Chapter 1 ALGEBRA 1.1 Simple expressions with two variables We have learnt about rectangle. Its area is l # b in which the letters 'l' and 'b' are variables. Variables follow the rules of four fundamental operations of numbers. Let us now translate a few verbal phrases into expressions using variables. Operation Addition Verbal phrase Sum of x and y Algebraic Expression x+y Subtraction Difference between a and b a - b ^if a > bh ^orh b - a (if b > a)  Multiplication product of x and y x # y (or) xy Division p divided by q p ' q (or) p q The following table will help us to learn some of the words (phrases) that can be used to indicate mathematical operations: Addition Subtraction The sum of the difference of increased by decreased by plus minus added to subtracted from more than less than Multiplication Division the product of the quotient of multiplied by divided by times the ratio of Example 1.1 Write the algebraic expressions for the following: 1) Twice the sum of m and n. 2) b decreased by twice a. 3) Numbers x and y both squared and added. 4) Product of p and q added to 7. 2 Algebra 5) Two times the product of a and b divided by 5. 6) x more than two-third of y. 7) Half a number x decreased by 3. 8) Sum of numbers m and n decreased by their product . 9) 4 times x less than sum of y and 6. 10) Double the sum of one third of a and m. 11) Quotient of y by 5 added to x. Solution: 1) 2^ m + nh 2) b - 2a 3) x2 + y2 4) 7 + pq 5) 2ab 5 x -3 2 ^ y + 6h - 4x y +x 5 6) 2 y + x 7) 9) 11) 3 8) (m + n) - mn 10) 2` 1 a + mj 3 Express each of the following as an algebraic expression (i) a times b. (ii) 5 multiplied by the sum of a and b. (iii) Twice m decreased by n . (iv) Four times x divided by y . (v) Five times p multiplied by 3 times q. Exercise 1.1 1. Choose the correct answer: (i) The sum of 5 times x, 3 times y and 7 (A) 5^ x + 3y + 7h (B) 5x + 3y + 7 (C) 5x + 3^ y + 7h (D) 5x + 3^7yh (ii) One half of the sum of numbers a and b (A) 1 ^a + bh (B) 1 a + b (C) 1 ^a - bh 2 2 2 (D) 1 + a + b 2 (iii) Three times the difference of x and y (A) 3x - y (B) 3 - x - y (C) xy - 3 3 (D) 3^ y - xh Chapter 1 (iv) 2 less than the product of y and z (A) 2 - yz (B) 2 + yz (C) yz - 2 (D) 2y - z (v) Half of p added to the product of 6 and q p 6q (A) + 6q (B) p + (C) 1 ^ p + 6qh (D) 1 ^6p + qh 2 2 2 2 2. Write the algebraic expressions for the following using variables, constants and arithmetic operations: (i) Sum of x and twice y. (ii) Subtraction of z from y. (iii) Product of x and y increased by 4 (iv) The difference between 3 times x and 4 times y. (v) The sum of 10, x and y. (vi) Product of p and q decreased by 5. (vii) Product of numbers m and n subtracted from 12. (viii) Sum of numbers a and b subtracted from their product. (ix) Number 6 added to 3 times the product of numbers c and d. (x) Four times the product of x and y divided by 3. 1.2 Simple Linear Equations Malar’s uncle presented her a statue. She wants to know the weight of that statue. She used a weighing balance to measure its weight. She knows her weight is 40kg . She finds that the weight of the statue and potatoes balance her weight. i.e., Weight of statue s Plus + Weight of potatoes 15 Equal = Malar’s weight 40            Table 1.1 Now we will think about a balance to find the value of s. 4 Algebra Take away 15 from both sides. Now the balance shows the weight of the statue. s + 15 = 40 (from Table 1.1) s + 15 – 15 = 40 – 15 (Taking away 15 from both the sides) s = 25 So the statue weighs 25 kg. The statement s + 15 = 40 is an equation. i.e., a statement in which two mathematical expressions are equal is called an equation. In a balance, if we take away some weight from one side, to balance it we must take away the same weight from the other side also. If we add some weight to one side of the balance, to balance it we must add the same weight on the other side also. Similarly, an equation is like a weighing balance having equal weights on each side. In an equation there is always an equality sign. This equality sign shows that value of the expression on the left hand side (LHS) is equal to the value of the expression on the right hand side (RHS). ¬ Consider the equation x + 7 = 15 Here LHS is x + 7 RHS is 15 We shall subtract 7 from both sides of the equation x + 7 - 7 = 15 - 7 (Subtracting 7 reduces the LHS to x) x = 8 (variable x is separated) 5 Chapter 1 ¬ Consider the equation n - 3 = 10 LHS is n - 3 RHS is 10 Adding 3 to both sides, we get n - 3 + 3 = 10 + 3 n = 13 (variable n is separated) ¬ Consider the equation 4m = 28 Divide both sides by 4 4m = 28 4 4 m=7 ¬ Consider the equation y =6 2 Multiply both sides by 2 y #2 = 6#2 2 y = 12 So, if we add (or subtract) any number on one side of an equation, we have to add (or subtract) the same number the other side of the equation also to keep the equation balanced. Similarly, if we multiply (or divide) both sides by the same non-zero number, the equation is balanced. Hence to solve an equation, one has to perform the arithmetical operations according to the given equations to separate the variable from the equation. Example 1.2 Solve 3p + 4 = 25 Solution: 3p + 4 - 4 = 25 - 4 (Subtracting 4 from both sides of the equation) 3p = 21 3p = 21 3 3 p=7 (Dividing both sides by 3) Example 1.3 Solve 7m - 5 = 30 Solution: 7m - 5 + 5 = 30 + 5 (adding 5 on both sides) 6 Algebra 7m = 35 7m = 35 7 7 m=5 (Dividing both sides by 7) While solving equations, the commonly used operation is adding or subtracting the same number on both sides of the equation. Instead of adding or subtracting a number on both sides of the equation, we can transpose the number. Transposing a number (i.e., changing the side of the number) is the same as adding or subtracting the number from both sides. While transposing a number we should change its sign. Let us see some examples of transposing. Example 1.4 Solve 2a - 12 = 14 Solution: Adding or subtracting on both sides Transposing 2a - 12 = 14 2a - 12 = 14 2a - 12 + 12 = 14 + 12 (adding 12 on Transpose (– 12) from LHS to RHS both sides) 2a = 14 + 12 (on transposing 2a = 26 - 12 becomes + 12 ) 2a = 26 (dividing both sides by 2) 2 2 2a = 26 2a = 26 2 2    a = 13 a = 13 (Dividing both sides by 2) Example 1.5 Solve 5x + 3 = 18 Solution: Transposing +3 from LHS to RHS 5x = 18 - 3 (on Transposing +3 becomes - 3 ) 5x = 15 5x = 15 5 5 (Dividing both sides by 5) x=3 7 Chapter 1 Example 1.6 Solve 2^ x + 4h = 12 Solution: Divide both sides by 2 to remove the brackets in the LHS. 2^ x + 4h 12 = 2 2 x+4 = 6 x = 6-4 (transposing +4 to RHS) x=2 Example 1.7 Solve - 3^m - 2h = 18 Solution: Divide both sides by ^- 3h to remove the brackets in the LHS. - 3^ m - 2h = 18 -3 -3 m - 2 =- 6 m =- 6 + 2 m =- 4 Example 1.8 Solve (3x + 1 ) – 7 = 12 Solution: (3x + 1) – 7 = 12 3x + 1 – 7 = 12 3x – 6 = 12 3x = 12 + 6 3x 3 = 18 3 x = 6 Example 1.9 Solve 5x + 3 = 17 - 2x Solution: 5x + 3 = 17 - 2x 8 (transposing - 2 to RHS) Algebra 5x + 2x = 17 - 3 (transposing + 3 to RHS and - 2x to LHS) 7x = 14 7x = 14 7 7 x=2 Example 1.10 Sum of three consecutive integers is 45. Find the integers. Solution: Let the first integer be x . & second integer = x + 1 Third integer = x+1+1 = x+2 Their sum = x + ^ x + 1h + ^ x + 2h = 45 3x + 3 = 45 3x = 42 x = 14 Hence, the integers are x = 14 x + 1 = 15 x + 2 = 16 Example 1.11 A number when added to 60 gives 75. What is the number? Solution: Let the number be x. The equation is 60 + x = 75 x = 75 - 60 x = 15 Example 1.12 20 less than a number is 80. What is the number? Solution: Let the number be x . The equation is x - 20 = 80 x = 80 + 20 x = 100 9 Chapter 1 Example 1.13 1 of a number is 63. What is the number? 10 Solution: Let the number be x . 1 ^ xh = 63 10 1 (x) # 10 = 63 # 10 10 x = 630 The equation is Example 1.14 A number divided by 4 and increased by 6 gives 10. Find the number. Solution: Let the number be x . The equation is x + 6 = 10 4 x = 10 - 6 4 x =4 4 x #4 = 4#4 4 \ the number is 16. Example 1.15 Thendral’s age is 3 less than that of Revathi. If Thendral’s age is 18, what is Revathi’ s age? Solution: Let Revathi’s age be x & Thendral’s age = x - 3 Given, Thendral’s age is 18 years & x - 3 = 18 x = 18 + 3 x = 21 Hence Revathi’s age is 21 years. 10 Algebra Exercise 1.2 1. Choose the correct answer. (i) If p + 3 = 9 , then p is (A) 12 (B) 6 (ii) If 12 - x = 8 , then x is (C) 3 (D) 27 (B) 20 (A) 4 q (iii) If = 7 , then q is 6 (B) 1 (A) 13 42 (iv) If 7^ x - 9h = 35 , then x is (C) - 4 (D) - 20 (C) 42 (D) 7 6 (B) - 4 (C) 14 (A) 5 (v) Three times a number is 60. Then the number is (A) 63 (B) 57 (C) 180 (D) 37 (D) 20 2. Solve : (i) x - 5 = 7 (ii) a + 3 = 10 (iv) b - 3 =- 5 (v) - x = 5 (viii) 14 - n = 10 (vii) 3 - x = 8 (iii) 4 + y =- 2 (vi) - x =- 7 (ix) 7 - m =- 4 (x) 20 - y =- 7 3. Solve : (i) 2x = 100 (iv) 51 = 17a (vii) - 7x = 42 (x) - 3x =- 18 (ii) 3l = 42 (iii) 36 = 9x (v) 5x =- 45 (vi) 5t =- 20 (viii) - 10m =- 30 (ix) - 2x = 1 4. Solve : (i) 1 x = 7 2 p (iv) =8 -7 5. Solve : (i) 3x + 1 = 10 (iv) 4a - 5 =- 41 y+3 (vii) = 14 5 (x) 11m = 42 + 4m (ii) a = 5 6 (v) - x = 2 5 (iii) n =- 8 3 (vi) - m =- 4 3 (ii) 11 + 2x =- 19 (iii) 4z - 3 = 17 (v) 3^ x + 2h = 15 (vi) - 4^2 - xh = 12 (viii) x + 5 = 7 3 (xi) - 3x =- 5x + 22 (ix) 6y = 21 - y (xii) 6m - 1 = 2m + 1 (xiii) 3x - 14 = x - 8 (xiv) 5x - 2x + 7 = x + 1 (xv) 5t - 3 = 3t - 5 11 Chapter 1 6. The sum of two numbers is 33. If one number is 18, what is the other number? 7. A number increased by 12 gives 25. Find the number. 8. If 60 is subtracted from a number, the result is 48. Find the number. 9. 5 times a number is 60. Find the number. 10. 3 times a number decreased by 6 gives 18. Find the number. 11. The sum of 2 consecutive integers is 75. Find the numbers. 12. Ram’s father gave him `70. Now he has `130. How much money did Ram have in the beginning? 13. 8 years ago, I was 27 years old. How old am I now? Solve: (i) y + 18 =- 70 (ii) - 300 + x = 100 (iii) t - 5 =- 6 (iv) 2x + 9 = 19 3 (v) 3x + 4 = 2x + 11 Fun game Ram asked his friends Arun, Saranya and Ravi to think of a number and told them to add 50 to it. Then he asked them to double it. Next he asked them to add 48 to the answer. Then he told them to divide it by 2 and subtract the number that they had thought of. Ram said that the number could now be 74 for all of them. Check it out if Arun had thought of 16, Saranya had thought of 20 and Ravi had thought of 7. Think of a number x Add 50 x+50 Double it 2x + 100  Add 48 2x + 148  Divide by 2 x + 74  Take away the number you thought of 74 12 Arun Saranya Ravi 16 20 7 Algebra 1. Algebra is a branch of Mathematics that involves alphabet, numbers and mathematical operations. 2. A variable or a literal is a quantity which can take various numerical values. 3. A quantity which has a fixed numerical value is a constant. 4. An algebraic expression is a combination of variables and constants connected by the arithmetic operations. 5. Expressions are made up of terms. 6. Terms having the same variable or product of variables with same powers are called Like terms. Terms having different variable or product of variables with different powers are called Unlike terms. 7. The degree of an expression of one variable is the highest value of the exponent of the variable. The degree of an expression of more than one variable is the highest value of the sum of the exponents of the variables in different terms 8. A statement in which two expressions are equal is called an equation. 9. An equation remains the same if the LHS and RHS are interchanged. 10. The value of the variable for which the equation is satisfied is called the solution of the equation. 13 Chapter 2 LIFE MATHEMATICS 2.1 Percent In the banners put up in the shops what do you understand by 25%, 20% ? Ramu’s mother refers to his report card to analyze his performance in Mathematics in standard VI. His marks in Maths as given in his report card are 17 / 25 , 36 / 50 , 75 / 100 , 80 / 100 , 22 / 25, 45 / 50 She is unable to find his best mark and his least mark by just looking at the marks. So, she converts all the given marks for a maximum of 100 (equivalent fractions with denominator 100) as given below: Unit Test 1 Monthly Test 1 Quarterly Exam Half - yearly Exam Unit Test 2 Monthly Test 2 68 100 72 100 75 100 80 100 88 100 90 100 14 Life Mathematics Now, all his marks are out of 100. So, she is able to compare his marks easily and is happy that Ramu has improved consistently in Mathematics in standard vi. Now let us learn about these special fractions. Try and help the duck to trace the path through the maze from ‘Start’ to ‘End’. Is there more than one path? No, there is only one path that can be traced from ‘Start’ to ‘End’. Total number of the smallest squares = 100 Number of shaded squares = 41 Number of unshaded squares = 59 Number of squares traced by the path = ____ Now, look at the table below and fill in the blanks: Shaded Portion Unshaded Portion Portion traced by the path 41 out of 100 59 out of 100 _____ out of 100 Ratio Fraction Percent 41 : 100 41 100 41% 59 : 100 59 100 59% ____ : 100 _____ % 100 The fraction with its denominator 100 is called a Percent. • • • • The word ‘Percent’ is derived from the Latin word ‘Percentum’, which means ‘per hundred’ or ‘hundredth’ or ‘out of 100’. Percentage also means ‘percent’. Symbol used for percent is % Any ratio x : y, where y = 100 is called ‘Percent’. 15 Chapter 2 To Express Percent in Different Forms: Pictorial Representation Shaded portion represented in the form of : Ratio 5 : 100 17 : 100 43 : 100 Fraction 5 100 17 100 43 100 Percent 5% 17% 43% Exercise 2.1 1) Write the following as a percent: (ii) 93 (iii) 11 divided by 100 100 2) Write the following percent as a ratio: (i) 20: 100 (i) 43% (ii) 75% (iii) 5% (iv) 1 100 (iv) 17 1 2 % (v) 33 1 3 % (iv) 70% (v) 82% (v) 100 100 3) Write the following percent as a fraction: (i) 25% (ii) 12 1 2 % (iii) 33% Think! Find the selling price in percentage when 25% discount is given, in the first shop. What is the reduction in percent given in the second shop? Which shop offers better price? 16 Life Mathematics 2.2 To Express a Fraction and a Decimal as a Percent We know that 5 = 5%, 1.2 = 1.2%, 175 = 175%. 100 100 100 To convert 5 to a percent 10 5 represented pictorially can be converted to a percent as shown below: 10 50 5 100 10 Multiply the numerator and denominator by 10 to make the denominator 100 5 # 10 = 50 = 50% 10 # 10 100 This can also be done by multiplying 5 by 100% 10 5 ` 10 # 100j % = 50% 50% of the circle is shaded. 25% of the circle is shaded. Try drawing circles with (i) 50%, (ii) 25% portion shaded in different ways. Do you know? Less than 1 and more than 100 can also be represented as a percent. 1% 2 120% 17 Chapter 2 (i) Fractions with denominators that can be converted to 100 Example 2.1 Express 3 as a percent 5 Solution: 1) 2 × ____ = 100 5 multiplied by 20 gives 100 2) 5 × 20 = _____ 3 # 20 = 60 = 60% 5 # 20 100 3 = 60% 5 3) 4 × 25 = _____ 4) 10 × _____ = 100 5) 1 × _____ = 100 Example 2.2 Express 6 1 as a percent 4 Solution: 6 1 = 25 4 4 4 multiplied by 25 gives 100 25 # 25 = 625 = 625% 4 # 25 100 (ii) Fractions with denominators that cannot be converted to 100 Example 2.3 Express 4 as a percent 7 Solution: Multiply by 100% 4 400 ` 7 # 100j % = 7 % = 57 1 % = 57.14% 7 Example 2.4 Express 1 as a percent 3 Solution: Multiply by 100% 1 100 ` 3 # 100j % = ` 3 j % = 33 1 3 %(or) 33.33% Example 2.5 There are 250 students in a school. 55 students like basketball, 75 students like football, 63 students like throw ball, while the remaining like cricket. What is the percent of students who like (a) basket ball? (b) throw ball? 18 Life Mathematics Solution: Total number of students = 250 (a) Number of students who like basket ball = 55 55 out of 250 like basket ball which can be represented as 55 250 55 Percentage of students who like basket ball = ` # 100j % 250 = 22% (b) Number of students who like throw ball = 63 63 out of 250 like throw ball and that can be represented as 63 250 Percentage of students who like throw ball = ` 63 # 100j % 250 = 126 % = 25.2% 5 22% like basket ball, 25.2% like throw ball. (iii) To convert decimals to percents Example 2.6 Express 0.07 as a percent Solution: Multiply by 100% (0.07 # 100) % = 7% Aliter: 0.07 = 7 = 7% 100 Example 2.7 Express 0.567 as a percent Solution: Multiply by 100% (0.567 # 100)% = 56.7% Aliter: 0.567 = 567 = 1000 = 56.7 100 567 10 # 100 = 56.7% Note: To convert a fraction or a decimal to a percent, multiply by 100%. 19 Chapter 2 Think! 1. 9 of your blood is water. What % of your blood is not water. 10 2. 2 of your body weight is muscle. What % of body is muscle? 5 About 2 of your body weight is water. Is muscle weight plus water weight 3 more or less than 100 %? What does that tell about your muscles? Exercise 2.2 1. Choose the correct answer: (i) 6.25 = (A) 62.5% (B) 6250% (C) 625% (D) 6.25% (B) 0.3% (C) 0.03% (D) 0.0003% (ii) 0.0003 = (A) 3% (iii) 5 = 20 (A) 25% (C) 0.25% (B) 1 % 4 (iv) The percent of 20 minutes to 1 hour is (A) 33 1 3 (C) 33 2 3 (B) 33 (D) 5% (D) none of these (v) The percent of 50 paise to Re. 1 is (C) 50 (B) 1 2 2. Convert the given fractions to percents (A) 500 (D) 20 ii) 9 iii) 5 1 i) 20 4 50 20 3. Convert the given decimals to percents iv) 2 3 v) 5 11 i) 0.36 iv) 3.05 v) 0.75 ii) 0.03 iii) 0.071 4. In a class of 35 students, 7 students were absent on a particular day. What percentage of the students were absent? 5. Ram bought 36 mangoes. 5 mangoes were rotten. What is the percentage of the mangoes that were rotten? 6. In a class of 50, 23 were girls and the rest were boys. What is the percentage of girls and the percentage of boys? 7. Ravi got 66 marks out of 75 in Mathematics and 72 out of 80 in Science. In which subject did he score more? 8. Shyam’s monthly income is `12,000. He saves `1,200 Find the percent of his savings and his expenditure. 20 Life Mathematics 2.3 To Express a Percent as a Fraction (or) a Decimal i) A percent is a fraction with its denominator 100. While expressing it as a fraction, reduce the fraction to its lowest term. Example 2.8 Express 12% as a fraction. Solution: 12 (reduce the fraction to its lowest terms) 100 3 25 12% = = Example 2.9 Percents that have easy fractions Express 233 1 3 % as a fraction. Solution: 50% = 1 2 1 25% = 4 1 1 33 % = 3 3 233 1 3 % = 700 % 3 700 = 7 = 3 # 100 3 = 2 13 Find more of this kind Example 2.10 Express 1 % as a fraction 4 Solution: 1%= 1 = 1 4 4 # 100 400 (ii) A percent is a fraction with its denominator 100. To convert this fraction to a decimal, take the numerator and move the decimal point to its left by 2 digits. Example 2.11 Express 15% as a decimal. Solution: 15% = 15 = 0.15 100 Example 2.12 Express 25.7% as a decimal. Solution: 25.7% = 25.7 100 = 0.257 21 Chapter 2 Math game - To make a triplet (3 Matching cards) This game can be played by 2 or 3 people. Write an equivalent ratio and decimal for each of the given percent in different cards as shown. 5% 1 : 20 0.05 33 1 % 3 1:3 0.33 Make a deck of 48 cards (16 such sets of cards) - 3 cards to represent one particular value - in the form of %, ratio and decimal. Shuffle the cards and deal the entire deck to all the players. Players have to pick out the three cards that represent the same value of percent, ratio and decimal and place them face up on the table. The remaining cards are held by the players and the game begins. One player chooses a single unknown card from the player on his left. If this card completes a triple (3 matching cards) the 3 cards are placed face up on the table. If triplet cannot be made, the card is added to the player’s hand. Play proceeds to the left. Players take turns to choose the cards until all triplets have been made. The player with the most number of triplets is the winner. To find the values of percents Colour 50% of the circle green and 25% of the circle red. 50% = 50 = 1 of the circle is to be coloured green. 100 2 Similarly, 25% = 25 = 25 = 1 100 100 4 1 of the circle is to be coloured red. 4 22 Life Mathematics Now, try colouring 1 of the square, green and 1 of the 2 4 square, red. Do you think that the green coloured regions are equal in both the figures? No, 50% of the circle is not equal to 50% of the square. Similarly the red coloured regions, 25% of the circle is not equal to 25% of the square. Now, let’s find the value of 50% of `100 and 50% of `10. What is 50% of `100? What is 50% of `10? 50% = 50 = 1 100 2 So, 1 of 100 = 1 # 100 = 50 2 2 50% = 50 = 1 100 2 1 of 10 = 1 # 10 = 5 2 2 50% of `100 = `50 50% of `10 = `5 Example 2.13 Find the value of 20% of 1000 kg. Solution: 20% of 1000 = = 20% of 1000 kg = 20 of 1000 100 20 # 1000 100 200 kg. Example 2.14 Find the value of 1 % of 200. 2 Solution: = = 1 2 of 200 100 1 # 200 2 # 100 1 # 200 = 1 200 1 % of 200 = 1 2 23 Chapter 2 Example 2.15 Find the value of 0.75% of 40 kg. Solution: 0.75% = 0.75% of 40 = = 0.75 100 0.75 # 40 100 3 = 0.3 10 0.75% of 40kg = 0.3kg. Example 2.16 In a class of 70, 60% are boys. Find the number of boys and girls. Solution: Total number of students = 70 Number of boys = 60% of 70 = 60 # 70 100 = 42 Number of boys = 42 Number of girls = Total students – Number of boys = 70 – 42 = 28 Number of girls = 28 Example 2.17 In 2010, the population of a town is 1,50,000. If it is increased by 10% in the next year, find the population in 2011. Solution: Population in 2010 = 1,50,000 Increase in population = 10 # 1, 50, 000 100 = Population in 2011 = = 24 15,000 150000 + 15000 1,65,000 Life Mathematics Exercise 2.3 1. Choose the correct answer: (i) The common fraction of 30 % is (A) 1 10 (B) 7 10 (C) 3 100 (D) 3 10 (C) 200 100 (D) 100 (B) 25 (C) 0.0025 (D) 2.5 (B) `20 (C) `30 (D) `300 (B) `7.50 (C) `5 (D) `100 (ii) The common fraction of 1 % is 2 (A) 1 (B) 1 2 200 (iii) The decimal equivalent of 25% is (A) 0.25 (iv) 10% of `300 is (A) `10 (v) 5% of `150 is (A) `7 2. Convert the given percents to fractions: iv) 2.5% iii) 1 % 4 3. Convert the given percents to decimals: i) 9% ii) 75% i) 7% ii) 64% iii) 375% iv) 0.03% v) 66 2 3 % v) 0.5% 4. Find the value of: i) 75% of 24 ii) 33 13 % of `72 iv) 72% of 150 v) 7.5% of 50kg iii) 45% of 80m 5. Ram spent 25% of his income on rent. Find the amount spent on rent, if his income is `25,000. 6. A team played 25 matches in a season and won 36% of them. Find the number of matches won by the team. 7. The population of a village is 32,000. 40% of them are men. 25% of them are women and the rest are children. Find the number of men and children. 8. The value of an old car is `45,000. If the price decreases by 15%, find its new price. 9. The percentage of literacy in a village is 47%. Find the number of illiterates in the village, if the population is 7,500. 25 Chapter 2 Think! 1) 2) Is it true? 20% of 25 is same as 25% of 20. The tax in a restaurant is 1.5% of your total bill. a) Write the tax % as a decimal. b) A family of 6 members paid a bill of ` 750. What is the tax for their bill amount? c) What is the total amount that they should pay at the restaurant? 2.4 Profit and Loss Ram & Co. makes a profit of `1,50,000 in 2008. Ram & Co. makes a loss of `25,000 in 2009. Is it possible for Ram & Co. to make a profit in the first year and a loss in the subsequent year? Different stages of a leather product - bag are shown below: Factory Wholesale Dealer Retailer Where are the bags produced? Do the manufactures sell the products directly? Whom does the products reach finally? Raja, the fruit stall owner buys fruits from the wholesale market and sells them in his shop. On a particular day, he buys apples, mangoes and bananas. 26 Life Mathematics Each fruit has two prices, one at each shop, as shown in the price list. The price at which Raja buys the fruit at the market is called the Cost Price (C.P.). The price at which he sells the fruit in his stall is called the Selling Price (S.P.). From the price list we can say that the selling price of the apples and the mangoes in the shop are greater than their respective cost price in the whole sale market. (i.e.) the shopkeeper gets some amount in addition to the cost price. This additional amount is called the profit. Selling Price of mango = Cost Price of mango + Profit Selling price – Cost price = Profit Profit = = Profit = i.e., Profit = Selling Price – Cost Price 15 – 10 `5 Selling Price – Cost Price In case of the apples, Selling price of apple > Cost price of apple, there is a profit. Profit = S.P. – C.P. = 8–6 Profit = `2 As we know, bananas get rotten fast, the shop keeper wanted to sell them without wasting them. So, he sells the bananas at a lower price (less than the cost price). The amount by which the cost is reduced from the cost price is called Loss. In case of bananas, Cost price of banana > selling price of banana, there is a loss. S.P. of the banana = C.P. of the banana – Reduced amount S.P. = C.P. – Loss Loss = C.P. – S.P. Loss = 3–2 Loss = `1 27 Chapter 2 So, we can say that • When the selling price of an article is greater than its cost price, then there is a profit. Profit = Selling Price - Cost Price • When the cost price of an article is greater than its selling price, then there is a loss. Loss = Cost Price – Selling Price • S.P = C.P + Profit • S.P = C.P - Loss. To find Profit / Loss % Rakesh buys articles for `10,000 and sells them for `11,000 and makes a profit of `1,000, while Ramesh buys articles for `1,00,000 and sells them for `1,01,000 and makes a profit of `1,000. Both of them have made the 1) Any fraction with its denominator 100 is called __________ 2) 1 = _________% 2 same amount of profit. Can you say 3) 35% = _________ ( in fraction) both of them are benefited equally? 4) 0.05 = _________% No. 5) 1 = __ % 4 To find who has gained more, we need to compare their profit based on their investment. We know that comparison becomes easier when numbers are expressed in percent. So, let us find the profit % Rakesh makes a profit of `1,000, when he invests `10,000. Profit of `1,000 out of `10,000 For each 1 rupee, he makes a profit of 1000 10000 Therefore for `100, profit = 1000 # 100 10, 000 Profit = 10% 28 Life Mathematics Ramesh makes a profit of `1000, when he invests `1,00,000. 1000 100000 Profit = 1000 # 100 = 1% 100000 Profit of 1000 out of 1,00,000 = So, from the above we can say that Rakesh is benefited more than Ramesh. So, Profit Percentage = Profit # 100 C.P Loss % is also calculated in the same way. Loss Percentage = Loss # 100 C.P. Profit Percentage or Loss Percentage is always calculated on the cost price of the article. Example 2.18 A dealer bought a television set for `10,000 and sold it for `12,000. Find the profit / loss made by him for 1 television set. If he had sold 5 television sets, find the total profit/loss Solution: Selling Price of the television set = `12,000 Cost Price of the television set = `10,000 S.P. > C.P, there is a profit Profit = S.P. – C. P. = 12000 – 10000 Profit = `2,000 Profit on 1 television set = `2,000 Profit on 5 television sets = 2000 × 5 Profit on 5 television sets = `10,000 Example 2.19 Sanjay bought a bicycle for `5,000. He sold it for `600 less after two years. Find the selling price and the loss percent. Solution: Cost Price of the bicycle = `5000 29 Chapter 2 Loss = `600 Selling Price = Cost Price – Loss = 5000 – 600 Selling Price of the bicycle Loss = `4400 = Loss # 100 C.P. = 600 # 100  5000 = 12% Loss = 12% Example 2.20 A man bought an old bicycle for `1,250. He spent `250 on its repairs. He then sold it for `1400. Find his gain or loss % Solution: Cost Price of the bicycle = `1,250 Repair Charges = `250 Total Cost Price Selling Price C.P. Loss = 1250 + 250 = `1,500 = `1,400 > S.P., there is a Loss = Cost Price – Selling Price = 1500 – 1400 = 100 Loss = `100 Loss = Loss # 100 C.P. = 100 # 100 1500 = 20 3 = 6 2 % (or) 6.67% 3 Loss = 6.67% 30 Life Mathematics Example 2.21 A fruit seller bought 8 boxes of grapes at `150 each. One box was damaged. He sold the remaining boxes at `190 each. Find the profit / loss percent. Solution: Cost Price of 1 box of grapes = `150 Cost Price of 8 boxes of grapes = 150 # 8 = `1200 Number of boxes damaged = 1 Number of boxes sold = 8 – 1 = 7 Selling Price of 1 box of grapes = `190 Selling Price of 7 boxes of grapes = 190 × 7 = `1330 S.P. > C.P, there is a Profit. Profit = Selling Price – Cost Price = 1330 – 1200 = 130 Profit = `130 Percentage of the profit = Profit # 100 C.P 130 = # 100 1200 = 10.83 Profit = 10.83 % Example 2.22 Ram, the shopkeeper bought a pen for `50 and then sold it at a loss of `5. Find his selling price. Solution: Cost price of the pen = `50 Loss = `5 31 Chapter 2 S.P. = C.P. – Loss = 50 – 5 = 45 Selling price of the pen = `45. Example 2.23 Sara baked cakes for the school festival. The cost of one cake was `55. She sold 25 cakes and made a profit of `11 on each cake. Find the selling price of the cakes and the profit percent. Solution: Cost price of 1 cake = `55 Number of cakes sold = 25 Cost price of 25 cakes = 55 × 25 = `1375 Profit on 1 cake = `11 Profit on 25 cakes = 11 × 25 = `275 S.P. = C.P. + Profit = 1375 + 275 = 1,650 = `1,650 Percentage of the profit = Profit # 100 C. P = 275 # 100 1375 = 20 Profit = 20 % Exercise 2.4 1. Choose the correct answer: i) If the cost price of a bag is `575 and the selling price is `625, then there is a profit of ` (A) 50 (B) 575 (C) 625 (D) none of these ii) If the cost price of the box is `155 and the selling price is `140, then there is a loss of ` (A) 155 (B) 140 (C) 15 32 (D) none of these Life Mathematics iii) If the selling price of a bag is `235 and the cost price is `200, then there is a (A) profit of `235 (B) loss of `3 (D) loss of `200 (C) profit of `35 iv) Gain or loss percent is always calculated on (A) cost price (B) selling price (C) gain (D) loss v) If a man makes a profit of `25 on a purchase of `250, then profit% is (A) 25 (B) 10 (C) 250 (D) 225 2. Complete the table by filling in the appropriate column: 3. C.P. S.P. Profit Loss ` ` ` ` 144 168 59 38 600 635.45 26599 23237 107.50 100 Find the selling price when cost price and profit / loss are given. i) Cost Price = `450 Profit = `80 ii) Cost Price = `760 Loss = `140 iii) Cost Price = `980 Profit = `47.50 iv) Cost Price = `430 Loss = `93.25 v) Cost Price = `999.75 Loss = `56.25 4. Vinoth purchased a house for `27, 50,000. He spent `2,50,000 on repairs and painting. If he sells the house for `33,00,000 what is his profit or loss % ? 5. A shop keeper bought 10 bananas for `100. 2 bananas were rotten. He sold the remaining bananas at the rate of `11 per banana. Find his gain or loss % 6. A shop keeper purchased 100 ball pens for `250. He sold each pen for `4. Find the profit percent. 7. A vegetable vendor bought 40 kg of onions for `360. He sold 36 kg at `11 per kg. The rest were sold at `4.50 per kg as they were not very good. Find his profit / loss percent. Choose one product and find out the different stages it crosses from the time it is produced in the factory to the time it reaches the customer. 33 Chapter 2 Think! Do you think direct selling by the manufacturer himself is more beneficial for the costumers? Discuss. Do it yourself 1. A trader mixes two kinds of oil, one costing `100 per Kg. and the other costing `80 per Kg. in the ratio 3: 2 and sells the mixture at `101.20 per Kg. Find his profit or loss percent. 2. Sathish sold a camera to Rajesh at a profit of 10 %. Rajesh sold it to John at a loss of 12 %. If John paid `4,840, at what price did Sathish buy the camera? 3. The profit earned by a book seller by selling a book at a profit of 5% is `15 more than when he sells it at a loss of 5%. Find the Cost Price of the book. 2.5 Simple Interest Deposit `10,000 now. Get `20,000 at the end of 7 years. Deposit `10,000 now. Get `20,000 at the end of 6 years. Is it possible? What is the reason for these differences? Lokesh received a prize amount of `5,000 which he deposited in a bank in June 2008. After one year he got back `5,400. Why does he get more money? How much more does he get? If `5,000 is left with him in his purse, will he gain `400? Lokesh deposited `5,000 for 1 year and received `5,400 at the end of the first year. When we borrow (or lend) money we pay (or receive) some additional amount in addition to the original amount. This additional amount that we receive is termed as Interest (I). 34 Life Mathematics As we have seen in the above case, money can be borrowed deposited in banks to get Interest. In the above case, Lokesh received an interest of `400. The amount borrowed / lent is called the Principal (P). In this case, the amount deposited - `5,000 is termed as Principal (P). The Principal added to the Interest is called the Amount (A). In the above case, Amount = Principal +Interest = `5000 + `400 = `5,400. Will this Interest remain the same always? Definitely not. Now, look at the following cases (i) If the Principal deposited is increased from `5,000 to `10,000, then will the interest increase? (ii) Similarly, if `5,000 is deposited for more number of years, then will the interest increase? Yes in both the above said cases, interest will definitely increase. From the above, we can say that interest depends on principal and duration of time. But it also depends on one more factor called the rate of interest. Rate of interest is the amount calculated annually for `100 (i.e.) if rate of interest is 10% per annum, then interest is `10 for `100 for 1 year. So, Interest depends on: Amount deposited or borrowed – Principal (P) Period of time - mostly expressed in years (n) Rate of Interest (r) This Interest is termed as Simple Interest because it is always calculated on the initial amount (ie) Principal. Calculation of Interest If ‘r’ is the rate of interest, principal is `100, then Interest = 100 # 1 # r 100 for 2 years = 100 # 2 # r 100 for 3 years = 100 # 3 # r 100 for n years = 100 # n # r 100 for 1 year 35 Chapter 2 So, Pnr 100 P+I I = A = A = A = Interest = Amount – Principal I = A–P P + Pnr 100 P`1 + nr j 100 The other formulae derived from I = r = n = P = Pnr are 100 100I Pn 100I Pr 100I rn Note: ‘n’ is always calculated in years. When ‘n’ is given in months \ days, convert it into years. Fill in the blanks Principal ` Interest ` 5,000 500 12,500 6,000 750 8,450 12,000 Amount ` 17,500 25,000 15,600 Example 2.24 Kamal invested `3,000 for 1 year at 7 % per annum. Find the simple interest and the amount received by him at the end of one year. Solution: Principal (P) Number of years (n) Rate of interest (r) = `3,000 = 1 = 7% 36 Life Mathematics Interest (I) = = Pnr 100 3000 # 1 # 7 100 I = `210 A = P+I = 3000 + 210 = `3,210 A Example 2.25 Radhika invested `5,000 for 2 years at 11 % per annum. Find the simple interest and the amount received by him at the end of 2 years. Solution: Principal (P) = `5,000 Number of years (n) = 2 years Rate of interest (r) = 11 % I = = Pnr 100 5000 # 2 # 11 100 = 1100 = `1,100 I Amount (A) = P+I = 5000 + 1100 = `6,100 A Example 2.26 Find the simple interest and the amount due on `7,500 at 8 % per annum for 1 year 6 months. Know this Solution: 12 months = 1 years P = `7,500 6 months = 6 year n = 1 yr 6 months 12 12 = 1 1 = 3 yrs 2 = 1 year 2 3 months = 3 year 12 = 1 year 4 = 1 6 yrs 2 r = 8% 37 Chapter 2 I = Pnr 100 = = 7500 # 3 # 8 2 100 7500 # 3 # 8 2 # 100 = 900 I = `900 A = P+I = 7500 + 900 = `8,400 Interest = `900, Amount = `8,400 Aliter: P = `7,500 n = 3 years 2 r = 8% A = P`1 + nr j 100 3 #8 e o 2 = 7500 1 + 100 = 7500 c1 + 3 # 8 m 2 # 100 = 7500` 28 j 25 = 300 # 28 = 8400 A = `8400 I = A–P = 8400 – 7500 = 900 I = `900 Interest = `900 Amount = `8,400 38 Life Mathematics Example 2.27 Find the simple interest and the amount due on `6,750 for 219 days at 10 % per annum. Solution: P = `6,750 Know this n = 219 days = 219 year = 3 year 5 365 365 days = 1 year 219 days = 219 year r = 10 % 365 = 3 year 5 73 days = 73 year 365 1 = year 5 I = Pnr 100 I = 6750 # 3 # 10 5 # 100 = 405 I = `405 A = P+I = 6750 + 405 = 7,155 A = `7,155 Interest = `405, Amount = `7,155 Example 2.28 Rahul borrowed `4,000 on 7th of June 2006 and returned it on 19th August2006. Find the amount he paid, if the interest is calculated at 5 % per annum. Solution: Number of days, P = `4,000 r = 5% June = July = August = Total number of days = n Know this days hath 24 (30 – 6) Thirty September, April, June 31 and November. All the rest have thirty - one 18 except February. 73 = 73 days 39 Chapter 2 = = A = 73 year 365 1 year 5 P`1 + nr j 100 = 4000 c1 + 1 # 5 m 5 # 100 = 4000`1 + 1 j 100 = 4000` 101 j 100 = 4,040 Amount = `4,040 Example 2.29 Find the rate percent per annum when a principal of `7,000 earns a S.I. of `1,680 in 16 months. Solution: P = `7,000 n = 16 months = 16 yr = 4 yr 12 3 I = `1,680 r = ? r = = = 100I Pn 100 # 1680 7000 # 4 3 100 # 1680 # 3 7000 # 4 = 18 r = 18 % Example 2.30 Vijay invested `10,000 at the rate of 5 % simple interest per annum. He received `11,000 after some years. Find the number of years. Solution: A = `11,000 P = `10,000 40 Life Mathematics r = 5% n = ? I = A–P = 11,000 – 10,000 = 1,000 I = `1000 n = = 100 I Pr 100 # 1000 10000 # 5 n = 2 years. A = P`1 + nr j 100 Aliter: = 10000 `1 + n # 5 j 100 11000 = 1+ n 11000 10000 11 10 11 # 20 10 = 20 20 + n 20 = 20 + n 22 = 20 + n 22 – 20 = n n = 2 years Example 2.31 A sum of money triples itself at 8 % per annum over a certain time. Find the number of years. Solution: Let Principal be `P. Amount = triple the principal = `3 P r = 8% n = ? 41 Chapter 2 I = A–P = 3P – P = 2P I = `2 P n = 100I Pr = 100 # 2P P#8 n = 25 years Number of years = 25 Aliter: Let Principal be `100 Amount = 3 # 100 = `300 I = A–P = 300 - 100 I = `200. n = 100I = 100 # 200 Pr 100 # 8 n = 200 = 25 8 Number of years = 25. Example 2.32 A certain sum of money amounts to `10,080 in 5 years at 8 % . Find the principal. Solution: A = `10,080 n = 5 years r = 8% P = ? A = P`1 + nr j 100 10080 = P`1 + 5 # 8 j 100 42 Life Mathematics 10080 = P` 7 j 5 10080 # 5 = P 7 7,200 = P Principal = `7,200 Example 2.33 A certain sum of money amounts to `8,880 in 6 years and `7,920 in 4 years respectively. Find the principal and rate percent. Solution: Amount at the end of 6 years = Principal + interest for 6 years = P + I6 = 8880 Amount at the end of 4 years = Principal + Interest for 4 years = P + I4 = 7920 I2 = 8880 – 7920 = 960 Interest at the end of 2 years = `960 Interest at the end of 1st year = 960 2 = 480 Interest at the end of 4 years = 480 # 4 = 1,920 P + I 4 = 7920 P + 1920 = 7920 P = 7920 - 1920 P = 6,000 Principal = `6,000 r = 100I = pn 100 # 1920 6000 # 4 r = 8% 43 Chapter 2 Exercise 2.5 1. Choose the correct answer: i) Simple Interest on `1000 at 10 % per annum for 2 years is (A) `1000 (B) `200 ii) If Amount = `11,500, (A) `500 (C) `100 (D) `2000 Principal = `11,000, Interest is (B) `22,500 (C) `11,000 (D) `11,000 (B) 1 yr 4 (C) 3 yr 4 (D) 1 yr (B) 3 yr 5 I = `1000, A is (C) 4 yr 5 (D) 2 yr 5 (C) `14000 (D) `1000 iii) 6 months = (A) 1 yr 2 iv) 292 days = (A) 1 yr 5 v) If P = `14000 (A) `15000 (B) `13000 2. Find the S.I. and the amount on `5,000 at 10 % per annum for 5 years. 3. Find the S.I and the amount on `1,200 at 12 1 2 % per annum for 3 years. 4. Lokesh invested `10,000 in a bank that pays an interest of 10 % per annum. He withdraws the amount after 2 years and 3 months. Find the interest, he receives. 5. Find the amount when `2,500 is invested for 146 days at 13 % per annum. 6. Find the S.I and amount on `12,000 from May 21 1999 to August 2 1999 at st nd 9 % per annum. 7. Sathya deposited `6,000 in a bank and received `7500 at the end of 5 years. Find the rate of interest. 8. Find the principal that earns `250 as S.I. in 2 1 years at 10 % per annum. 2 9. In how many years will a sum of `5,000 amount to `5,800 at the rate of 8 % per annum. 10. A sum of money doubles itself in 10 years. Find the rate of interest. 44 Life Mathematics 11. A sum of money doubles itself at 12 1 % per annum over a certain period of 2 time. Find the number of years. 12. A certain sum of money amounts to `6,372 in 3 years at 6 % Find the principal. 13. A certain sum of money amounts to `6,500 in 3 years and `5,750 in 1 1 years 2 respectively . Find the principal and the rate percent. 14. Find S.I. and amount on ` 3,600 at 15% p.a. for 3 years and 9 months. 15. Find the principal that earns ` 2,080 as S.I. in 3 1 years at 16% p.a. 4 Think! 1) Find the rate percent at which, a sum of money becomes 9 times 4 in 2 years. 2) If Ram needs `6,00,000 after 10 years, how much should he invest now in a bank if the bank pays 20 % interest p.a. 1. A fraction whose denominator is 100 or a ratio whose second term is 100 is termed as a percent. 2. Percent means per hundred, denoted by % 3. To convert a fraction or a decimal to a percent, multiply by 100. 4. The price at which an article is bought is called the cost price of an article. 5. The price at which an article is sold is called the selling price of an article. 6. If the selling price of an article is more than the cost price, there is a profit. 45 Chapter 2 7. If the cost price of an article is more than the selling price, there is a loss. 8. Total cost price = Cost Price + Repair Charges / Transportation charges. 9. Profit or loss is always calculated for the same number of articles or same units. 10. Profit = Selling Price – Cost Price 11. Loss = Cost Price – Selling Price 12. Profit% = Profit # 100 C.P. 13. Loss% = Loss # 100 C.P. 14. Selling Price = Cost Price + Profit 15. Selling Price = Cost Price - Loss 16. Simple interest is I = Pnr 100 17. A = = = P+I P + Pnr 100 P`1 + nr j 100 18. I = A–P 19. P = 20. r = 21. n = 100I nr 100I Pn 100I Pr 46 Measurements MEASUREMENTS 3.1 Trapezium A trapezium is a quadrilateral with one pair of opposite sides are parallel. The distance between the parallel sides is the height of the trapezium. Here the sides AD and BC Fig. 3.1 are not parallel, but AB || DC. If the non - parallel sides of a trapezium are equal ( AD = BC ), then it is known as an isosceles trapezium. +A = +B ; +C = +D Here AC = BD + A + + D = 180° ; + B + + C = 180° Fig. 3.2 Area of a trapezium ABCD is a trapezium with parallel sides AB and DC measuring ‘a’ and ‘b’. Let the distance between the two parallel sides be ‘h’. The diagonal BD divides the trapezium into two triangles ABD and BCD. Area of the trapezium = area of D ABD + area of D BCD = 1 # AB # h + 1 # DC # h = = 2 2 1 # h 6 AB + DC @ 2 1 # h 6 a + b @ sq. units 2 Fig. 3.3 ` Area of a trapezium = 1 # height # (sum of the parallel sides) sq. units 2 Example 3.1 Find the area of the trapezium whose height is 10 cm and the parallel sides are 12 cm and 8 cm of length. 47 Chapter 3 Solution Given: h = 10 cm, a = 12 cm, b = 8 cm Area of a trapezium = 1 # h^a + bh 2 1 = # 10 # ^12 + 8h = 5 # ^20h 2 \ Area of the trapezium = 100 sq. cm2 Example 3.2 The length of the two parallel sides of a trapezium are 15 cm and 10 cm. If its area is 100 sq. cm. Find the distance between the parallel sides. Solution Given: a = 15 cm, b = 10 cm, Area = 100 sq. cm. Area of the trapezium = 100 1 h^a + bh = 100 2 1 # h # ^15 + 10h = 100 2 h # 25 = 200 h = 200 = 8 25 ` the distance between the parallel sides = 8 cm. Example 3.3 The area of a trapezium is 102 sq. cm and its height is 12 cm. If one of its parallel sides is 8 cm. Find the length of the other side. Solution Given: Area = 102 cm2, h = 12 cm, a = 8 cm. Area of a trapezium = 1 h^ a + bh = 1 # 122# ^8 + bh = 2 6 (8 + b) = 102 102 102 102 8 + b = 17    & b = 17 – 8 = 9 ` length of the other side = 9 cm By paper folding method: In a chart paper draw a trapezium ABCD of any measure. Cut and take the trapezium separately. Fold the trapezium in such a way that DC lies on AB and crease it on the middle to get EF. 48 Measurements E Fig. 3.4 EF divides the trapezium in to two parts as shown in the Fig. 4.40 (ii) From D draw DG = EF. Cut the three parts separately. Arrange three parts as shown in the Fig. 3.4 (iii) The figure obtained is a rectangle whose length is AB + CD = a + b and breadth is 1 ( height of trapezium ) = 1 h 2 2 ` Area of trapezium = area of rectangle as shown in Fig. 4.40 (iii) = length # breadth = ^a + bh` 1 hj 2 1 = h^a + bh sq. units 2 Exercise 3.1 1. Choose the correct answer. i) The area of trapezium is ____________ sq. units (B) 1 h (a + b) (C) h^a - bh (D) 1 h (a – b) (A) h^a + bh 2 2 ii) In an isosceles trapezium (A) non parallel sides are equal (B) parallel sides are equal (C) height = base (D) parallel sides = non parallel sides iii) The sum of parallel sides of a trapezium is 18 cm and height is 15 cm. Then its area is (B) 115 cm 2 (C) 125 cm 2 (D) 135 cm 2 (A) 105 cm 2 iv) The height of a trapezium whose sum of parallel sides is 20 cm and the area 80 cm2 is (A) 2 cm (B) 4 cm (C) 6 cm (D) 8 cm 49 Chapter 3 2. Find the area of a trapezium whose altitudes and parallel sides are given below: i) altitude = 10 cm, parallel sides = 4 cm and 6 cm ii) altitude = 11 cm, parallel sides = 7.5 cm and 4.5 cm iii) altitude = 14 cm, parallel sides = 8 cm and 3.5 cm 3. The area of a trapezium is 88 cm2 and its height is 8 cm. If one of its parallel side is 10 cm. Find the length of the other side. 4. A garden is in the form of a trapezium. The parallel sides are 40 m and 30 m. The perpendicular distance between the parallel side is 25 m. Find the area of the garden. 5. Area of a trapezium is 960 cm2. The parallel sides are 40 cm and 60 cm. Find the distance between the parallel sides. 3.2 Circle In our daily life, we come across a number of objects like wheels, coins, rings, bangles,giant wheel, compact disc (C.D.) What is the shape of the above said objects? ‘round’, ‘round’, ‘round’ Yes, it is round. In Mathematics it is called a circle. Now, let us try to draw a circle. Take a thread of any length and fix one end tightly at a point O as shown in the figure. Tie a pencil (or a chalk) to the other end and stretch the thread completely to a point A. Holding the thread stretched tightly, move the pencil. Stop it when the pencil again reaches the point A. Now see Fig. 3.5 the path traced by the pencil. Is the path traced by the pencil a circle or a straight line? ‘Circle’ Yes, the path traced by the point, which moves at a constant distance from a fixed point on a given plane surface is called a circle. Parts of a Circle The fixed point is called the centre of the circle. The constant distance between the fixed point and the moving point is called the radius of the circle. i.e. The radius is a line segment with one end point at the centre and the other end on the circle. It is denoted by ‘ r ’. A line segment joining any two points on the circle is called a chord. 50 Fig. 3.6 Measurements Diameter is a chord passing through the centre of the circle. It is denoted by ‘ d ’. The diameter is the longest chord. It is twice the radius.(i.e. d = 2r ) The diameter divides the circle into two equal parts. Each equal part is a semicircle. Think it: How many diameters can a circle have ? The plural of radius is “radii”. All the radii of a circle are equal. Circumference of a circle: Can you find the distance covered by an athlete if he takes two rounds on a circular track. Since it is a circular track, we cannot use the ruler to find out the distance. So, what can we do ? Take a one rupee coin.Place it on a paper and draw its outline. Remove the coin. Mark a Fig. 3.7 point A on the outline as shown in the Fig. 3.7 Take a thread and fix one end at A. Now place the thread in such a way that the thread coincides exactly with the outline. Cut the other end of the thread when it reaches the point A. Length of the thread is nothing but the circumference of the coin. Fig. 3.8 So, the distance around a circle is called the circumference of the circle, which is denoted by ‘C’. i.e., The perimeter of a circle is known as its circumference. Take a bottle cap or a bangle or any other circular objects and find the circumference. If possible find the relation between the circumference and the diameter of the circular objects. Relation between diameter and circumference of the circle Draw four circles with radii 3.5 cm, 7 cm, 5 cm, 10.5 cm in your note book. Measure their circumferences using a thread and the diameter using a ruler as shown in the Fig. 3.9 given below. 51 Chapter 3 Fig. 3.9 Fill in the missing values in table 3.1 and find the ratio of the circumference to the diameter. Diameter Circumference (C) (d ) Ratio ` C j d Circle Radius 1 3.5 cm 7 cm 22 cm 22 = 3.14 7 2 7 cm 14 cm 44 cm 44 = 22 = 3.14 14 7 3 5 cm 10 cm ---- ---- 4 10.5 cm 21 cm ---- ---- Table 3.1 What do you infer from the above table?. Is this ratio ` C j approximately d the same? Yes ! C d = 3.14 & C = ^3.14h d So, can you say that the circumference of a circle is always more than 3 times its diameter ? Yes ! In all the cases, the ratio C is a constant and is denoted by the Greek letter r d (read as ‘pi’ ). Its approximate value is 22 or 3.14. 7 so, C = r & C = rd units d where d is the diameter of a circle. We know that the diameter of a circle is twice the radius r . i.e., d = 2r . 52 Measurements from the above formula, C = rd = r^2r h & C = 2rr units. The value of r is calculated by many mathematicians. Babylonians : r = 3 Greeks : r = 22 or 3.14 7 62838 Aryabhata : r = (or) 3.1416 2000 Now, we use r = 22 or 3.14 7 Archemides : 3 1 < r < 3 10 7 71 Example 3.4 Find out the circumference of a circle whose diameter is 21 cm. Solution Circumference of a circle = rd = 22 # 21 7 Here r = 22 7 = 66 cm. Example 3.5 Find out the circumference of a circle whose radius is 3.5 m. Solution Circumference of a circle = 2rr = 2 # 22 # 3.5 7 = 2 × 22 × 0.5 = 22 m Example 3.6 A wire of length 88 cm is bent as a circle. What is the radius of the circle. Solution Length of the wire = 88 cm Circumference of the circle = Length of the wire 2rr = 88 2 # 22 # r = 88 7 r = 88 # 7 = 14 cm 2 # 22 ` radius of a circle is 14 cm. Example 3.7 The diameter of a bicycle wheel is 63 cm. How much distance will it cover in 20 revolutions? 53 Chapter 3 Solution When a wheel makes one complete revolutions, Distance covered in one rotation = Circumference of wheel ` circumference of the wheel = rd units = 22 # 63 cm 7 = 198 cm For one revolution, the distance covered = 198 cm ` for 20 revolutions, the distance covered = 20 × 198 cm = 3960 cm = 39 m 60 cm [100 cm = 1 m] Example 3.8 A scooter wheel makes 50 revolutions to cover a distance of 8800 cm. Find the radius of the wheel. Solution Distance travelled = Number of revolutions # Circumference Distance travelled Number of revolutions 2rr = 8800 50 i.e., 2rr = 176 Circumference = 2 # 22 # r = 176 7 r = 176 # 7 2 # 22 r = 28 cm ` radius of the wheel = 28 cm. Example 3.9 The radius of a cart wheel is 70 cm. How many revolution does it make in travelling a distance of 132 m. Solution Given: r = 70 cm, Distance travelled = 132 m. ` Circumference of a cart wheel = 2rr = 2 # 22 # 70 7 = 440 cm 54 Measurements Distance travelled = Number of revolutions # Circumference ` Number of revolutions = Distance travelled Circumference = 132 m 440 cm cm (1 m = 100 cm, 132 m = 13200 cm) 13200 = 440 cm = 30 ` Number of revolutions = 30. Example 3.10 The circumference of a circular field is 44 m. A cow is tethered to a peg at the centre of the field. If the cow can graze the entire field, find the length of the rope used to tie the cow. Solution Length of the rope = Radius of the circle Circumference = 44 m (given) i.e., 2rr = 44 2 # 22 # r = 44 7 ` r = 44 # 7 = 7 m 2 # 22 Fig. 3.10 ` The length of the rope used to tie the cow is 7 m. Example 3.11 The radius of a circular flower garden is 56 m. What is the cost of fencing it at `10 a metre ? Solution Length to be fenced = Circumference of the circular flower garden Circumference of the flower garden = 2rr = 2 # 22 # 56 = 352 m 7 ` Length of the fence = 352 m Cost of fencing per metre = `10 ` cost of fencing 352 m = `10 # 352 = `3520 ` Total cost of fencing is `3520. 55 Chapter 3 Example 3.12 The cost of fencing a circular park at the rate of `5 per metre is `1100. What is the radius of the park. Solution Cost of fencing = Circumference # Rate Cost of fencing Rate 1100 i.e., 2rr = 5 2rr = 220 ` Circumference = ` 2 # 22 # r = 220 7 r = 220 # 7 2 # 22 = 35 m ` Radius of the park = 35 m. Activity - Circular Geoboard Take a square Board and draw a circle. Fix nails on the circumference of the circle. ( See fig ) Using rubber band, form various diameters, chords, radii and compare. Exercise 3.2 1. Choose the correct answer: i) The line segment that joins the centre of a circle to any point on the circle is called (A) Diameter (B) Radius (C) Chord (D) None ii) A line segment joining any two points on the circle is called (A) Diameter (B) Radius (C) Chord iii) A chord passing through the centre is called (D) None (A) Diameter (B) Radius (C) Chord iv) The diameter of a circle is 1 m then its radius is (D) None (A) 100 cm (B) 50 cm (C) 20 cm v) The circumference of a circle whose radius is 14 cm is (D) 10 cm (A) 22 cm (B) 44 cm (C) 66 cm 56 (D) 88 cm Measurements 2. Fill up the unknown in the following table: radius ( r ) diameter ( d ) circumference (c) (i) 35 cm ----- ----- (ii) ----- 56 cm ----- (iii) ----- ----- 30.8 cm 3. Find the circumference of a circle whose diametre is given below: (i) 35 cm (ii) 84 cm (iii) 119 cm (iv) 147 cm 4. Find the circumference of a circle whose radius is given below: (i) 12.6 cm (ii) 63 cm (iii) 1.4 m (iv) 4.2 m 5. Find the radius of a circle whose circumference is given below: (i) 110 cm (ii) 132 cm (iii) 4.4 m (iv) 11 m 6. The diameter of a cart wheel is 2.1 m. Find the distance travelled when it completes 100 revolutions. 7. The diameter of a circular park is 98 m. Find the cost of fencing it at `4 per metre. 8. A wheel makes 20 revolutions to cover a distance of 66 m. Find the diameter of the wheel. 9. The radius of a cycle wheel is 35 cm. How many revolutions does it make to cover a distance of 81.40 m? 10. The radius of a circular park is 63 m. Find the cost of fencing it at `12 per metre. Area of a circle Consider the following A farmer levels a circular field of radius 70 m. What will be the cost of levelling? What will be the cost of polishing a circular table-top of radius 1.5 m ? How will you find the cost ? To find the cost what do you need to find actually? The region enclosed by the circumference of a Area or perimeter ? circle is a circular region. Area, area, area Yes. In such cases we need to find the area of the circular region. So far, you have learnt to find the area of triangles and quadrilaterals that made up of straight lines. But, a cirlce is a plane figure made up of curved line different from other plane figures. 57 Chapter 3 So, we have to find a new approach which will make the circle turn into a figure with straight lines. Take a chart paper and draw a circle. Cut the circle and take it separately. Shade one half of the circular region. Now fold the entire circle into eight parts and cut along the folds (see Fig. 3.11). Fig. 3.11 Arrange the pieces as shown below. Fig. 3.12 What is the figure obtained? These eight pieces roughly form a parallelogram. Similarly, if we divide the circle into 64 equal parts and arrange these, it gives nearly a rectangle. (see Fig. 3.13) Fig. 3.13 What is the breadth of this rectangle? The breadth of the rectangle is the radius of the circle. i.e., breadth b = r ..... (1) What is the length of this rectangle ? As the whole circle is divided into 64 equal parts and on each side we have 32 equal parts. Therefore, the length of the rectangle is the length of 32 equal parts, which is half of the circumference of a circle. 58 Measurements ` length l = 1 [circumference of the circle] 2 1 = 62rr @ = rr 2 ` l = rr ..... (2) Area of the circle = Area of the rectangle (from the Fig. 4.50) = l # b = (rr) # r (from (1) and (2)) = rr2 sq. units. ` Area of the circle = rr2 sq. units. Example 3.13 Find the area of a circle whose diameter is 14 cm Solution Diameter d = 14 cm radius r = d = 14 = 7 cm So, 2 Area of circle = rr 2 2 = 22 # 7 # 7 7 = 154 sq. cm ` Area of circle = 154 sq. cm Example 3.14 A goat is tethered by a rope 3.5 m long. Find the maximum area that the goat can graze. Solution Radius of the circle = Length of the rope ` radius r = 3.5 m = 7 m 2 maximum area grazed by the goat = rr2 sq. units. = 22 # 7 # 7 7 2 2 77 = 38.5 sq. m = 2 ` maximum area grazed by the goat is 38.5 sq. m. 59 Fig. 3.14 Chapter 3 Example 3.15 The circumference of a circular park is 176 m. Find the area of the park. Solution Circumference = 176 m (given) 2rr = 176 2 # 22 # r = 176 7 r = 176 # 7 44 ` r = 28 m Area of the park = rr2 = 22 # 28 # 28 7 = 22 # 4 # 28 = 2464 sq. m. Example 3.16 A silver wire when bent in the form of a square encloses an area of 121 sq. cm. If the same wire is bent in the form of a circle. Find the area of the circle. Solution Let a be the side of the square Area of the square = 121 sq. cm. (given) (11 # 11 = 121) a2 = 121 & a = 11 cm Perimeter of the square = 4a units = 4 # 11 cm = 44 cm Length of the wire = Perimeter of the square = 44 cm The wire is bent in the form of a circle The circumference of the circle = Length of the wire ` circumference of a circle = 44 cm 2rr = 44 22 `2# # r = 44 7 r = 44 # 7 44 r = 7 cm ` Area of the circle = rr2 = 22 × 7 cm × 7 cm 7 Area of the circle = 154 cm2. 60 Measurements Example 3.17 When a man runs around circular plot of land 10 times, the distance covered by him is 352 m. Find the area of the plot. Solution Distance covered in 10 times = 352 m Distance covered in one time = 352 m = 35.2 m 10 The circumference of the circular plot = Distance covered in one time ` circumference = 35.2 m 2rr = 35.2 2 # 22 # r = 35.2 7 r = 35.2 # 7 44 = 0.8 # 7 = 5.6 m Area of the circular plot = rr2 = 22 # 5.6 # 5.6 7 = 22 # 0.8 # 5.6 = 98.56 m2 ` Area of circular plot = 98.56 m2 Example 3.18 A wire in the shape of a rectangle of length 37 cm and width 29 cm is reshaped in the form of a circle. Find the radius and area of the circle. Solution Length of the wire = perimeter of the rectangle = 2 [ length + breadth ] = 2 [37 cm + 29 cm] = 2 # 66 cm = 132 cm. Since wire is bent in the form of a circle, The circumference of the circle = The length of the wire ` Circumference of a circle = 132 2rr = 132 22 2# # r = 132 7 r = 132 # 7 = 21 44 ` radius of the circle = 21 cm Area of the circle = rr2 = 22 # 21 # 21 = 22 # 3 # 21 7 ` Area of the circle = 1386 sq. cm. 61 Chapter 3 Exercise 3.3 1. Find the area of the circles whose diameters are given below: (i) 7 cm (ii) 10.5 cm (iii) 4.9 m (iv) 6.3 m 2. Find the area of the circles whose radii are given below: (take r = 22 ) 7 (take r = 22 ) 7 3. The diameter of a circular plot of ground is 28 m. Find the cost of levelling the ground at the rate of `3 per sq. m. (i) 1.2 cm (ii) 14 cm (iii) 4.2 m (iv) 5.6 m 4. A goat is tied to a peg on a grass land with a rope 7 m long. Find the maximum area of the land it can graze. 5. A circle and a square each have a perimeter of 88 cm. Which has a larger area? 6. A wheel goes a distance of 2200 m in 100 revolutions. Find the area of the wheel. 7. A wire is in the form of a circle of radius 28 cm. Find the area that will enclose, if it is bent in the form of a square having its perimeter equal to the circumference of the cirlce. 8. The area of circular plot is 3850 m2. Find the radius of the plot. Find the cost of fencing the plot at `10 per metre. 9. The radius of a circular ground is 70 m. Find the distance covered by a child walking along the boundary of the ground. 10. The area of a circular field is 154 m2. Find the time taken by an athlate to complete 2 rounds if she is jogging at the rate of 5 km/hr. 11. How many circles of radius 7 cm can be cut from a paper of length 50 cm and width 32 cm. 3.3 Area of the path way In our day - to - day life we go for a walk in a park, or in a play ground or even around a swimming pool. Can you represent the path way of a park diagrammatically ? Have you ever wondered if it is possible to find the area of such paths ? Can the path around the rectangular pool be related to the mount around the photo in a photo frame ? Can you think of some more examples? In this section we will learn to find • Area of rectangular pathway • Area of circular pathway 62 Measurements Area of rectangular pathway (a) Area of uniform pathway outside the rectangle Consider a rectangular building. A uniform flower garden is to be laid outside the building. How do we find the area of the flower garden? The uniform flower garden including the building is also a rectangle in shape. Let us call it as outer rectangle. We call the building as inner rectangle. Let l and b be the length and breadth of the building. Fig. 3.15 ` Area of the inner rectangle = l b sq. units. Let w be the width of the flower garden. Fig. 3.16 What is the length and breadth of the outer rectangle ? The length of the outer rectangle (L) = w + l + w = ^l + 2wh units The breadth of the outer rectangle (B) = w + b + w = ^b + 2wh units ` area of the outer rectangle = L # B = ^l + 2wh^b + 2wh sq. units Now, what is the area of the flower garden ? Actually, the area of the flower garden is the pathway bounded between two rectangles. ` Area of the flower garden = (Area of building and flower garden) –   (Area of building) Generally, Area of the pathway = (Area of outer rectangle) – (Area of inner rectangle) i.e. Area of the pathway = (l + 2w) (b + 2w) – lb. Example 3.19 The area of outer rectangle is 360 m2. The area of inner rectangle is 280 m2. The two rectangles have uniform pathway between them. What is the area of the pathway? Solution Area of the pathway = (Area of outer rectangle) – (Area of inner rectangle) 63 Chapter 3 = (360 – 280) m2 = 80 m2 ` Area of the pathway = 80 m2 Example 3.20 The length of a building is 20 m and its breadth is 10 m. A path of width 1 m is made all around the building outside. Find the area of the path. Solution Inner rectangle (given) l = 20 m b = 10 m Area = l × b Area = 20 m × 10 m = 200 m2 Outer rectangle width, w =1m L = l + 2w = 20 + 2 = 22 m B = b + 2w = 10 + 2 = 12 m Area = (l + 2w) (b + 2w) Area = 22 m # 12 m = 264 m2 Area of the path = (Area of outer rectangle) – (Area of inner rectangle) = ( 264 – 200 ) m2 = 64 m2 ` Area of the path = 64 m2 Example 3.21 A school auditorium is 45 m long and 27 m wide. This auditorium is surrounded by a varandha of width 3 m on its outside. Find the area of the varandha. Also, find the cost of laying the tiles in the varandha at the rate of `100 per sq. m. Solution Fig. 3.17 Inner (given) rectangle = 45 m l = 27 m b Area = 45m # 27 m = 1215 m2 Outer rectangle Width, w = 3 m L = l + 2w = 45 + 6 = 51 m B = b + 2w = 27 + 6 = 33 m Area = 51m # 33 m 64 = 1683 m2 Measurements (i) Area of the verandha = (Area of outer rectangle) – (Area of inner rectangle) = (1683 – 1215) m2 = 468 m2 ` Area of the verandha = 468 m2 (or) 468 sq. m. (ii) Cost of laying tiles for 1 sq. m = `100 Cost of laying tiles for 468 sq. m = `100 # 468 = `46,800 ` Cost of laying tiles in the verandha = `46,800 (b) Area of uniform pathway inside a rectangle A swimming pool is built in the middle of a rectangular ground leaving an uniform width all around it to maintain the lawn. If the pathway outside the pool is to be grassed, how can you find its cost ? If the area of the pathway and cost of grassing per sq. unit is known, then the cost of grassing the pathway can be found. Fig. 3.18 Here, the rectangular ground is the outer rectangle where l and b are length and breadth. ` Area of the ground (outer rectangle) = l b sq. units If w be the width of the pathway (lawn), what will be the length and breath of the swimming pool ? The length of the swimming pool = l-w-w = l - 2w The breadth of the swimming pool = b-w-w = b - 2w ` Area of the swimming pool (inner rectangle) = ^l - 2wh^b - 2wh Sq. units Area of the lawn = Area of the ground – Area of the swimming pool. Generally, Area of the pathway = (Area of outer rectangle) – (Area of inner rectangle) = lb – (l – 2w) (b – 2w) 65 Chapter 3 Example 3.22 The length and breadth of a room are 8 m and 5 m respectively. A red colour border of uniform width of 0.5 m has been painted all around on its inside. Find the area of the border. Solution Fig. 3.19 Outer (given)rectangle l =8m b =5m Area = 8m # 5 m = 40 m2 Inner rectangle width, w = 0.5 m L = l - 2w = (8 – 1) m = 7 m B = b - 2w = (5 – 1) m = 4 m Area = 7m × 4 m = 28 m2 Area of the path = (Area of outer rectangle) – (Area of inner rectangle) = (40 – 28) m2 = 12 m2 ` Area of the border painted with red colour = 12 m2 Example 3.23 A carpet measures 3 m # 2 m. A strip of 0.25 m wide is cut off from it on all sides. Find the area of the remaining carpet and also find the area of strip cut out. Solution Outer rectangle Inner rectangle carpet before cutting the strip carpet after cutting the strip width, w = 0.25 m l =3m b =2m L = l - 2w = (3 – 0.5) m Area = 3m # 2m = 2.5 m 2 =6m B = b - 2w = (2 – 0.5) m = 1.5 m Area = 2.5m # 1.5m = 3.75 m2 The area of the carpet after cutting the strip = 3.75 m2 66 Measurements Area of the strip cut out = (Area of the carpet) –              (Area of the remaining part) = (6 – 3.75) m2 = 2.25 m2 ` Area of the strip cut out = 2.25 m2 Note: If the length and breadth of the inner rectangle is given, then the length and breadth of the outer rectangle is l + 2w , b + 2w respectively where w is the width of the path way. Suppose the length and breadth of the outer rectangle is given, then the length and breadth of the inner rectangle is l - 2w , b - 2w respectively. Fig. 3.20 Exercise 3.4 1. A play ground 60 m # 40 m is extended on all sides by 3 m. What is the extended area. 2. A school play ground is rectangular in shape with length 80 m and breadth 60 m. A cemented pathway running all around it on its outside of width 2 m is built. Find the cost of cementing if the rate of cementing 1 sq. m is `20. 3. A garden is in the form of a rectangle of dimension 30 m # 20 m. A path of width 1.5 m is laid all around the garden on the outside at the rate of `10 per sq. m. What is the total expense. 4. A picture is painted on a card board 50 cm long and 30 cm wide such that there is a margin of 2.5 cm along each of its sides. Find the total area of the margin. 5. A rectangular hall has 10 m long and 7 m broad. A carpet is spread in the centre leaving a margin of 1 m near the walls. Find the area of the carpet. Also find the area of the un covered floor. 6. The outer length and breadth of a photo frame is 80 cm , 50 cm. If the width of the frame is 3 cm all around the photo. What is the area of the picture that will be visible? Circular pathway Concentric circles Circles drawn in a plane with a common centre and different radii are called concentric circles. Circular pathway Fig. 3.21 A track of uniform width is laid around a circular park for walking purpose. Can you find the area of this track ? 67 Chapter 3 Yes. Area of the track is the area bounded between two concentric circles. In Fig. 4.59, O is the common centre of the two circles.Let the radius of the outer circle be R and inner circle be r. The shaded portion is known as the circular ring or the circular pathway. i.e. a circular pathway is the portion bounded between two concentric circles. Fig. 3.22 width of the pathway, w = R – r units i.e., w = R – r & R = w + r units r = R – w units. The area of the circular path = (area of the outer circle) – (area of the inner circle) = rR2 - rr2 = r^ R2 - r2h sq. units ` The area of the circular path = r^ R2 - r2h sq. units = r^ R + r h^ R - r h sq. units Example 3.24 The adjoining figure shows two concentric circles. The radius of the larger circle is 14 cm and the smaller circle is 7 cm. Find The area of the larger circle. (i) (ii) The area of the smaller circle. (iii) The area of the shaded region between two circles. Solution Fig. 3.23 i) Larger circle ii) Smaller circle R = 14 r = 7 area = rR2 area = rr2 = 22 # 14 # 14 = 22 # 7 # 7 7 7 = 22 # 28 = 22 # 7 = 616 cm2 = 154 cm2 iii) The area of the shaded region = (Area of larger circle) – (Area of smaller circle) = (616 – 154) cm2 = 462 cm2 Example 3.25 From a circular sheet of radius 5 cm, a concentric circle of radius 3 cm is removed. Find the area of the remaining sheet ? (Take r = 3.14 ) 68 Measurements Solution Given: R = 5 cm, r = 3 cm Area of the remaining sheet = r^ R2 - r2h = 3.14 (52 – 32) = 3.14 (25 – 9) = 3.14 × 16 = 50.24 cm2 Aliter: Outer circle R = 5 cm Area = rR2 sq. units = 3.14 # 5 # 5 = 3.14 # 25 = 78.5 cm2 Area of the remaining sheet = = = ` Area of the remaining sheet = Inner circle r = 3 cm Area = rr2 sq. units = 3.14 # 3 # 3 = 3.14 # 9 = 28.26 cm2 (Area of outer circle) – (Area of inner circle) (78.5 – 28.26) cm2 50.24 cm2 50.24 cm2 Example 3.26 A circular flower garden has an area 500 m2. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden (Take r = 3.14) Solution Given, area of the garden = 500 m2 Area covered by a sprinkler = rr2 = 3.14 # 12 # 12 = 3.14 # 144 = 452 .16 m2 Fig. 3.24 Since, the area covered by a sprinkler is less than the area of the circular flower garden, the sprinkler cannot water the entire garden. Example 3.27 A uniform circular path of width 2 m is laid out side a circular park of radius 50 m. Find the cost of levelling the path at the rate of `5 per m2 (Take r = 3.14) 69 Chapter 3 Solution Given: r = 50 m, w = 2 m, R = r + w = 50 + 2 = 52 m Area of the circular path = r^ R + r h^ R - r h = 3.14 # ^52 + 50h^52 - 50h = 3.14 # 102 # 2 = 3.14 # 204 Fig. 3.25 = 640.56 m2 The cost of levelling the path of area 1 sq m = `5 The cost of levelling the path of 640.56 m2 = `5 # 640.56 = `3202.80 ` the cost of levelling the path = `3202.80 Exercise 3.5 1. A circus tent has a base radius of 50 m. The ring at the centre for the performance by an artists is 20 m in radius. Find the area left for the audience. (Take r = 3.14) 2. A circular field of radius 30 m has a circular path of width 3 m inside its boundary. Find the area of the path (Take r = 3.14) 3. A ring shape metal plate has an internal radius of 7 cm and an external radius of 10.5 cm. If the cost of material is `5 per sq. cm, find the cost of 25 rings. 4. A circular well has radius 3 m. If a platform of uniform width of 1.5 m is laid around it, find the area of the platform . (Take r = 3.14) 5. A uniform circular path of width 2.5 m is laid outside a circular park of radius 56m. Find the cost of levelling the path at the rate of `5 per m2 (Take r = 3.14) 6. The radii of 2 concentric circles are 56 cm and 49 cm. Find the area of the pathway. 7. The area of the circular pathway is 88 m2. If the radius of the outer circle is 8 m, find the radius of the inner circle. 8. The cost of levelling the area of the circular pathway is ` 12,012 at the rate of ` 6 per m2. Find the area of the pathway. 70 Measurements Figure Area Forumula 1 × height × sum of 2 1 × h × (a + b) sq. 2 Perimeter of the circle = 2 × r × radius 2 r r units Area of the circle = r × radius × radius r r2 sq. units parallel sides Area of the pathway i) area of the rectangular pathway ii) area of the circular pathway 71 units Area of outer rectangle – Area of inner rectangle Area of outer circle – Area of inner circle = r (R 2 – r2) sq. units = r (R + r) (R – r) sq. units Chapter 4 GEOMETRY 4.1 Triangle: Revision A triangle is a closed plane figure made of three line segments. In Fig. 4.1 the line segments AB, BC and CA form a closed figure. This is a triangle and is denoted by D ABC. This triangle may be named as D ABC or D BCA or Fig. 4.1 D CAB. The line segments forming a triangle are the three sides of the triangle. In Fig.4.1 AB , BC and CA are the three sides of the triangle. The point where any two of the three line segments of a triangle intersect is called the vertex of the triangle. In Fig. 4.1 A,B and C are the three vertices of the D ABC. When two line segments intersect, they form an angle at that point. In the triangle in Fig. 4.1 AB and BC intersect at B and form an angle at that vertex. This angle at B is read as angle B or + B or + ABC. Thus a triangle has three angles + A, + B and + C. In Fig. 4.1 D ABC has AB, BC, CA Sides : Angles : + CAB, + ABC,+ BCA Vertices : A, B, C The side opposite to the vertices A, B, C are BC, AC and AB respectively. The angle opposite to the side BC, CA and AB is + A, + B and + C respectively. A triangle is a closed figure made of three line segments. It has three vertices, three sides and three angles. 72 Geometry 4.2 Types of Triangles Based on sides A triangle is said to be Equilateral, when all its sides are equal. Isosceles, when two of its sides are equal. Scalene, when its sides are unequal. Based on angles A triangle is said to be Right angled, when one of its angle is a right angle and the other two angles are acute. Obtuse - angled, when one of its angle is obtuse and the other two angles are acute. Acute - angled, when all the three of its angles are acute. The sum of the lengths of any two sides of a triangle is always greater than the length of the third side. 4.3 Angle sum property of a triangle: Activity 1 Draw any triangle ABC on a sheet of paper and mark the angles 1, 2 and 3 on both sides of the paper as shown in Fig. 4.2 (i). Is it possible to form a triangle whose sides are 7cm, 5cm and 13cm? Fig. 4.2 Cut a triangle ABC in a paper. Fold the vertex A to touch the side BC as shown in the Fig. 4.2 (ii) Fold the vertices B and C to get a rectangle as shown in the Fig. 4.2 (iii) Now you see that + 1, + 2 and + 3 make a straight line. 73 Chapter 4 From this you observe that + 1 + + 2 + + 3 = 180° + A + + B + + C = 180° The sum of the three angles of a triangle is 180° Activity 2 Draw a triangle. Cut on the three angles. Re arrange them as shown in Fig. 4.2 (ii). You observe that the three angles now constitute one angle. This angle is a straight angle and so has measure 180° Fig. 4.3 The sum of the three angles of a triangle is 180° Think it. 1. Can you have a triangle with the three angles less than 60°? 2. Can you have a triangle with two right angles? 4.4 Exterior angle of a triangle and its property Activity 3 Fig. 4.4 Draw a triangle ABC and produce one of its sides, say BC as shown in Fig. 4.4 (i) observe the angles ACD formed at the point C. This angle lies in the exterior of T ABC formed at vertex C. + BCA is an adjacent angle to + ACD. The remaining two angles of the triangle namely + A nd + B are called the two interior opposite angles. Now cut out (or make trace copies of) + A and + B and place them adjacent to each other as shown in Fig. 4.4 (ii) You observe that these two pieces together entirely cover + ACD. 74 Geometry From this we conclude that the exterior angle of a triangle is equal to the sum of the two interior opposite angles. The relation between an exterior angle and its two interior angles is referred to as the exterior angle property of a triangle. Draw a triangle ABC and produce one of its sides BC as shown in Fig. 4.4 (i) + ACD formed at the point C. Now take a protractor and measure + ACD, + A and + B. Find the sum + A + + B. and compare it with the measure of + ACD. Do you observe that + ACD = + A + B? Example 4.1 In the given figure find the value of x. Solution + CAB + + ABC ++ BCA 40° + x + x 40° + 2x 2x 2x = = = = = 180° 180° 180° 180° – 40° 140° (Since sum of the three angles of a triangle is 180°) x = 140 = 70° 2 The value of x = 70°. Example 4.2 Two angles of a triangle are 40° and 60°. Find the third angle. Solution + RPQ + + PQR ++ QRP = 180° x + 40° + 60° x + 100° x ` The third angle x = = = = = (Since sum of the three angles of a triangle is 180°) 180° 180° 180° – 100° 80° 80° Example 4.3 In the given figure, find the measure of + A. Solution + CAB + + ABC + + BCA = 180° (Since sum of the three angles of a 2x + 120° + x = 180° triangle is 180°) 75 Chapter 4 3x + 120° = 3x = 3x = x= = ` +A = 180° 180° – 120° 60° 60 3 20° 2x = 2 × 20° = 40° Example 4.4 In the given figure. Find the value of x. Solution In the figure exterior angle = + ABD = 110°. Sum of the two interior opposite angle = + BCA + + CAB = x + 50° x + 50° = 110° (Since the sum of the two interior opposite x = 110° – 50° angle is equal to the exterior angle) = 60° ` The value of x is 60°. Example 4.5 In the given figure find the values of x and y. Solution In the give figure, Exterior angle = + DCA = 130° 50° + x = 130° (Since sum of the two interior opposite x = 130° – 50° angle is equal to the exterior angle) = 80° In D ABC, + A + + B + + C = 180° (Since sum of three angles of a triangle is 180°) 50° + x + y = 180° 50° + 80° + y = 180° 130° + y = 180° y = 180° – 130° = 50° ` The values of x = 80° and y = 50°. 76 Geometry Aliter: + ACB + + DCA = 180° (Since sum of the adjacent angles on a line is 180°) y + 130° = 180° y = 180° – 130° = 50° In D ABC, + A + + B + + C = 180° (Since sum of the three angles of a triangle is 180°) 50° + x + y = 180° 50° + x + 50° = 180° 100° + x = 180° x = 180° – 100° = 80° Example 4.6 Three angles of a triangle are 3x + 5°, x + 20°, x + 25°. Find the measure of each angle. Solution Sum of the three angles of a triangle = 180° 3x + 5° + x + 20° + x + 25° = 180° 5x + 50° = 180° 5x = 180° – 50° 5x = 130° x = 130 5 = 26° 3x + 5° = (3 × 26°) + 5° = 78° + 5° = 83° x + 20° = 26° + 20° = 46° x + 25° = 26° + 25° = 51° ` The three angles of a triangle are 83°, 46° and 51°. 77 Chapter 4 Exercise 4.1 1. Choose the correct answer. i) The sum of the three angles of a triangle is (A) 90° (B) 180° (C) 270° (D) 360° ii) In a triangle, all the three angles are equal, then the measure of each angle is 0 0 0 0 (B) 45 (C) 60 (D) 90 (A) 30 iii) Which of the following can be angles of a triangle? 0 0 0 0 0 0 0 0 0 (A) 50 , 30 , 105 (B) 36 , 44 , 90 (C) 70 , 30 , 80 0 iv) Two angles of a triangle are 40° and 60°, then the third angle is 0 0 0 0 (A) 20 (B) 40 (C) 60 (D) 80 0 v) In T ABC, BC is produced to D and + ABC = 50 , 0 + ACD = 105 , then + BAC will be equal to (A) 75° (B) 15° (C) 40° (D) 55° 2. State which of the following are triangles. 0 0 0 (i) + A = 25 + B = 35 + C = 120 0 0 0 (ii) + P = 90 + Q = 30 + R = 50 0 0 0 (iii) + X = 40 + Y = 70 + Z = 80 3. Two angles of a triangle is given, find the third angle. 0 (i) 75 , 45 0 0 (ii) 80 , 30 0 0 (iii) 40 , 90 0 0 (iv) 45 , 85 0 4. Find the value of the unknown x in the following diagrams: 78 0 (D) 45 , 45 80 0 Geometry 5. Find the values of the unknown x and y in the following diagrams: 6. Three angles of a triangle are x + 5°, x + 10° and x + 15° find x. 79 Chapter 4 1. The sum of the three angles of a triangle is 180°. 2. In a triangle an exterior angle is equal to the sum of the two interior opposite angles. 80 Practical Geometry Practical Geometry 5.1 Construction of triangles In the previous class, we have learnt the various types of triangles on the basis of their sides and angles. Now let us recall the different types of triangles and some properties of triangle. Classification of triangles No. CLASSIFICATION OF TRIANGLES on the basis SIDES on the basis ANGLES 1 2 Name of Triangle Figure Equilateral triangle Note Three sides are equal Isosceles triangle Any two sides are equal 3 Scalene triangle Sides are unequal 4 Acute angled triangle All the three angles are acute (less than 90°) 5 Obtuse angled triangle Any one of the angles is obtuse (more than 90°) 6 Right angled triangle Any one of the angles is right angle (90°) 81 Chapter 5 Some properties of triangle 1. The sum of the lengths of any two sides of a triangle is greater than the third side. 2. The sum of all the three angles of a triangle is 180°. To construct a triangle we need three measurements in which at least the length of one side must be given. Let us construct the following types of triangles with the given measurements. (i) Three sides (SSS). (ii) Two sides and included angle between them (SAS). (iii) Two angles and included side between them (ASA). (i) To construct a triangle when three sides are given (SSS Criterion) Example 5.1 Construct a triangle ABC given that AB = 4cm, BC = 6 cm and AC = 5 cm. Solution Rough Diagram Given measurements AB = 4cm BC = 6 cm AC = 5 cm. Steps for construction Step 1 : Draw a line segment BC = 6cm Step 2 : With ‘B’ as centre, draw an arc of radius 4 cm above the line BC. Step 3 : With ‘C’ as centre, draw an arc of 5 cm to intersect the previous arc at ‘A’ Step 4 : Join AB and AC. Now ABC is the required triangle. 82 Practical Geometry By using protector measure all the angles of a triangle. Find the sum of all the three angles of a triangle. Rough Figure m P 2c 1. A student attempted to draw a triangle with given measurements PQ = 2cm, QR = 6cm, PR = 3 cm. (as in the rough figure). First he drew QR = 6cm. Then he drew an arc of 2cm with Q as centre and he drew another arc of radius 3 cm with R as centre. They could not intersect each to get P. 3c m Q 6 cm (i) What is the reason? (ii) What is the triangle property in connection with this? R The sum of any two sides of a triangle is always greater than the third side. Draw the bisectors of the three angles of a triangle. Check whether all of them pass through a same point. This point is incentre. (ii) To construct a triangle when Two sides and an angle included between    them are given. (SAS Criterion) Example 5.2 Construct a triangle PQR given that PQ = 4 cm, QR = 6.5 cm and + PQR = 60°. Solution Given measurements PQ = 4 cm QR = 6.5 cm + PQR = 60° 83 Chapter 5 Steps for construction Step 1 : Draw the line segment QR = 6.5 cm. Step 2 : At Q, draw a line QX making an angle of 600 with QR. Step 3 : With Q as centre, draw an arc of radius 4 cm to cut the line (QX) at P. Step 4 : Join PR. PQR is the required triangle. Construct a triangle with the given measurements XY = 6cm, YZ = 6cm and + XYZ = 70°. Measure the angles of the triangle opposite to the equal sides. What do you observe? (iii) To construct a triangle when two of its angles and a side      included between them are given. (ASA criterion) Example 5.3 Construct a triangle XYZ given that XY = 6 cm, + ZXY = 30° and + XYZ = 100°. Examine whether the third angle measures 50°. Solution Given measurements XY = 6 cm + ZXY = 30° + XYZ = 100° 84 Practical Geometry Step 1 : Draw the line segment XY = 6cm. Step 2 : At X, draw a ray XP making an angle of 30° with XY. Step 3 : At Y, draw another ray YQ making an angle of 100° with XY. The rays XP and YQ intersect at Z. Step 4 : The third angle measures 50° i.e + Z = 50°. Construct a triangle PQR given that PQ = 7 cm, + Q = 70°, + R = 40°. Hint: Use the Angle Sum Property of a triangle. Exercise : 5.1 I. Construct the triangles for the following given measurements. 1. Construct 3 PQR, given that PQ = 6cm, QR = 7cm, PR = 5cm. 2. Construct an equilateral triangle with the side 7cm. Using protector measure each angle of the triangle. Are they equal? 3. Draw a triangle DEF such that DE = 4.5cm, EF = 5.5cm and DF = 4.5cm. Can you indentify the type of the triangle? Write the name of it. II. Construct the triangles for the following given measurements. 4. Construct 3 XYZ, given that YZ = 7cm, ZX = 5cm, + Z = 50°. 5. Construct 3 PQR when PQ = 6cm, PR = 9cm and + P = 100°. 6. Construct 3 ABC given that AB = 6 cm, BC = 8 cm and + B = 90° measure length of AC. III. Construct the triangles for the following given measurements. 7. Construct 3 XYZ, when + X = 50°, + Y = 70° and XY = 5cm. 8. Construct 3 ABC when + A = 120°, + B = 30° and AB = 7cm. 9. Construct 3 LMN, given that + L = 40°, + M = 40° and LM = 6cm. Measure and write the length of sides opposite to the + L and + M. Are they equal? What type of Triangle is this? 85 Chapter 6 DATA HANDLING 6.1 Mean, Median and Mode of ungrouped data Arithmetic mean We use the word ‘average’ in our day to day life. Poovini spends on an average of about 5 hours daily for her studies. In the month of May, the average temperature at Chennai is 40 degree celsius. What do the above statements tell us? Poovini usually studies for 5 hours. On some days, she may study for less number of hours and on other days she may study longer. The average temperature of 40 degree celsius means that, the temperature in the month of May at chennai is 40 degree celsius. Some times it may be less than 40 degree celsius and at other times it may be more than 40 degree celsius. Average lies between the highest and the lowest value of the given data. Rohit gets the following marks in different subjects in an examination. 62, 84, 92, 98, 74 In order to get the average marks scored by him in the examination, we first add up all the marks obtained by him in different subjects. 62 + 84 + 92 + 98 + 74 = 410. and then divide the sum by the total number of subjects. (i.e. 5) The average marks scored by Rohit = 410 = 82. 5 This number helps us to understand the general level of his academic achievement and is referred to as mean. ` The average or arithmetic mean or mean is defined as follows. Mean = Sum of all observations Total number of observations 86 Data Handling Example 6.1 Gayathri studies for 4 hours, 5 hours and 3 hours respectively on 3 consecutive days. How many hours did she study daily on an average? Solution: Total number of study hours Number of days for which she studied. = 4 + 5 + 3 hours 3 12 = 3 Average study time = = 4 hours per day. Thus we can say that Gayathri studies for 4 hours daily on an average. Example 6.2 The monthly income of 6 families are ` 3500, ` 2700, ` 3000, ` 2800, ` 3900 and ` 2100. Find the mean income. Solution: Average monthly income = Total income of 6 familes Number of families = ` 3500 + 2700 + 3000 + 2800 + 3900 + 2100 6 18000 = ` 6 = ` 3,000. Example 6.3 The mean price of 5 pens is ` 75. What is the total cost of 5 pens? Solution: Total cost of 5 pens Number of pens Total cost of 5 pens = Mean # Number of pens Mean = = ` 75 # 5 = ` 375 Median Consider a group of 11 students with the following height (in cm) 106, 110, 123, 125, 115, 120, 112, 115, 110, 120, 115. The Physical EducationTeacher Mr. Gowtham wants to divide the students into two groups so that each group has equal number of students. One group has height lesser than a particular height and the other group has student with height greater than the particular height. 87 Chapter 6 Now, Mr. Gowtham arranged the students according to their height in ascending order. 106, 110, 110, 112, 115, 115,115, 120, 120, 123, 125 The middle value in the data is 115 because this value divides the students into two equal groups of 5 students each. This value is called as median. Median refers to the value 115 which lies in the middle of the data.Mr. Gowtham decides to keep the middle student as a referee in the game. Median is defined as the middle value of the data when the data is arranged in ascending or descending order. Find the actual Find the median of the following: distance between 40, 50, 30, 60, 80, 70 your school and Arrange the given data in ascending order. house. Find the median of the place. 30, 40, 50, 60, 70, 80. Here the number of terms is 6 which is even. So the third and fourth terms are middle terms. The average value of the these terms is the median. (i.e) Median = 50 + 60 = 110 = 55. 2 2 (i) When the number of observations is odd, the middle number is the median. (ii) When the number of observations is even, the median is the average of the two middle numbers. Example 6.4 Find the median of the following data. 3, 4, 5, 3, 6, 7, 2. Solution: Arrange the data in ascending order. 2, 3, 3, 4, 5, 6, 7 The number of observation is 7 which is odd. ` The middle value 4 is the median. Example 6.5 Find the median of the data 12, 14, 25, 23, 18, 17, 24, 20. Solution: Arrange the data in ascending order 88 In highways, the yellow line represents the median. Data Handling 12, 14, 17, 18, 20, 23, 24, 25. The number of observation is 8 which is even. ` Median is the average of the two middle terms 18 and 20. Median = 18 + 20 = 38 = 19 2 2 Example 6.6 Find the median of the first 5 prime numbers. Solution: The first five prime numbers are 2, 3, 5, 7, 11. The number of observation is 5 which is odd. ` The middle value 5 is the median. Mode Look at the following example, Mr. Raghavan, the owner of a ready made dress shop says that the most popular size of shirts he sells is of size 40 cm. Observe that here also, the owner is concerned about the number of shirts of different sizes sold. He is looking at the shirt size that is sold, the most. The highest occurring event is the sale of size 40 cm. This value is called the mode of the data. Mode is the variable which occurs most frequently in the given data. Mode of Large data Putting the same observation together and counting them is not easy if the number of observation is large. In such cases we tabulate the data. Example 6.7 Following are the margin of victory in the foot ball matches of a league. 1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 2, 3, 2, 3, 1, 1, 2, 3, 2, 6, 4, 3, 2, 1, 1, 4, 2, 1, 5, 3, 4, 2, 1, 2. Find the mode of this data. Solution: Margin of victory 1 2 3 4 5 6 Tally Marks |||| |||| |||| |||| |||| |||| || |||| || || Total Table 6.1 89 Number of Matches 9 14 7 5 3 2 40 Chapter 6 Now we quickly say that ‘2’ is the mode. Since 2 has occurred the more number of times, then the most of the matches have been won with a victory margin of 2 goals. Example 6.8 Find the mode of the following data. 3, 4, 5, 3, 6, 7 Find the mode of the transport in your place. Solution: 3 occurs the most number of times. ` Mode of the data is 3. Example 6.9 Find the mode of the following data. 2, 2, 2, 3, 3, 4, 5,5, 5, 6,6, 8 Solution: 2 and 5 occur 3 times. ` Mode of the data is 2 and 5. Example 6.10 Find the mode of the following data Find the mode 90, 40, 68, 94, 50, 60. of the flower. Solution: Here there are no frequently occurring values. Hence this data has no mode. Example 6.11 The number of children in 20 families are 1, 2, 2, 1, 2, 1, 3, 1, 1, 3 1, 3, 1, 1, 1, 2, 1, 1, 2, 1. Find the mode. Solution: Number of Children 1 2 3 Tally Marks |||| |||| || |||| ||| Total Number of Families 12 5 3 20 Table 6.2 12 families have 1 child only, so the mode of the data is 1. 90 Data Handling Exercise: 6.1 1. Choose the correct answer: i) The arithmetic mean of 1, 3, 5, 7 and 9 is (A) 5 (B) 7 (C) 3 (D) 9 ii) The average marks of 5 children is 40 then their total mark is (A) 20 (B) 200 (C) 8 (D) 4 (C) 30 (D) 10 (C) 7 (D) 14 (C) 7 (D) 2 iii) The median of 30,50, 40, 10, 20 is (A) 40 (B) 20 iv) The median of 2, 4, 6, 8, 10, 12 is (A) 6 (B) 8 v) The mode of 3, 4, 7, 4, 3, 2, 4 is (A) 3 (B) 4 2. The marks in mathematics of 10 students are 56, 48, 58, 60, 54, 76, 84, 92, 82, 98. Find the range and arithmetic mean 3. The weights of 5 people are 72 kg, 48 kg, 51 kg, 69 kg, 67 kg. Find the mean of their weights. 4. Two vessels contain 30 litres and 50 litres of milk separately. What is the capacity of the vessels if both share the milk equally? 5. The maximum temperature in a city on 7 days of a certain week was 34.8°C, 38.5°C, 33.4°C, 34.7°C, 35.8°C, 32.8°C, 34.3°C. Find the mean temperature for the week. 6. The mean weight of 10 boys in a cricket team is 65.5 kg. What is the total weight of 10 boys. 7. Find the median of the following data. 6, 14, 5, 13, 11, 7, 8 8. The weight of 7 chocolate bars in grams are 131, 132, 125, 127, 130, 129, 133. Find the median. 9. The runs scored by a batsman in 5 innings are 60, 100, 78, 54, 49. Find the median. 10. Find the median of the first seven natural numbers. 11. Pocket money received by 7 students is given below. ` 42, ` 22, ` 40, ` 28, ` 23, ` 26, ` 43. Find the median. 12. Find the mode of the given data. 3, 4, 3, 5, 3, 6, 3, 8, 4. 91 Chapter 6 13. Twelve eggs collected in a farm have the following weights. 32 gm,40 gm, 27 gm, 32 gm, 38 gm, 45 gm, 40 gm, 32 gm, 39 gm, 40 gm, 30 gm, 31 gm, Find the mode of the above data. 14. Find the mode of the following data. 4, 6, 8, 10, 12, 14 15. Find the mode of the following data. 12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16, 15, 17, 13, 16, 16, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14. 1. Average lies between the highest and the lowest value of the given data. 2. Mean = sum of all the observations total number of observations 3. Median is defined as the middle value of the data, when the data is arranged in ascending or descending order. 4. Mode is the variable which occurs most frequently in the given data. 92 Answers Answers Unit 1 Exercise 1.1 1. (i) B (ii) A (iii) D (iv) C 2. (i) x + 2y (ii) y – z (iii) xy + 4 (v) A (iv) 3x – 4y (if 3x > 4y) or 4y – 3x (if 4y > 3x) (v) 10 + x + y (vi) pq – 5 (viii) ab – (a + b) (ix) 3cd + 6 (vii) 12 – mn (x) 4xy 3 Exercise 1.2 1. (i) B (ii) A (iii) C 2. (i) x = 12 (ii) a = 7 (iii) y = – 6 (iv) b = – 2 (vi) x = 7 (iv) C (v) D (v) x=–5 (ix) m = 11 (x) y = 27 (iv) a = 3 (v) x=–9 (vii) x = –5 (viii) n = 4 3. (i) x = 50 (ii) l = 14 (iii) x = 4 (vi) t = – 4 (vii) x = – 6 (viii) m = 3 4. (i) x = 14 (ii) a = 30 (iii) n = – 24(iv) p = – 56 (v) x = – 10 (ii) x = – 15 (iii) z = 5 (ix) x = - 1 (x) 2 x=6 (vi) m = 12 5. (i) x = 3 (iv) a = – 9 (v) x=3 (ix) y = 3 (x) m=6 (vi) x = 5 (vii) y = 67 (viii) x = 6 (xi) x = 11 (xii) m = 1 (xiii) x = 3 (xiv) x = – 3 (xv) 2 6. 15 7. 13 11.37, 38 12. 60 8. 108 9. 12 10. t=–1 8 13. 35 Unit - 2 Exercise 2.1 1. (i) 20% (ii) 93% (iii) 11% (iv) 1% (v) 100% 2. (i) 43 : 100 (ii) 75 : 100 (iii) 5 : 100 (iv) 35:200 (v) 100 : 300 3. (i) 25 100 (ii) 25 200 (iii) 33 100 (iv) 70 (v) (iv) A (v) C 100 82 100 Exercise 2.2 1. (i) C (ii) C (iii) A 93 Answers 2. (i) 100% (ii) 18% (iii) 525% (iv) 66.67% (v) 45.45% 3. (i) 36% (ii) 3% (iii) 7.1% (iv) 305% (v) 75% 4. 20% 5. 13.89% 6. Girls 46%; Boys 54 % 7. He got more marks in Science. 8. Savings 10%; Expenditure 90% Exercise 2.3 1. (i) B (ii) B (iii) A (ii) 3 (iii) 3. (i) 0.07 (ii) 0.64 (iii) 3.75 (iv) 0.0003 (v) 0.005 4. (i) 18 (ii) ` 24 (iii) 36 m (iv) 108 2. (i) 9 100 4 1 400 5. ` 6250 6. 9 matches 8. ` 38250 9. 3975 illiterates (iv) C (v) B (iv) 1 (v) 2 40 3 (v) 3.75 kg 7. 12,800 men; 11,200 children Exercise 2.4 1. (i) A (ii) C (iv) A (v) B (iii) C 2. Profit = ` 24, Loss = ` 21; Profit = ` 35.45, Loss = ` 3362, Loss = ` 7.50 3. (i) ` 530 (iv) ` 336.75 4. Profit 10% (ii) ` 620 (iii) ` 1027.50 (v) ` 943.50 5. Loss 12% 6. Profit 60% 7. Profit 15% Exercise 2.5 1. (i) B (iv) C 2. ` 2,500; ` 7,500 5. ` 2,630 (ii) A (iii) A (v) A 3. ` 450; ` 1,650 6. ` 216; ` 12,216 7. 5% 94 4. ` 2,250 8. ` 1,000 Answers 9. 2 years 12. ` 5,400 10. 10% 11. 8 years 13. ` 5,000; 10% 14. S.I. = 2025; ` 5,625 15. ` 4,000 Unit - 3 Exercise 3.1 1. (i) B (ii) A 2. (i) 50 cm2 (ii) 66 cm2 (iii) 80.5 cm2 3. 12 cm 4. 875 m2 (iii) D (iv) D 5. 19.2 cm Exercise 3.2 1. (i) B (ii) C (iii) A (iv) D (v) D 2. (i) d = 70 cm, c = 220 cm (ii) r = 28 cm, c = 176 cm (iii) r = 4.9 cm, d = 9.8 cm 3. (i) 110 cm (ii) 264 cm (iii) 374 cm (iv) 462 cm 4. (i) 79.2 cm (ii) 396 cm (iii) 8.8 m (iv) 26.4 m 5. (i) 17.5 cm (ii) 21 cm 6. 660 m 7. ` 1232 (iii) 0.7 m (iv) 1.75 m 8. 1.05 m 9. 37 10. ` 4,752 Exercise 3.3 1. (i) 38.5 cm2 (iii) 18.865 m2 2. (i) 4.525 cm2 (iii) 55.44 m2 (ii) 86.625 cm2 (iv) 124.74 m2 (ii) 616 cm2 (iv) 98.56 m2 3. ` 1848 4. 154 m2 5. circle has larger area 6. 38.5 m2 7. 1936 cm2 8. r = 35, ` 2200 9. 440 m 10. 63.36 Second 11. 10 Exercise 3.4 1. 636 m2 2. ` 1152 3. ` 1590 4. 375 cm2 5. 40 m2, 30 m2 6. 3256 cm2 95 Answers Exercise 3.5 1. 6594 m2 2. 536.94 m2 3. ` 24,050 4. 21.195 m2 5.` 4494 6. 2310 cm2 8. 2002 m2 7. 6 m Unit - 4 Exercise 4.1 1. (i) B (ii) C (iii) C (iv) D (v) D 2. (i) + A = 25°, + B = 35°, + C = 120° 3. (i) 60° (ii) 70° (iii) 50° (iv) 50° 4. (i) 70° (ii) 60° (iii) 40° (iv) 30° (v) 65°, 65° (vi) 60°, 60°, 60° 5. (i) y = 60°, x = 70° (ii) y = 80°, x = 50° (iii) y = 70°, x = 110° (iv) x = 60°, y = 90° (v) y = 90°, x = 45° (vi) x = 60°, y = 50° 6. x = 50°. Unit - 6 Exercise 6.1 1. (i) A (ii) B (iii) C (iv) C (v) B 2. Range is 50; A.M. = 70.8 3. 61.4 kg. 4. 40 litres 5. 34.9°C 6. 655.0 kg 7. 8 12. 3 8. 130 gram 13. 32 gm and 40 gm 9. 60 10. 4 14. no mode 96 15. 15 11. ` 28 Answers 97