Oasis School Mathematics-8 47 Example 2 If the cost of 2 5 of a piece of land is Rs. 5,00,000, what is the cost of 3 4 piece of the same land? Solution: The cost of 2/5 of a piece of land = Rs. 5,00,000 The cost of 1 piece of land = Rs. 500000 2 / 5 = Rs. 500000×5 2 = Rs. 1250000 The cost of 3 4 of a piece of land = Rs. 3 4 × 12,50,000 = Rs.9,37,500 Hence, the required cost of 3 4 of a piece of the same land is Rs. 9,37,500. Example 3 48 men are employed to construct a building in 75 days. They wanted the building to be ready in 60 days. How many more men should be employed to construct the building in 60 days? Solution: In 75 days, 48 men can construct a building Alternative method Fraction of land Cost 2 5 3 4 5,00, 000 x As the area of land and its cost are direct variations then, 3 4 2 5 = x 5,00,000 or, 3 4 × 5 2 = x 5,00,000 or, 15 8 = x 5,00,000 x = 15 × 5,00,000 5,00,000 = 9, 37, 500 Hence, the cost of 3 4 part of land is Rs. 9, 37, 500. More men are required to construct the building in 1 day. Less men are required to construct the building in 60 days.
48 Oasis School Mathematics-8 In 1 day, 48 × 75 men can construct the building In 60 days, 48 × 75 60 men can construct the building ∴ Men required to construct the building = 48 × 75 60 = 60. Number of additional men = (60 – 48)men = 12 men Example 4 A hostel has provisions for 45 students for 30 days. If 15 more students joined the hostel after 6 days, how long will the provision last? Solution: Here, Total number of students = 45 + 15 = 60 Remaining number of days = 30 – 6 = 24 For 45 students remaining provision lasts for 24 days. For 1 student remaining provision lasts for 24 × 45 days. 60 students have provision for 24 × 45 60 days = 6 × 3 days = 18 days. Hence, the remaining provision will last for 18 days. Example 5 20 men can do a piece of work in 12 days working 9 hours a day. In how many days 30 men can do the same work working 6 hours a day. (a) What are the variations used in the above question? (b) Prepare the table according to the above question. (c) Solve the above question, using variation. Alternative method Here, the total number of students = 45 + 15 = 60 Remaining number of days = 30 – 6 = 24 Let us suppose the remaining provision will last for x days. Now, Student Days 45 24 60 x As the number of students and days are indirect variations, x 24 = 45 60 or, x = 45 60 × 24 = 18 days. Alternative method Days Men 75 60 48 48+x Since, the men and days are indirect proportion, 48 + x 48 = 75 60 or, 48 + x 48 = 5 4 or, 4(48 + x) = 240 or, 192 + 4x = 240 or, 4x = 240 – 192 or, 4x = 48 ∴ x = 48 4 = 12
Oasis School Mathematics-8 49 Solution: Men Working hours Days 20 9 12 30 6 x (a) Variations used in the above questions are men, working hours and days. (b) Let's suppose 30 men can do a work in x days working 9 hours a day. (c) Menanddays are indirectvariations.Similarly,hours anddays are alsoindirectvariations, x 12 = 9 6 × 20 30 x = 9 × 20 × 12 6 × 30 = 12 ∴ Required number of days = 12 days. Exercise 5.1 1. (a) If the cost of 20 pens is Rs 640, what is the cost of 12 pens? (b) If the cost of 5m cloth is Rs. 1,200, what is the cost of 15 m cloth? (c) If a dozen oranges cost Rs. 120, find the cost of 4 oranges. (d) If the cost of 50 kg of sugar is Rs. 5,000, what is the cost of 2 quintals of sugar? 2. (a) Lochan earns Rs. 2,16,000 in 3 months! How much does he earn in 1 year? (b) A worker earns Rs. 54,000 in 6 months. How much does he earn in 9 months? (c) Cost of 18 kg rice is Rs. 900. How much rice can be bought with Rs. 500? (d) If the cost of 25 pens is Rs. 246, how many pens can be bought for Rs. 984? (e) A car consumes 8 litres of petrol in covering a distance of 120 km. How many kilometers will it go with 15 litres of petrol? 3. (a) If 1 4 part of a property is Rs. 48600, find the 2 5 part of it. (b) If the cost of 1 4 part of the land is Rs. 80,000, find the cost of 1 2 part of the land. (c) If 1 3 part of a money is Rs. 40,000, what is 1 4 part of it? 4. (a) If 4 men can do a piece of work in 30 days, in how many days could 12 men do the same work? (b) 12 workers can do a piece of work in 20 days. How many workers should be added to complete the work in 16 days?
50 Oasis School Mathematics-8 (c) 15 men can do a piece of work in 80 days. How long will it take to finish the work if 5 men are added? (d) 6 taps take 15 minutes to fill up a water tank. How many minutes will 10 taps take to fill up the same tank? 5. (a) A garrison of 1500 men has provisions for 60 days. How long would the food last if the garrison is reduced to 900 men ? (b) 150 students of a hostel have food enough for 45 days. 25 students leave the hostel after 10 days. How long will the remaining food last ? (c) A garrison of 1200 men has food for 60 days. How long does it last if 300 men were added after 10 days? 6. (a) 15 men reap 45 hectares of field in 6 days. How many hectares will 9 men reap in 4 days? (b) 40 men can do a piece of work in 72 days if they work 8 hours a day. How many men are required to finish the same work in 40 days working 9 hours a day? 7. 10 men can complete a work, working 5 hours per day in 8 days. In how many days 5 men can complete the same work working 8 hours per day? (a) What are variations in the above questions? (b) Prepare the table according to the above questions. (c) Using direct and indirect variation, solve the above question. 8. 20 men can dig a field in 18 days, working 8 hours a day. (a) Identify whether the variations men and days, working hours and days are dicrect or indirect variations? (b) In how many days, can 16 men dig the same field working 9 hours a day? Prepare the table. (c) Using direct and indirect variation solve the question of (b) Project Work 1. Collect 4 pairs of variation which are in direct proportion and 2 pairs of variation which are in indirect proportion. Present them with suitable example. Answer 1. (a) Rs. 384 (b) Rs. 3600 (c) Rs. 40 (d) Rs. 20,000 2. (a) Rs. 8,64,000 (b) Rs. 81,000 (c) 10 kg (d) 100 pens (e) 225 km 3. (a) Rs. 77760 (b) Rs. 1,60,000 (c) Rs. 30,000 4. (a) 10 days (b) 3 (c) 60 days (d) 9 mins. 5. (a) 100 days (b) 42 days (c) 40 days 6. (a) 18 hectares (b) 64 men (c) 28 days. 7. (a), (b) Consult your teacher (c) 10 days 8. (a), (b) Consult your teacher (c) 28 days.
Oasis School Mathematics-8 51 6.1 Flash back • If S.P > C.P there is profit. So, profit = S.P – C.P • If S.P< C.P, there is loss. So, loss = C.P – S.P • Overhead expenses like transportation charge, maintenance, repair and other expenditure are always added on cost price to get net price. Machine is bought for Rs. 50,000. If transportation charge of Rs. 6,000 is paid then net C.P is equal to Rs. 50,000 + Rs. 6000 = Rs. 56,000. • Profit percent = Profit C.P × 100% • Loss percent = Loss C.P × 100% A machine is bought for Rs. 5000 and sold it for Rs. 6000. Then Profit = S.P– C.P = Rs. 6000 – Rs. 5000 = Rs. 1000 Again, Profit percent = Profit C.P × 100% = 1000 5000 × 100% = 20% Profit = Profit% of C.P 6.2 Calculation of C.P. or S.P. if Profit and Loss are Given I. To find out S.P. when C.P. and profit or loss per cent are given As we know that, profit = P% of C.P. Then, S.P. = C.P. + Profit Profit = Profit percent of C.P. and Loss = Loss percent of C.P. Unit 6 Profit and Loss You know! • Cost price (C.P) and selling price (S.P) • Profit percent and loss percent
52 Oasis School Mathematics-8 Then, S.P. is calculated as follows S.P = C.P. + profit % of C.P.………… (i) or, S.P. = 100 + P% 100 × C.P. ………… (ii) When L% is given, S.P. = C.P. – Loss, S.P. = C.P. – L% of C.P ………… (iii) or, S.P. = 100 – L% 100 × C.P. ………… (iv) II. To find out C.P. when S.P. and profit or loss percent are given (i) When profit P% is given, we can use same relation to obtain C.P. i.e. S.P. = C.P. + P % of C.P. …………………….. (i) or, directly we can obtain C.P by using the relation, C.P = S.P. × 100 100 + P% ….. (ii) (ii) When loss % is given, we can use the relation S.P. = C.P. – L % of C.P. ……………………. (iii) or, directly we can obtain C.P. by using the relation C.P. = S.P. × 100 100 – L% Remember ! Loss = Loss % of C.P. S.P. = C.P.–L% of C.P. S.P. = 100–L% 100 ×C.P. C.P. = S.P×100 100–L% Profit = Profit % of C.P. S.P. = C P + P% of C.P. S. P. = 100+P% 100 ×C.P. C. P. = S.P.×100 100+P% .....…. (iv) Worked Out Examples Example 1 A man buys a motorbike for Rs. 1,50,000 and after some days he sells it for Rs. 1,30,000. What is his gain or loss percentage? Solution: Here, C.P. = Rs. 1,50,000 S.P. = Rs. 1, 30,000 Since, C.P. > S.P. Loss = C.P. – S.P.
Oasis School Mathematics-8 53 = 1,50,000 – 1,30,000 = 20,000 We have, loss per cent = Loss C.P. × 100% = 20,000 1,50,000 × 100% = 2 15 × 100% = 13 1 3 % Example 2 A man bought 100 apples at Rs. 1,800. 20 of them were rotten and he sold remaining at the rate of Rs. 25 per piece. Find his profit or loss percent. Solution: Here, C.P. of 100 apples = Rs. 1800 Number of rotten apples = 20 Remaining apples = 100 – 20 = 80 S.P. 80 apples = Rs.80 × 25 = Rs. 2,000. We have, Profit = S.P. – C.P. = Rs.2000 – Rs.1,800 = Rs. 200 Profit per cent = Profit C.P. × 100% = 200 1800 × 100% = 100 9 = 11 1 9 %. Example 3 An article bought for Rs. 1,000 is sold at a profit of 10%. Find the selling price. Solution: Here, C.P. = Rs. 1000 Profit = 10%, S.P. = ? Now, Profit = 10% of C.P. = 10 100 × Rs. 1000 • In this method, assume C.P. as 100. • S.P. as 100 + profit per cent if profit per cent is given. • S.P. as 100 – loss percent if loss per cent is given.
54 Oasis School Mathematics-8 = Rs. 100 ∴ S.P. = C.P. + Profit = Rs. 1000 + Rs. 100 = Rs. 1100 Example 4 By selling a radio for Rs. 1,500, a trader gains 20%. Find the cost price of the radio. Solution: Here, S.P. = Rs. 1,500 Profit % = 20%, C.P. = ? Let, C.P. = x S.P. = C.P.+ P% of C.P. or, 1500 = x + 20% of x or, 1500 = x + 20 100 × x or, 1500 = x + x 5 or, 1500 = 6x 5 or, 6 x = 1500 × 5 or, x = 1500×5 6 = Rs. 1,250 Hence, the required C.P. is Rs. 1,250. Alternative method-I Here C. P. = Rs. 1000, P% = 10%, S.P. = ? We have, S.P. = 100 + P% 100 × C.P. = ( 100 + 10 100 ) × 1000 = 110 100 × 1,000 = Rs 1,100. Since there is the profit of 10 % When C.P. is Rs.100, S.P. is Rs.110. When, C.P. is Re.1, S.P. is Rs. 110 100 . When C.P. is Rs. 1000, S.P. is Rs. 110 100 × 1000 ∴ S.P. = Rs. 110 100 × 1,000 S.P. = Rs. 1,100. Alternative method-II Alternative method-I Here, S.P.=Rs. 1500, P%=20%, C.P.=? S.P. = 100 + S.P. (100 + P%) = 100 × Rs.1500 120 = Rs 1,250. Since, there is the profit of 20% If S.P is Rs. 120, then C.P. = Rs. 100 If S.P. is Re 1, then C.P. = Rs.100 120 If S.P. is Rs. 1500, then C.P. = Rs. 100 120 × 1,500 Hence, C.P. = Rs. 100 120 × 1,500 = Rs. 1,250 Alternative method-II
Oasis School Mathematics-8 55 Exercise 6.1 1. Answer the following questions. (a) Write the formula to calculate profit if profit, C.P. and S.P. are given. (b) Write the formula to calculate profit if profit percent is given. (c) Write the formula to calculate loss if loss percent is given. (d) Write the relation among S.P., C.P. and loss percent. (e) Write the relation among S.P., C.P. and profit percent. 2. Find the profit or loss percent in each of the following cases. (a) C.P. = Rs. 200, S.P. = Rs. 250 (b) C.P. = Rs. 800, S.P. = Rs. 900. (c) C.P. = Rs. 750, S.P. = Rs. 600. (d) C.P. = Rs. 1,600, S.P. = Rs. 1,800 3. (a) If C.P. = Rs. 800, profit = Rs. 25, find S.P. (b) If S.P. = Rs. 1,285, profit = Rs. 85, find C.P. 4. (a) Arun buys a table fan for Rs. 400 and sells it for Rs. 425. Find his profit and profit percent. (b) A man buys an article for Rs. 4,000 and sells it for Rs. 3500. Find his (i) loss, and (ii) loss per cent. (c) Asmita bought an article at Rs. 2,000 and she sold it at Rs. 1,750. Find her loss and loss per cent. 5. (a) A man bought a mobile set for Rs. 12,000. What should be its selling price to gain Rs. 1250? (b) A man bought an article for Rs. 15,000. What is the selling price of the article, if the loss is Rs. 2,250. (c) Sampada sold an article for Rs. 3,000 at a profit of Rs. 750. What is the cost price of the article? 6. (a) Suman buys an old scooter for Rs. 95,000 and spends Rs. 5,000 on its repair. If he sells the scooter for Rs. 1,05,000, find his gain or loss per cent. (b) Salman sold a mobile set at Rs. 12,000 after spending Rs. 1,500 for its repairing. If he had paid Rs. 10,000 for it, find his profit or loss per cent. 7. (a) A shop-keeper bought 200 eggs at Rs. 5 each. 50 eggs were broken during transportation and then he sold the remaining eggs at Rs. 8 each. Find, his gain or loss per cent. (b) A fruit-seller purchased 1,000 mangoes at Rs. 15 each. 300 of them were rotten and he sold the rest at Rs. 20 each. Find his profit or loss per cent. 8. Find the selling price in each of the following cases. (a) C.P. = Rs. 1000; profit = 25% (b) C.P. = Rs. 5000, loss = 20% 9. Find the cost price when (a) S.P. = Rs. 2200, profit = 10% (b) S.P. = Rs. 2400, loss = 20%
56 Oasis School Mathematics-8 10. (a) A shopkeeper bought a radio for Rs. 1800 and sold it at 20% profit. Find his amount of profit. (b) A watch is bought at Rs. 500 and sold at a loss of 15%. Find the amount of loss . 11. (a) Aadhya bought a mobile set for Rs. 12000 and sold at a profit of 15%. Find the selling price. (b) Lakpa bought a television for Rs. 20,000 and sold it at the loss of 10%, find the selling price of TV. (c) At what price an article should be sold to make a profit of 20%, if it was initially paid Rs. 4000. 12. (a) By selling an article for Rs. 900, a man gains 20%. Find his cost price. (b) By selling a pen for Rs. 40, a man makes a profit of 25%. How much does it cost him? (c) A man sold an article at Rs. 1,250 at the gain of 25%. Find its cost price. 13. A laptop is bought at Rs. 90,000. (a) What is its selling price if it is sold at the profit of Rs. 12,000? (b) What is the profit percent in this case? (c) What will be selling price, if he wants to make the profit of 20%? 14. A mobile set is sold at Rs. 60,000. (a) What is its cost price if it is sold at the profit of Rs. 8000. (b) What is the profit percent in this case? (c) What is the selling price if it is sold at the profit of 20%? 15. A television is sold at Rs. 72,000. (a) What is cost price, if it is sold at the profit of 20%? (b) At what price should it be sold to make the profit of Rs. 4000? (c) What is the profit percent in this case? Answer 1. Consult your teacher. 2. (a) Profit% = 25% (b) Profit% = 12.5% (c) Loss% = 20% (d) Loss% = 12.5% 3. (a) Rs. 825 (b) Rs. 1200 4. (a) Rs. 25, 6 1 4 % (b) (i) Rs. 500, 12 1 2 % (c) Rs. 250 , 12 1 2 % 5. (a) Rs. 13,250 (b) Rs. 12, 750 (c) Rs. 2250 6. (a) 5% (b) 4 8 23% 7. (a) 20% profit (b) 6 2 3 % loss 8. (a) Rs. 1250 (b) Rs. 4000 9. (a) Rs. 2000 (b) Rs. 3000 10. (a) Rs. 360 (b) Rs. 75 11. (a) Rs. 13800 (b) Rs. 18000 (c) Rs. 4800 12. (a) Rs. 750 (b) Rs. 32 (c) Rs. 1000 13. (a) Rs. 1,02,000 (b) 13.33% (c) Rs. 1,08,000 14. (a) Rs. 52,000 (b) 15.38% (c) Rs. 62,400 15. (a) Rs. 60,000 (b) Rs. 64,000 (c) 6.67%
Oasis School Mathematics-8 57 Project Work Go to a nearby shop of your community. Select any 5 items available in the shop. Ask the cost price of each item with shopkeeper. Ask the selling price of each item. Then identify which item is more profitable on selling. Present in the classroom with calculation. 6.3 Discount Discount Read the following question and answer the questions given below. • Price of a machine is tagged as Rs. 5000. Shopkeeper reduced the price by Rs. 500 while selling. At what price shopkeeper sold the machine? • What is the tagged price (marked price) of the article? • What amount of discount is given to the customer? • Establish the relation among marked price discount and selling price. Let's observe one more example, Tagged price of watch is Rs. 15,000. Discount of 10% is offered, then list the market price, discount and selling price. • In the above example, Marked price (MP) = Rs. 15000 Discount = 10% of Rs. 15,000 = 10 100 × 15,000 = Rs. 1500. Selling price = Rs. 15,000 – 1500 = Rs. 13,500 Marked price We can observe the price tag which is kept along with the article while purchasing goods from the shop or department store. The price of the goods so kept using the tag or the list or mark is called marked price or the quoted price or the labeled price or the listed price. It is generally fixed to be either greaterthan or equal to the selling price. i.e. M.P. ≥ S.P. • In the above example, Marked price (MP) = Rs. 5000 Discount = Rs. 500 Selling price = Rs. 4500 Relation among them is Selling price = marked price – Discount.
58 Oasis School Mathematics-8 Discount The concession or deduction over the marked price of an article at the time of selling is called discount. It is normally expressed in terms of percentage of marked price. Hence, if the discount is d%, that means. Discount = d% of M.P. Selling price The price of the article after allowing discount from its M.P. is called its net price or the selling price. Hence, relating above three terms we have. M.P. – Discount = S.P. As we know that the discount is always provided on the marked price of an article, the discount per cent is calculated by the following formula: Discount percent Discount M.P. × 100% Discount M.P. × 100% Worked Out Examples Example 1 The marked price of an article is Rs. 500. Find the selling price after a discount of Rs. 50. Solution: Here, M.P. = Rs. 500 Discount = Rs. 50 ∴ S.P. = M.P. – Discount = Rs. 500 – Rs. 50 = Rs. 450. Example 2 If a discount of 10% is offered on the goods whose M.P. is Rs.1,000, find its S.P. Solution: • Discount = Discount % of M.P. • S.P. = M.P. – Discount = M.P. – Discount % of M.P. • Discount = M.P. – S.P • Discount Percent = M.P. – S.P = Discount M.P. × 100% Remember !
Oasis School Mathematics-8 59 Here, M.P. = Rs. 1,000 Discount = 10% of M.P. = 10% of Rs. 1,000 = 10 100 × Rs. 1,000 = Rs. 100 ∴ S.P. = M.P. – Discount = Rs. 1,000 – Rs. 100 = Rs. 900 Example 3 Alina sold a book for Rs. 135 after a discount of 10% on M.P.. Find its M.P. Solution: Let, M.P. be x, then discount = 10% of x = 10 100 × x = x 10 Now, S.P. = Rs. 135 We have, M.P. – Discount = S.P. or x – x 10 = Rs. 135 or, 9x 10 = Rs. 135 or, 9x = Rs 135 × 10 ∴ x = Rs. 135 × 10 9 = Rs. 150. ∴ M.P. = Rs. 150 Example 4 The marked price of an article is Rs. 260. If the shopkeeper allows some discount and sells it for Rs. 234, find the discount and discount per cent. Solution: Here, M.P. = Rs. 260, S.P. = Rs. 234 Discount = ? Discount percent = ? We have, Discount = M.P. – S.P. = Rs. 260 – Rs. 234 = Rs. 26. Again, Discount percent = Discount M.P. × 100 % = 26 260 × 100% = 10%. Hence Discount = Rs. 26 and Discount per cent = 10 % Example 5 Marked price of an article is Rs. 10,000. Find the cost price if a profit of Rs. 200 is made after the discount of 10%. Solution: Given, M.P. = Rs. 10,000 Discount% = 10% We have, Discount = Discount % of M.P. = 10% of 10,000
60 Oasis School Mathematics-8 = 10 100 × 10,000 = Rs. 1,000 Again, we have, S.P. = M.P. – Discount = 10,000 – 1,000 = Rs. 9,000 Profit = Rs. 200 We have, C.P. = S.P. – Profit = Rs. (9,000 – 200) = Rs. 8,800 Exercise 6.2 1. (a) What is the formula to calculate discount if discount percent and MP are given? (b) Write the relation of MP, SP and discount. (c) Write the formula to calculate discount percent. (d) Write the formula to calculate SP if MP and discount percent are given. 2. Find discount and discount percent in the following cases. (a) M. P. = Rs. 600, S.P. = Rs. 480 (b) M. P. = Rs. 5000, S.P. = Rs. 4,000 (c) M. P. = Rs. 5000, S.P = Rs. 4,500 3. Find the selling price in the following cases. (a) M.P. = Rs. 200, Discount = 20% (b) M.P. = Rs. 3,600, Discount = 5% (c) M.P. = Rs. 4,000, Discount = 15% 4. Find the marked price in the following cases. (a) S.P. = Rs. 360, Discount = 10% (b) S.P. = Rs. 1,900, Discount = 5% (c) S.P. = Rs. 950, Discount = 5% 5. (a) The marked price of a mobile is Rs. 15,000 and it is sold for Rs. 12,000. Find the amount of discount and its percentage. (b) What should be the discount percent so that a watch marked Rs. 6,000 is sold for Rs. 4,500? 6. (a) An article marked Rs. 4,500 is sold at a discount of 10%. Find the selling price of the article. (b) The marked price of an article is Rs. 700 and it is sold at a cash discount of 10%. Find its selling price. 7. (a) After a discount of 12%, a radio was sold for Rs.4400. Find its marked price. (b) After allowing a discount of 8% on marked price, an article is sold for Rs. 2,300. Find its marked price.
Oasis School Mathematics-8 61 (c) An article is sold at Rs. 2,400 after a discount of 4%. Find its marked price. 8. (a) Marked price of a watch is Rs. 1,000. It is sold allowing a discount of 15%. If the profit is Rs. 10, find the cost price of the watch. (b) Ram marked the price of an article as Rs. 400. He sold it allowing a discount of 20% at a loss of Rs. 60. Find the cost price of the article. 9. Marked price of an article is Rs. 15,000. (a) What is discount amount if it is sold at the discount of 20% (b) Find the selling price of the article. (c) If there is a profit of Rs. 2500, find its cost price 10. Marked price of an article is Rs. 12,000. (a) It is sold at the discount of 10%, find the discount amount. (b) Find the selling price. (c) If it is sold at the profit of Rs. 800, find the cost price. 11. An article is sold with the discount of 20% (a) Find the marked price if the discount amount is Rs. 3000. (b) Find the selling price. (c) If it is sold at the loss of Rs. 2000, find its cost price. Project Work • Visit a department store or fancy shop. Select any 10 items available there. List out their marked price. Ask the selling price of each item with shopkeeper and find the discount percent in each item. Answer 1. Consult your teacher. 2. (a) Rs. 120, 20% (b) Rs. 1000, 20% (c) Rs. 500, 10% 3. (a) Rs. 160 (b) Rs. 3420 (c) Rs. 3400 4. (a) Rs. 400 (b) Rs. 2000 (c) Rs. 1000 5. (a) Rs. 3000, 20% (b) Rs. 1500, 25% 6. (a) Rs. 4050 (b) Rs. 630 7. (a) Rs. 5000 (b) Rs. 2500 (c) Rs. 2500 8. (a) Rs. 840 (b) Rs. 380 9. (a) Rs. 3000 (b) Rs. 12000 (c) Rs. 9,500 10. (a) Rs. 1200 (b) Rs. 10,800 (c) Rs. 10,000 11. (a) Rs. 15,000 (b) 12,000 (c) Rs. 14,000
62 Oasis School Mathematics-8 7.1 Flack back Paridhi deposited Rs. 10,000 in a bank. After 1 year, Rs. 600 was added and her money rose to Rs.10,600. The extra amount of money Rs. 600 paid by the bank is the interest. Rs. 10,000 deposited is the principal and Rs. 10,600 is the amount. In the above example, • Rs. 600 is the interest. • Rs. 10,000 is the principal • Rs. 10,600 is the amount. Let's define some basic terms of simple interest. Principal: The initial amount money borrowed or invested or deposited is called principal or sum. It is denoted by P. Interest: The additional money paid by the borrower for having used another person's money, is called interest. It is denoted by I. Rate of Interest : The interest on Rs. 100 in 1 year is known as rate percent per annum. It is denoted by R i.e. if the interest of Rs. 100 in 1 year is Rs. 12 then, Rate (R) = 12% p.a. Amount : The total sum of money paid with interest is called amount. It is denoted by A. Thus, Amount (A) = principal (P) + simple interest (I) i.e. A = P + I 7.2 Calculation of Simple Interest Let, principal deposited be Rs. P for T years, rate be R% per year, and the interest be Rs. I. By the definition of the rate of interest, Unit 7 Simple Interest You know! • Concept of percentage • Terms related to interest
Oasis School Mathematics-8 63 Interest on Rs. 100 for 1 year = Rs. R Interest on Re 1 for 1 year = Rs. R 100 Interest on Rs. P for 1 year = Rs. R 100 × P The interest on Rs. P for T years = Rs. R 100 × P × T = Rs. P × T × R 100 ∴ Simple interest (I) = P × T × R 100 .............. (i) There are four quantities involved in the calculation of simple interest, namely principal (P), rate per cent (R), time (T) and interest (I). So, if we are given any three, we can find the fourth. To find any one of P, R, T, it is convenient to rearrange the formula as follows. P = I × 100 R × T ....................... (ii) R = I × 100 P × T ....................... (iii) T = I × 100 P × R ....................... (iv) Again, we have, A = P + I, where A = amount = P + P × T × R 100 = 100P + P × T × R 100 A = P(100 + T × R) 100 P = A × 100 100 + T × R ....................... (v) Worked Out Examples Example 1 Find the interest on Rs. 3,600 at the rate of 12% p.a. for 3 years. Solution: Here, Principal (P) = Rs. 3,600 Rate (R) = 12% p.a. Remember ! I = P × T × R 100 A = P + I R = 100 × I P × T T = 100 × I P × R P = 100 × I T × R P = 100 × A 100 + T × R
64 Oasis School Mathematics-8 Time (T) = 3 years We have, Interest (I) = P × T × R 100 = 3,600 × 3 × 12 100 = Rs. 1,296 Example 2 In what time will Rs. 3000 amount to Rs. 3360 at 6% p.a. ? Solution: Here, Principal (P) = Rs. 3,000 Amount (A) = Rs. 3,360 Rate(R) = 6% p.a We have, Interest (I) = A – P = Rs. (3,360 – 3,000) = Rs. 360 Time (T) = ? We have, T = I × 100 P × R = 360 × 100 3,000 × 6 = 2 years Hence, the required time is 2 years. Example 3 Find the rate if the interest per rupee per month is 1 paisa. Solution: Here, Principal (P) = Re. 1 Interest (I) = 1 paisa = Re. 1 100 . Time (T) = 1 month = 1 12 year Rate (R) = ? We have, R = 100 × I P × T = 100 × 1 × 1 100 1 12 = 12 % ∴ Rate (R) = 12 % p.a. Alternative method Here, Interest of Re. 1 in 1 month is 1 paisa. Interest of Rs. 100 in 1 month is 100 paisa. Interest of Rs. 100 in 12 months is 1200 paisa. i.e. Interest of Rs. 100 in 1 year is Rs. 12. ∴ Rate (R) = 12 %
Oasis School Mathematics-8 65 Example 4 In how many years will a sum of money double itself at 10% per annum simple interest. Solution: Let, Principal (P) = Rs. x. Then, Amount (A) = Rs. 2x Rate (R) = 10% p.a. Time (T) = ? Now, Interest (I) = A – P = Rs. 2x – Rs. x = Rs. x We have, T = 100 × I P × R = 100 × x x × 10 = 10 years Hence, the required time is 10 years. Example 5 What sum will amount to Rs. 1500 in 5 years at 10% per year simple interest ? Solution: Here, Amount (A) = Rs. 1,500 Time (T) = 5 years Rate (R) = 10% p.a. Principal (P) = ? We have, P = 100 × A 100 + T × R = 100 × 1500 100 + 5 × 10 = 1,50,000 150 = Rs. 1,000 So, the required sum is Rs. 1,000. Example 6 A sum of money doubles itself in 20 years at a certain rate. In how many years will it treble itself at the same rate of interest? Solution: Here, in the first case, Principal (P) = Rs. x (suppose) Then, Amount (A) = Rs. 2x, Time (T) = 20 years Rate (R) = ? Now, Interest (I) = A – P
66 Oasis School Mathematics-8 = Rs. 2x – Rs. x = Rs. x We have, R = 100 × I P × T = x × 100 x × 20 = 5% p.a. Again in the second case, Amount = Rs. 3x Principal (P) = Rs. x Rate (R) = 5% p.a. (by question) Time (T) = ? Now, I = A – P = Rs. 3x – Rs. x = Rs. 2x We have, T = 100 × I P × R = 2x × 100 x × 5 = 40 years Hence, the sum will treble itself in 40 years. Example 7 Aslam deposited Rs. 4500 in a bank for 3 years at the interest rate of 10% p.a. How much would he get after 3 years if he paid 5% tax on the interest. Solution: Here, Principal (P) = Rs. 4,500 Time (T) = 3 years Rate (R) = 10 % We have, Interest (I) = P × T × R 100 = 4500 × 3 × 10 100 = Rs.1,350 Tax on interest = 5% of 1,350 = 5 100 × 1,350 = Rs. 67.50 Net interest = Rs. 1,350 – Rs. 67.50 = Rs. 1282.50 Amount = P + Net interest = 4,500 + 1,282.50 = Rs. 5,782.50 Hence, the sum will be Rs. 5,782.50
Oasis School Mathematics-8 67 Exercise 7.1 1. (a) If P be the principal T be the time in year and R be the rate of interest, write their relation with interest (I). (b) If x be the principal and y be the interest, what is the amount in terms of x and y? 2. Find the simple interest and amount in the given cases. (a) P = Rs. 1,600, T = 3 years, R = 5% p.a. (b) P = Rs. 1,2000, T = 6 months, R = 10% p.a. (c) P = Rs. 2,400, T = 73 days, R = 15% p.a. 3. Find the principal and amount in the following cases. (a) I = Rs. 200, T = 5 years, R = 10% p.a. (b) I = Rs.1,200, T=2 years, 6 months, R = 20% p.a. 4. Find the time in the following cases. (a) P = Rs.6,500, I = Rs. 3120, R = 6% p.a. (b) P = Rs. 2,500, I = Rs. 750, R = 5% p.a. (c) P = Rs. 8,000, A = Rs. 10,000, R = 6 1 4 % p.a. 5. Find the rate of interest in the following cases: (a) P = Rs. 6,000, I = Rs. 2,100, T = 5 years (b) P = Rs. 1,600, A = Rs. 2,832, T = 5 1 2 years (c) Interest per rupee per month is 1.5 paisa. 6. Find the principal in the following cases. (a) A = Rs. 1,500, T = 2 years, R = 10% p.a. (b) A = Rs. 4,000, T = 6 years, R = 10% p.a. 7. (a) Umesh borrowed Rs. 30,000 from a bank. The bank charged him 12% p.a. simple interest. How much should he pay at the end of 3 years ? (b) Bikalpa deposits Rs. 4,800 in a bank which pays 6% interest per annum. Find the amount he would get at the end of 1 year 9 months. (c) Sankalpa borrows Rs. 25,000 from a bank at 10% per annum. What amount must he pay to the bank if he repays the loan in 6 months.? 8. (a) In how many years, Rs. 1,250 invested at 6% p.a. will yield Rs. 225 as simple interest ?
68 Oasis School Mathematics-8 (b) In what time will Rs. 10,000 amount to Rs. 15,250 at 7 1 2 % p.a. simple interest? (c) In how many years will a sum of money double itself at 5% p.a. simple interest ? 9. (a) At what rate percent will Rs. 1,250 yield an interest of Rs. 100 in 2 years? (b) At what rate percent p.a. will Rs. 1,600 amount to Rs. 2,000 in 5 years? (c) At what rate percent p.a. will the simple interest on Rs. 75 for 9 months be Rs.4.50 ? (d) At whatrate percent per annum will a given sum of money treble itself in 25 years ? 10. (a) What sum amounts to Rs. 3,600 in 2 years at the rate of 10% p.a.? (b) What sum amounts to Rs. 480 in 2 years at 15% per annum? (c) A received from B a certain sum of money at 10% per annum simple interest. After 73 days, A returned Rs. 5,100 to B and cleared the debt. Find the sum borrowed by A. 11. A sum of money doubles itself in 10 years at a certain rate. In how many years will it treble itself at the same rate of interest? 12. (a) Write the relation among principal (P), rate of interest (R), time (T) and interest (I). (b) What is the interest of Rs. 10,000 for 2 years at the interest rate of 6% p.a.? (c) How much money would be collected in 2 years? 13. Lakpa deposited Rs. 5,000 in a bank for 3 years at the rate of 10% p.a. (a) How much interest will he get after 3 years? (b) How much tax does he have to pay, if the tax on interest is 5%? (c) How much money is collected in the bank after 3 years? Project Work • Visit any bank or finance company nearby your home. Ask their interest rate. Suggest which bank or finance company is appropriate to take the loan. Give suggestion with calculation. Answer 1. Consult your teacher. 2. (a) Rs. 240, Rs. 1840 (b) Rs. 600, Rs. 12,600 (c) Rs. 72, Rs. 2472 3. (a) Rs. 400, Rs. 600 (b) Rs. 2400, Rs. 3600 4. (a) 8 yrs. (b) 6 yrs (c) 4 yrs. 5. (a) 7% (b) 14% (c) 18% 6. (a) Rs. 1250 (b) Rs. 2500 7. (a) Rs. 40800 (b) Rs. 5304 (c) Rs. 26, 250 8. (a) 3 yrs (b) 7 yrs. (c) 20 yrs. 9. (a) 4% (b) 5% (c) 8% (d) 8% 10. (a) Rs. 3000 (b) Rs. 369.23 (c) Rs. 5000 11. 20 years 12. (a) Consult your teacher (b) Rs. 1200 (c) Rs. 60 (d) Rs. 11,140 13. (a) Rs. 1500 (b) Rs. 75 (c) Rs. 6425
Oasis School Mathematics-8 69 Miscellaneous Exercise 1. Study the following questions and answer them: (a) Which digits are in binary number system and quinary number system? (b) Convert the 100011012 into decimal number. (c) Convert the decimal number thus obtained into quinary number. Show the relation of binary and quinary number. 2. Answer the following questions. (a) What type of number are called rational number? (b) Is 0.66666...... a rational number? Give reason. (c) Why 2.14374218.........is an irrational number? Give reason. 3. (a) Which of the following statement is not true? (i) 2 3 is a rational number. (ii) 3 is an irrational number. (iii) -3 4 is an irrational number. (b) 2 is a natural number. Is it an rational number also? Give reason. (c) 5.23417213..........is an irrational number. Why? 4. Distance of the sun from the earth is 149,180,000 km. (a) Write this fact in Scientific notation. (b) Speed of the light in vacuum is 2.98 × 105 km per sec. (c) Using the relation distance = speed × time, calculate the time taken by light to reach to the earth from the sun. 5. Mass of the earth is 5.977 × 1024 kg. and the mass of the sun is 1.989 × 1030 kg. (a) By how many times the sun is heavier than the earth? (b) If 1.989 × 1030 represents one solar mass, then what is the mass of the earth in solar mass. (c) Write both the result of (a) and (b) in Scientific notation. 6. (a) In the ratio, 5:4, what do 5 and 4 present? (b) The income of Pasang and Shakar is the ratio of 5:4. If the income of Pasang is Rs. 20,000, what is the income of Shankar? (c) If the income of Pasang increased by Rs. 4,000 and that of Shankar decreased by Rs. 2000; what is the ratio of their income? Ans: [(a) Consult your teacher (b) Rs. 16,000 (c) 12:7] 7. (a) If the antecedent of a ratio is 5 and the consequent is 2, find the ratio.
70 Oasis School Mathematics-8 (b) If the ratio of the ages of father and his son is in the given ratio, after 10 years. The ratio of their ages is 2:1, find their present age. (c) What is the ratio of their ages before 10 years? Ans: [(a) Consult your teacher (b) 50 yrs, 20 yrs, (c) 4:1] 8. If a, b, c, d are proportion, (a) What is the relation among them? What is 'a' and 'd' called? What is 'b' and 'c' called? (b) Check whether the numbers 3, 11, 23 are in proportion or not? (c) If not, what should be added to each number such that the resulting numbers are in proportion. Ans: [(a) Consult your teacher, (b) No, (c) 1] 9. (a) What type of proportions are called direct proportion? Give an example. (b) The cost of 2 5 part of a land is Rs. 6,80,000. What is the cost of 7 10 part of the land? (c) Also find the cost of the land. Ans: [(a) Consult your teacher, (b) Rs. 11,90,000, (c) Rs. 17,00,000] 10. Study the given table and answer the questions given below. (a) What type of variations are days and men? Why? (b) What type of variations are work and days? Why? (c) Using variations calculate the value of x. Ans: [(a), (b) Consult your teacher, (c) 44 4 9 days] 11. (a) On which condition, there is profit and there is loss? (b) The cost of one cartoon cooking oil which contains 10 packet of oil is Rs. 2500, what is the cost of 6 packets of oil? (c) If 1 packet of oil is sold at Rs. 285, what is the total profit on selling 10 packets of oil? (d) Calculate the profit percent Ans: [(a) Consult your teacher (b) Rs. 1500, (c) Rs. 350, (d) 14%] 12. The cost of 80 litres of milk is Rs. 8,000. (a) What is the cost of 100 litres of milk? Men Work Days 20 60 30 x 3 5 2 3
Oasis School Mathematics-8 71 (b) A man bought 100 litres of milk 5 litres of milk is wasted by leakage. He sold remaining milk at the rate of Rs. 120 per litre, what is the total SP of the milk? (c) Find his total profit or loss. (d) Find the profit or loss percent. Ans: [(a) Rs. 10,000, (b) Rs. 11,400, (c) Rs. 1,400, (d) 14%] 13. An article is bought at Rs. 15,000. (a) What is the profit if it is sold at Rs. 18,000. (b) Find the profit percent in this case. (c) If the profit of 50% is to be made, what is the selling price of the article? Ans: [(a) Rs. 3,000, (b) 20%, (c) Rs. 22,500] 14. A mobile set is sold at Rs. 60,000 at the profit of 20%; (a) Write the relation among CP, SP and profit percent. (b) Using this relation, find the cost price (CP). (c) What should be its selling price if the profit of 10% is to be made? Ans: [(a) Rs. 50,000, (b) S.P.= C.P + profit% of CP, (c) Rs. 55,000] 15. Tagged price of television is Rs. 30,000. Shopkeeper has offered a discount of 10%. (a) Calculate the discount amount. (b) Calculate the selling price of the shirt. (c) If the shopkeeper made the profit of 20% on selling the television, find its cost price. (d) By how much tagged price is more than the cost price? Ans: [(a) Rs. 3,000, (b) Rs. 27,000, (c) Rs. 22,500, (d) Rs. 7,500] 16. (a) Which of the following relation is not true? (i) Discount = Discount% of MP (ii) S.P. = MP – Discount (iii) Discount = Discount% of SP (b) If an article is sold at Rs. 2500 with the discount of Rs. 350, what is its marked price. (c) What is its discount percent? (d) If the article is bought at Rs.2000 and sold it without discount, what would be the profit percent. Ans: [(a) Consult your teacher, (b) Rs. 2850, (c) 12.28%, (c) 42.5%] 17. (a) Write the formula to calculate Rate Percent, if Principal (P), Time (T) and Interest (I) are given. (b) A sum of money doubles itself in 10 years, what is the interest rate? (c) At the same rate, how many years will it take to be treble? Ans: [(a) Consult your teacher, (b) 10 years, (c) 20 years]
72 Oasis School Mathematics-8 Assessment Test Paper Full marks - 30 Attempt all the questions. 1. (a) What are the digits used in binary number system and quinary number system? (1) (b) Convert 250 into binary number system. (2) (c) Convert (10101101)2 into decimal number system. (2) 2. (a) In the fraction 3 4 , are 3 and 4 both integers? (1) (b) What type of number 3 4 is? (1) (c) Convert 3 4 into decimal. (1) (d) What is the nature of that decimal? (1) (e) Draw out the conclusion from the result of (b) and (d). (1) 3. (a) Multiply 10–6 with 1010 using the law of indices. (1) (b) Using this result multiply 7 × 10–6 and 1.5 × 1010. (1) (c) Using above result, Evaluate: 7 × 10–6 × 1.5 × 1010 2 × 10³ . (2) 4. (a) In the ratio 2 : 3, what do 2 and 3 represent? (1) (b) The ages of two girls are in the ratio 2 : 3. If the age of elder one is 15 years, what is the age of younger one? (2) (c) After how many years ratio of their ages will be 4 : 5? (2) 5. (a) Identify whether the number of articles and its cost are in direct proportion or in indirect proportion. (1) (b) Then what is the cost of 10 articles, if the cost of an article is 600. (2) (c) If the marked price of the article is Rs. 900. (2) (d) If the discount of 20% is given, what is its selling price? (1) 6. (a) Write the relation of P, T, R and I. (1) (b) If a man deposited Rs. 6000 for 2 years at the rate of 10% p.a, how much interest does he receive in 2 years? (2) (c) How much money is collected in the bank in 2 years? (1)
Oasis School Mathematics-8 73 Estimated Teaching Hours 15 Contents: • Area of triangle and quadrilateral • Area of circle. Expected learning outcomes: At the end of this unit, students will be able to develop the following competencies • Find the area of plane figures like triangle and quadrilateral • Find the area of circle. Teaching materials: Graph sheet, models of triangle, different types of quadrilateral and circle. Mensuration
74 Oasis School Mathematics-8 Unit 8 Perimeter and Area Flash Back Let's try to understand the given examples: Santosh wishes to enclose his home by a boundary wall. What is the length of the wall? Aadhya walks 3 times around the park in the morning. How much distance does she cover? In the first case, the length of boundary is the perimeter. Similarly, in the second case the distance covered by Aadhya is three times the perimeter. Again, how to calculate the perimeter of given figures? Perimeter of figure (i) is AB + BC + AC and perimeter of figure (ii) is AB + BC + CD + DE + EF + FA. Hence, the length of the outer boundary of a closed figure is called its perimeter. Area Square having length of one edge is 1cm, 1cm² is called unit square or one centimeter square. It has an area of 1 square centimeter (cm²). A rectangle of sides 7cm by 4 cm can be divided into 28 squares of each 1 cm². Since the area of 1 square is 1 cm2 , and the figure contains 28 squares area of this figure = 28 cm2 = 7 × 4 cm2 = l × b B A C E D C A B F (i) (ii) You know! • The perimeter of a plane figure
Oasis School Mathematics-8 75 Hence, area of a rectangle = l × b. In square, all sides are equal so its area is A = l². SN Name Figure Area Perimeter 1. Triangle 1 2 × base (b)×height (h) = 1 2 × BC × AD AB+BC+AC = c+a+b and Semiperimeter (S) = a+b+c 2 2. Equilateral Triangle 3 4 a2 3a 3. Scalene Triangle s(s-a) (s-b) (s-c) P = a +b + c s = a+b+c 2 4. Rectangle length(l) × breadth (b) = BC × AB 2 (l + b) 5. Parallelogram base(b) × height (h) = BC × AE 2(a+b) 6. Square (side)2 = (l) 2 4l 7. Rhombus 1 2× product of diagonals = 1 2 dl ×d2 = 1 2 AC × BD Or, base×height = BC × AE AB+BC+CD+AD Square A = l² l l l A = l l × b Rectangle l b A B D C b hheight base c a A B C c a b A B C A B D C b b l height b b(base) A D C B E a a l l l l A D B C A B D C E d1 d2
76 Oasis School Mathematics-8 8. Quadrilateral 1 2 × diagonal × (sum of perpendiculars) = 1 2 × d × (p1 + p2 ) AB+BC+CD+AD 9. Trapezium 1 2 × height×(sum of //sides) = 1 2 × h(a + b) = 1 2 ×AE (AD+BC) AB+BC+CD+AD 10. Kite 1 2 × product of diagonals = 1 2 × d1 × d2 = 1 2 × AC × BD AB + BC + CD + AD Worked Out Examples Example 1 Find the area of the following figure: Solution: (i) In the given figure, ABCD is a rhombus where, d1 = AC = 12 cm d2 = BD = 18 cm ∴ Area of rhombus = 1 2 × d1 × d2 = 1 2 ×12×18 cm² = 108 cm² (ii) In the given figure, ABCD is a parallelogram. where, DC = 12 cm, AE = 8 cm ∴ Area of parallelogram = base × height = DC ×AE = 12 × 8 = 96 cm². Example 2 Find the area of shaded region. Solution: (i) Here, The area of the parallelogram ABCD = base × height A B C N M D d P1 P2 A B D h b h C E a A D C B d1 d2 D C B 12cm 18cm A A D E B C 8cm 12cm A F D E C B 14cm 8cm
Oasis School Mathematics-8 77 = BC × DE = 14 × 8 = 112 cm2 And the area of the triangle BCF = 1 2 × base × height = 1 2 × BC × DE = 1 2 × 14 × 8 = 7 × 8=56 cm2 ∴ Area of the shaded region = Area of parallelogram ABCD – Area of ∆BCF = 112 – 56 = 56 cm2 Example 3 A rectangular field is 30m long and 10m broad. Find (a) its area and (b) its cost of plastering the field at Rs. 20 per square metre, (c) Its perimeter, (d) Cost of fence he field at the rate of Rs. 35 per meter. Solution: Here, length of rectangular field (l) = 30 m, Breadth of rectangular field (b) = 10m (a) ∴ Area of rectangular field = l × b = 30 × 10 = 300 m2 = 300 sq.m. (b) Here, cost of plastering 1 sq.m. area = Rs. 20. ∴ Cost of plastering 300 sq.m area = Rs. 20 ×300 = Rs. 6000. (c) We have, perimeter = 2 (l + b) = 2 (30 + 10)m = 80 m (d) Cost to fence the field perimeter = Rs. 25 Cost to fence the field = Rs. 80 × 25 = Rs. 2000 Exercise 8.1 1. Find the area of the following triangles. (a) (b) (c) (d) 2. (a) Calculate the area of a triangle whose base is 8 cm and height is 9 m. (b) Calculate the perimeter and semi-perimeter of a triangle whose sides are 10 cm, 24 cm and 26 cm. 3. Find the area of the following rectangles: (a) (b) (c) 6 cm 6 cm A B C P Q R X Y Z D E F 5 cm 6 cm 8 cm 8 cm 6 cm 8 cm 6 cm 4 cm 4 cm 6 cm 10.5 cm 3.5 cm
78 Oasis School Mathematics-8 4. (a) Find the area of a rectangle if its length and breadth are 25 cm and 12 cm respectively. (b) The area of a rectangle is 64 cm2 and its length is 16 cm. Find the breadth of the rectangle. (c) The area of a rectangle is 450 cm2 and its breadth is 15 cm. Find its length. (d) The area of a rectangle is 540 cm2 and its length is 27 cm. Find its breadth and perimeter. 5. Find the area of the following squares. (a) (b) (c) 6. Find the area and perimeter of the following parallelograms. (a) (b) (c) 7. Find the area of the following rhombuses. (a) (b) (c) BD = 6 cm, AC = 5 cm SQ = 4.6 cm, PR = 3.5 cm MO = 8 cm, PN = 10 cm 8. Find the area of the following quadrilaterals. (a) (b) (c) BD = 4 cm, AQ = 3 cm SQ = 33 cm, PX = 19 cm, GE = 44 cm, DM = 20 cm, and CP = 2 cm and RY = 11 cm and FN = 15 cm 4 cm 5 cm 16 cm P Q R S T 2 cm 9 cm 3 cm 4 cm A E B D C 5 cm 6 cm 6 cm 4.2 cm 4.2 cm P S R Q A P M D S P B Q N C R O A D B C S P Q R G N Y Q M X P D E F
Oasis School Mathematics-8 79 9. Find the area of the following trapeziums: (a) (b) (c) 10. Find the area of the following kites. (a) (b) (c) AC = 15 cm SQ = 10 cm WY = 8.6 cm BD = 8 cm PR = 8 cm XZ = 3.5 cm 11. Find the area of shaded regions in the following figures. (a) (b) (c) 12. A rectangular field is 20 m long and 12 m broad. (a) Find its area. (b) Find the cost of harvesting the vegetable on the field at Rs. 10 per square metre. (c) Find its perimeter (d) Find the cost of fencing the field at the rate of Rs. 20 per m. 13. The perimeter of a square room is 60 m. (a) Find its length. (b) Find the area of its floor. (c) Find the total cost for carpeting the floor at Rs. 20 per sq. m. 14. Study the given figure and answer the questions given below. (a) What do the parts DC and AF represent? (b) Calculate the area of parallelogram ABCD and area of triangle EDC. (c) Find the area of shaded part. 6 cm 10 cm A D B C P S Q R P X O Y M N E 9 cm 13 cm 40 cm 30 cm 40 cm 28 cm 9 cm B C S P R Q W Y Z X D A A B 20cm 6 cm D E C F D 15cm 8cm 4 cm C B A E P S Q R A E D B C 5 cm14 cm 15cm 10cm 6cm A E B D F C 12 cm 30 cm
80 Oasis School Mathematics-8 15. The shape of a field is as shown in the figure. (a) What is the shape of given field? (b) Calculate the area of given field. (c) Find the area of non-shaded part and the area of shaded part. (d) Find the cost to harvest vegetable on the shaded part at the rate of Rs. 30 per m2 . Answer 1. (a) 18cm2 (b) 14.98cm2 (c) 24cm2 (d) 16 3 cm2 2. (a) 36cm2 (b) 60cm, 30cm 3. (a) 24cm2 (b) 24cm2 (c) 36.75cm2 4. (a) 300cm2 (b) 4cm (c) 30cm (d) 20cm, 94cm 5. (a) 16cm2 (b) 25cm2 (c) 128cm2, 6. (a) 24cm2 , (b) 18cm2 , (c) 25.2cm2 , 7. (a) 15cm2 (b) 8.05cm2 (c) 40cm2 8. (a) 10cm2 (b) 495cm2 (c) 770cm2 9. (a) 72cm2 (b) 455cm2 (c) 306cm2 10. (a) 60cm2 (b) 40cm2 (c) 15.05cm2 11. (a) 60cm2 (b) 16cm2 (c) 145cm2 12. (a) 240cm2 (b) Rs. 2400 (c) 64 m (d) Rs. 180 13. (a) 15m (b) 225m2 (c) Rs. 4500 14. (a) Consult your teacher (b) 360 cm2 , 180 cm2 (c) 180 cm2 15. (a) Consult your teacher (b) 5000 m2 , 3000m2 (d) Rs. 60,000 Project Work 1. Collect some triangular object available in the surroundings. Take the measurement of each side and find their area. 2. Draw the structure of the yard nearby your home. Measure its different dimensions and find its area. S M P Q R 50 cm 80 cm 120 cm
Oasis School Mathematics-8 81 8.2 Circle Review: Observe the given figure properly and answer the questions given below. • What is the given figure called? • What is the length of outer boundary of above figure called? • What is the line AOB called? • What is the meaning of circumference of the circle? • What does the diameter of a circle mean? • What is the ratio of circumference and diameter of a circle called? • Is it same for every circle? • What is the meaning of π? • What is the approximate value? • What is the formula to calculate the circumference of a circle? • The ratio of circumference to the diameter of a circle is always constant. • This constant value is called pi (π). • The approximate value of π is 22 7 = 3.1416. ........... • Length of outer boundary of a circle is called circumference of the circle. • Formula to calculate the circumference of the circle is 2πr. Remember ! Activity Relation between diameter and circumference of a circle It can be verified by experiments that, for every circle, circumference diameter = π (constant) where π = 22 7 or 3.14 or 3.1416 Now, draw three circles with radii 2 cm, 3 cm and 4 cm respectively. Measure the circumference (C) of each circle by pieces of string. Also measure its diameter (d) and tabulate them as follows. A • B O
82 Oasis School Mathematics-8 2 cm 3 cm 4 cm Fig. (i) Fig. (ii) Fig. (iii) Figures Circumference (c) Diameter (d) Circumference (C)Diameter (d) i. 12.56 cm 4 cm 12.56 4 = 3.14 ii. 18.84 cm 6 cm 18.84 6 = 3.14 iii. 25.12 cm 8 cm 25.12 8 = 3.14 From above experiment, we state below the circumference diameter formula. C d = π, C = πd Also, C = π(2r) = 2πr. [∵ d = 2r] Area of a circle Draw a circle. Divide the circle into 16 equal parts. Shade 8 sectors by two different colour as shown in the figure. Cut all 16 sectors and arrange them as shown. As we know the, circumference of a circle is 2πr. If we arrange the sectors as above it is some how like a rectangle, whose length is πr and the breadth is r. Now, area of second figure = πr × r = πr² So, the area of circle = πr² • d = 2r , π = 22 7 • circumference = 2πr • Area = πr² Remember !
Oasis School Mathematics-8 83 Worked Out Examples Example 1 Find the area of the circle having radius 42 cm. Solution: Here, radius (r) = 42 cm We have, area of a circle (A) = πr2 = 22 7 × (42)2 = 22 7 × 42 × 42cm2 = 5544cm2 . Example 2 If the area of a circle is 154 cm2 , find its radius and circumference. Solution: Here, Area of a circle (A) = 154 cm² Radius (r) = ? We have, Area of circle = πr² or, 154 = πr2 or, r2 = 154 π = 154 × 7 22 = 49 cm² ∴ r = 49 = 7 cm Hence, radius (r) = 7 cm Again, circumference of a circle (C) = 2πr = 2 × 22 7 × 7cm = 44cm. Example 3 If the circumference of a circle is 88cm, find its radius and area. Solution: Here, Circumference (C) = 88 cm We have, 2πr = 88cm or, 2 × 22 7 × r = 88cm or, 44r = 88 × 7 or, r = 88 × 7 44 ∴ r = 14cm Again, Area of a circle (A) = πr2 = 22 7 × (14)2 cm2 = 22 7 × 14 × 14cm2 = 616 cm2 .
84 Oasis School Mathematics-8 Example 4 Find the area of the non-shaded part. Solution: As there are four quarter circle, which are shaded, So one full circle is shaded there. Radius of the circle inside square (r) = 14 2 cm. We have, area of a circle (A) = πr² = 22 7 × (7)² cm² = 154 cm² Area of square = l² = (14)² = 196 cm² ∴ Area of the non-shaded part = 196cm² – 154 cm² = 42 cm² Exercise 8.2 1. Find the area of the following circles. (a) (b) (c) (d) 2. Find the area of the following circles. (a) Radius (r) = 14 cm (b) Radius (r) = 35 cm (c) Diameter (d) = 5.6 cm (d) Diameter (d) = 4.2 cm 3. Find the radius and area of each of the following circles. (a) Circumference (C) = 88 cm (b) Circumference (C) = 440 cm (c) Circumference (C) = 308 cm 4. Find the radius and circumference of circles having (a) Area (A) = 154 cm2 (b) Area (A) = 616 cm2 5. (a) Area of a wheel is 308cm2 . Find. (i) its radius (ii) its circumference (iii) distance covered by it in 10 revolutions. (b) Area of circular wheel is 616cm2 . Find, (i) its radius (ii) its circumference (iii) distance covered by it in 100 revolutions. 14cm r = 7 cm d = 70 cm d = 42 cm
Oasis School Mathematics-8 85 6. Find the area of the shaded region of the given figures. (a) (b) (c) (d) 7. (a) A cow is tied on the pole in the centre of meadow by a rope of length 7m. Find the area of a meadow over which cow can graze? (b) A man runs 20 times around the circular meadow. Altogether he runs 3 km 520 m. Find the area of that meadow. 8. A circular pond having radius 28 m is surrounded by a path of uniform width 3m as shown in the figure. (a) Find the area of the pond. (b) Find the area of the pond and path around it. (c) Find the area of path around it. (d) Find the cost to pave the stone on the path at the rate of 150 per m2 . 9. Radius of a wheel of the cycle is 35 cm. Find: (a) Distance covered by the wheel in one revolution. (b) If it makes 80 revolution per minute, find the distance covered by the wheel in one minute. (c) Speed of the cycle in km per hour. 3 m 28 m Project Work Draw a circle in a hard paper. Divide it into 16 equal parts. Cut all the parts. Arrange them to form a rectangle. Find the area of circle. Answer 1. (a) 154 cm2 (b) 616 cm2 (c) 3850cm2 (d) 1386 cm2 2. (a) 616 cm2 (b) 3850 cm2 (c) 24.64 cm2 , 17.6 cm (d) 13.86cm2 3. (a) 14 cm, 616 cm2 (b) 70 cm, 15400 cm2 (c) 49 cm, 7546 cm2 4. (a) 7 cm, 44 cm (b) 14 cm, 88cm 5. (a) (i) 9.9 cm ii) 62.2 cm (iii) 622 cm (b) (i) 14 cm (ii) 88 cm (iii) 8800 cm 6. (a) 42 cm2 (b) 707.14 cm2 (c) 23.625 cm2 (d) 168 cm2 7. (a) 154 m2 (b) 2464 m2 8. (a) 2464 m2 (b) 3020.28 m2 (c) 556.28 m2 (d) Rs. 83,442.85 9. (a) 3.85 m (b) 308m (c) 18.48 km/hr.
86 Oasis School Mathematics-8 Miscellaneous Exercise 1. (a) Write the formula to calculate the area of quadrilateral. (b) If a diagonal of quadrilateral is 12cm and the length of perpendicular drawn from the opposite vertices to the diagonal are 8cmand 9cm. Show this information in diagram. (c) Find the area of quadrilateral. [Ans: (a), (b) consult your teacher (c) 102cm²] 2. A quadrilateral has its all side equal opposite sides are parallel. (a) Which special type of quadrilateral is this? (b) If its two diagonals are 12cm and 16cm, find its area. (c) If one side of this quadrilateral is 16cm, find the height of the quadrilateral. [Ans: (a) consult you teacher, (b) 96cm², (c) 6cm] 3. (a) Which type of special quadrilateral is shown in the figure? (b) Does BC represent the hight of the figure? Give reason. (c) Calculate the area of given figure. [Ans: (a) consult your teacher, (b) consult your teacher, (c) 156cm²] 4. In the given figure, ABCD is a parallelogram. (a) What is the area of parallelogram ABCD. (b) What is the area of ∆FDC? (c) What is the area of shaded part in the figure? (d) If CE = 4cm, what is the area of the figure ABED? [Ans: (a) 320cm² ,(b) 160cm², (c) 160cm, (d) 352cm²] 5. Study the given figure and answer the questions given below. (a) Find the area of square. (b) Is the area of non shaded part equal to the area of circle having radius 7cm? Given reason. (c) Find the area of non shaded part. (d) Find the area of shaded part. [Ans: (a) 196cm², (b) consult your teacher ,(c) 154 cm², (d) 42cm²] A B D C 12 cm 16 cm 10 cm A F B D C E 16cm 20cm 14 cm
Oasis School Mathematics-8 87 6. A circular meadow having area 616 ft² is surrounded by metallic wire 3 times. (a) Find the radius of the medow. (b) Find the circumference of the meadow. (c) Find the length of wire to surround it 3 times. (d) Cost of wire if the cost per feet is Rs. 9. [Ans: (a) 14ft (b) 88ft (c) 264 ft (d) Rs. 2376] 7. Diameter of a circular pond is 14m. It is surrounded by a path of uniform width 3m. (a) Draw a figure to represent above question. (b) Find the area of the pond. (c) Find the area of pond and path. (d) Find the area of path around the pond. [Ans: (a) consult your teacher (b) 616m² (c) 908.28m² (d) 292.28m²] 8. The structure of the plot of land is as shown in the figure. It has circular pond at the middle (a) Find the area of the whole plot. (b) Find the area of the pond. (c) Find the area of remaining part except pond. (d) Find the cost to grow the grass on the remaining part at the rate of Rs. 35 per square feet. [Ans: (a) 980 ft², (b) 616ft², (c) 364 ft², (d) Rs. 12,740] 9. Observe the given figure and answer the questions given below. (a) Which parallelogram is there in the figure? (b) Does AF represent the height of both triangle ABC and parallelogram EBCD? (c) Find the area of triangle ABC and parallelogram EBCD. [Ans: (a), (b) Consult your teacher, (c) 80cm2 , 160 cm2 ] 30 ft 40 ft 28 ft 20 cm A F B E D C 8 cm
88 Oasis School Mathematics-8 Objective Questions Choose the correct alternatives. 1. If 'a', 'b' and 'c' be the three sides of triangle. Then its semi perimeter 's' is (i) a + b + c (ii) abc (iii) a + b + c 2 2 If 'x' be a side of an equilateral triangle, then its area is (i) 3x (ii) 3x² 4 (iii) 3x 2 3. If the base of a parallelogram is 20cm and the height is 12cm then its area is (i) 240 cm² (ii) 120 cm² (iii) 60 cm² 4. If one pair of opposite sides of a quadrilateral are parallel then it is a (i) trapezium (ii) parallelogram (iii) rhombus 5. Which one of the following represents area of the quadrilateral ABCD (i) 1 2 BD × (AD + BC) (ii) 1 2 BD × (AB + DC) (iii) 1 2 BD × (AE + CF) 6. If two diagonals of a rhombus are 15cm and 20cm, then its area is (i) 300cm² (ii) 150cm² (iii) 600 cm² 7. The area of given figure is (i) 80cm² (ii) 40cm² (iii) 160cm² 8. Formula to calculate the area of a circle is (i) 2πr (ii) πr² (iii) π 9. The area of given data is (i) 616cm² (ii) 308cm² (iii) 154cm² 10. The area of the shaded part of the given figure is (i) 462cm² (ii) 616cm² (iii) 154cm² 11. π is the ratio of (i) the diameter and the radius (ii) circumference and the radius (iii) circumference and the diameter B A D C E F A D B C E 6cm 10cm 5cm 28cm P O 7cm 14cm
Oasis School Mathematics-8 89 Assessment Test Paper Full marks - 23 Attempt all the questions 1. (a) What is the name of given figure? Explain with is features. (1) (b) Write the formula to calculate area of given figure. (1) (c) If AB = 10cm, DC = 16cm, AM = 8cm, find the area of given figure. (2) 2. (a) Write the formula to calculate the area of a circle and the area of a square. (1) (b) In the given figure, find the diameter, radius and area of circle (2) (c) Find the area of square and circle. (2) (d) Find the area of shaded part. (1) 3. (a) Write the formula to calculate the area of a parallelogram and the area of a triangle. (1) (b) Using this formula calculate the area ofparallelogram ABCD and area of ΔEBC. (2) (c) Find the area of shaded part (1) 4. (a) Write the formula to calculate the circumference of a circle. (1) (b) Using this formula calculate the circumference of the wheel of a cycle whose diameter is 28 cm. (1) (c) If the wheel revolves 100 revolutions per minute, find the distance covered by the wheel in one minute. (2) (d) Find the speed of the cycle in km per hour. (1) 5. In the given figure, ABCD is a quadrilateral. (a) What do the parts AC, BM and DN represent? (1) (b) If AC = 10cm, BM = 4 cm, and DN = 6cm, find the area of quadrilateral ABCD. (2) (c) If AB = AD and BC = DC, then what type of special quadrilateral is ABCD? (1) A D B C M A B 20cm 10cm D E C E B A D C N M
90 Oasis School Mathematics-8 Estimated Teaching Hours 30 Contents: • Indices • Factorisation • HCF and LCM • Rational Expressions • Simultaneous Equations • Quadratic Equations Expected Learning outcomes At the end of this unit, students will be able to develop the following competencies. • State and use the properties of indices to solve the simple problems related to it. • Factorise simple algebraic expressions (Binomial andTrinomial) • Find H.C.F and L.C.M of algebraic expressions • Simplify the simple rational expressions • Solve two simultaneous equations using different methods • Solve the quadratic equations. Teaching Materials Flash cards, Chart paper, A4 size paper, etc. Algebra
Oasis School Mathematics-8 91 9.1 Indices Let us consider an algebraic term axb . Here, 'a' is coefficient, 'x' is the base and 'b' is the index (or power) of the base. An index of a base means how many times the base is multiplied. The plural form of index is called indices. Laws of Indices If m and n are two positive integers, then (i) am × an = am+n [Product law] (ii) am ÷ an = am–n [Quotient law] (iv) (am)n = amn [Power law] (v) (ab)m = ambm (vi) ( a b) m = am bm (vii) a0 = 1 [Law of zero index] (viii) n am = a [Root law of index] Worked Out Examples Example 1 Find the products in their exponential form. (i) 3 × 3 × 3 ×3 (ii) (–2) × (–2) × (–2) × (–2) × (–2) (iii) 2 5 × 2 5 × 2 5 Solution: (i) 3 × 3 × 3 × 3 = 34 (ii) (–2) × (–2) × (–2) × (–2) × (–2) = (–2)5 (iii) 2 5 × 2 5 × 2 5 = ( 2 5 ) 3 am bn bn am m n Unit 9 Indices You know! • 2 × 2 × 2 × 2 × 2 = 25 • 2 × 2 × 2 × 2 × 2 × 2 2 × 2 = 24 • 23 × 22 × 24 25 29 25 = = 24
92 Oasis School Mathematics-8 Example 2 Find the value of the following: (i) 32 × 53 (ii) 36 3 2 (iii) ( 8 125) 1 3 Solution: (i) Here, 32 × 53 = 3 × 3 × 5 × 5 × 5 = 9 × 125 = 1125 (ii) Here, 36 3 2 = 62 × 3/2 = 63 = 6 × 6 × 6 = 216 (iii) Here, ( 8 125) 1 3 = ( 23 53 ) 1 3 = 23 × 1 3 53 × 1 3 = 2 5 [∵ ( a b) n = an bn] Example 3 Simplify by using laws of indices. The answer should contain only positive indices. (i) 24 ÷ 26 (ii) a6 × a4 a² (iii) ax–y × ay–z × az–x Solution: (i) Here, 24 ÷ 26 = 24 26 = 24–6 = 2–2 [∵ am an = am–n] = 1 2² = 1 4 [∵ a–m = 1 am ] (ii) Here, a6 × a4 a² = a6+4 a² [∵ am × an = am+n] = a10 a² = a10–2 = a8 . [∵ am an = am–n] (iii) Here, ax–y × ay–z × az–x = ax–y+y–z+z–x [∵ am × an = am+n] = ax–x–y+y–z+z = a0 = 1 Example 4 Evaluate: Solution: 5 9 8 27 9 4 1 3 1 2 × ( ) ÷ ( ) − 5 9 8 27 9 4 5 9 2 3 3 2 1 3 1 2 3 1 3 2 1 2 × ( ) ÷ ( ) = × ( ) ÷ ( ) − × − ( ) x = × ÷ ( ) = × ÷ ( ) = × × = − 5 9 2 3 3 2 5 9 2 3 2 3 5 9 2 3 3 2 5 9 1 1
Oasis School Mathematics-8 93 Example 5 Simplify: (xa – b)a + b × (xa + c)b – c × (xc + a)c – a Solution: x(a –b) (a + b) × x (b+c) (b – c) × x(c+a) (c–a) = xa² – b² × xb² – c² × xc² – a² = xa² – b + b² – c² + c² – a² = x° = 1 Exercise 9.1 1. Express the product in their exponent form: (a) 6 × 6 × 6 × 6 (b) (–b) × (–b) × (–b) × (–b) × (–b) (c) ( 3 5 ) × ( 3 5 ) × ( 3 5 ) (d) x 4 × x 4 × x 4 2. Express the following exponent form into the product form. (a) (2x)3 (b) (– 2 3 ) 2 (c) ( x a ) 3 3. (a) What power of x yields unit value? (b) What index of m will yield 1 m² ? (c) What is the power of 6 so that its value becomes 1 36 ? (d) If ax = 1, what is the value of x? (e) If x = 2, what is the value of 2x ? (f) If x = 2, what is the value of 4 2x ? 4. Simplify by using laws of indices. Answer should contain only positive indices. (a) 34 × 35 (b) a6 × a–5 (c) 45 ÷ 43 (d) c6 ÷ c3 (e) (43 ) 4 × (44 )2 (f) (54 )3 × (52 )3 (g) (p q) 5 × (p q) 2 (h) ( 3 5 ) 2 × ( 3 5 ) 4 (i) (– 1 3 ) 3 × (– 1 3 ) 4 × (– 1 3 ) 5 (j) a² × a3 a4 (k) (4p)3 × (4p)² (4p)4 (l) 4² × 9² × 49 2³ × 7 × 3² (m) 8p2 × 5p3 (n) 4r5 × 4r4 (o) 9x2 y2 × 7x3 y4 5. Simplify: (a) bx-y × by-z× bz-x (b) ax²–y²× ay²–z² × az²–x² (c) (–4x²y)³ × (3x–3y)² (d) (–2ab²c–1)³× (3a²b³c)–1 (e) (2x²y³z-1)²× (3x²y³z)–3 (f) (3x²y³z–1) × (4x–2y3 z2 )–2 (g) ( a³b–5 a²b–3 ) –5
94 Oasis School Mathematics-8 6. (a) If x = 1, y = 2 and z = 3, find the value of each of the following: (i) x–2 (ii) 27xºy6 z–3 (iii) (x5 y)–4 (iv) xy yz zx (b) If a = 2, b = 0, c = 1, d = –1, find the value of (i) a–1.b–2 (ii) b³c3 d2 (iii) ca . ab .db (iv) ab .cd.da 7. Evaluate: (a) 82/3 (b) 323/5 (c) 272/3 (d) (27)–2/3 (e) (–125)–2/3 (f) ( 243 32 ) 2/5 (g) ( 1 216 ) – 1 3 (h) ( 5 4 ) –5 (i) (625)–1/4 (j) ( 1 16 ) 0.25 (k) (40.5)2 (l) ( 9 16) 1 2 × ( 32 243) 3 5 (m) (20)³ × (30)² (40)² × (10)³ (n) 8. Simplify: (a) 3x+1 – 3x 2 × 3x (b) 2x+1 – 2x 3 × 2x (c) 5x+1 – 5x 4 × 5x (d) ax+2 – ax+1 ax+1 –ax 9. Simplify: (a) 128x6 y–3 64x–3y³ (b) (8x6 y–3) 1/3 2xy (c) (4xy)² (2x–3y–3)³ (d) 81x6 y–5z² (3x²y³z)³ 10. Simplify: (a) ax+y ax+2y × ay–z a2x+z (b) x3a+3b x2a–2b × x3a+b x2a+b (c) xa(b + c) × x–b(c+a) × x-c(a + b) (d) (ya+b)a–b × (yb–c)b+c × (yc+a)c-a (e) ( za zb ) a²+ab+b² . ( zb zc ) b²+bc+c² . ( zc za ) c²+ca+a² (f) xa²–b² x2b²–a² × xb²–2a² xb² (g) (xa+b)a–b × (xb+c)b–c (xa+c)a–c (h) ( xa x–b ) a–b × ( xb x–c ) b–a × ( xc x–a ) c–a 26 × 39 × 512 23 × 33 × 125 3 Answer 1. (a) 64 (b) (-b)5 (c) (3/5)3 (d) (x/4)3 2. (a) 2x×2x×2x (b) (-2/3)×(-2/3)×(-2/3) (c) x/a×x/a×x/a 3. (a) 0 (b) (-2) (c) -2 (d) 0 (e) 1 (f) 1 4. (a) 39 (b) a (c) 42 (d) c3 (e) 420 (f) 518 (g) (p/q)7 (h) (3/5)6 (i) (-1/3)12 (j) a (k) 4p (l) 126 (m) 40p5 (n) 16r9 (o) 63x5 y6 5. (a) 1 (b) 1 (c) -192y5 (d) -8/3 ab3 /c4 (e) 4/27x2 y3 z5 (f) 3x6 /16y3 z5 (g) b10/a5 6. (a) (i) 1 (ii) 64 (iii) 1/16 (iv) 24 (b) (i) 0 (ii) 0 (iii) 1 (iv) 1 7 4 8 9 1 9 1 25 9 4 6 1024 3125 1 5 1 2 . ( ) ( ) ( ) ( ) ( ) ( ) ( ) (h) ( ) ( ) ( ) a b c d e f g i j k 4 2 9 9 2 750 8 1 1 3 1 9 2 2 9 6 2 (l) ( ) ( ) . ( ) ( ) / ( ) ( ) . ( ) ( ) ( ) m n a b c d a a x y b x y c x y d y z a a b x c x d e f x z a b bc 11 11 14 1 2 2 5 2 3 10 1 1 1 1 ( ) . ( ) ( ) ( ) ( ) ( ) ( ) ( ) + + x g 3b2 ( ) 1 (h) 1
Oasis School Mathematics-8 95 Unit 10 Factorisation Flash Back Look and learn the following. 5 × 3 = 15, 5 and 3 are the factors of 15. Similarly, 6 × x = 6x, 6 and x are the factors of 6x. (a+b) (a–b) = a²–b², (a+b) and (a–b) are the factors of a²–b². Thus, the product of 6 and x is 6x, the term 6x is called a multiple of 6 and x, whereas 6 and x are the factors of 6x. The process of finding two or more expressions whose product is equal to the given expression is called factorisation. For example: Factorisation of 2a + 4ab = 2a(1 + 2b) Method of factorisation of algebraic expression: Factorisation of a polynomial having a monomial factors Steps: Find by inspection, find the common from the given terms and take common. For example: Factorise: 2x2 y + 6xy2 + 4xy Solution: By inspection, we find that each term of given expression 2x2 y + 6xy2 + 4xy is exactly divisible by 2xy. So, 2xy is a common factor. Here, 2x2 y + 6xy2 + 4xy = 2xy (x + 3y + 2) When the common factor is a polynomial. In this case, we take out the common factor and use distributive property. Look and learn For example : Factorise: (a) x(a + b) + y(a + b) + z(a + b) Solution: Here, x(a + b) + y(a + b) + z(a + b)=(a + b) (x + y + z) If first term is '–', then it is usual to take '–' as common. You know! • Factors and Multiples
96 Oasis School Mathematics-8 (b) ax + by + ay + bx Solution: Here, ax + by + ay + bx = ax + ay + bx + by = a(x + y) + b(x + y) = (x + y) (a + b) Example 1 Factorise : 2a²b + 3ab² Solution: 2a²b + 3ab² = ab(2a + 3b) Example 3 Factorise: 2x + 2y + xy + y² Solution: 2x + 2y + xy + y² = 2(x + y) + y (x + y) = (x + y) (2 + y) 1. Factorise (a) 2x + 6y (b) 15x²y + 25xy2 (c) 3a² – 15ab (d) 6x4 y – 18x3 y2 (e) – 10 x³y²z³ – 15 x²yz (f) – 21x4 a²c³ + 14 x³ac² 2. Factorise (a) 4a + 3ab + 5a² (b) 3x²y – 6xy + 9xy² (c) 4x³y7 – 2xy6 + 6y9 (d) –abx – acy – adz (e) 4x2 y – 6xy2 + 10xy 3. Factorise (a) a(x + y) + b(x + y) (b) a2 (a + 1) + 1(a + 1) (c) y(a – b) – x(a – b) (d) x(a – b) + y(b – a) (e) x(x – 2y) + y(2y – x) 4. Factorise (a) (a – 2) (x+ 4) + (a – 2) (x + 5) (b) (x – y)2 + (x – y) (c) (a – b) – (a – b)2 (d) p(a – b) + q(a – b) + r(a – b) (e) x(a + b + c) + y(a + b + c) + z(a + b + c) Worked Out Examples Example 2 Factorise: x(a – b) + y (a – b) + z (a – b) Solution: x(a – b) + y(a – b) + z (a – b) = (a – b) (x + y + z) Example 4 Simplify : 7 × 16 + 7 × 14 Solution: 7 × 16 + 7 × 14 = 7 (16 + 14) = 7 × 30 = 210 Exercise 10.1