the point P rotates about O through angle 90° in the anticlockwise direction. In figure (ii), ∠POP' = 90° and OP = OP' i.e. the point P rotates about O through an angle 90° in the clockwise direction. In figure (iii), OP = OP' and ∠POP'= 180° i.e. the point P rotates about O through a 180° in both direction producing the same image. In figure, (iv), OP = OP' and ∠POP' = 270° i.e. the point P rotates about O through an angle 270° in anticlockwise direction. In figure (v), OP = OP', ∠POP' = 360°, the point P rotates about O through an angle of 360° in both direction producing the same image. Thus, a rotation is defined when its centre, the angle of rotation and the direction of the rotation are given. The direction of rotation can be clockwise or anticlockwise direction. Rotation is a rule which shifts each point of an object in the same direction through a certain angle about a fixed point. Figure (i) Figure (ii) Figure (iii) Figure (iv) Figure (v)

98 Oasis School Mathematics-8 I. Rotation of a Line PQ, about the Point O Through 90° in Anticlockwise Direction Steps: • Join OP and OQ. • Taking O as centre and OP as radius, draw an arc in anticlockwise direction. • At O, draw ∠P'OP = 90° and OP = OP' meeting the arc through P at P'. • Similarly, taking O as centre and OQ as radius draw an arc in anticlockwise direction. • At O, draw ∠Q'OQ = 90° and OQ = OQ', meeting the arc drawn through Q at Q'. • Join P'Q'. Hence, P'Q' is the image of PQ. II. Rotation of a ∆ABC about the Point O Through 60° in Clockwise Direction Let ABC be a triangle and O is the point. Steps: • Join AO, BO and CO. • Taking O as centre and OA as radius, draw an arc in clockwise direction. • At O, draw ∠A'OA = 60° and OA = OA' meeting the arc drawn through A at A'. • Similarly, taking O as centre and OB and OC as radius, draw arcs in clockwise direction respectively. • At O, draw ∠B'OB = 60° and OB = OB' and ∠C'OC = 60° and OC = OC' meeting the arcs drawn through B and C at B' and C' respectively. • Join A'B', B'C' and A'C'. Hence, ∆A'B'C' is the image of ∆ABC.

99 Oasis School Mathematics-8 99 6.3 Properties of Rotation The following are the properties of rotation. (i) Area of object and its image are equal. (ii) The centre of rotation is the invariant point. (iii) Each point of an object turns through an equal angular displacement in the same direction. (iv) The perpendicular bisector of the line segment joining a point of the object and its corresponding image passes through the centre of rotation. 6.4 Rotation Using Co-ordinates Just as in reflection, the images of the objects can easily be obtained by the rotation through a given angle using co-ordinates. Look and learn the following illustrations. (c) Rotation through 180° in anti-clockwise and clockwise direction. (a) Rotation through 90° (or –270°) in anticlockwise direction about origin. P(3,4) +900 P'(–4, 3) Symbolically, (b) Rotation through 90° in clockwise direction about origin. P(3, 4) – 900 P' (4, –3) Symbolically, P(x, y) R(O, –900 ) P' (y, –x) P(x, y) R(O,+900 ) P'(–y, x) Anticlockwise Clockwise Clockwise Anticlockwise P(3, 4) R(O,±1800 ) P'(–3, –4) Symbolically, P(x, y) R(O,±1800 ) P'(–x, –y)

100 Oasis School Mathematics-8 • R(O, + 90°) is equivalent to R (O, – 270°) • R(O, –90°) is equivalent to R(O, +270°) • R (O, + 180°) is equivalent to R(O, – 180°) Remember ! Worked Out Examples Example: 1 Find the image of the point A(–4, 6) under the rotation about the origin through the angle (i) +90° (ii) –90° (iii) 180° (iv) 270° Solution: We have, (i) P(x, y) R(O,+900 ) P' (–y, x) ∴ A(–4, 6) R(O,+900 ) A'(–6,–4) (ii) P(x, y) R(O,–900 ) P'(y , –x) ∴ A(–4, 6) R(O,–900 ) A'(6, 4) (iii) P(x, y) R(0,1800 ) P'(–x, –y) ∴ A (–4, 6) R(O, 1800 ) A'(4, –6) (iv) P(x, y) R(O,2700 ) P'(y, –x) ∴ A(–4, 6) R(O,2700 ) A(6, 4) Example: 2 A(1, 3), B(6, 4), C(7, 2) are the vertices of a triangle ABC. Rotate the ∆ABC about origin through 90° in anticlockwise direction. Present the ∆ABC and its image in the graph. Solution: We have, R(O, + 90°) : P(x, y) → P'(–y, x) ∴ A(1, 3) R(O,+900 ) A'(–3, 1) B(6, 4) R(O,+900 ) B'(–4, 6) and C(7, 2) R(O,+900 ) C'(–2, 7) Hence, ∆ABC and ∆A'B'C' are presented on the graph.

101 Oasis School Mathematics-8 101 Exercise 6.2 1. Copy the figures given below in your exercise book. Rotate them about O through the given angle in the given direction. Draw the image formed by rotation. (a) (b) (c) 2. Draw a ∆ABC in which AB = BC = AC = 4.5 cm. Take any point O outside of ∆ABC and rotate it about O through 60° in anticlockwise direction. 3. (a) Find the image of the following points underthe rotation through +900 about origin. (i) (4, 3) (ii) (2, –3) (iii) (-5, -4) (iv) (-3, 2) (b) Find the image of the following points under the rotation through –90º about the origin. (i) (–2, 4) (ii) (1, 3) (iii) (5, –3) (iv) (-1, -4) (c) Find the image of the following points under the rotation through 180° about the origin. (i) (1, –5) (ii) (–3, –2) (iii) (2, 6) (iv) (–1, 2) 4. (a) Rotate the ∆ABC with verities. A(1, 2), B(4, 5) and C(6, 1) through +90° about origin. Draw both the figure on the graph paper. (b) ABC is a triangle with vertices A(–1, 4), B(5, 2) and C(3, 7). Find the images of the verticesA, B and C of the triangleABC underthe rotation about the origin through (i) 180° (ii) 270°. Plot both the triangles on the same graph paper. (c) ABCD is a quadrilateral with vertices A(2, 4), B(–1, 2), C(3, –1) and D(4, 1). Find the images of the vertices A, B, C and D of the quadrilateral ABCD under the rotation about the origin through (i) +90° (ii) –90°. Present the quadrilaterals ABCD and A'B'C'D' on the same graph paper. 1. Consult your teacher. 2. Consult your teacher. 3. (a) (i) (–3, 4) (ii) (3, 2) (iii) (4, –5) (iv) (–2, –3) (b) (i) (4, 2) (ii) (3, –1) (iii) (–3 –5) (iv) (–4, 1) (c) (i) (–1, 5) (ii) (3, 2) (iii) (–2, –6) (iv) (1, –2) 4. (a) A'(–2, 1), B'(–5, 4) C'(–1, 6) (b) (i) A' (1, -4) B' (-5, -2), C' (–3, –7) (ii) A' (4,1), B' (2, -5), C'(7, -3) (c) (i) A' (–4, 2), B' (–2, –1), C' (1, 3), D' (–1, 4) (ii) A' (4, -2), B' (2,1), C' (-1,-3), D' (1,–4) Answer A B C O • Rotation through +90 O • Rotation through –600 P Q R 3 O • Rotation through 1800 A B C 5

102 Oasis School Mathematics-8 6.5 Translation or Displacement Each point of an object (or a geometrical figure) shifted along the same distance and same direction is called translation or displacement. The displacement of an object (or geometrical figure) has magnitude as well as direction. Hence it is a vector quantity. Example: Let AB be given line and a → be translation vector. Now, A and B are shifted to A' and B' such that AA' = a → . Then AB is said to be translated to A'B' by translation vector a → . Translation vector: The fixed value by which the geometrical figure will be displaced in definite direction of translation is called translation vector. Properties (i) Object and image are congruent. (ii) The line segments joining the object points and its corresponding image points are parallel and equal in length. Translation using co–ordinates: Any point P(x, y) moves a units parallel to x–axis and b units parallel to y–axis, then new co-ordinates of P will be (x + a, y + b). We mark the point P' on the plane whose co–ordinates are (x + a, y + b). Thus, if the translation vector T = ( a b), the point P(x, y) is translated to P'(x + a, y + b) Symbolically, P(x, y) T ( ( a b P' (x + a, y + b) Worked Out Examples Example: 1 Translate ∆ABC given in the figure by the given vector. Here, ∆ABC is displacing by given vector a → . From A, draw AA' || a → such that AA'= a → From C, draw BB' || a → such that BB' = a → From C, draw CC' || a → such that CC' = a → Join A', B' and C' Hence, ∆A'B'C' is the displaced image of ∆ABC. a → A B A' B' Translation vector Image P'(x + a, y + b) P(x, y) b a

103 Oasis School Mathematics-8 103 Example: 2 ABC is a triangle with vertices A(–2, 3), B(–2, 5) and C(2, 3). Translate the ∆ABC by translation vector T = ( ( 4 3 . Present the ∆ABC and its image on the graph. Solution: We have, P(x, y) T ( ( a b P' (x + a, y + b) A(–2, 5) T ( ( 4 3 A'(–2 + 4, 5 + 3) = A' (2, 8) B(–2, 3) T ( ( 4 3 B'(–2 + 4, 3 + 3) = B' (2, 6) and C(2, 3) T ( ( 4 3 C'(2 + 4, 3 + 3) = C' (6, 6) Hence, ∆A'B'C' is the translated image of ∆ABC. Exercise 6.3 1. Copy the figures given below in your copy and transform them by the translation vector in the direction and magnitude of it. Shade the image so formed. 2. Find the image of the following points (i) A(2, 3), (ii) B(6, –3), (iii) C(3, 0), (iv) D(–4, –6) when they are translated by the vectors (a) T = ( ( 4 3 (b) T =( ( -3 5 . 3. A point P(4, 5) is translated to the point P'(6, 8). Find the translation vector by which P is translated to P'. 4. ABC is a triangle with vertices A(3, 7), B(2, 2) and C(6, 1). Translate the ∆ABC by the vector (i) T = ( ( 2 3 (ii) T = ( ( -3 -4 . Present the ∆ABC and its image on the graph. 1. Consult your teacher. 2. a) (i) A' (6,6) (ii) B' (10,0) (iii) C' (7, 3) (iv) D' (0,–3) b) A' (-1, 8), B' (3, 2), C' (0, 5), D' (-7, -1) 3. ( 2 3 ) 4. (i) A'(5,10), B' (4,5) C' (8,4) (ii) A' (0,3), B' (-1,-2) C' (3, –3) Answer Y Y' X' X O A A' B' C' B C (c) S P R O a → (b) A B C a → (a) A B a →

104 Oasis School Mathematics-8 7.1 Bearing Bearing is the method of finding the distance between the two places in degree. There are two types of bearing. I. The compass bearing II. The three figure bearing I. The compass bearing The compass bearing is the old method of locating the direction. In the given figure NOS and EOW represent north-south and east west direction respectively. O is the point of reference for the direction. We use line ON as the base line. Now, we have simple compass card having 8 directions as shown in the figure alongside. Where, NE - North East direction SE - South East direction SW - South West direction NW - North West direction Look at these two examples: The direction of P is N 30° E. The direction of Q is S 20° E. Note : The bearing are always measured from N or S but not from E and W. O NE E SE S N NW W SW S Q 200 W O E N N P W E S 30 O Unit 7 Bearing and Scale Drawing

105 Oasis School Mathematics-8 105 II. The three digits bearing The direction of any point in terms of an angle expressed in three digits as measured in clockwise direction with reference to the north line is called the three digit bearing. Here, the bearing of P from N is 060°. The bearing of Q from N is 150° Worked Out Examples Example: 1 From the given figure, write down the compass bearing of P, NE and SW. Solution: Bearing of point P is N 15° E. Bearing of NE is N 45°E. Bearing of SW is S 45° W. Example: 2 Write down the three digit bearing of the given points from O. (a) (b) (c) Solution: Here, (a) The bearing of A from O is 055°. (b) The bearing of B from O is 140°. (c) The bearing of C from O is (360° - 25°) = 335°. P N E Q O 300 600 S W Note : ∠NAB is the bearing of B from A. Reflex ∠N'BA is the bearing of A from B. N A N' B N P W E S 15º O SW SE NW NE O N A 550 N O B 1400 N O C 250

106 Oasis School Mathematics-8 Example: 3 Write in three digits bearing. (a) N60° E (b) S 30° W Solution: From the figure, S30°W = (180°+ 30°) N 60° E = 060° = 210° Example: 4 If the bearing of B from A is 60°, what is the bearing of A from B? Solution: Here, The bearing of B from A is 060°. i.e. ∠NAB = 60° Since, NA||N'B ∠NAB + ∠N'BA = 180° [∵ Sum of co-interior angles is 180°] 60° + ∠N'BA = 180° or, ∠N'BA = 180° – 60° = 120° ∴ The bearing of A from B = reflex ∠N'BA = (360° – 120°) = 240° Exercise 7.1 1. Write down the compass bearing of the following figure: 2. Draw the diagrams to show each of the following bearings. (a) N50°E (b) S 30°E (c) S 60°W (d) N 80°W N S W O 300 N A O 600 A 600 B N' N 300 N S A B S A 450 O 600 O N S W SW S N E NW N E S W (a) (b) (c) (d) (e) C O O O

107 Oasis School Mathematics-8 107 3. From the given figure, write down the three digit bearing of NE, E, SE, S, SW, W, and NW. 4. Write the three digit bearing of given points from O. 5. Sketch the following bearing. (a) 075° (b) 145° (c) 215° (d) 305° 6. Write the three digit bearing of: (a) N60°E (b) S70°W (c) S50°E (d) N30°W 7. In each ofthe given figures, bearing ofBfrom A is given. Find the bearing of A from B. 8. From the given map of Nepal, using three digit bearing, write down the bearing of: (a) Kathmandu from Pokhara. (b) Gorkha from Nepalganj. (c) Jumla from Dhangadi. (d) Biratnagar from Janakpur. O E SW SE NW NE S N W N E O 1500 O N D 400 O N C O N B 1600 A O (a) N (b) (c) (d) (e) 800 N A N N N N A B A B B B A 2300 3100 400 1150 N' N' N' (a) (b) (c) (d) Kathmandu

108 Oasis School Mathematics-8 1. (a) S300 E (b) S450 W (c) N600 W (d) S450 W (e) N450 W 2. Consult your teacher. 3. 0450 , 0900 , 1350 , 1800 , 2250 , 2700 , 3150 4. (a) 0800 , (b) 2000 , (c) 3200 (d) 2700 (e) 1500 5. Consult your teacher. 6. (a) 0600 , (b) 2500 (c) 1300 , (d) 3300 7. (a) 2950 (b) 2200 (c) 500 (d) 1300 8. Consult your teacher Answer 7.2 Scale Drawing The figures or maps can be drawn by using different scales. It is impossible to show lengths and distance which are too great or too small in a sheet of paper. In this case, we have to enlarge smaller figure and reduce the larger figure by taking a suitable scale. Scale factor In the given fig. (i) and (ii), we can see that each side of second figure is twice that of first figure. Here each side of first figure is enlarged twice. ∴ AB A'B' = 3 cm 6 cm = 1 2 or, BC B'C' = 3 cm 6 cm = 1 2 or, AC A'C' = 3 cm 6 cm = 1 2 Thus, first figure is the scale drawing of second figure to the scale 1:2. Thus, the ratio 1:2 is called the scale factor. Note : The scale factor 1:100 means 1 unit length in drawing is 100 units of length in actual size. i.e. Length in drawing Actual length = 1 100 = scale factor. A A C 3 cm Figure (i) 3 cm 3 cm Figure (ii) 6 cm 6 cm 6 cm C' A' B'

109 Oasis School Mathematics-8 109 Worked Out Examples Example: 1 What is the actual distance between two places which is represented by 1.5 cm on a map which is drawn to the scale 1:60,00,000? Solution: Here, Scale = 1: 6000000 i.e. 1 cm on a map represents 6000000 cm. = 6000000 1000 m = 60000 100 m. = 60 km. ∴ 1.5 cm on a map represents = 60 × 1.5 km = 90 km. ∴ Distance between the two places = 90 km Example: 2 The actual length and breadth of a rectangular field is 100 m and 80 m respectively, which is drawn on the scale 1:800. Find the length and breadth of the field on the drawing. Solution: Here, 800 cm is represented by 1 cm i.e. 800 100 m is represented by 1 cm or, 8 m is represented by 1 cm length of the field = 100 m 8 m is represented by 1 cm 1 m is represented by 1 8 cm 80 m is represented by 1 8 × 80 cm = 10 cm 100 m is represented by 1 8 × 100 cm = 12.5 cm. Alternative method We have, scale = length in map actual length 1:800 = length in map/100 m or, length in map = 100 800 m = 1 8 m = 1 8 × 100 cm = 12.5 cm. Again, Scale = Length in map Actual length 1:800 = Length in map 80 m ∴ Length in map = 80 800 m = 1 10 m = 1 10 ×100 cm = 10 cm.

110 Oasis School Mathematics-8 Example: 3 If an aeroplane flies 600 km in the bearing of 030° and 800 km in the bearing of 120°, find the distance between the two places and the bearing of the starting place from the last place using the scale 1 cm = 100 km. Solution: In the given figure, A is the starting place, last place is B, AB = 10 cm. (By measurement) therefore, distance between the places (AB)=10 cm = 1000 km. [∵ 1 cm = 100 km] From the given figure ∠ABN2 = 113° (By measurement). Hence, bearing of the starting place from the last place = 360° – 113° = 247°. Exercise 7.2 1. Given figure shows the position of students in a class. If the scale is 1:150. Find the actual distance between: (a) Ramesh and Shyam. (b) Goma and Bidhya. (c) Ramesh and Goma. (d) Shyam and Bidhya. 2. If the scale in a map is 1:1500. Find the actual distance between the given places if the distance in the map is: (a) 4 cm (b) 8 cm (c) 6 cm (d) 15 cm (e) 29 cm 3. The scale in a map is 1:5000. Find the distance between two places in the map if the actual distance between the two places is: (a) 0.8 km (b) 600 m (c) 1500 m (d) 3 km (e) 4 km 4. (a) Find the actual height of a tree whose height in a map is 5 cm and is represented by a scale 1: 600. (b) What is the actual distance between two places which is represented by 25 mm. on a map which is drawn to the scale of 1 mm = 2 km? 8 cm 300 1200 N A B N N2 1 6 cm Ramesh Goma Shyam Bidhya

111 Oasis School Mathematics-8 111 5. The figure alongside shows the outline sketch of a house plan. Draw a scale of 1 cm to represent 1 m. (a) Find the actual length and breadth of bedroom, kitchen and toilet. (b) Find the area of the garage. 6. Find the actual distance between the given places from the given scale. (a) Kathmandu and Dhangadi (b) Pokhara and Jumla (c) Gorkha and Janakpur 7. The figure or map of Nepal is drawn on the scale of 1 cm is equal to 40 km showing the major trade centers. Find out the actual distance between: (a) Dhangadhi and Nepalgunj. (b) Bhaktapur and Birgunj. (c) Jiri and Biratnagar. (d) Birgunj and Biratnagar. 8. If an aeroplane flies 400 km in the bearing of 030° and then 300 km in the bearing of 120°, find the distance between the two places and the bearings of the starting place from the last place using the scale 1 cm = 100 km. 9. A motor is at S, a place 200 km east to T. Travelling a distance of 300 km to the due west of S, it reached at U. Express this information by the drawing and find (a) the distance from T to U. (b) the bearing of T from U. T U N 2cm S 3cm Using scale : 1cm = 100km Kitchen Toilet Common Room Bedroom Garage 3 cm 3 cm 3 cm 3 cm 4 cm 4 cm 3 cm 3 cm 1cm

112 Oasis School Mathematics-8 1. Consult your teacher. 2. (a) 60m (b) 120m (c) 90m (d) 225m (e) 435m 3. (a) 16cm (b) 12cm (c) 30cm (d) 60cm (e) 80cm 4. (a) 30m (b) 50km 5. (a) Bedroom 4m, 3m Kitchen - 4m, 2m, Toilet-4m, 1m. (b) 9m2 6. Consult your teacher 7. (a) 152km (b) 88km (c) 152 km. (d) 72km 8. 500km and 2470 9. (a) 100km (b) 0900 Answer Assessment Tes t Paper Attempt all the questions: Group A [4 × 1 = 4] 1. (a) Draw a compass bearing of 0450 . (b) In the given figure, write the compass bearing of SE (c) Find the bearing of B from A in the given figure. (d) Find the actual height of a tower whose height in a map is 3.5cm and represented by scale 1:800. Group B [4×2=8] 2. Draw a triangle ABC with the vertices A (3,1), B (5,2) and C (2, 5) on graph paper. Reflect ∆ABC on Y-axis. Also draw ∆A'B'C' on the same graph paper. 3. Plot ∆PQR with the vertices P (-4, -1), Q(2,1) and R (-1, 5) on a graph paper then plot the image ∆P'Q'R' after rotating through +900 about origin. W N S SE 45° E N N1 B A 120° Full marks : 12

113 Oasis School Mathematics-8 113 Sets Contents • Operations on Sets - Union of Sets - Intersection of Sets - Differences of Sets - Complement of Sets • Cardinality of Sets Expected Learning Outcomes At the end of this unit, students will be able to develop the following competencies: • Find the union and intersection of two sets and show these relations in the Venn diagram. • Find the difference of two sets and complement of a set and show these relations in Venn diagram. • Find the cardinal number of sets and use the cardinality relations to solve problems related to sets. Teaching Materials • Tracing paper, flash card, chart paper, etc. Estimated Teaching Hours5

114 Oasis School Mathematics-8 8.1 Review A collection of well-defined objects is called a set. The term well-defined means that we must be able to tell whether a particular object belongs to the set or not. There must be any repeated objects in a set. (i) The set of boys of class VIII in a school (ii) The set of vowels in the English alphabet The objects of a set are called its members or elements. Set notation: The sets are usually denoted by capital letters A, B, C etc. Elements are enclosed inside the braces { } and separated with comma (,). Example: A = {2, 3, 5, 7, 11} 2 is an element of A, then 2∈A. 4 is not an element of A, then 4 ∉A. Note: Say Write 2 belongs to A 2 ∈ A. 4 does not belong to A 4 ∉ A. Specification of Sets The common method to represent the set are- (i) Listing method (ii) Description method (iii) Set builder method Let us represent the set of prime numbers less than 15 by three different methods: Method Way to specify Example Listing method All the elements are listed within curly braces separating each element by comma. A = { a, e, i, o, u} Description method The common property of elements are described by words with suitable sentence. A set of vowel letters. Set builder method A rule or a statement or a formula is written to represent the sets. A = {x : x is a vowel letter of English alphabet} Unit 8 Sets

115 Oasis School Mathematics-8 115 Types of Sets Null set: A set having no element is called null set or an empty set or void set. It is denoted by φ (phi) or { }. Example: A set of triangle having 4 sides. Singleton Set: A set having only one element as its member is called singleton set or unit set. Example: {1}, {0}, etc. are singleton sets. Finite set: A set having finite number of elements is called a finite set. Example: A = {x : x is a prime number less than 20} is a finite set. Infinite set: A set having an infinite number of elements is called an infinite set. Example: A = {1, 2, 3, 4, ……} is an infinite set. • A set of male students in Padmakanya college is a null set. • A set of the highest peak of the world is a singleton set. • A set of even numbers from 10 to 30 is a finite set. • A set of natural numbers is an infinite set. • φ is a null set and {φ} is a unit set. Remember ! Set Relations The set relations between two sets can be understood with the help of the given table. Type Nature Example Overlapping sets Sets having at least one common element A = {1, 2, 3, 4} and B = {3, 4, 5, 6} Disjoint sets Sets having no common element A = {p, q, r, s} and B = {x, y, z} Equal sets Sets containing same elements A = {a, b, c, d} B = {a, b, c, d} then A = B. Equivalent sets Sets containing equal number of elements A = {1, 2, 3, 4} B = {w, x, y, z} Subsets: In this case, A may be equal to B. Let A = {1, 2, 3} and B = { 1, 2, 3, 4, 5}. Here every element of set A is also the element of set B. A is subset of B. Hence, set A is said to be a subset of set B, if every element of set A is also an element of set B. It is denoted by A ⊂ B which is read as 'A is a subset of B or A is contained in B'.

116 Oasis School Mathematics-8 Note: • If A = {a, b} and B = {a, b, c}, then A is the proper subset of B., i.e. A ⊂ B. • Every set is a subset of itself. i.e. A ⊆ A, B ⊆ B. • Null set is a subset of every set i.e. φ ⊂ A, φ ⊂ B. Number of subsets of a given set Let us observe the following examples, • Consider the null set { } or φ. Since every set is a subset of itself, the subset of the null set = { }.i.e. number of subsets of a null set = 1 = 2°. • Consider the set A = {1}. Since the null set is a subset of every set and a set is also a subset of itself, its subsets are { } and {1}. ∴ Number of subsets of this sets = 2 = 2¹ • Consider the set A = {1, 2} of two elements. Its subsets are { }, {1}, {2}, and {1, 2}. So, the number of subsets = 4 = 2² . • Consider the set A = {1, 2, 3}. Its subsets are { }, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3} and {1,2,3}. ∴ Number of subsets of this set = 8 = 2³. The above information obtained can be written in the tabular as below. In general, the number of subsets of set having 'n' elements = 2n. Set No. of elements No. of subsets { } 0 1 = 20 { 1 } 1 2 = 21 { 1, 2 } 2 4 = 22 { 1, 2, 3 } 3 8 = 23 Exercise 8.1 1. Write the following sets in listing method (a) A = a set of prime numbers less than 10. (b) B = a set of first three days of the week. (c) C = {x : x is an even number less than 10}. 2. Write the following sets in the set-builder form (a) A = {2, 3, 5, 7} (b) B = {1, 3, 5, 7, 9} (c) W = {a set of whole numbers less than 5.}

117 Oasis School Mathematics-8 117 3. Write the following sets in description method (a) A = {1, 3, 5, 7, 9, 11, 13} (b) B = {x : x is a natural number less than 5}. (c) C = { 5, 10, 15, 20}. 4. Determine whether the following sets are disjoint or overlapping. (a) A = {1, 2, 3}, B = {a, b, c} (b) C = {4, 5, 6, 7}, D = {6, 7, 8, 9} (c) E = {4, 5, 6, 7, 8}, F = {7, 8, 9, 10, 11} (d) P = {x, y}, Q = {a, b, c, d} (e) F(8) = the set of factors of 8, F(7) = the set of factors of 7. 5. Identify whether the following sets are null set or unit set. (a) { } (b) A = {men over 200 years old} (c) D = {0} (d) E = the set of men with height more than 15 feet (e) {φ} 6. Identify whether the following sets are finite or infinite. (a) A set of prime numbers less than 20 (b) A set of natural numbers (c) A set of odd numbers from 1 to 100 7. Which of the following pairs of sets are equal? (a) A = {1, 2, 3, 4}, B = {4, 3, 2, 1} (b) C = {4, 5, 6, 7}, D = {a, b, c, d} (c) A = a set of prime numbers less than 8, B = {2, 3, 5, 7} (d) P = a set of odd numbers less than 8, R = { 1, 3, 5, 7} 8. Which of the following pairs of sets are equivalent? (a) A = {1,2,3,4}, B = {6,7,8,9} (b) C = {Umesh, Dolendra,Prabhat}, D={Manita, Alina, Gita} (c) A = {1,2,3,4,5}, B = {4,5,6,7} 9. Let A = {0, 1, 2, 3, 4, 5}. State whether the following statements are true or false. (a) {2} ⊂ A......... (b) {o} ⊆ A........ (c) φ ⊆A..... (d) {1, 3} ⊆ A... (e) A ⊆ A.... (f) 2 ⊆ A.... 10. State whether the following statements are true or false. (a) No set is a proper subset of itself. (b) The universal set is the subset of every set. (c) Null set is subset of every set.

118 Oasis School Mathematics-8 11. If A = {0, 1, 2}, which of the following are proper subsets of A? (a) {0, 1} (b) {2, 1, 0} (c) {2} (d) {a, b, c} (e) {4, 5} 12. Write down all possible subsets of the following sets. (a) {a} (b) {a, b} (c) {2, 4, 6} 13. Find the number of subsets of the following sets. (a) { p } (b) {x, y, z} (c) { 1, 2} 1. (a) A = {2, 3, 5, 7} (b) B = {Sunday, Monday, Tuesday} (c) C= {2, 4, 6, 8} 2. (a) A = {x: x is a prime number less than 10} (b) B = {x : x is an odd number less then 10} (c) W = {x : x is a whole numbers less then 5} 3. (a) A = a set of odd number less then 15 (b) B = a set of natural numbers less then 5 (c) C = a set of first 4 multiples of 5 4. (a) disjoint set (b) overlapping sets (c) overlapping sets (d) disjoint sets (e) overlapping sets 5. (a) null set (b) null set (c) unit set (d) null set (e) unit set 6. (a) finite set (b) infinite set (c) finite set 7. (a) equal sets (b) unequal sets (c) equal sets (d) equal sets 8. (a) equivalent sets (b) equivalent sets (c) not equivalent sets 9. (a) true (b) false (c) true (d) true (e) true (f) false 10. (a) true (b) false (c) true 11. a and c are the proper subsets. 12. (a) φ, {a} (b) φ, {a}, {b}, {a, b} (c) φ, {2}, {4}, {6}, {2,4}, {2, 6}, {4, 6}, {2,4,6} 13. (a) 2 (b) 8 (c) 4 Answer 8.2 Operations on Sets Four operations on sets are - (i) Union of sets (ii) Intersection of sets (iii) Difference of sets (iv) Complement of a set Union of sets The union of two sets A and B is a new set of whose elements are either in A or in B or in both of them. It is denoted by A ∪ B which is read as 'A union B' or 'A cup B' Symbolically, it is defined as A ∪ B = { x : x belongs to at least one of the sets A or B} = {x : x ∈ A or x ∈ B} To form A ∪ B from two given sets A and B, first take the elements of A and then take elements of B which are not in A. For example,

119 Oasis School Mathematics-8 119 (i) If A = {a, b, c} and B = {b, c, d, e}, then A ∪ B = {a, b, c} ∪ {b, c, d, e} = {a, b, c, d, e}. Here, b and c are common elements of A and B. Its Venndiagram is given alongside. (ii) If A = {a, b, c} and B = {d, e}, then A ∪ B = {a,b,c} ∪ {d, e} = {a, b, c, d, e} Here, A and B have no common element. Its Venn-diagram is given alongside. (iii) If A = {1, 4, 5}, B = {1, 4, 5, 8, 10} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then A ∪ B = {1, 4, 5} ∪ {1, 4, 5, 8, 10} = {1, 4, 5, 8, 10} Here, A is subset of B. Thus, we have A ∪ B = B. Its Venn-diagram is given alongside. Intersection of sets The intersection of two sets A and B is the set of all elements common to both A and B. The intersection of two sets A and B is denoted by A ∩ B which is read as "A intersection B" or "A cap B". Thus, A ∩ B = {x : x ∈ A and x ∈ B} In order to form A ∩ B from the two given sets A and B, we take all those elements which are common to A and B. If there is no common element between A and B, then A ∩ B is the null set and A and B are called disjoint sets. For example, (i) If A = {a, b, c, d, e, f} and B = {b, d, f, h, j}, then A ∩ B = {a, b, c, d, e, f} ∩ {b, d, f, h, j} = {b, d, f} Here, b, d, f are common elements to both A and B. A∩B is shaded b h a c d e f j A B U a b d e A A∪B is shaded B c U U A∪B is shaded B A 7 8 1 4 5 10 2 9 6 3 U a b d e A A∪B is shaded B c U

120 Oasis School Mathematics-8 (ii) If A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8} then A ∩ B = {1, 3, 5, 7, 9} ∩ {2, 4, 6, 8} = φ There is no common element between A and B. So A and B are disjoint sets. 1 2 6 4 8 5 7 3 9 A B U Difference of Sets The difference of two sets A and B is the set of all elements of A only, which are not in B. The difference of two sets A and B is denoted by A – B which is read as "A difference B" or "A minus B". Symbolically, it is defined as, A – B = {x : x ∈ A and x ∉ B} = elements of A, which are not in B Similarly, B – A = {x : x ∈ B and x ∉ A} = elements of B, which are not in A In order to form A– B, we take all those elements of A which are not in B. This means that A – B is the new set which is formed by taking elements of A only. For examples, (i) If A = {b, c, d, e} and B ={a, b, c}, then A–B = {b, c, d, e}–{a, b, c}={d, e}and B – A = {a, b, c} – {b, c, d, e} = {a}. Their Venn-diagrams are given alongside. Complement of a Set Let U = {x:x is an odd number less than 15} and A = {1, 3, 5, 7, 9}, then U = {1,3,5,7,9,11,13}. Here, two elements 11 and 13 from U do not belong to set A. ∴ {11, 13} is the complement of set A. The complement of a set A with respect to universal set U is the set of all those elements of U only, which are not in A. The complement of a set A is denoted by A' or A or Ac . Symbolically, it is defined as U – A = {x : x ∈ U and x ∉ A}. Its Venn-diagram is given below. The shaded portion represents the complement of a set A. d b d b c c a a e e A B A B U U A–B is shaded B–A is shaded

121 Oasis School Mathematics-8 121 For example, (i) If U = {1, 2, 3, 4, 5, 6, 7} and A = {1, 2, 3, 4}, then = U – A = {1, 2, 3, 4, 5, 6, 7} – {1, 2, 3, 4} = {5, 6, 7}. Its Venn-diagram is given alongside. Also, (A) = U – A = {1, 2, 3, 4, 5, 6, 7} – {5, 6, 7} = {1, 2, 3, 4} = A ∴ (A) = A. This shows the complement of a complement set is given set. Its Venn-diagram is given alongside. (ii) If U = {1, 2, 3, 4, 5, 6, 7}, A = {1, 2, 3} and B = {3, 4, 5}, then A ∪ B = { 1, 2, 3} ∪ {3, 4, 5} = {1, 2, 3, 4, 5}. Thus, A ∪ B is the set of all the elements which belong to either A or B or both A and B. ∴ (A ∪ B) = U – (A ∪ B) = {1, 2, 3, 4, 5, 6, 7} – {1, 2, 3, 4, 5} = {6, 7}. Thus, it is the set of all those elements which belong to neither A nor B. Worked Out Examples Example: 1 If U = {x : x is a natural number less than 13}, A = {x : x is a prime number less than 13}, B = {x : x is an even number less than 13} and C = {x : x is a natural number less than 6}, find; (i) A ∪ B (ii) A ∪ C. Show it in Venn-diagrams. Solution: Here, U = {x : x is a natural number less than 13} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} 1 2 3 4 5 7 U A U A A is shaded 1 2 3 4 5 7 A U A (A) is shaded 6 6 3 1 4 2 5 6 7 A B U A∪B is shaded

122 Oasis School Mathematics-8 A = {x : x is a prime number less than 13}={2, 3, 5, 7, 11} B = {x : x is an even number less than 13}={2, 4, 6, 8, 10, 12} and C = {x : x is natural number less than 6} ={ 1, 2, 3, 4, 5} (i) A ∪ B = {2, 3, 5, 7, 11} ∪ {2, 4, 6, 8, 10, 12} = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12} Its Venn-diagram is given alongside. (ii) A ∪ C = {2, 3, 5, 7, 11} ∪{1, 2, 3, 4, 5} = {1, 2, 3, 4, 5, 7, 11}. Its Venn-diagram is given alongside. Example: 2 If A = {factors of 24} and B = {factors of 36}, find (i) A ∪ B (ii) A ∩ B. Show it also in Venn-diagram. Solution: Here, A = factors of 24 = {1, 2, 3, 4, 6, 8, 12, 24} B = factors of 36 = {1, 2, 3, 4, 6, 9, 12, 18, 36} (i) A ∪ B = {1, 2, 3, 4, 6, 8, 12, 24} ∪{1, 2, 3, 4, 6, 9, 12, 18, 36} = { 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}. Its Venn-diagram is given alongside. (ii) A ∩ B = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {1, 2, 3, 4, 6, 9, 12, 18, 36} = {1, 2, 3, 4, 6,12}. Its Venn-diagram is given alongside. Example: 3 If U = set of natural numbers less than 10, A = {1, 2, 3, 4, 5}, B = set of odd numbers less than 10, find- (i) A – B (ii) B–A (iii) A ∩ B. Also show them in the Venn diagram. Solution: U = {1, 2, 3, 4, 5, 6, 7, 8, 9} A = {1, 2, 3, 4, 5} B = {1, 3, 5, 7, 9} (i) A–B = {1, 2, 3, 4, 5} – {1, 3, 5, 7, 9} = {2, 4} A B U A∪B is shaded 11 12 10 7 8 4 1 9 6 5 3 2 A C U A∪C is shaded 12 11 1 7 4 6 8 9 2 5 3 A B U A∪B is shaded 8 9 36 18 24 12 1 2 4 3 6 A B U A∩B is shaded 1 2 4 6 3 3 18 24 36 8 12 A B U A–B is shaded 2 1 7 9 8 6 3 4 5

123 Oasis School Mathematics-8 123 Illustration in Venn diagram. (ii) B–A = {1, 3, 5, 7, 9} – {1, 2, 3, 4, 5} = {7, 9} Illustration in the Venn diagram. (iii) A ∩ B = {1, 2, 3, 4, 5} ∩ {1, 3, 5, 7, 9} = {1, 3, 5} A ∩ B = U – (A∩B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 5} = {2, 4, 6, 7, 8, 9} Illustration in the Venn diagram. Example: 4 From the given Venn diagram, find (i) A (ii) B (iii) A ∩ B (iv) A ∪ B (v) A–B (vi) B–A (vii) A ∪ B Solution: Here, from the Venn diagram, (i) A = {a, e, i, o, u} (ii) B = {p, q, r, o, u} (iii) A ∩ B = {o, u} (iv) A ∪ B = {a, e, i, o, u, p, q, r} (v) A–B = {a, e, i} (vi) B–A = {p, q, r} (vii) A ∪ B = {s, t}. Example: 5 Shade the following sets separately in the adjoining Venn-diagrams. (i) A – B (ii) B – C (iii) A – (B ∩ C) (iv) A (v) A ∪ B Solution: A B U B–A is shaded 2 1 7 9 6 8 3 4 5 A B U 1 7 9 3 5 1 2 4 8 6 3 5 A∩B is shaded A B U 1 p r 3 q 5 a e i s t o u A B C U A B C U (i) A – B (ii) B – C A–B is shaded B–C is shaded A–(B∩C) is shaded (iii) A–(B∩C) A B C U

124 Oasis School Mathematics-8 Example: 6 If A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, C = {4, 6, 9, 10}, verify that (i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) Solution: (i) B ∩ C = {4, 5, 6, 7, 8} ∩ {4, 6, 9, 10} = {4, 6} ∴ A ∪ (B ∩ C) = {1, 2, 3, 4, 5} ∪ {4, 6} = {1, 2, 3, 4, 5, 6} .................. (i) Again, A ∪ B = {1, 2, 3, 4, 5} ∪ {4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8} and A ∪ C = {1, 2, 3, 4, 5} ∪ {4, 6, 9 10} = {1, 2, 3, 4, 5, 6, 9, 10} ∴ (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8} ∩ {1, 2, 3, 4, 5, 6, 9, 10} = {1, 2, 3, 4, 5, 6} ....................... (ii) From (i) and (ii), we get, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). Hence, verified. (ii) B ∪ C = {4, 5, 6, 7, 8} ∪ {4, 6, 9, 10} = {4, 5, 6, 7, 8, 9, 10} ∴ A ∩ (B ∪ C) = {1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8, 9, 10} = {4, 5}............ (iii) Again, A ∩ B = {1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8} = {4, 5} And A ∩ C = {1, 2, 3, 4, 5} ∩ {4, 6, 9, 10} = {4} ∴ (A ∩ B) ∪ (A ∩ C) = {4, 5} ∪ {4} = {4, 5} ...................... (iv) From (iii) and (iv), we get, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). Hence, verified. A B C U A B C U (iv) A (v) A ∪ B A∪B is shaded A is shaded

125 Oasis School Mathematics-8 125 Exercise 8.2 1. What does the shaded portion represent in each of the given venn-diagrams? A B U A B U A U A B U A B U A U A A A B B B B B U U (a) (d) (g) (b) (e) (h) (c) (f) (i) U 2. (a) If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 3, 5, 7} and B = {2, 4, 6, 8}, find (i) A ∪ B (ii) A ∩ B. (iii) A ∪ U (iv) B ∩ U. (b) If A = {a, e, i, o, u} and B = {r, a, e, s}, find (i) A ∩ B. (ii) A ∪ B. (c) If P = {1, 2, 3, 4, 5}, Q = {1, 2, 4, 6, 7}, find (i) P – Q. (ii) Q – P. (d) If A = {factors of 8}, and B = {factors of 12}, find (i) A ∪ B (ii) A ∩ B. (e) If U = {1, 2, 3, 4, 5, ………….., 10}, A = {factors of 2}, B = {factors of 3}, find (i) A ∪ B (ii) A ∩ B (iii) A ∪ B (iv) A ∩ B . 3. From the given Venn diagram, find the following sets (i) A (ii) B (iii) A ∪ B (iv) A ∩ B (v) A–B (vi) B–A (vii) A (viii) B 4. (a) If U = {x : x is a positive integer less than 20}; A = {x : x is a prime number less than 20} and B = {x : x is divisible by 3, x ≤ 15}, find (i) A ∪ B (ii) A ∪ B (iii) A ∩ B (iv) A ∩ B (v) A ∪ B (vi) A – B (vii) B – A. Show all these relations in the Venn diagram. A B U 3 7 5 2 4 6 8 1 9

126 Oasis School Mathematics-8 (b) If U = {x : x is a positive integer less than 10}, A = {x : x is a prime number less than 10} and B = {x : x is an odd number less than 10}, find (i) A (ii) A (iii) B (iv) B (v) A ∪ B (vi) A ∩ B Also draw Venn-diagram to represent the above sets. 5. Show the following set relations in the Venn diagram. (a) A–B (b) B–A (c) A (d) B (e) A – B (f) A ∪ B (g) A ∩ B (h) A–(A ∩ B) (i) (A ∪ B) – A 6. What does the shaded part of each figure represent? (a) (d) (b) (e) (c) (f) 7. Draw Venn-diagram separately like the adjoining figure and shade the portion of the following sets. (a) A ∪ B (b) A ∩ B (c) A ∩ C (d) A ∪ (B ∪ C) (e) (A∩B)∩C (f) U ∩ C 8. From the adjoining Venn-diagram, write down the elements of the following sets. (a) U (b) A (c) C (d) A ∪ B (e) B ∩ C (f) A ∪ (B ∪ C) (g) (A ∩ B) ∩ C 9. If A = {2, 3, 5, 7, 11}, B = {3, 6, 9, 11, 15, 17} and C = {5, 6, 7, 11}, then verify that- (a) A ∪ (B∩C) = (A ∪ B) ∩ (A ∪ C) (b) A∩(B ∪ C) = (A∩B) ∪ (A∩C) A C B U A C B U 3 4 6 8 10 7 11 9 12 1 2 5

127 Oasis School Mathematics-8 127 Answer 1. (a) A∪B (b) A∩B (c) A– B (d) B –A (e) A ∪ B (f) A ∩ B (g) A – B (h) A (i) B – A 2. (a) (i) {2, 3, 4, 5, 6, 7, 8} (ii) {2} (iii) {1, 2, 3, ..... 9} (iv) {2, 4, 6, 8} (b) (i) {a, e} (ii) {a, e, i, o, u, r, s} (c) (i) {3, 5} (ii) {6, 7} (d) (i) {1, 2, 3, 4, 6, 8, 12} (ii) {1, 2, 4} (e) (i) {1, 2, 3} (ii) {1} (iii) {4,5,6,7,8,9,10} (iv) {2, 3, 4, 5, 6, 7, 8, 9, 10} 3. (i) {2, 3, 5, 7} (ii) {2, 4, 6, 8} (iii) {2, 3, 4, 5, 6, 7, 8} (iv) {2} (v) {3, 5, 7} (vi) {4, 6, 8} (vii) {1, 4, 6, 8, 9} (viii) {1, 3, 5, 7, 9} 4. (a) (i) {2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 19} (ii) {1, 4, 8, 10, 14, 16, 18} (iii) {3} (iv) {1, 2, 4, 5, 6, 7, 8, 9, 10 .... 19} (v) {1, 2, 3, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 18, 19} (vi) (2, 5, 7, 11, 13, 17, 19} (vii) {6, 9, 12, 15} (b) (i) {1, 4, 6, 8, 9} (ii) {2, 3, 5, 7} (iii) {2, 4, 6, 8} (iv) {1, 3, 5, 7, 9} (v) {4, 6, 8} (vi) {1, 2, 4, 6, 8, 9} 5. (i) U A B (A ∪ B) – A U A B A–(A ∩ B) (h) (g) A∩B U A B 6. (a) A∩B∩C (b) A∩B (c) A∪B∪C (d) A (e) A∪C (f) A∪B∪C 7. U A B C U A B C U A B C (a) (b) (c) A ∪ B A ∩ B A ∩ C

128 Oasis School Mathematics-8 8.3 Cardinality of Two Sets Let A = {a, b, c, d}, B = { 1, 2 }, C = { 0 } and φ are some sets. There are four elements in set A, two elements in set B: In short, we write n(A) = 4, n(B) = 2, n(C) = 1 and n(φ) = 0, which are called cardinal numbers of sets. Hence, the number of elements present in the set is called cardinal number of sets. Let 'A' and 'B' be any two overlapping sets, such that n(A) = a, n(B) = b and n(A∩B) = x. Then, from the Venn–diagram, n(A∪B) = a – x + x + b – x n(A∪B) = a + b – x n(A∪B) = n(A) + n(B) –n(A∩B) If two sets are disjoint, then n(A∩B) = 0 n(A∪B) = n(A) + n(B) Some important results on cardinality of two sets (i) n(A∪B) = n(A) + n(B) – n(A∩B) (ii) n(A∪B) = n(A) + n(B) if A and B are disjoint sets. (iii) n0 (A) = n(A) – n(A∩B) (iv) n0 (B) = n(B) – n(A∩B) (v) n(A∪B) = n0 (A) + n0 (B) + n(A∩B) (vi) n(A∪B) = n(U) – n(A∪B) (vii) n0 (A) = n(A∪B) – n(B) (viii) n0 (B) = n(A∪B) – n(A) U A B (a-x) x (b-x) U n(A∪B) A B n(A) n0 (A) n(B) n0 (B) n(A∩B) Note: n0 (A) = n(A–B) and n0 (B) = n(B – A) 8. (a) {1, 2, 3, ...12} (b) {1, 2, 3, 4, 5} (c) {1, 2, 5, 8, 10, 12} (d) {1, 2, 3, 4, 5, 6, 8} (e) {1, 2, 8} (f) {1, 2, 3, 4, 5, 6, 8, 10, 12} (g) {1, 2} (d) (e) (f) U A B C U A B C U A B C A ∪ B ∪ C (A ∩ B) ∩ C U ∩ C

129 Oasis School Mathematics-8 129 Worked Out Examples Example: 1 If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}, find (i) n(A) (ii) n(B) (iii) n(A∩B) (iv) n(A∪B) (v) n(A–B) (vi) n(B–A) Solution: Here, A = { 1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8} (i) n(A) = number of elements in set A = 5 (ii) n(B) = number of elements in set B = 5 (iii) A ∩ B = {1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8} = {4, 5} ∴ n(A ∩ B) = number of elements common to both sets A and B = 2 (iv) A ∪ B = {1, 2, 3, 4, 5} ∪ {4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8} ∴ n(A ∪ B) = number of elements in set A ∪ B = 8 (v) A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7, 8} = {1, 2, 3} ∴ n(A – B) = number of elements in set A only = 3 (vi) B – A = {4, 5, 6, 7, 8} – {1, 2, 3, 4, 5} = {6, 7, 8} ∴ n(B – A) = number of elements in set B only = 3. Example: 2 In the given Venn diagram "A" and "B" are the subsets of universal set U. Then find (i) n(A ∩ B) (ii) n(A) (iii) n(B) (iv) n(A–B) (v) n(B–A) (vi) n(A ∪ B) (vii) n(A ∪ B) Solution: From the Venn diagram (i) n(A ∩ B) = 5 (ii) n(A) = 12 + 5 = 17 (iii) n(B) = 5 + 9 = 14 (iv) n(A–B) = 12 (v) n(B – A) = 9 (vi) n(A ∪ B) = 12 + 5 + 9 = 26 (vii) n(A ∪ B) = 6 U A B 12 5 9 6

130 Oasis School Mathematics-8 Example: 3 In the given Venn diagram, if n(U) = 200, find (i) n(A ∩ B) (ii) no (A) (iii) no (B) (iv) n(A ∪ B). Solution: Given, n(U) = 200 From the Venn diagram n(U) = 120 – x + x + 100 – x + 40 or, 200 = 260 – x or, x = 260 – 200 or, x = 60 ∴ (i) n(A ∩ B) = x = 60 (ii) no (A) = 120 – x = 120 – 60 = 60 (iii) no (B) = 100–x = 100–60 = 40 (iv) n(A ∪ B) = n(U)–40 = 200–40 = 160. Example: 4 If n(A) = 40, n(B) = 60 and n(A ∪ B) = 80, using Venn diagram, find the value of n(A ∩ B). Solution: Given, n(A) = 40 n(B) = 60 n(A ∪ B) = 80 n(A ∩ B) = ? Let, n(A ∩ B) = x From the Venn diagram, n(A ∪ B) = 40 – x + x + 60 – x or, 80 = 100 – x or, x = 100 – 80 ∴ n(A ∩ B) = 20 U A B 120-x x 100-x 40 U A B 40–x x 60–x

131 Oasis School Mathematics-8 131 Example: 5 In a class of 150 students, 70 study English, 90 study Mathematics and 50 study both. Find the number of students who study neither of them. Solution: Let 'E' and 'M' be the set of students who study English and Mathematics respectively. Here, n(U) = 150 n(E) = 70 n(M) = 90 n(E ∩ M) = 50 n(E ∪ M) = ? We have, n(E ∪ M) = n(E)+n(M)–n(E ∩ M) = 70 + 90 – 50 = 110. Again we have, n(E ∪ M) = n(U)–n(E ∪ M) = 150–110 = 40 Hence, 40 students study neither subjects. Example: 6 In a group of students 45 speak either Nepali or English or both, 22 can speak Nepali only and 12 can speak English only. How many students can speak both Nepali and English? Solution: Let, 'N' and 'E' be the set of students who speak Nepali and English respectively. Here, n0 (N) = 22 n0 (E) = 12 n(N∪E) = 45 n(N∩E) = ? We have, n(N∪E) = n0 (N) + n0 (E) + n(N∩E) 45 = 22 + 12 + n(N∩E) or, 45 = 34 + n(N∩E) or, n(N∩E) = 45 – 34 = 11 ∴ 11 students can speak both Nepali and English. From the Venn diagram, n(U) = 20+50+40+x 150 = 110 + x or, x = 150 – 110 or, x = 40 Alternative method U M 70–50 90–50 = 20 = 40 x 50 E Let, n(E ∪ M) = x Alternative method Here, n0 (N) = 22, n0 (E) = 12, n(A∪B) = 45 Let, n(N∩E) = x From the Venn diagram, 22 + x + 12 = 45 or, 34 + x = 45 or, x = 45 – 34 Hence, 11 students can speak both Nepali and English. N E 22 x 12 N E no(N) = 22 no(E) = 12 n(N∩E) = 11

132 Oasis School Mathematics-8 Example: 7 In a group of 50 students, 25 play cricket, 30 play football and 8 play neither game. Find the number of students who play both cricket and football. Solution: Let, 'C' and 'F' be the sets of students who play cricket and football respectively. Here, n(U) = 50 n(C) = 25 n(F) = 30 n(C∪F) = 8 n(C∩F) = ? We have, n(C∪F) = n(U) – n(C∪F) = 50 – 8 = 42 Again, We have, n(C∪F) = n(C) + n(F) – n(C∩F) 42 = 25 + 30 – n(C∩F) or, n(C ∩ F) = 25 + 30 – 42 = 13 ∴ 13 students play both cricket and football. Exercise 8.3 1. (a) If A = {w, x, y, z}, find n(A). (b) If B = {a, b, c, d, e}, find n(B). 2. From the given Venn diagram, find the cardinal number of following sets. (i) n(A) (ii) n(B) (iii) n0 (A) (iv) n0 (B) (v) n(A ∩ B) (vi) n(A ∪ B) (vii) n(A∪B) (viii) n (A) (ix) n (B) 3. (a) If A = {2,3,5,7} and B={1,2,3,4,6} be two subsets of the universal set U={1, 2, 3, 4, 5, 6, 7, 8}, find (i) n(A) (ii) n(B) (iii) n(A – B) (iv) n(B – A) (b) If P = {2, 4, 6, 8} and Q = {a set of prime numbers less than 10} be the subsets of universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, find (i) n(P) (ii) n(Q) (iii) n(P∩Q) (iv) no (P) (v) no (Q) (vi) n(P∪Q) (vii) n(P ∪ Q). Alternative method Let, n(C∩F) = x From, the Venn diagram, 25 – x+ x + 30 – x + 8 = 50 or, 63 – x = 50 or, x = 63 – 50 or, x = 13 ∴ 13 students play both cricket and football. U C F 25 – x 30 – x 8 x U A B a j c b k l m d e o p n x h f i

133 Oasis School Mathematics-8 133 4. If n(A) = 12, n(B) = 14, find the maximum value of n(A∪B) and n(A∩B) 5. (a) If n(A) = 40, n(B) = 60 and n(A∩B) = 10, find the value of n(A∪B). (b) n(A) = 80, n(B) = 70, n(A∪B) = 100, find n(A∩B). (c) If n0 (A) = 35, n0 (B) = 45, n(A∩B) = 20, find (i) n(A) (ii) n(B) (iii) n(A∪B). 6. (a) If n(U) = 50, n(A) = 15, n(B) = 30 and n(A∩B) = 7, find (i) n(A∪B) (ii) n(A – B) (iii) n(B – A) (iv) n(A ∪ B) . (b) If n(P) = 35, n(Q) = 45, n(P∪Q) = 60, find the n(P∩Q). 7. If n(U) = 200, n(M) = 160, n(N) = 150, n(M ∪ N) = 30, find the value of (i) n(M∩N) (ii) n0 (M) (iii) n0 (N) (iv) Show the above information in the Venn diagram. 8. (a) In a group of 90 people, 70 like mango, 50 like both mango and orange. Each people like at least one of the two fruits,. Using Venn diagram find, (i) the number of people who like orange. (ii) the number of people who like only one fruit. (b) In a survey conducted among 350 people in a community, 200 people are found influenced by materialism and 120 people are found influenced by spiritualism. If 120 persons are found influenced by both, (i) Show the above information in the Venn diagram. (ii) Find the number of people who are influenced by neither of them. 9. (a) In a school, all students play either football or volleyball or both of then 300 play football, 250 play volleyball and 110 play both the games. Draw a Venn diagram to find, (i) the number of students playing football only. ii) the number of students playing volleyball only. (iii) the total number of students. (b) In a class of 50 students, 20 students play football and 16 students play hockey. It is found that 10 students play both the games. Find the number of students who play (i) at least one game (ii) neither of them. (c) Out of 30 students of class VIII, 15 like to play cricket, 12 like to play football, 6 do not like to play any game. (i) Draw the Venn diagram to illustrate this information. (ii) How many students like to play both the games? (iii) How many students like to play cricket only? (iv) How many students like to play football only?

134 Oasis School Mathematics-8 10. 40 students of a class study either Mathematics or English or both. 12 students take both subjects and 17 study English only. How many students study (i) Mathematics only? (ii) English ? 1. (a) 4 (b) 5 2. (i) 9 (ii) 8 (iii) 5 (iv) 4 (v) 4 (vi) 3 (vii) 13 (viii) 7 (ix) 8 3. (a) (i) 4 (ii) 5 (iii) 2 (iv) 3 (b) (i) 4 (ii) 4 (iii) 1 (iv) 3 (v) 3 (vi) 7 (viii) 2 4. 26 and 12 5. (a) 90 ((b) 50 (c) (i) 55 (ii) 65 (iii) 100 6. (a) (i) 38 (ii) 8 (iii) 23 (iv) 12 (b) 20 7. (i) 40 (ii) 120 (iii) 110 8. (a) (i) 70 (ii) 40 b. (ii) 150 9. (a) (i) 190 (ii) 140 (iii) 440 (b) (i) 26 (ii) 24 (c) (ii) 3 (iii) 12 (iv) 9 10. (i) 11 (ii) 29 Answer Full marks : 20 Assessment Test Paper Group - A [6 × 1 = 6] 1. (a) If A = a set of prime numbers less than 10 and B = {2, 3, 5, 7}, identify whether A and B are equal sets or not. (b) If A = {a, b, c}, find the number of subsets of A. 2. (a) What does the shaded part of given Venn diagram represent? (a) (b) 3. (a) If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}, find n(A∪B). (b) Write the formula that shows the relation among n(A), n(B), n(A∪B) and n(A∩B). Group - B [ 3×2=6] 4. (a) Let A = {2, 3, 5, 7, 11, 13}, B = {2, 4, 6, 8, 10, 12, 14}, find (i) A∪B (ii) A∩B. (b) Draw a Venn diagram to show the relation (A∩B)∩C. (c) If A = {p, q, r}, B = { q, r, s}, find (i) A–B and B–A. Group- C [2 × 4 = 8] 5. (a) In the given Venn diagram, if n(∪) = 150, find the value of (i) x, (ii) no (A), (iii) no (B) (b) In a group of 100 people, 70 like tea and 80 like coffee. If 60 like both, find the number of people who do not like both. U A B U A B U A B 70–x 80–x 20 x

135 Oasis School Mathematics-8 135 Arithmetic Contents • Number System of Different Bases • Integers • Rational and Irrational Numbers • Ratio and Proportion • Percentage • Unitary Method • Simple Interest Expected Learning Outcomes At the end of this unit, students will be able to develop the following competencies: • Convert decimal number into binary and quinary numbers and vice-versa. • Add or subtract the binary numbers and quinary numbers. • Use the four fundamental operations on integers. • Simplify the problems on integers. • Identify rational and irrational numbers. • Operate on irrational numbers. Teaching Materials: • Chart paper, A4 size paper, flash cards, etc. Estimated Teaching Hours 49

136 Oasis School Mathematics-8 9.1 Review In Hindu-Arabic system of numeration, ten basic symbols are used to represent the large numbers. They are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. The numbers are expressed in powers of 10. This number system is 10 based number system or decimal number system. For example, 24 = 2 × 101 + 4 × 100 386 = 3 × 102 + 8 × 101 + 6 × 100 4512 = 4 × 103 + 5 × 102 + 1 × 101 + 2× 100 9.2 Binary number system This is another system of numeration. This system uses only two digits 0 and 1. In this system, numbers are expressed in terms of powers of 2. So it is base-two system. For example, 10012 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 8 + 0 + 0 + 1 = 9 11102 = 1 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 = 16 + 8 + 4 + 2 + 0 = 30. ∴ 10012 = 9 and 11102 = 30 binary number decimal number binary number decimal number Unit 9 Number System of Different Bases

137 Oasis School Mathematics-8 137 Place value chart Decimal numeration Base two grouping Binary numeration 27 26 25 24 23 22 21 20 0 0 0 1 1 12 2 1 0 102 3 1 1 112 4 1 0 0 1002 5 1 0 1 1012 6 1 1 0 1102 7 1 1 1 1112 8 1 0 0 0 10002 9 1 0 0 1 10012 10 1 0 1 0 10102 In the binary system, we see from the above table that the even numbers have zero (0) in ones digit and the odd numbers have one (1) in ones digit. I. Conversion of Binary System (2 base system) into decimal number system (10 base system): To convert a binary number system into decimal system, expand the given binary number by giving the place value of each digit in power of 2. For example, 10012 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 1 × 8 + 0 × 4 + 0 × 2 + 1 × 1 = 8 + 0 + 0 + 1 = 9 1000012 = 1 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 1 × 32 + 0 × 16 + 0 × 8 + 0 × 4 + 0 × 2 + 1 × 1 = 32 + 0 + 0 + 0 + 0 + 1 = 33 II. Conversion of decimal numbers into binary numbers: To convert a decimal number into a binary number system, follow the given steps. For example: Convert 30 into a binary number. 2 2 2 2 2 30 7 Remainder 0 1 1 1 1 15 3 1 0 Writing all the remainders from the bottom to the top 11110. ∴ 30 = 111102 • Divide the given number successively by 2 until the quotient is zero. • List the remainder obtained in each successive division in a column. • Write all the remainders from the bottom to the top. Which is the required binary number.

138 Oasis School Mathematics-8 9.3 Quinary Number System The number system which uses the number 5 as the base is called quinary number system. Under the quinary number system, the numbers are expressed in powers of 5. The five basic symbols used in the quinary number system are 0, 1, 2, 3 and 4. For example, 135 = 1 × 51 + 3 × 50 = 1 × 5 + 3 × 1 = 8 ∴ 135 = 810 2335 = 2 × 5² + 3 × 5¹ + 3 × 5º = 2 × 25 + 3 × 5 + 3 × 1 = 50 + 15 + 3 = 68 ∴ 2335 = 6810 Let's see the given table Decimal numbers Quinary numbers 0 0 1 15 2 25 3 35 4 45 5 105 6 115 7 125 8 135 9 145 10 205 Place value chart Base 5 system 54 53 52 51 50 Base-10 system 235 2 3 13 1425 1 4 2 47 12305 1 2 3 0 190 I. Conversion of quinary number system (base-5 system) into decimal number system (base - 10 system) To convert a quinary number system into decimal number system, we should expand the given number by giving the place value of each digit in powers of 5. Find the product and then the sum of all numbers. For example, 1235 = 1 × 5² + 2 × 5¹ + 3 × 5° = 1 × 25 + 2 × 5 + 3 × 1 = 25 + 10 + 3 = 38 ∴ 1235 = 3810

139 Oasis School Mathematics-8 139 and 11435 = 1 × 5³ + 1 × 5² + 4 × 5¹ + 3 × 5° = 1 × 125 + 1 × 25 + 4 × 5 + 3 × 1 = 125 + 25 + 20 + 3 = 173 ∴ 11435 = 17310. II. Conversion of decimal numbers into quinary numbers To convert decimal numbers into quinary number system • Divide the given number successively by 5 until the quotient is zero. • List the remainder obtained in each successive division in a column. • Write all the remainders from the bottom to the top. Which is the required quinary number. For example: Convert 258 into quinary number system. Here, Writing the remainder from the bottom to the top 2013. ∴ 258 = 20135 Worked Out Examples Example: 1 Display the numbers 10012 , 111012 , 1000112 and 10000012 in place value chart in binary system. Solution: Place value chart in binary system 26 25 24 23 22 21 20 10012 1 0 0 1 111012 1 1 1 0 1 1000112 1 0 0 0 1 1 10000012 1 0 0 0 0 0 1 5 5 5 5 258 10 Remainder 3 1 0 2 51 2 0

140 Oasis School Mathematics-8 Example: 2 Convert the following binary numbers into decimal system. (i) 10012 (ii) 10000012 Solution: (i) 10012 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 1 × 8 + 0 × 4 + 0 × 2 + 1 × 1 = 8 + 0 + 0 + 1 = 9 (ii) 10000012 = 1 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20 = 1 × 64 + 0 × 32 + 0 × 16 + 0 × 8 + 0 × 4 + 0 × 2 + 1 × 1 = 64 + 0 + 0 + 0 + 0 + 0 + 1 = 65 Example: 3 Convert 59 into binary (base two) number system. Solution: 2 2 2 2 2 2 59 14 Remainder 1 1 0 1 1 1 29 7 3 1 0 ∴ 59 = 1110112 Example: 4 Display the numbers 23015 , 123145 and 2304105 in place value chart of quinary number system. Solution: 55 54 53 52 51 50 23015 2 3 0 1 12345 1 2 3 1 4 2304105 2 3 0 4 1 0 Example: 5 Convert the following quinary numbers into decimal numbers (i) 1005 (ii) 23405 (iii) 11345 Solution: (i) 1005 = 1 × 5² + 0 × 5¹ + 0 × 50 = 1 × 25 + 0 × 5 + 0 × 1 = 25 + 0 + 0 = 25

141 Oasis School Mathematics-8 141 (ii) 23405 = 2 × 53 + 3 × 52 + 4 × 51 + 0 × 50 = 2 × 125 + 3 × 25 + 4 × 5 + 0 × 1 = 250 + 75 + 20 + 0 = 345 iii) 11345 = 1 × 53 + 1 × 52 + 3 × 51 + 4 × 50 = 1 × 125 + 1 × 25 + 3 × 5 + 4 × 1 = 125 + 25 + 15 + 4 = 169 Example: 6 Convert the following decimal numbers into quinary numbers. (i) 145 (ii) 9999 Solution: (i) Successive remainders are taken from bottom to top 1040. ∴ 145 = 10405 Checking : 10405 = 1 × 5³ + 0 × 5² + 4 × 5¹ + 0 × 50 = 125 + 0 + 20 + 0 = 145 (ii) 5 5 5 5 5 5 9999 1999 399 Remainder 4 4 4 4 0 3 79 15 3 0 ∴ 9999 = 3044445 Example: 7 Convert 21345 into binary number system. Solution: Lets convert 21345 into decimal system. Now, 21345 = 2 × 53 + 1 × 52 + 3 × 51 + 4 × 50 = 250 + 25 + 15 + 4 = 294 ∴ 21345 = 29410 .....................(i) 5 5 5 5 145 5 Remainder 0 4 0 1 29 1 0

142 Oasis School Mathematics-8 Again, let's convert 29410 into binary number. ∴ 29410 = 1001001102 .................(ii) From (i) and (ii) ∴ 21345 = 1001001102 .....................(i) Example: 8 Convert 110102 into quinary number system. Solution: Let's convert 110102 into decimal number system. Now, 110102 = 1 × 24 + 1 × 2³ + 0 × 2² + 1 × 2¹ + 0× 20 = 1 × 16 + 1 × 8 + 0 × 4 + 1 × 2 + 0 × 1 = 16 + 8 + 0 + 2 + 0 = 26 ∴ 110102 = 2610 ..........................(i) Again, lets convert 2610 into quinary number system. ∴ 2610 = 1015 ...........................(ii) From (i) and (ii) 110102 = 1015 . Exercise 9.1 1. Display the following binary numbers in place value chart. (a) 1101002 (b) 1000100112 (c) 11110012 Remainder 0 1 1 0 0 1 0 0 1 2 2 2 2 2 2 2 2 2 294 147 73 36 18 9 4 2 1 0 5 5 5 26 1 Remainder 1 0 1 5 0

143 Oasis School Mathematics-8 143 2. Convert each of the binary number (i.e. base 2 system) into decimal number system. (a) 102 (b) 1002 (c) 11112 (d) 10012 (e) 101002 (f) 101012 (g) 111102 (h) 11110102 (i) 10000012 (j) 11111112 3. Convert the following decimal numbers into binary numbers. (a) 45 (b) 105 (c) 355 (d) 400 (e) 529 (f) 512 (g) 190 (h) 101 (i) 723 4. Display the following quinary numbers in place value chart. (a) 423015 (b) 1230015 (c) 23101235 5. Convert the following quinary numbers into decimal number (i.e. base 10 number). (a) 1005 (b) 2105 (c) 4235 (d) 200015 (e) 32345 6. Convert the following decimal numbers into quinary numbers. (a) 150 (b) 500 (c) 1024 (d) 5054 (e) 8085 7. Express the following base two numbers into base five numbers. (a) 111102 (b) 1111112 (c) 1011012 8. Convert the following base five numbers into base two numbers. (a) 10125 (b) 44445 (c) 31045 (d) 400025 1. Consult your teacher 2. (a) 2 (b) 4 (c) 15 (d) 9 (e) 20 ( f) 21 (g) 30 (h) 122 (i) 65 (j) 127 3. (a) 1011012 (b) 11010012 (c) 1011000112 (d) 1100100002 (e) 10000100012 (f) 10000000002 (g) 101111102 (h) 11001012 (i) 10110100112 4. (a) Consult your teacher 5. (a) 25 (b) 55 (c) 113 (d) 1251 (e) 444 6. (a) 11005 (b) 40005 (c) 130445 (d) 1302045 (e) 2243205 7. (a) 1105 (b) 2235 (c) 1405 8. (a) 100001002 (b) 10011100002 (c) 1100101002 (d) 1001110001102 Answer

144 Oasis School Mathematics-8 9.4 Addition and Subtraction of Binary Numbers a. Addition of Binary Numbers In decimal system, if the sum of the digits becomes 10 or more, we start to take carry over 1 ten, 2 tens, 3 tens, etc. to the higher places. But, in binary (base 2) system, if the sum of the digits becomes 2 or more, we start to take carry 1 two, 2 twos, 3 twos, etc. to the higher places. Binary number system has only two digits 0 and 1. Let's remember the basic rule 0 + 0 = 0 1 + 0 = 1 1 + 1 = 10 i.e. write 0 in its place and carry over 1 1 + 1 + 1 = 10 + 1 = 11, i.e write 1 in its place and carry over 1. Example, Add : 1012 + 112 Steps 1 0 12 +0 1 1 2 10 0 0 1 1 • 1 + 1 = 10, set down 0 in its place, carry over 1. • 1+ 0 + 1 = 10, set down 0, carry over 1. • 1 + 1 = 10 b. Subtraction of binary numbers Let's remember the basic rule for subtraction. 0 – 0 = 0 1 – 1 = 0 1 – 0 = 1 10–1 = 1 We can borrow from higher places if it is necessary. Let's be clear with the help of given example. Subtract : 11012 – 1112 Here, Steps 1 1012 – 1112 1 1 0 0 10 10 • 1 – 1 = 0, set 0 down. • Borrow 1. then 10–1 = 1, set 1 down • Again, borrow 1, 10–1=1, set 1 down