Ratio and Proportion PDF

4 Ratio and Proportion Let’s study. Ratio Properties of ratios Operations on equal ratios Theorem of equal ratios Continued proportion k method...

4 Ratio and Proportion Let’s study. Ratio Properties of ratios Operations on equal ratios Theorem of equal ratios Continued proportion k method Let’s recall. In earlier standards, we have learnt about ratio and proportion. We have also solved examples based on it. Let us discuss following example. Ex. The rawa ladoo prepared by Vimal are tasty, for which she takes 1 bowl of ghee, 3 bowls of rawa and 2 bowls of sugar. 3 Here proportion of rawa and sugar is 3 : 2 or 2. If 12 units of rawa is used, how many units of sugar are required ? Let the number of bowls of sugar required be x. 3 12 \ from above information, 2 = \ 3x = 24 \x=8 x That is for preparation of ladoo, with 12 units of rawa requires 8 units of sugar. Alternatively we can solve the above example in the following way. 3k 3 3k bowls of rawa, 2k bowls of sugar is required because = 2 2k If 3k = 12 then k = 4 \ 2k = 2 × 4 = 8 bowls of sugar is required. Let’s learn. Ratio and proportion The concept of ratio of two numbers can be extended to three or more numbers. Let us see the above example of ladoos. The proportion of ghee, rawa and sugar is 1 : 3 : 2. Here proportion of ghee and rawa is 1 : 3 and that of rawa and sugar is 3 : 2, This means the proportion of ghee, rawa and sugar is 1 : 3 : 2. Let us take k bowls of ghee, 3k bowls of rawa and 2k bowls of sugar. Hence for 12 bowls of rawa, how much quantity of ghee and sugar is required can be found as follows. Now 3k = 12 \ k = 4 and 2k = 8. \ 4 bowls of ghee and 8 bowls of sugar is required. 57  The same concept can be extended for proportion of 4 or more entities. If a, b, c, d are in the ratio 2 : 3 : 7 : 4 then let us assume that the numbers are 2m, 3m, 7m, 4m. From the given information, value of m can be determined. For example if the sum of these four numbers is 48, we find these numbers. 2m + 3m + 7m + 4m = 16 m = 48 \ m=3 \ 2m = 6, 3m = 9, 7m = 21, 4m = 12 \ required numbers are 6, 9, 21, 12 Ex (1) The proportion of compounds of nitrogen, phosphorous and potassium in certain fertilizer is 18 : 18 : 10. Here compound of nitrogen is 18%, compound of phosphorous is 18% and that of potassium is 10%. Remaining part is of other substances. Find the weight of each of the above compounds in 20 kg of fertilizer. Solution : Let the weight of nitrogen compound in 20 kg of fertilizer be x kg. 18 x 18 × 20 ∴ = ∴ x= = 3.6 100 20 100 \ weight of nitrogen compound is 3.6 kg. The percentage of phosphorous compound is also 18%. \ Weight of compound of phosphorous is 3.6 kg. If we assume the weight of potassium compound y kg then 10 = y \ y = 2 \ weight of potassium compound is 2 kg. 100 20 Direct proportion A car covers a distance of 10 km consuming 1 litre of petrol. It will cover a distance of 20 ´ 10 = 200 km consuming 20 litre of petrol. Consuming 40 litre of petrol, it will cover a distance of 40 ´ 10 = 400 km. Let us write this information in tabular form. Petrol : x litre 1 20 40 Distance : y km 10 200 400 x 1 20 1 40 1 x = = =k y 10 200 10 400 10 y The ratio of consumption of petrol (in litre) and distance covered by the car (in kilometres), is constant. In such case, it is said that the two quantities are in direct proportion or in direct variation. 58  Inverse proportion A car takes two hours to cover a distance of 100 km at the speed of 50 km/hr. A bullock- cart travels 5 km in 1 hour. To cover a distance of 100 km at the speed of 5 km/hr, the bullock-cart takes 20 hours. We know that, Speed ´ time = distance By using the relation let us put the above information in a tabular form. Vehicle Speed/hr (x) Time (y) x´y x´y=k Car 50 2 100 Bullock-cart 5 20 100 Hence, we see that, the product of speed of the vehicle and time is constant. In such a case it is said that the quantities are in inverse proportion or in inverse variation. Let’s recall. Properties of ratio a (1) Ratio of numbers a and b is written as a : b or. a is called the predecessor (first term) b and b is called successor (Second term). (2) In the ratio of two numbers, if the second term is 100 then it is known as a percentage. (3) The ratio remains unchanged, if its terms are multiplied or divided by non-zero number. e.g.. 3 : 4 = 6:8 = 9:12, Similarly 2:3:5 = 8:12:20. If k is a non-zero number, then a : b = ak : bk a : b : c = ak : bk : ck (4) The quantities taken in the ratio must be expressed in the same unit. (5) The ratio of two quantities is unitless. For example The ratio of 2 kg and 300 g is not 2 : 300 , but it is 2000 : 300 as (2 kg = 2000 gm) i.e. 20 : 3 Ex (1) The ratio of ages of Seema and Rajashree is 3 : 1. The ratio of ages of Rajashree and Atul is 2 : 3. Then find the ratio of ages of Seema, Rajashree and Atul. Solution : Seema's age : Rajashree's age = 3 : 1 Rajashree's age : Atul's age = 2 : 3 Second term of first ratio should be the same as the first term of second ratio. 59  Hence to get the continuous ratio, multiplying each term of the first ratio by 2. We get 3:1 = 6:2. Seema's age 6 Rajashree's age 2 = , = Rajashree's age 2 Atul's age 3 \ Seema's age : Rajashree's age : Atul's age 6 : 2 : 3. Ex (2) The length of a rectangular field is 1.2 km and its breadth is 400 metre. Find the ratio of length to breadth. Solution : Here the length is in kilometer and breadth is in meter. In order to find the ratio of length to breadth, they must be expressed in same unit. Hence we convert kilometre to meter. 1.2 km= 1.2 ´1000 = 1200 m \ ratio of 1200 m , to 400 m is 1200 3 = , that is 3 : 1 400 1 Ex (3) The ratio of expenditure and income of Mahesh is 3 : 5. Find the percentage of expenses to his income. Solution : The ratio of expenditure to income is 3 : 5. To convert it into percentage, convert second term into 100. 3 3 × 20 60 = = \ Expenditure = 60 = 60% \ Mahesh spends 60% of his income. 5 5 × 20 100 Income 100 Ex (4) The ratio of number of mango trees to chikoo trees in an orchard is 2 : 3. If 5 more trees of each type are planted the ratio of trees would be 5 : 7. Then find the number of mango and chickoo trees in the orchard. Solution : The ratio of trees is 2 : 3. Let the number of mango trees = 2x and chikoo trees = 3x From given condition, 2 x + 5 = 5 3x + 5 7 14x + 35 = 15x + 25 \ x = 10 \ number of mango trees in the orchard = 2x = 2 ´10= 20 and number of chikoo trees = 3x = 3 ´10= 30 60 Ex (5) The ratio of two numbers is 5 : 7. If 40 is added in each number, then the ratio becomes 25 : 31, Find the numbers. Solution : Let the first number be 5x and and second number be 7x. From the given condition, 5 x + 40 = 25 7 x + 40 31 31( 5 x + 40 ) = 25 ( 7 x + 40 ) 155 x + 1240 = 175 x + 1000 1240 −11000 = 175 x − 155 x 240 = 20 x x = 12 \ first number = 5 ´ 12 = 60 and second number = 7 ´ 12 = 84 \ given numbers are 60 and 84. Practice set 4.1 (1) From the following pairs of numbers, find the reduced form of ratio of first number to second number. (i) 72, 60 (ii) 38,57 (iii) 52,78 (2) Find the reduced form of the ratio of the first quantity to second quantity. (i) 700 `, 308 ` (ii) 14 `, 12 `. 40 paise. (iii) 5 litre, 2500 ml (iv) 3 years 4 months, 5 years 8 months (v) 3.8 kg, 1900 gm (vi) 7 minutes 20 seconds, 5 minutes 6 seconds. (3) Express the following percentages as ratios in the reduced form. (i) 75 : 100 (ii) 44 : 100 (iii) 6.25% (iv) 52 : 100 (v) 0.64% (4) Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required? (5) Convert the following ratios into percentage. 7 (i) 15 : 25 (ii) 47 : 50 (iii) (iv) 546 (v) 7 10 600 16 (6) The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha's birth her mothers age was 27 year. Find the present ages of Abha and her mother. (7) Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4? (8) The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana's present age ? 61  Let’s learn. Comparison of ratios a c The numbers a, b, c, d being positive, comparison of ratios , can be done b d using following rules : a c a c (i) If ad > bc then > (ii) If ad < bc then < b d b d a c (iii) If ad = bc then = b d Compare the following pairs of ratios. 4 7 13 7 Ex (1) , Ex (2) , 9 8 8 5 Solution : 4 ´ 8 ? 7 ´ 9 13 × 5 , ? 8× 7 32 < 63 \ 4 7 < 65 ? 56 9 8 65 > 56 \ 13 > 7 8 5 a -1 a +1 Ex (3) If a and b are integers and a < b, b > 1 then compare ,. b -1 b +1 Solution : a 1 ab − b + a − 1 − ab − b + a + 1 2 (a − b) = < 0.......... (2) b2 − 1 b2 − 1 2a − 2b a −1 a +1 = 2 − < 0.....from (1) & (2) b −1 b −1 b +1 2 ( a − b ).......... (1) a −1 a +1 = 2 < b −1 b −1 b +1 62  3  a4  Ex (4) If a : b = 2 : 1 and b : c = 4 : 1 then find the value of  2 2 .  32b c  a 2 \a=2b b=4 \ b = 4c Solution : b =1 c 1 a = 2 b = 2 ´ 4c = 8c \a=8c Now substituting the values a = 8 c, b = 4c 3 ( 8c ) 3    a 2 2  = 4 4    32b c   32 × 42 × c 2 × c 2    3  8 × 8 × 8 × 8 × c4  =  32 ×16 × c 2 × c 2    = (8)3 3  a4  \  2 2  = 512  32b c  Practice set 4.2 a ak (1) Using the property = , fill in the blanks substituting proper numbers in the b bk following. 5.... 35.... 9 4.5........ (i) = = = (ii) = = = 7 28.... 3.5 14.... 42 3.5 (2) Find the following ratios. (i) The ratio of radius to circumference of the circle. (ii) The ratio of circumference of circle with radius r to its area. (iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm. (iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its perimeter to area. (3) Compare the following pairs of ratios. 5 3 3 5 63 5 17 (i) , (ii) , (iii) , 18 121 3 7 5 7 125 80 45 9.2 3.4 (iv) 48 , 27 (v) , 5.1 7.1 (4) (i) ABCD is a parallelogram. The ratio of ÐA and ÐB of this parallelogram is 5 : 4. Find the measure of Ð B. (ii) The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages. 63  (iii) The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle. (iv) The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers. (v) If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers. 3  a3  a2 (5*) If a : b = 3 : 1 and b : c = 5 : 1 then find the value of (i)  15b 2 c  (ii)   7bc a (6*) If 0.04 × 0.4 × a = 0.4 × 0.04 × b then find the ratio b. (7) (x + 3) : (x + 11) = (x - 2) : (x + 1) then find the value of x. Let’s learn. Operations on equal ratios Using the properties of equality, we can perform some operations on ratios. Let's study them. Let us learn some properties of the equal ratios, if a, b, c, d, are positive integers. a c b d (I) Invertendo : If = then = b d a c a c = b d \ a × d = b × c \ b×c = a×d b×c a×d \ = a × c a × c...(dividing both sides by a ´ c ) b d = a c a c b d \ If b = d then a = c. This property is known as Invertendo. a c a b (II) Alternando : If = then = b d c d a c = b d a×d = b×c \ a×d b×c =...(dividing both sides by c ´ d) c×d c×d a b = c d a c a b If = then =. This property is known as Alternando. b d c d 64  a c a+b c+d (III) Componendo : If= then = b d b d a c = b d a c + 1 = + 1...(adding 1 to both sides) b d a+b c+d = b d a c a+b c+d If = then =. This property is known as Componendo. b d b d a c a −b c −d (IV) Dividendo : If = then = b d b d a c \ = b d a c \ − 1 = − 1...(subtracting 1 from both sides) b d a −b c −d \ b = d a c a −b c −d If = then =. This property is known as Dividendo. b d b d a c a ca + b c + d (V) Componendo-Dividendo : If = then = = , a ¹ b, c ¹ d b d b da − b c − d a c a+b c+d If = b d \ =...(using componendo)....(1) b d a+b c+d a −b c −d b = d \ =...(using dividendo)....(2) b d a −b c −d a+b c+d = \ =...from (1) and (2) b d a −b c −d a ca + b = ca ++ db c + d If = a −then =. This property is known as Componendo-dividendo. b d b ca −− db c − d General form of Componendo and Dividendo a c If = then a + b = c + d...(performing componendo once) b d b d a + 2b c + 2d...(performing componendo twice) = b d a + mb c + md Generally =... (performing componendo m times)...(I) b d a c a − mb c − md Similarly if = then =...(performing dividendo m time)...(II) b d b d a c and if = then a + mb = c + md...[dividing (I) by (II)] b d a − mb c − md 65  Remember this ! a c b d (Invertendo) a c a+b c+d If = then = If = then = (Componendo) b d a c b d b d a c (Alternando) (Dividendo) If = then ac = db - If ba = dc then a b b = c d d - b d a+b c+d If ba = dc then = (Componendo-Dividendo) a −b c−d Solved Examples : a 5 a + 7b Ex (1) If = then find the ratio =.? b 3 7b Method I Method II a 5 a b a 5 Solution : If = then = = k , = b 3 5 3 b 3...(using alternando) a 5 \ a = 5k, b = 3k \ 7b = 21 \ a + 7b = 5k + 7 × 3k a + 7b = 5 + 21 7b 7 × 3k \ 7b 21...(using componendo) 5k + 21k a + 7b 26 = = 21k \ 7b 21 26k 26 = = 21k 21 Ex. (2) If ba = 74 then find the ratio 5ab- b. Method I Method II a 7 a 7 Solution : b = 4 b = 4 a b 5a = 5 # 7 \ 7 = 4...(using alternando) b 4 a b Let 7 = 4 = m = 35 4 5a − b 35 − 4 \ a = 7m, b = 4m =...(using dividendo) b 4 5 (7m) - 4m 5a - b 31 5a - b = \ b = 4m b 4 35m - 4m = 4m 31 = 4 66 Ex. (3) If ba = 73 then find the value of the ratio aa +- 22bb. Solution : Method I : Method II : \ a 7 b =3 Let a = 7m, b = 3m 1 7 m + 2 × 3m \ 2ab = 67...(multiplying both sides by 2 ) \ aa +- 22bb = 7 m − 2 × 3m a + 2b 7 + 6 \ = (using componendo = 7m + 6m a − 2b 7 − 6 - dividendo) 7m - 6m 13m 13 \ aa +- 22bb = 13 1 = m = 1 a b 5a + 3b Ex (4) If = then find the value of the ratio. 3 2 7 a − 2b Solution : Method I Method II a b a b = = 3 2 3 2 \ a = 3........ (using Alternando) Let a b = = t. b 2 3 2 Now dividing each term of 5a + 3b by b. \ by substituting a = 3t and b = 2t, 7 a − 2b 5a 3b 5  a  + 3 5a + 3b 5(3t ) + 3(2t ) (t ¹ 0) +   = b b = b 7a − 2b 7(3t ) − 2(2 t) 7 a 2b a − 7  − 2 b b b 15t + 6t = 3 21t − 4 t 5  + 3 =   2 21t 3 = 7  − 2 17t 2 21 15 = +3 17 = 2 21 −2 2 15 + 6 = 21 − 4 21 = 17 67  x 4 4 − Ex (5) If y = 5 then find the value of the ratio. 4 + Solution : x 4 = y 5 4x 16...(multiplying both sides by 4) = y 5 \ 4x + y 16 + 5...(using componendo-dividendo) = 4x − y 16 − 5 \ 4x + y = 21 4x − y 11 \ 4x − y = 11 4x + y 21 \ 3x 2 + y 2 Ex (6) If 5x = 4y then find the value of the ratio 3 x 2 − y 2. Solution : x 4 = y 5 x2 16 = y 2 25 \ 3x 2 48...(multiplying both sides by 3) = y 2 25 \ 3x 2 + y 2 48 + 25...(using componendo-dividendo) = 3x − y 2 2 48 − 25 \ 3x 2 + y 2 = 73 3x − y 2 2 23 \ Let’s learn. Application of properties of equal ratios To solve some types of equations, it is convenient to use properties of equal ratios rather than using other methods. 3x 2 + 5 x + 7 3x 2 + 4 x + 3 Ex (1) Solve the equation. = 10 x + 14 8x + 6 3x 2 + 5 x + 7 3x 2 + 4 x + 3 Solution : = 10 x + 14 8x + 6 2 ( 6 x + 10 x + 14 = 6 x2 + 8x + 6 ) ( )....(multiplying both sides by 2) 10 x + 14 8x + 6 68  (6x 2 + 10 x + 14 ) − (10 x + 14 ) = (6x 2 + 8 x + 6 ) − (8 x + 6 )...(using dividendo) 10 x + 14 8x + 6 6x2 6x2 ∴ = 10 x + 14 8 x + 6 This equation is true for x = 0 \ x = 0 is a solution of the given equation. 1 1 ¹ 0 then x2 ¹ 0, \ dividing by 6x2, 10 x + 14 = 8 x + 6 If x \ 8 x + 6 = 10 x + 14 \ 6 − 14 = 10 x − 8 x \ −8 = 2 x \ x = −4 \ x = −4 or x = 0 are the solutions of the given equation. Ex (2) Solve. x+7 + x−2 5 = x+7 − x−2 1 ( x + 7 + x − 2) + ( x + 7 − x − 2) 5 +1 Solution : =...(using componendo-dividendo) ( x + 7 + x − 2 ) − ( x + 7 − x − 2 ) 5 −1 \ 2 x+7 = 6 2 x−2 4 \ x+7 3 = x−2 2 \ x+7 9 =...(squaring both sides of the equation) x−2 4 \ 4 x + 28 = 9 x − 18 \ 28 + 18 = 9 x − 4 x \ 46 = 5 x \ 46 =x 5 46 \ x= 5 is the solution of the given equation. 69  Activity : Take 5 pieces of card paper. Write the following statements, one on each paper. a+b c+d a b a ac c c-a a rc (i) b = d (ii) = (iii) = (iv) = (v) = rd c d b bd d d -b b a c a, b, c, d are positive numbers and = is given. Which of the above statements are true b d or false, write at the back of each card, if false explain why. Practice set 4.3 a 7 (1) If = then find the values of the following ratios. b 3 5a + 3b 2a 2 + 3b 2 a 3 - b3 7 a + 9b (i) 5a − 3b (ii) 2a 2 − 3b 2 (iii) b3 (iv) 7 a − 9b 15a 2 + 4b 2 47 (2) If = then find the values of the following ratios. 15a 2 − 4b 2 7 a 7 a − 3b b 2 − 2a 2 b 3 − 2a 3 (i) b (ii) 7a + 3b (iii) 2 (iv) b + 2a 2 b 3 + 2a 3 3a + 7b 4 3a 2 − 7b 2 (3) If = then find the value of the ratio 3a 2 + 7b 2. 3a − 7b 3 (4) Solve the following equations. x 2 + 12 x − 20 x 2 + 8 x + 12 10 x 2 + 15 x + 63 2 x + 3 (i) = (ii) = 3x − 5 2x + 3 5 x 2 − 25 x + 12 x −5 (2 x + 1) 2 + (2 x − 1) 2 17 4x +1 + x + 3 4 (iii) = (iv*) = (2 x + 1) 2 − (2 x − 1) 2 8 4x +1 − x + 3 1 (4 x + 1) 2 + (2 x + 3) 2 61 (3 x − 4)3 − ( x + 1)3 61 (v) = (vi) = 4 x 2 + 12 x + 9 36 (3 x − 4)3 + ( x + 1)3 189 Activity : In the following activity, the values of a and b can be changed. That is by changing a : b we can create many examples. Teachers should give lot of practice to the students and encourage them to construct their own examples. 5a 2 + 2b 2 3a 2a − b = ----- = ----- 2a + b = ----- 5a 2 − 2b 2 4b a 2 + b2 a 3 a b2 = --- = = ----- b 4 2b 70  Let’s learn. Theorem on equal ratios a a+c c If ba = dc then = = This property is called the theorem of equal ratios. b b+d d Prrof : Let ba = dc = k. \ a = bk and c = dk a + c bk + dk k ( b + d ) \ b+d = b+d = b+d =k \ a = c = a + c b d b+d We know that, ba = bl al al cm al + cm \ If ba = dc = k , then = = =k bl dm bl + dm a c e If = = =........ (finite terms) and if l, m, n are non-zero numbers b d f al + cm + en +... then each ratio = (finite terms) is the general form of the above bl + dm + fn +... theorem. Use your brain power ! In a certain gymnasium, there are 35 girls and 42 boys in the kid's section, 30 girls and 36 boys in the children's section and 20 girls and 24 boys in the teens' section. What is the ratio of the number of boys to the number of girls in every section ? For physical exercises, all three groups gathered on the ground. Now what is the ratio of number of boys to the number of girls ? From the answers of the above questions, did you verify the theorem of equal ratios ? Ex (1) Fill in the blanks in the following statements. a b 4a + 9b x y z 5x − 3 y + 4 z (i) = = (ii) 3 = 5 = 4 =............... 3 7.......... a b 4a + 9b 4a + 9b 4a + 9b Solution : (i) = = = = 3 7 4 × 3 + 9 × 7 12 + 63 75 (ii) x y z 5 × x −3 × y 4 × z = = = = = 3 5 4 5 × 3 −3 × 5 4 × 4 5 x −3 y 4 z ∴ = = = 15 −15 16 5x − 3 y + 4 z = ----(by the theorem of equal ratio) 15 − 15 + 16 5x − 3 y + 4 z = 16 71  a b c Ex (2) If ( x − 2 y + 3z ) = ( y − 2 z + 3x) = ( z − 2 x + 3 y ) and x + y + z ¹ 0 a+b+c then prove that each ratio = 2( x + y + z ) a b c Solution : Let ( x − 2 y + 3z ) = ( y − 2 z + 3x) = ( z − 2 x + 3 y ) = k. \ by theorem of equal ratios a+b+c k= ( x − 2 y + 3z ) + ( y − 2 z + 3x) + ( z − 2 x + 3 y ) a+b+c = 2x + 2 y + 2z a+b+c = 2( x + y + z ) a b c a+b+c ∴ = = = x − 2 y + 3z y − 2 z + 3x z − 2x + 3y 2( x + y + z ) y z x a b c Ex (3) If = = then prove that z + x = x + y = y + z. b+c −a c + a −b a +b−c Solution : By invertendo, we get b+c −a c + a −b a +b−c = = y z x b+c −a c + a −b a +b−c Now let = = = k. y z x \ by theorem of equal ratios (c + a − b ) + ( a + b − c ) (a + b − c) + (b + c− a) (b + c − a ) + (c + a − b) k= k= k= z+x x+ y y+z 2a 2b 2c =.....(I) = = z+x x+ y.......(II) y+z.....(III) 2a 2b 2c \ = = z+x x+ y y+z a b c \ = = z+x x+ y y+z 14 x 2 − 6 x + 8 7 x − 3 Ex (4) Solve : = 10 x 2 + 4 x + 7 5 x + 2 Solution : By observation, we see that multiplying by 2x the predecessor and the successor of right hand side, we get two terms of the predecessor and the successor of the left hand side. But before multiplying, we must ensure that x ¹ 0. 72  14 x 2 − 6 x + 8 8 7 x − 3 −3 If x = 0 then 10 x 2 + 4 x + 7 = 7 and = 5x + 2 2 8 −3 \ 7 = 2 Which is a contradiction. \ x¹0 \ multiplying predecessor and successor of RHS by 2x. 14 x 2 − 6 x + 8 2 x(7 x − 3) = =k 10 x 2 + 4 x + 7 2 x(5 x + 2) \ 14 x2 − 6 x + 8 = 14 x2 − 6 x = k 2 2 10 x + 4kx + 7 10 x + 4 x 14 x 2 − 6 x + 8 − 14 x 2 + 6 x 8 \ = =k 10 x 2 + 4 x + 7 − 10 x 2 − 4 x 7 8 \ k= 7 \ 7x − 3 8 = 5x + 2 7 \ 49 x − 21 = 40 x + 16 \ 49 x − 40 x = 16 + 21 37 \ 9 x = 37 \ x= 9 Practice set 4.4 (1) Fill in the blanks of the following x y 3x + 5 y 7 x − 9 y a b c a − 2b + 3c...... (i) = = 7 3....... =...... (ii) 3 = 4 = 7 =...... = 6 − 8 + 14 (2) 5 m -n =3m +4n then find the values of the following expressions. m2 + n2 3m + 4n (i) m 2 − n 2 (ii) 3m − 4n (3) (i) If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal y−z z−x x− y then show that, = =. a (b − c) b(c − a ) c(a − b) x y z (ii) If = = and x+y +z ¹ 0 then show that the value of 3x − y − z 3 y − z − x 3z − x − y each ratio is equal to 1. 73  a+b (iii) If ax + by = bx + az = ay + bz and x+y +z ¹ 0 then show that. x+ y x+z y+z 2 y+z z+x x+ y x y z (iv) If = = then show that = =. a b c b+c −a c + a −b a +b−c 3x − 5 y x + 5 z y − z x (v) If = = then show that every ratio = y. 5z + 3 y y − 5x x − z 16 x 2 − 20 x + 9 4 x − 5 5 y 2 + 40 y − 12 y + 8 (4) Solve. (i) = (ii) = 8 x 2 + 12 x + 21 2 x + 3 5 y + 10 y 2 − 4 1 + 2 y Let’s learn. Continued Proportion Let us consider the ratios 4 : 12 and 12 : 36. They are equal ratios. In the two ratios, the successor (second term) of the first ratio is equal to the predecessor (first term) of the second ratio. Hence 4, 12, 36 are said to be in continued proportion. If a = b then a, b, c are in continued proportion. b c a b If ac = b2, then dividing both sides by bc we get =. b c \ if ac = b2 , then a, b, c are in continued proportion. When a, b, c are in continued proportion then b is known as Geometric mean of a and c or Mean proportional of a and c. Hence all the following statements convey the same meaning. a b \ (1) = (2) b2= a c (3) a, b, c are in continued proportion. b c (4) b is the geometric mean of a and c. (5) b is the mean proportional of a and c. We can generalise the concept of continued proportion. a b c d e If = = = = then a, b, c, d, e and f are said to be in continued proportion. b c d e f Ex (1) If x is the geometric mean of 25 and 4, then find the value of x. Solution : x is the geometric mean of 25 and 4. \ x2 = 25 ´ 4 \ x2 = 100 \ x = 10 74 Ex (2) If 4 a2b, 8 ab2, p are in continued proportion then find the value of p. Solution : From given information, 4 a2b, 8 ab2, p are in continued proportion. ∴ 4a 2b 8ab 2 = 8ab 2 p 8ab 2 × 8ab 2 p = = 16b3 4a b2 Ex (3) Which number should be subtracted from 7, 12 and 18 such that the resultant numbers are in continued proportion? Solution : Let x be subtracted from 7, 12 and 18 such that resultant numbers are in contin- ued proportion. (7-x), (12-x), (18 - x) are in continued proportion. Tally \ (12-x)2 = (7-x) (18 - x) (7-x) = 7-(-18) = 25 \ 144-24 x + x = 126 - 25x + x 2 2 (12-x) = 12 - (-18) = 30 \ -24 x +25x = 126 - 144 (18 - x) = 18 - (-18) = 36 \ x = -18 302 = 900 and 25 ´ 36 = 900 25, 30, 36 are in continued proportion \ If -18 is subtracted from 7, 12, 18 the resultant numbers are in continued proportion. k - method The k-method is used to solve examples based on equal ratios, i.e. equal proportions. In this simple method every equal ratio is assumed to be equal to k. a c 5a − 3c 7 a − 2c Ex (1) If = d then show that = b 5b − 3d 7b − 2d a c Solution : Let = d = k \ a =bk, c = dk b Substituting values of a and c in both sides, 5a − 3c 5(bk ) − 3(dk ) k (5b − 3d ) LHS = = = =k 5b − 3d 5b − 3d (5b − 3d ) RHS = 7a − 2c = 7(bk ) − 2(dk ) = k (7b − 2d ) = k 7b − 2 d 7b − 2d 7b − 2d \ LHS = RHS. 5a − 3c 7 a − 2c \ = 5b − 3d 7b − 2d 75  (a + b) 2 (b + c) 2 Ex (2) If a, b, c are in continued proportion then show that, ab = bc. a b Solution : a, b, c are in continued proportion. Let = = k. b c \ b =ck, a = bk = ck ´ k = ck2 Substituting values of a and b. (a + b) 2 (ck 2 + ck ) 2 c 2 k 2 (k + 1) 2 (k + 1) 2 LHS = ab = (ck 2 )(ck ) = c2k 3 = k RHS = (b + c) = (ck + c) = c (k2+ 1) = (k + 1) 2 2 2 2 2 bc (ck )c ck k (a + b) 2 (b + c) 2 \ LHS = RHS. \ ab = bc Ex (3) If a, b, c are in continued proportion Ex (4) Five numbers are in continued proportion. The first term is 5 and the last a a 2 + ab + b 2 then show that = 2 term is 80. Find these numbers. c b + bc + c 2 Solution : Let the numbers in continued Solution : a, b, c are in continued proportion. proportion be a, ak, ak2, ak3, ak4. a b \b= Here a = 5 and ak4 = 80 c a b \ 5 ´ k4 = 80 Let, = =k \b = ck and a =ck2 b c \ k4 =16 a ck 2 2 LHS = = = k c c \k =2 24 = 16 \ a 2 + ab + b 2 ak = 5 ´ 2 = 10 ak2 = 5 ´ 4 = 20 RHS = b 2 + bc + c 2 (k 2c)2 + k 2c(ck ) + (ck )2 ak3 = 5 ´ 8 = 40 ak4 = 5 ´ 16 = 80 = (ck )2 + (ck )(c) + c 2 \ the numbers are 5, 10, 20, 40, 80. k c +k c +c k 4 2 3 2 2 2 = c2k 2 + c2k + c2 c 2 k 2 (k 2 + k + 1) = 2 2 c (k + k + 1) = k2 \ LHS = RHS \ a = a 2 + ab + b2 2 2 c b + bc + c 76  Practice set 4.5 (1) Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion? (2) If (28-x) is the mean proportional of (23-x) and (19-x) then find the value of x. (3) Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers. (4) If (a + b + c) (a - b + c) = a2 + b2 + c2 show that a, b, c are in continued proportion. a b (5) If = and a, b, c > 0 then show that, b c (i) (a + b + c) (b - c) = ab - c2 (ii) (a2 + b2) (b2 + c2)= (ab + bc)2 a 2 + b2 a + c (iii) = ab b x + y x2 − y 2 (6) Find mean proportional of , x− y x2 y 2 Activity : Observe the political map of India from a Geography textbook. Study the scale of this map. From the given scale find the straight line distances between various cities like (i) New Delhi to Bengaluru (ii) Mumbai to Kolkata, (iii) Jaipur to Bhubaneshwar. Problem set 4 (1) Select the appropriate alternative answer for the following questions. (i) If 6 : 5 = y : 20 then what will be the value of y ? (A) 15 (B) 24 (C) 18 (D) 22.5 (ii) What is the ratio of 1 mm to 1 cm ? (A) 1 : 100 (B) 10 : 1 (C) 1 : 10 (D) 100 : 1 (iii*) The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age ? (A) 3 : 2 (B) 2 : 3 (C) 4 : 3 (D) 3 : 4 77  (iv) 24 Bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get ? (A) 8 (B) 15 (C) 12 (D) 9 (v) What is the mean proportional of 4 and 25 ? (A) 6 (B) 8 (C) 10 (D) 12 (2) For the following numbers write the ratio of first number to second number in the reduced form. (i) 21, 48 (ii) 36, 90 (iii) 65, 117 (iv) 138, 161 (v) 114, 133 (3) Write the following ratios in the reduced form. (i) Radius to the diameter of a circle. (ii) The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm. (iii) The ratio of perimeter to area of a square, having side 4 cm. (4) Check whether the following numbers are in continued proportion. (i) 2, 4, 8 (ii) 1, 2, 3 (iii) 9, 12, 16 (iv) 3, 5, 8 a, b, c are in continued proportion. If a = 3 and c = 27 then find b. (5) (6) Convert the following ratios into percentages.. 5 22 5 144 (i) 37 : 500 (ii) (iii) 30 (iv) (v) 8 16 1200 (7) Write the ratio of first quantity to second quantity in the reduced form. (i) 1024 MB, 1.2 GB [(1024 MB = 1 GB)] (ii) 17 Rupees, 25 Rupees 60 paise (iii) 5 dozen, 120 units (iv) 4 sq.m, 800 sq.cm (v) 1.5 kg, 2500 gm (8) If a = 2 then find the values of the following expressions. b 3 4a + 3b 5a 2 + 2b 2 (i) (ii) 2 3b 5a − 2b 2 a 3 + b3 7b − 4a (iii) (iv) 7b + 4a b3 (9) If a, b, c, d are in proportion, then prove that 11a 2 + 9ac a 2 + 3ac (i) = 11b 2 + 9bd b 2 + 3bd a 2 + 5c 2 a (ii*) = b b2 + 5d2 a 2 + ab + b 2 c 2 + cd + d 2 (iii) 2 = a − ab + b 2 c 2 − cd + d 2 78 (10) If a, b, c are in continued proportion, then prove that a a − 2b b a −b (i) = (ii) = a + 2b a − 4c b+c a−c (11) Solve : 12 x 2 + 18 x + 42 = 2 x + 3 2 18 x + 12 x + 58 3 x + 2 2x − 3y z − y x + 3z x (12) If = = then prove that every ratio = y. 3z + y z − x 2 y − 3x by + cz cz + ax ax + by x y z (13*) If = = then prove that a b =c. = b2 + c2 c2 + a 2 a 2 + b2 79  5 Linear Equations in Two Variables Let’s study. Introduction Linear equation in two variables Solving simultaneous equations Simultaneous equations Word problems based on simultaneous equations Let’s recall. Ex. Solve the following equations. x 4 (1) m+3=5 (2) 3y + 8 = 22 (3) =2 ( 4) 2p = p + 3 9 m= y= x= p= (5) Which number should be added to 5 (6) Which number should be subtracted to obtain 14 ? from 8 to obtain 2 ? + 5 = 14 8- =2 x + 5 = 14 8-y=2 x= y= In all above equations, degree of the variable is 1. These are called as Linear equations. Let’s learn. Linear equations in two variables Find two numbers whose sum is 14. Using variables x and y for the two numbers, we can form the equation x + y = 14. This is an equation in two variables. We can find many values of x and y satisfying the condition. e.g. 9 + 5 = 14 7 + 7 = 14 8 + 6 =14 4 + 10 = 14 (-1) + 15= 14 15 + (-1) = 14 2.6 + 11.4 =14 0 + 14 = 14 100 + (-86) = 14 (-100) + (114) = 14 + = 14 + = 14 Hence, above equation has many solutions like (x = 9, y = 5); (x = 7, y = 7); (x = 8, y = 6) etc. 80 Conventionally, the solution x = 9, y = 5 is written as an ordered pair (9, 5) where 9 is the value of x and 5 is the value of y. To satisfy the equation x + y = 14 , we can get infinite or- dered pairs like (9,5), (7,7), (8,6), (4,10), (10,4), (-1,15), (2.6, 11.4), … etc. All of these are the solutions of x + y = 14. Consider second example. Find two numbers such that their difference is 2. Let the greater number be x and the smaller number be y. Then we get the equation x - y = 2 For the values of x and y, we can get following equations. 10 - 8 = 2 9-7=2 8-6=2 (-3) - (-5) = 2 5.3-3.3=2 15 - 13 = 2 100 - 98 = 2 - = 2 - = 2 Here if we take values x = 10 and y = 8, then the ordered pair (10, 8) satisfies the above equation. Here we cannot write as (8, 10) because (8, 10) will imply x = 8 and y = 10 and it does not satisfy the equation x - y = 2. Therefore, note that, the order of numbers in the pair indicating solution is very important. Now let us write the solutions of x- y = 2 in the form of ordered pairs. (7, 5), (-2, -4), (0, -2), (5.2, 3.2), (8, 6) etc. There are infinite solutions. Find the solution of 4m - 3n = 2. Construct 3 different equations and find their solutions. Now, observe the first two equations. x + y = 14........ I x - y = 2........ II Solution of equation I : (9, 5), (7, 7), (8, 6)... Solutions of Equation II : (7, 5), (-2, -4), (0, -2), (5.2, 3.2), (8, 6)... (8, 6) is the only common solution of both the equations. This solution satisfies both the equa- tions. Hence it is the unique common solution of both the equations. Remember this ! When we consider at the same time two linear equations in two variables those equations are called Simultaneous equations. 81 Activity : On the glasses of following spectacles, write numbers such that 29 13 (i) Their sum is 42 and difference is 16 (ii) Their sum is 37 and difference is 11 (iii) Their sum is 54 and difference is 20 (iv) Their sum is... and difference is... Let’s recall. x+y = 5 and 2x + 2y = 10 are two equations in two variables. Find five different solu- tions of x+y = 5, verify whether same solutions satisfy the equation 2x + 2y = 10 also. Observe both equations. Find the condition where two equations in two variables have all solutions in common. Let’s learn. Elimination method of solving simultaneous equations By taking different values of variables we have solved the equations x + y = 14 and x - y = 2. But every time, it is not easy to solve by this method, e.g. , 2x + 3y = -4 and x - 5y = 11. Try to solve these equations by taking different values of x and y. By this method observe that it is not easy to obtain the solution. Therefore to solve simultaneous equations we use different method. In this method, we eliminate one of the variables to obtain equations in one variable. We can solve and find the value of one of the two variables and then substituting this value in one of the given equations we can find the value of the other variable. Study the following example to understand this method. 82 Ex (1) Solve x + y = 14 and x - y = 2. Solution : By adding both the equations we get an equation in one variable x+y = 14.........I + x-y = 2.........II 2x + 0 = 16 Substituting x =8 in the equation (I) 2x = 16 x + y = 14 x = 8 \ 8 + y =14 \ y =6 Here (8, 6) is the solution of first equation. Let us check, whether it satisfies the second equation also. x - y = 8 - 6 =2 is true. \ (8,6) is the solution for both the equations. Hence (8, 6) is the solution of simultaneous equations x + y = 14 and x - y = 2. Ex (2) Sum of the ages of mother and son is 45 years. If son's age is subtracted from twice of mother's age then we get answer 54. Find the ages of mother and son. It becomes easy to solve a problem if we make use of variables. Solution : Let the mother's today's age be x years and son's today's age be y years. From the first condition x+y =45.........I From the second condition 2x-y = 54.........II Adding equations (I) and (II) 3x+0 = 99 3x = 99 x = 33 Substituting x = 33 in equation (I), 33+y = 45 y = 45-33 y = 12 Verify that x=33 and y = 12 is the solution of second equation. Today's age of mother = 33 and today's age of son = 12. 83  General form of linear equation in two variables The general form of a linear equation in two variables is ax + by + c= 0 where a, b, c are real numbers and a and b are non-zero at the same time. In this equation the index of both the variables is 1. Hence it is a linear equation. Ex (1) Solve the following equations Ex (2) Solve the following simultaneous 3x + y = 5........... (I) equations. 2x + 3y = 1........(II) 3x – 4y – 15 = 0........... (I) Solution : To eliminate one of the variables, y +x + 2 = 0........(II) we observe that in both equations, not a sin- Solution : Let us write the equations by gle coefficient is equal or opposite number. shifting constant terms to RHS Hence we will make one of them equal. 3x – 4 y = 15........(I) Multiply both sides of the equation (I) by 3. x+ y = -2............. (II) \ 3x ´ 3 + 3 ´ y= 5 ´ 3 To eliminate y, multiply second equation by \ 9x + 3y = 15.......(III) 4 and add to equation (I). 2x + 3y = 1.......(II) 3x – 4 y = 15 Now subtracting eqn (II) from eqn (III) + 4x + 4y = -8 9x + 3y = 15 7x = 7 + 2x + 3y = 1 x = 1 - - - Substituting x = 1 in the equation (II). 7x = 14 x+y = -2 x = 2 \1 + y = -2 Substituting x = 2 in one of the equations. 2x + 3 y = 1 \ y = -2 -1 \ 2 ´ 2 + 3y = 1 \ y = -3 \ 4 + 3y = 1 (1, -3) is the solution of the above equations. \ 3y = -3 Verify that it satisfies equation (I) also. \ y = -1 Verify that (2, -1) satisfies the second equation. Use your brain power! 3x - 4y - 15 = 0 and y + x + 2 = 0. Can these equations be solved by eliminating x ? Is the solution same? 84  Let’s learn. Substitution method of solving simultaneous equations There is one more method to eliminate a variable. We can express one variable in terms of other from one of the equations. Then substituting it in the other equation we can eliminate the variable. Let us discuss this method from following examples. Ex (1) Solve 8x + 3y = 11 ; 3x – y = 2 Ex (2) Solve. 3x – 4 y= 16 ; 2x – 3y = 10 Solution : 3x-4y=16..........(I) Solution : 8x + 3y = 11.................. (I) 2x – 3y = 10.............(II) 3x – y = 2.......................(II) Writing x in terms of y from equation (I). In Equation (II), it is easy to 3x – 4 y = 16 express y in terms of x. 3x = 16 + 4y 3x – y = 2 16 + 4 y x= 3 3x – 2 = y Substituting this value of x in equation (II) Substituting y = 3x -2 in equation (I). 2 x − 3 y = 10 8x + 3y = 11  16 + 4 y  2  − 3 y = 10 \ 8x + 3(3x-2) = 11  3  32 + 8 y \ 8x + 9x -6 = 11 3 − 3 y = 10 \ 17x – 6 = 11 32 + 8 y − 9 y = 10 3 \ 17x = 11 + 6 = 17 32 + 8y – 9y =30 \x=1 32 – y = 30 \y=2 Now, substituting this value of x in the Now, substituting y = 2 in equation (I) equation y = 3x – 2. 3x -4y =16 y=3´1–2 \ 3x -4´2 = 16 \y=1 \ 3x -8 = 16 \ 3x = 16 + 8 \ (1, 1) is the solution of \ 3x = 24 the given equations \ x = 8 \ x = 8 and y = 2 \ (8, 2) is the solution of the given equations. 85  Practice set 5.1 (1) By using variables x and y form any five linear equations in two variables. (2) Write five solutions of the equation x + y = 7. (3) Solve the following sets of simultaneous equations. (i) x + y = 4 ; 2x – 5y = 1 (ii) 2x + y = 5; 3x-y = 5 (iii) 3x - 5y=16; x-3y=8 (iv) 2y - x=0; 10x + 15y = 105 (v) 2x + 3y+4 = 0; x- 5y = 11 (vi) 2x - 7y = 7; 3x + y = 22 Let’s learn. Word problems based on simultaneous equations While solving word problems, converting the given information into mathematical form is an important step in this process. In the following flow-chart, the procedure for finding solutions of word problems is given. Steps Example Read the given word problem Sum of two numbers is 36. If 9 is carefully and try to subtracted from 8 times the first number, understand it. we get second number. Find the numbers. From given information, Let first number = x use proper variables for given quantities and second number = y Sum of the numbers 36 \x + y = 36 Form the mathematical state- 8 times first number = 8x ments from above variables 8 times first number – 9 = 8x – 9 \ Second number = y = 8x - 9 x+y = 36 x + y = 36 Use suitable method + 8x – y =9 \ 5 + y = 36 to solve the equations \ 9x = 45 \ y = 36 -5 \ x =5 \ y = 31 Find the solution x = 5, y = 31 Substituting the values of the 31 + 5 = 36..........(I) variables in both the equations, verify your result. 31 = 8 ´ 5 – 9..............(II) Write the answer. \ the numbers are 5 and 31. 86  Word Problems Now we will see various types of word problems. (1) Problems regarding age. (2) Problems regarding numbers. (3) Problems based on fractions. (4) Problems based on money transactions. (5) Problems based on geometrical properties (6) Problems based on speed, distance, time. Ex (1) Sum of two numbers is 103. If greater number is divided by smaller number then the quotient is 2 and the remainder is 19. Then find the numbers. Solution : Step 1 : To understand the given problem. Step 2 : Use proper variables for given quantities. Also note the rule dividend = divisor ´ quotient + remainder. Let the greater number be x and the smaller number be y Step 3 : Given information : Sum of the numbers = 103 x + y = 103 is the first equation. By dividing greater number by smaller numbers quotient is 2 and remainder is 19. x = 2 ´ y + 19...(dividend = divisor ´ quotient + remainder) x – 2y = 19 is the second equation. Step 4 : Let us find the solution of the equations. x + y = 103................(I) x – 2y = 19................(II) Subtracting eqn. (II) from eqn. (I) x + y = 103 x – 2y = 19 – + - 0 + 3y = 84 \ y = 28 Step 5 : Substituting value of y in equation x + y = 103. \ x + 28 = 103 \ x = 103 – 28 \ x = 75 Step 6 : Given numbers are 75 and 28. 87 Ex (2) Salil's age is 23 years more than half of the Sangram's age. Five years ago, the sum of their ages was 55 years. Find their present ages. Solution : Let Salil's present age be x and Sangram's present age be y. y Salil's age is 23 years more than half of the Sangram's age \ x = + 2 Five years ago Salil's age = x – 5. Five years ago Sangram's age = y - 5 The sum of their ages five years ago = 55 + = 55 Finding the solution by solving equations 2x = y + 46 2x – y = 46.............(I) (x - 5) + (y-5) = 55 x + y = 65............(II) Adding equation (I) and (II) Substituting x = 37 in equation (II) 2x – y = 46 x+y = 65 + x + y = 65 \ 37 + y = 65 \ 3x = 111 \ y = 65 -37 \ x = 37 \ y = 28 Salil's present age is 37 years and Sangram's present age is 28 years. Ex (3) A two digit number is 4 times the sum of its digits. If we interchange the digits, the number obtained is 9 less than 4 times the original number. Then find the number. Solution : Let the units place digit in original number be x, and tens place be y. Digit in Digit in Number Sum of tens place units place the digits For original number y x 10y + x y+x Number obtained by x+y interchanging the digits x y 10x + y From first condition, 10y + x = 4 (y + x) \10y + x = 4y + 4x \ x – 4x + 10y – 4y = 0 \ -3x + 6y = 0 \ -3x = -6y \x = 2y.....(I) 88  From second condition, 10x + y = 2(10y+x)-9 10x+y = 20y + 2x-9 10x-2x+y-20y = -9 8x – 19y = -9.............(II) x = 2y..............(I) Substituting x = 2y in equation (II). 16y – 19y = -9...............(I) \ – 3y = -9 \ y = 3 Substituting y = 3 in equation (I). x–2y = 0 x -2 x 3 = 0 \ x–6=0 \x=6 Original two digit number : 10y + x = 10 ´ 3 + 6 = 36 Ex (4) The population of a certain town was 50,000. In a year, male population was increased by 5% and female population was increased by 3%. Now the population became 52020. Then what was the number of males and females in the previous year? Solution : Let the number of males in previous year be x, number of females be y. By first condition + = 50000.......(I) Male population increased by 5% \ number of males = x Female population increased by 3% \ number of females = y. From second condition x+ y = 52020 x+ y = 5202000.......(II) Multiplying equation (I) by 103 x+ y = 5150000.......(III) Subtracting equation (III) from equation (II). 2x = 5202000 - 5150000 2x = 52000 \ number of males = x = \ number of females = y = 89  Activity I : There are instructions written near the arrows in the following diagram. From this information form suitable equations and write in the boxes indicated by arrows. Select any two equations from these boxes and find their solutions. Also verify the solutions. By taking one pair of equations at a time, how many pairs can be formed ? Discuss the solutions for these pairs. Sum of my length If the breadth is and breadth is 36. subtracted from twice the length, the answer is 27. I am a rectangle. My length is x units and My breadth is breadth is y units. 5 times of the The difference 7 length. between my length and breadth is 6 units. Practice set 5.2 (1) In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of these notes together is 350 rupees. Number of 5 rupee notes are less by 10 than number of 10 rupee notes. Then find the number of 5 rupee and 10 rupee notes. (2) The denominator of a fraction is 1 less than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is 3 : 5. Find the fraction. (3) The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their today's ages. (4) The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is140. Then find the number of lions and peacocks in the zoo. (5) Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was Rs. 4500 and after 10 years his monthly salary became 5400 rupees, then find his original salary and yearly increment. (6) The price of 3 chairs and 2 tables is 4500 rupees and price of 5 chairs and 3 tables is 7000 rupees, then find the total price of 2 chairs and 2 tables. 90 (7) The sum of the digits in a two-digits number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number. (8*) InD ABC, the measure of angle A is equal to the sum of the measures of Ð B and Ð C. Also the ratio of measures of Ð B and Ð C is 4 : 5. Then find the measures of angles of the triangle. (9*) Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part 1 is equal to 3 of the larger part. Then find the length of the larger part. (10) In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong ? Problem set 5 (1) Choose the correct alternative answers for the following questions. (i) If 3x + 5y = 9 and 5x + 3y = 7 then What is the value of x + y ? (A) 2 (B) 16 (C) 9 (D) 7 (ii) 'When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26.' What is the mathematical form of the statement ? (A) x - y = 8 (B) x + y = 8 (C) x + y = 23 (D) 2x + y = 21 (iii) Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay's age ? (A) 20 (B) 15 (C) 10 (D) 5 (2) Solve the following simultaneous equations. (i) 2x + y = 5 ; 3x - y = 5 (ii) x - 2y = -1 ; 2x- y = 7 (iii) x + y = 11 ; 2x - 3y = 7 (iv) 2x + y = -2 ; 3x - y = 7 (v) 2x - y = 5 ; 3x + 2y = 11 (vi) x - 2y = -2 ; x + 2y = 10 (3) By equating coefficients of variables, solve the following equations. (i) 3x - 4y=7; 5x + 2y=3 (ii) 5x + 7y=17 ; 3x - 2y=4 (iii) x - 2y= -10; 3x - 5y= -12 (iv) 4x + y=34 ; x + 4y= 16 (4) Solve the following simultaneous equations. x y x y x y (i) + =4; − = 1 (ii) + 5 y = 13 ; 2 x + = 19 3 4 2 4 3 2 2 3 5 4 (iii) + = 13 ; − = −2 x y x y 91 (5*) A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number. (6) The total cost of 8 books and 5 pens is 420 rupees and the total cost of 5 books and 8 pens is 321 rupees. Find the cost of 1 book and 2 pens. (7*) The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves rupees 200, find the income of each. (8*) If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 8 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle. (9*) The distance between two places A and B on road is 70 kilometers. A car starts from A and the other from B. If they travel in the same direction, they will meet after 7 hours. If they travel towards each other they will meet after 1 hour, then find their speeds. (10*) The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number. Activity : Find the fraction. Numerator x Denominator y If numerator is multiplied by 3 and If numerator is increased by 8 and 3 is subtracted from the denomina- denominator is doubled then the 18 1 tor then the fraction obtained is. resulting fraction is. 11 2 Equation I Equation II 11x - 6y + 18 = 0 x-y+8=0 \ Given fraction = Verify the answer obtained. 92  6 Financial Planning Let’s study. Introduction to financial planning Savings and Investments Tax structure Computation of Income -tax Let’s discuss. Anagha : Shall we buy computer ? Mother : Ok, let's buy it next year. Anagha : Mamma, why not now ? Mother : Anagha, you don't know how expensive it is ! Anagha : You mean we will have to save up for it, right ? Mother : Yes, thats the thing. We often hear such conversations. Everyone requires money to meet a variety of needs. That is why after spending on the necessities of the present, everyone tries to save money to make provisions for the future needs. That is what we call 'Saving' money. In order to protect our savings or even to make them grow, we keep them as fixed deposits or buy immovable properties such as a house, land etc. That is what we call 'investment'. Every investor, first spends the amount required to meet primary necessities and saves the remaining amount. One also uses these savings to make a carefully considered investments. This is called financial planning. The main purpose of financial planning is protection and growth of the wealth. Financial planning is useful for making provisions for the predictable and unpredictable expenses that each of us has to meet in our life. Predictable expenses Unpredictable expenses (1) Education of children and other expenses (1) Natural disasters for them (2) Capital for a profession or business (2) Medical expenses for a family member. (3) Buying a vehicle (3) Loss due to an accident (4) Buying or building a house. (4) Sudden death (5) Old age requirements. The above considerations make it quite clear why financial planning is a must. However some important points must be kept in mind as we plan our finances. 93  Let’s learn. Savings (1) It is for our own benefit to keep our savings safe and to make them grow. Our savings remain safe in a bank or in a post office. Money saved in a bank is also useful for cashminus transactions. This way, we do not have to carry large amounts of cash or worry about losing it or getting it stolen. (2) If the money we get or earn is in the form of cash and we keep it as it is, without investing it, its value diminishes with time. For example, if today your can buy two pencils for ten rupees, a few years hence, you may be able to buy only one for that amount. (3) If the amount invested is used for expanding a business, to start an industry or other such purposes, it contributes to the growth of the national production. (4) If some part of the income is spent for a socially useful cause everyone benefits from it in the long run. (5) After spending on necessities it is beneficial to limit spending on luxuries and to save, instead for education, medical treatment etc.. Let’s discuss. M u fu tua nd l N fun SS ds Shares Bo nd nce Insura s P acc igmi ou s nt Saving Recurr s en Postal Life itiz t deposit or ctmene Insurance (PLI) n i Se nveshem i sc deposit Fixed ing R.D. Pr iva s PPF Po Saving te st O k Nation accoun an t sec ffic a ry e B bank d lize Sala unt tor eposits c o ac Savings & Investments ‍ bserve the above picture, which shows some modes of investment. Discuss them. Find O out other modes of investment and write them in the blank spaces in the picture. 94  Let’s learn. Investments Investments are of many types. Investors often favour institutions like banks and postal departments for investing their money because it is safe there. There is a certain risk in investing money in shares, mutual funds, etc. That is because this money is invested in a business or industry and if that incurs a loss, the investor suffers the loss too. On the other hand, if it makes a profit the money is safe and there is the opportunity to get a dividend too. An investor must take two important points into account when making an investment namely the risk and the gain. It is possible to make big gains by taking greater risk. However it must be kept in mind that the greater risk can also lead to greater loss. Study the following examples based on income and investment. Ex(1) Shamrao's income in 2015-16 after paying all taxes is Rs. 6,40,000. He pays Rs. 2000 per month for insurance and 20% of his annual income into his provident fund. He puts aside Rs. 500 per month for emergencies. How much money does he have for yearly spending ? Solution : (i) Annual income = 6,40,000 rupees (ii) Insurance premium = 2000 ´ 12 = 24,000 rupees 20 (iii) Contribution to provident fund = 6,40,000 ´ = 1,28,000 rupees 100 (iv) Amount put aside for emergency = 500 ´ 12 = 6000 rupees \ Total planned expenditure = 24,000 + 1,28,000 + 6,000 = 1,58,000 rupees \ Amount available for yearly expenses = 6,40,000 - 1,58,000 = 4,82,000 rupees Ex(2) Mr. Shah invested Rs. 3,20,000 in a bank at 10% compound interest. He also invested Rs. 2,40,000 in mutual funds. At market rates he got Rs. 3,05,000 after 2 years. How much did he gain ? Which of his investments was more profitable ? Solution : (i) We shall first calculate the compound interest on the money invested in the bank. Compound interest = Amount - Principal That is, I = A - P n  r  = P 1 +  − P  100   r  n  = P  100  − 1 1 +     10  2   + = 3,20,000  100  1 − 1    95  = 3,20,000 (1.1) − 1 2 = 3,20,000 [1.21 − 1] = 3,20,000 ´ 0.21 = 67,200 rupees Mr. Shah invested Rs. 3,20,000 in the bank and got Rs. 67,200 as interest. Let us see percentage of interest obtained on the investment. 100 ´ 67200 Percentage of interest = = 21 3, 20, 000 \ The investment in the bank gave a profit of 21%. (ii) The amount Mr. Shah got at the end of 2 years from the mutual fund = 3,05,000 rupees \ The gain from the mutual fund = 3,05,000 - 2,40,000 = 65,000 rupees 65000 ×100 \ Percentage gain = = 27.08 2, 40, 000 The investment in the mutual fund yielded a profit of 27.08%. It is clear that Mr. Shah's investment in the mutual fund was more profitable. Ex(3) Mr. Shaikh invested Rs. 4,00,000 in a glass industry. After 2 years he received Rs. 5,20,000 from the industry. Putting aside the original investment, he invested his gains in a fixed deposit and in shares in the ratio 3 : 2. How much amount did he invested originally in each of the schemes ? Solution : Mr. Shaikh's profit at the end of 2 years = 5,20,000 - 4,00,000 = 1,20,000 rupees Amount invested in the fixed deposit = 3 ´ 1,20,000 5