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Exercise 1.1 : Solutions of Questions on Page Number : 14 Q1 : Using appropriate properties find: (i) (ii) Answer : (i) (ii) (By commutativity) Q2 : Write the additive inverse of each of the following: (i) (ii) (iii) (iv) (v) Answer : (i) Additive inverse = (ii) Additive inverse = (iii) Additive inverse = (iv) Additive inverse (v) Additive inverse Q3 : Verify that – ( – x) = x for. (i) (ii) Answer : (i) The additive inverse of is  as This equality represents that the additive inverse of is or it can be said that i.e., -…

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Exercise 14.1 Question 1:Draw a circle with circle of radius 3.2 and write down it’s procedures. Answer: Procedure: (a) For the required radius of 3 cm open the compass (b) Make a point with a sharp pencil where we want the centre of circle to be. (c) Name it O (d) Place the pointer of compasses on 0. (e) Turn the compasses slowly to draw the circle. Hence, it is the required circle. Question 2: With the same centre 0, draw two circles of radii 4 cm and 2.5 cm Answer: Steps of construction: (a) Marks a point ‘0’ with…

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Exercise 13. 1 Q1. Observe your surroundings in school or home and list some symmetrical objects. Ans: CD, Glass, Paper sheet, Bucket, Bathtub Q2. Find out the mirror line,for the figure given below. Ans: In the given figure, line l2 is the mirror line. The left and right part of the object is same when folded about the line l2. Q3. From given figure, identify the shapes of the objects and check if they are symmetrical in nature. Also, draw the symmetry line. Ans: (a) Symmetry (b) Symmetry (c) Asymmetry (d) Symmetry (e) Symmetry (f) Symmetry The symmetry line for…

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EXERCISE-12.1 Question 1:The number of girls and boys studying in a class are 25 and 20 respectively. Calculate the ratio of number of girls to that of boys? Calculate the ratio of number of girls to that of the total students in the class? Solution: Given; No. of girls = 25 No. of boys = 20 Total number of students studying the class = 25 + 20 = 45 Ratio of no. of girls to boys = 25⁄20 = 5⁄4 Ratio of girls to total students in the class = 25⁄45 = 5⁄9 Question 2: The strength of a class…

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Exercise 1 Question 1: Let the side of an equilateral triangle be m. Show the perimeter of the triangle with the help of m. Answer Perimeter = sum of three sides = m+m+m = 3m Question 2:Let the side of a regular hexagon be denoted by m. Show the perimeter of the hexagon with the help of m. (All sides are equal in regular hexagon) Answer Side of regular hexagon = m Perimeter = sum of all sides = 6m Question 3:  In the given figure a cube, 3 – D cube is shown. It has six identical squares as…

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Exercise 10.1 Q.1 Find the perimeter of each of the following figures (a) Ans. Perimeter is the distance covered along the boundary of a closed figure. Perimeter = Sum of the distances or lengths of all the sides = PQ + QR + RS + SP. = 3 cm + 2 cm + 1 cm + 4 cm. = (3 + 2 + 1 + 4) cm = 10 cm The perimeter of the given figure is 10 cm. (b) Ans. Perimeter = Sum of the distances or lengths of all the sides = EF + FG + GH +…

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Exercise 1 Question 1: Total no. of animals in five villages are as follows: Village V = 120 Village W = 60 Village X = 80 Village Y = 50 Village Z = 90 Prepare a table of the animals with the help of a symbol 1 in ques which represent 20 animals. Answer the given questions: a) How many symbols represent animals of village X? b) Name the village which has min no. of animals? c) Name the village which has max no. of animals? d) Arrange the name of villages in ascending order according to the number of…

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Class 6 Maths Chapter -8 Decimals Exercise 8.1 1. Using the given diagram write the data in form of numbers in the table given below: Hundreds (100) Tens(10) Ones(1) Tenths(⅒)  –  –  –  –  –  –  –  – Answer: Hundreds (100) Tens(10) Ones(1) Tenths(⅒) Number 0 3 1 2 31.2 1 1 0 4 110.4 2. Write the given decimals in the place value table. (a) 20.4 (b) 0.4 (c) 18.5 (d) 302.1 Answer: Hundreds (100) Tens(10) Ones(1) Tenths(⅒) Number (a) 0 2 0 4 20.4 (b) 0 0 0 4 0.4 (c) 0 1 8 5 18.5 (d) 3…

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Class 6 Maths Chapter -7 Fractions Exercise 2.1 1) Solve the following (a) 2-⅗ Soln: To solve ,we have to make both numbers in fraction form ⇒2⁄1−⅗ Lets take LCM of 1,5 = 1*5 =5 ⇒ 2⁄1∗5⁄5  =  10⁄5 Since both numbers has the denominator as 5, we can solve the equation now, The solution is ⇒10⁄5 −3⁄5 =7⁄5 (b) 4+7⁄8 To solve we have to make both numbers in fraction form ⇒4⁄1+7⁄8 Lets take LCM of 1,8 = 1*8 =8 ⇒4⁄1∗8⁄8=32⁄8 Since both numbers has the denominator as 8, we can solve the equation now, The solution is ⇒32⁄8+7⁄8=39⁄8…

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