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Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability Probability Exercise 23 – Selina Concise Mathematics Class 8 ICSE Solutions Question 1. A die is thrown, find the probability of getting: (i) a prime number (ii) a number greater than 4 (iii) a number not greater than 4. Solution: A die has six numbers : 1, 2, 3, 4, 5, 6 ∴ Number of possible outcomes = 6 Question 2. A coin is tossed. What is the probability of getting: (i) a tail? (ii) ahead? Solution: On tossing a coin once, Number of possible outcome = 2 Question 3. A coin is tossed twice. Find the probability of getting: (i) exactly one head (ii) exactly one tail (iii) two tails (iv) two heads Solution: Question 4. A letter is chosen from the word ‘PENCIL’ what is the probability that the letter chosen is a consonant? Solution: Total no. of letters in the word ‘PENCIL = 6 Total Number of Consonant = ‘PNCL’ i.e. 4 Question 5. A b

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling Data Handling Exercise 22A – Selina Concise Mathematics Class 8 ICSE Solutions Question 1. Arrange the following data as an array (in ascending order): (i) 7, 5, 15, 12, 10, 11, 16 (ii) 6.3, 5.9, 9.8, 12.3, 5.6, 4.7 Solution: (i) Ascending order = 5, 7, 10, 11, 12, 15, 16 (ii) Ascending order = 4.7, 5.6, 5.9, 6.3, 9.8, 12.3 Question 2. Arrange the following data as an array (descending order): (i) 0 2, 0, 3, 4, 1, 2, 3, 5 (ii) 9.1, 3.7, 5.6, 8.3, 11.5, 10.6 Solution: (i) Descending order = 5, 4, 3, 3, 2, 2, 1, 0 (ii) Descending order = 11.5, 10.6, 9.1, 8.3, 5.6, 3.7 Question 3. Construct a frequency table for the following data: (i) 6, 7, 5, 6, 8, 9, 5, 5, 6, 7, 8, 9, 8, 10, 10, 9, 8, 10, 5, 7, 6, 8. (ii) 3, 2, 1, 5, 4, 3, 2, 5, 5, 4, 2, 2, 2, 1, 4, 1, 5, 4. Solution: (i) (ii) Question 4. Following are the marks obtained by 30

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacity

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacity Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacity (Cuboid, Cube and Cylinder) Surface Area, Volume and Capacity Exercise 21A – Selina Concise Mathematics Class 8 ICSE Solutions Question 1. Find the volume and the total surface area of a cuboid, whose : (i) length = 15 cm, breadth = 10 cm and height = 8 cm. (ii) l = 3.5 m, b = 2.6 m and h = 90 cm, Solution: (i) Length =15 cm, Breadth = 10 cm, Height = 8 cm. Volume of a cuboid = Length x Breadth x Height = 15 x 10 x 8 =1200 cm3. Total surface area of a cuboid 2 (l x b + b x h + h x l) = 2 (15 x 10 + 10 x 8 + 8 x 15) = 2(150 + 80 +120) = 2 x 350 = 700 cm2 (ii) Length = 3.5 m Breadth = 2.6 m, Height = 90 cm =   m = 0.9 m. Volume of a cuboid = l x b x h = 3.5 x 2.6 x 0.9 = 8.19 m3 Total surface area of a cuboid = 2(l x b + b x h + h x l) = 2 (3.5 x 2.6 + 2.6 x 0.9 + 0.9 x 3.5) = 2 (910 + 2.34

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon Area of Trapezium and a Polygon Exercise 20A – Selina Concise Mathematics Class 8 ICSE Solutions Question 1. Find the area of a triangle, whose sides are : (i) 10 cm, 24 cm and 26 cm (ii) 18 mm, 24 mm and 30 mm (iii) 21 m, 28 m and 35 m Solution: (i) Sides of ∆ are a = 10 cm b = 24 cm c = 26 cm (ii) Sides of ∆ are a = 18 mm b = 24 mm c = 30 mm (iii) Sides of ∆ are a = 21 m b = 28 m c = 35 m Question 2. Two sides of a triangle are 6 cm and 8 cm. If height of the triangle corresponding to 6 cm side is 4 cm ; find : (i) area of the triangle (ii) height of the triangle corresponding to 8 cm side. Solution: Question 3. The sides of a triangle are 16 cm, 12 cm and 20 cm. Find : (i) area of the triangle ; (ii) height of the triangle, corresponding to the largest side ; (iii) height of the triangle,

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D Representing 3-D in 2-D Exercise 19 – Selina Concise Mathematics Class 8 ICSE Solutions Question 1. If a polyhedron has 8 faces and 8 vertices, find the number of edges in it. Solution: Faces = 8 Vertices = 8 using Eulers formula, F + V – E = 2 8 + 8 – E = 2 -E = 2 – 16 E= 14 Question 2. If a polyhedron has 10 vertices and 7 faces, find the number of edges in it. Solution: Vertices = 10 Faces = 7 Using Eulers formula, F + V – E = 2 7 + 10 – E = 2 -E = -15 E = 15 Question 3. State, the number of faces, number of vertices and number of edges of: (i) a pentagonal pyramid (ii) a hexagonal prism Solution: (i) A pentagonal pyramid Number of faces = 6 Number of vertices = 6 Number of edges = 10 (ii) A hexagonal prism Number of faces = 8 Number of vertices = 12 Number of edges = 18 Question 4. Verily Euler’s formula

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 18 Constructions (Using ruler and compass only) Constructions Exercise 18A – Selina Concise Mathematics Class 8 ICSE Solutions Question 1. Given below are the angles x and y. Without measuring these angles, construct : (i) ∠ABC = x + y (ii) ∠ABC = 2x + y (iii) ∠ABC = x + 2y Solution: (i) Steps of Construction : 1. Draw a line segment BC of any suitable length. 2. With B as centre, draw an arc of any suitable radius. With the same radius, draw arcs with the vertices of given angles as centres. Let these arcs cut arms of the arc x at points P and Q and arms of angle y at points R and S. 3. From the arc, with centre B, cut DE = PQ arc of x and EF = RS arc of y 4. Join BF and produce upto point A. Thus ∠ABC = x + y (ii) Steps of Construction : Proceed in exactly the same way as in part(i) takes DE = PQ = arc of x. EF = PQ = arc of x and FG = RS