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COMPARING QUANTITIES, , 117, , CHAPTER, , Comparing Quantities, , 8, , 8.1 Recalling Ratios and Percentages, We know, ratio means comparing two quantities., A basket has two types of fruits, say, 20 apples and 5 oranges., Then, the ratio of the number of oranges to the number of apples = 5 : 20., 5, 1, The comparison can be done by using fractions as,, =, 20 4, 1, The number of oranges is th the number of apples. In terms of ratio, this is, 4, 1 : 4, read as, “1 is to 4”, OR, 20 4, = which means, the number of apples, 5 1, is 4 times the number of oranges. This comparison can also be done using percentages., , Number of apples to number of oranges =, , By unitary method:, Out of 25 fruits, number of oranges are 5., So out of 100 fruits, number of oranges, , There are 5 oranges out of 25 fruits., So percentage of oranges is, 5, , 4, , =, , 20, , = 20%, 25 4 100, [Denominator made 100]., , Since, , ×, , OR, =, , 5, 25, , × 100 = 20., , contains only apples and oranges,, , So,, percentage of apples + percentage of oranges = 100, or, percentage of apples + 20 = 100, or, percentage of apples = 100 – 20 = 80, Thus the basket has 20% oranges and 80% apples., Example 1: A picnic is being planned in a school for Class VII. Girls are 60% of the, total number of students and are 18 in number., The picnic site is 55 km from the school and the transport company is charging at the rate, of ` 12 per km. The total cost of refreshments will be ` 4280., 2021–22
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118, , MATHEMATICS, , Can you tell., 1. The ratio of the number of girls to the number of boys in the class?, 2. The cost per head if two teachers are also going with the class?, 3. If their first stop is at a place 22 km from the school, what per cent of the total, distance of 55 km is this? What per cent of the distance is left to be covered?, Solution:, 1. To find the ratio of girls to boys., Ashima and John came up with the following answers., They needed to know the number of boys and also the total number of students., Ashima did this, , John used the unitary method, , Let the total number of students, , There are 60 girls out of 100 students., 100, There is one girl out of, students., 60, So, 18 girls are out of how many students?, , be x. 60% of x is girls., Therefore, 60% of x = 18, , 60, × x = 18, 100, 18 × 100, or, x =, = 30, 60, , OR, , Number of students =, , 100, × 18, 60, , = 30, , Number of students = 30., , So, the number of boys = 30 – 18 = 12., 18 3, Hence, ratio of the number of girls to the number of boys is 18 : 12 or, = ., 12 2, 3, is written as 3 : 2 and read as 3 is to 2., 2, 2. To find the cost per person., Transportation charge = Distance both ways × Rate, = ` (55 × 2) × 12, = ` 110 × 12 = ` 1320, Total expenses = Refreshment charge, + Transportation charge, = ` 4280 + ` 1320, = ` 5600, Total number of persons =18 girls + 12 boys + 2 teachers, = 32 persons, Ashima and John then used unitary method to find the cost per head., For 32 persons, amount spent would be ` 5600., 5600, The amount spent for 1 person = `, = ` 175., 32, 3. The distance of the place where first stop was made = 22 km., 2021–22
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COMPARING QUANTITIES, , 119, , To find the percentage of distance:, Ashima used this method:, , John used the unitary method:, , 22 22 100, =, ×, = 40%, 55 55 100, , Out of 55 km, 22 km are travelled., , She is multiplying, , OR, , 100, =1, 100, and converting to, the ratio by, , Out of 1 km,, , 22, km are travelled., 55, , 22, × 100 km are travelled., 55, That is 40% of the total distance is travelled., , Out of 100 km,, , percentage., , Both came out with the same answer that the distance from their school of the place where, they stopped at was 40% of the total distance they had to travel., Therefore, the percent distance left to be travelled = 100% – 40% = 60%., , TRY THESE, In a primary school, the parents were asked about the number of hours they spend per day, 1, in helping their children to do homework. There were 90 parents who helped for hour, 2, 1, to 1 hours. The distribution of parents according to the time for which,, 2, they said they helped is given in the adjoining figure ; 20% helped for, 1, more than 1 hours per day;, 2, 1, 1, 30% helped for hour to 1 hours; 50% did not help at all., 2, 2, Using this, answer the following:, (i) How many parents were surveyed?, (ii) How many said that they did not help?, 1, (iii) How many said that they helped for more than 1 hours?, 2, , EXERCISE 8.1, 1. Find the ratio of the following., (a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour., (b) 5 m to 10 km, (c) 50 paise to ` 5, 2. Convert the following ratios to percentages., (a) 3 : 4, (b) 2 : 3, 3. 72% of 25 students are interested in mathematics. How many are not interested, in mathematics?, 4. A football team won 10 matches out of the total number of matches they played. If, their win percentage was 40, then how many matches did they play in all?, 5. If Chameli had ` 600 left after spending 75% of her money, how much did she have, in the beginning?, 2021–22
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120, , MATHEMATICS, , 6. If 60% people in a city like cricket, 30% like football and the remaining like other, games, then what per cent of the people like other games? If the total number of, people is 50 lakh, find the exact number who like each type of game., , 8.2 Finding the Increase or Decrease Per cent, We often come across such information in our daily life as., (i) 25% off on marked prices, (ii) 10% hike in the price of petrol, Let us consider a few such examples., Example 2: The price of a scooter was ` 34,000 last year. It has increased by 20%, this year. What is the price now?, Solution:, Sunita used the unitary method., 20% increase means,, ` 100 increased to ` 120., So, ` 34,000 will increase to?, , Amita said that she would first find, the increase in the price, which is 20% of, ` 34,000, and then find the new price., 20, , × 34000, 100, = ` 6800, New price = Old price + Increase, = ` 34,000 + ` 6,800, = ` 40,800, , 20% of ` 34000 = `, , OR, , Increased price = `, , 120, 100, , × 34000, , = ` 40,800, , Similarly, a percentage decrease in price would imply finding the actual decrease, followed by its subtraction the from original price., Suppose in order to increase its sale, the price of scooter was decreased by 5%., Then let us find the price of scooter., Price of scooter = ` 34000, Reduction = 5% of ` 34000, 5, × 34000 = ` 1700, =`, 100, New price = Old price – Reduction, = ` 34000 – ` 1700 = ` 32300, We will also use this in the next section of the chapter., , 8.3 Finding Discounts, Discount is a reduction given on the Marked Price, (MP) of the article., This is generally given to attract customers to buy, goods or to promote sales of the goods. You can find, the discount by subtracting its sale price from its, marked price., So, Discount = Marked price – Sale price, 2021–22
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COMPARING QUANTITIES, , Example 3: An item marked at ` 840 is sold for ` 714. What is the discount and, discount %?, Solution:, , Discount = Marked Price – Sale Price, = ` 840 – ` 714, = ` 126, Since discount is on marked price, we will have to use marked price as the base., On marked price of ` 840, the discount is ` 126., On MP of ` 100, how much will the discount be?, Discount =, , 126, × 100% = 15%, 840, , You can also find discount when discount % is given., Example 4: The list price of a frock is ` 220., A discount of 20% is announced on sales. What is the amount, of discount on it and its sale price., Solution: Marked price is same as the list price., 20% discount means that on ` 100 (MP), the discount is ` 20., By unitary method, on `1 the discount will be `, , 20, ., 100, , 20, × 220 = ` 44, 100, The sale price = (` 220 – ` 44) or ` 176, , On ` 220, discount = `, , Rehana found the sale price like this —, A discount of 20% means for a MP of ` 100, discount is ` 20. Hence the sale price is, ` 80. Using unitary method, when MP is ` 100, sale price is ` 80;, 80, When MP is ` 1, sale price is `, ., Even though the, 100, discount was not, 80, × 220 = ` 176., Hence when MP is ` 220, sale price = `, found, I could find, 100, the sale price, directly., , TRY THESE, 1. A shop gives 20% discount. What would the sale price of each of these be?, (a) A dress marked at ` 120, (b) A pair of shoes marked at ` 750, (c) A bag marked at ` 250, 2. A table marked at ` 15,000 is available for ` 14,400. Find the discount given and, the discount per cent., 3. An almirah is sold at ` 5,225 after allowing a discount of 5%. Find its marked price., 2021–22, , 121
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122, , MATHEMATICS, , 8.3.1 Estimation in percentages, Your bill in a shop is ` 577.80 and the shopkeeper gives a discount of 15%. How would, you estimate the amount to be paid?, (i) Round off the bill to the nearest tens of ` 577.80, i.e., to ` 580., 10, (ii) Find 10% of this, i.e., `, × 580 = ` 58 ., 100, 1, (iii) Take half of this, i.e., × 58 = ` 29 ., 2, (iv) Add the amounts in (ii) and (iii) to get ` 87., You could therefore reduce your bill amount by ` 87 or by about ` 85, which will be, ` 495 approximately., 1. Try estimating 20% of the same bill amount., 2. Try finding 15% of ` 375., , 8.4 Prices Related to Buying and Selling (Profit and Loss), For the school fair (mela) I am going to put a stall of lucky dips. I will charge ` 10 for one, lucky dip but I will buy items which are worth ` 5., So you are making a profit of 100%., No, I will spend ` 3 on paper to wrap the gift and tape. So my expenditure is ` 8., This gives me a profit of ` 2, which is,, , 2, 8, , × 100% = 25% only., , Sometimes when an article is bought, some additional expenses are made while buying or, before selling it. These expenses have to be included in the cost price., These expenses are sometimes referred to as overhead charges. These may include, expenses like amount spent on repairs, labour charges, transportation etc., , 8.4.1 Finding cost price/selling price, profit %/loss%, Example 5: Sohan bought a second hand refrigerator for ` 2,500, then spent ` 500 on, its repairs and sold it for ` 3,300. Find his loss or gain per cent., Solution: Cost Price (CP) = ` 2500 + ` 500 (overhead expenses are added to give CP), = ` 3000, Sale Price (SP) = ` 3300, As SP > CP, he made a profit = ` 3300 – ` 3000 = ` 300, His profit on ` 3,000, is ` 300. How much would be his profit on ` 100?, Profit =, , 300, 30, × 100% = % = 10%, 3000, 3, , 2021–22, , P% =, , P, × 100, CP
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COMPARING QUANTITIES, , 123, , TRY THESE, 1. Find selling price (SP) if a profit of 5% is made on, (a) a cycle of ` 700 with ` 50 as overhead charges., (b) a lawn mower bought at ` 1150 with ` 50 as transportation charges., (c) a fan bought for ` 560 and expenses of ` 40 made on its repairs., Example 6: A shopkeeper purchased 200 bulbs for ` 10 each. However 5 bulbs, were fused and had to be thrown away. The remaining were sold at ` 12 each. Find the, gain or loss %., Solution: Cost price of 200 bulbs = ` 200 × 10 = ` 2000, 5 bulbs were fused. Hence, number of bulbs left = 200 – 5 = 195, These were sold at ` 12 each., The SP of 195 bulbs = ` 195 × 12 = ` 2340, CP is ` 10, He obviously made a profit (as SP > CP)., Profit = ` 2340 – ` 2000 = ` 340, On ` 2000, the profit is ` 340. How much profit is made on ` 100? Profit, 340, × 100% = 17%., 2000, Example 7: Meenu bought two fans for ` 1200 each. She sold one at, a loss of 5% and the other at a profit of 10%. Find the selling price of, each. Also find out the total profit or loss., , =, , Solution: Overall CP of each fan = ` 1200. One is sold at a loss of 5%., This means if CP is ` 100, SP is ` 95., Therefore, when CP is ` 1200, then SP = `, Also second fan is sold at a profit of 10%., It means, if CP is ` 100, SP is ` 110., Therefore, when CP is ` 1200, then SP = `, , 95, × 1200 = ` 1140, 100, , 110, × 1200 = ` 1320, 100, , Was there an overall loss or gain?, We need to find the combined CP and SP to say, whether there was an overall profit or loss., , Total CP = ` 1200 + ` 1200 = ` 2400, Total SP = ` 1140 + ` 1320 = ` 2460, Since total SP > total CP, a profit of ` (2460 – 2400) or ` 60 has been made., , TRY THESE, 1. A shopkeeper bought two TV sets at ` 10,000 each. He sold one at a profit 10%, and the other at a loss of 10%. Find whether he made an overall profit or loss., , 2021–22, , SP is ` 12
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124, , MATHEMATICS, , 8.5 Sales Tax/Value Added Tax/Goods and Services Tax, The teacher showed the class a bill in which the following heads were written., Bill No., , Date, Menu, , S.No., , Item, , Quantity, , Rate, , Amount, , Bill amount, + ST (5%), Total, Sales tax (ST) is charged by the government on the sale of an item. It is collected by the, shopkeeper from the customer and given to the government. This is, therefore, always on, the selling price of an item and is added to the value of the bill. There is another type of tax, which is included in the prices known as Value Added Tax (VAT)., From July 1, 2017, Government of India introduced GST which stands for Goods and, Services Tax which is levied on supply of goods or services or both., , Example 8: (Finding Sales Tax) The cost of a pair of, roller skates at a shop was ` 450. The sales tax charged was, 5%. Find the bill amount., Solution: On ` 100, the tax paid was ` 5., 5, × 450, 100, = ` 22.50, Bill amount = Cost of item + Sales tax = ` 450 + ` 22.50 = ` 472.50., , On ` 450, the tax paid would be = `, , Example 9: (Value Added Tax (VAT)) Waheeda bought an air cooler for ` 3300, including a tax of 10%. Find the price of the air cooler before VAT was added., Solution: The price includes the VAT, i.e., the value added tax. Thus, a 10% VAT, means if the price without VAT is ` 100 then price including VAT is ` 110., Now, when price including VAT is ` 110, original price is ` 100., 100, × 3300 = ` 3000., Hence when price including tax is ` 3300, the original price = `, 110, Example 10: Salim bought an article for ` 784 which included GST of 12% . What is, the price of the article before GST was added?, Solution: Let original price of the article be ` 100. GST = 12%., Price after GST is included = ` (100+12) = ` 112, When the selling price is ` 112 then original price = ` 100., When the selling price is ` 784, then original price = ` 100 × 784 = ` 700, 12, 2021–22
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COMPARING QUANTITIES, , THINK, DISCUSS AND WRITE, 1. Two times a number is a 100% increase in the number. If we take half the number, what would be the decrease in per cent?, 2. By what per cent is ` 2,000 less than ` 2,400? Is it the same as the per cent by, which ` 2,400 is more than ` 2,000?, , EXERCISE 8.2, 1. A man got a 10% increase in his salary. If his new salary is ` 1,54,000, find his, original salary., 2. On Sunday 845 people went to the Zoo. On Monday only 169 people went. What, is the per cent decrease in the people visiting the Zoo on Monday?, 3. A shopkeeper buys 80 articles for ` 2,400 and sells them for a profit of, 16%. Find the selling price of one article., 4. The cost of an article was ` 15,500. ` 450 were spent on its repairs. If it is, sold for a profit of 15%, find the selling price of the article., 5. A VCR and TV were bought for ` 8,000 each. The shopkeeper made a, loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss, percent on the whole transaction., 6. During a sale, a shop offered a discount of 10% on the, marked prices of all the items. What would a customer, have to pay for a pair of jeans marked at ` 1450 and, two shirts marked at ` 850 each?, 7. A milkman sold two of his buffaloes for ` 20,000 each., On one he made a gain of 5% and on the other a loss of, 10%. Find his overall gain or loss. (Hint: Find, CP of each), 8. The price of a TV is ` 13,000. The sales tax charged on, it is at the rate of 12%. Find the amount that Vinod will, have to pay if he buys it., 9. Arun bought a pair of skates at a sale where the discount given was 20%. If the, amount he pays is ` 1,600, find the marked price., 10. I purchased a hair-dryer for ` 5,400 including 8% VAT. Find the price before VAT, was added., 11. An article was purchased for ` 1239 including GST of 18%. Find the price of the, article before GST was added?, , 8.6 Compound Interest, You might have come across statements like “one year interest for FD (fixed deposit) in, the bank @ 9% per annum” or ‘Savings account with interest @ 5% per annum’., , 2021–22, , 125
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126, , MATHEMATICS, , Interest is the extra money paid by institutions like banks or post offices on money, deposited (kept) with them. Interest is also paid by people when they borrow money., We already know how to calculate Simple Interest., Example 10: A sum of ` 10,000 is borrowed at a rate of interest 15% per annum for 2, years. Find the simple interest on this sum and the amount to be paid at the end of 2 years., Solution: On ` 100, interest charged for 1 year is ` 15., 15, × 10000 = ` 1500, 100, Interest for 2 years = ` 1500 × 2 = ` 3000, Amount to be paid at the end of 2 years = Principal + Interest, = ` 10000 + ` 3000 = ` 13000, , So, on ` 10,000, interest charged =, , TRY THESE, Find interest and amount to be paid on ` 15000 at 5% per annum after 2 years., My father has kept some money in the post office for 3 years. Every year the money, increases as more than the previous year., We have some money in the bank. Every year some interest is added to it, which is, shown in the passbook. This interest is not the same, each year it increases., Normally, the interest paid or charged is never simple. The interest is calculated on the, amount of the previous year. This is known as interest compounded or Compound, Interest (C.I.)., Let us take an example and find the interest year by year. Each year our sum or, principal changes., Calculating Compound Interest, A sum of ` 20,000 is borrowed by Heena for 2 years at an interest of 8% compounded, annually. Find the Compound Interest (C.I.) and the amount she has to pay at the end of, 2 years., Aslam asked the teacher whether this means that they should find the interest year by, year. The teacher said ‘yes’, and asked him to use the following steps :, 1. Find the Simple Interest (S.I.) for one year., Let the principal for the first year be P1. Here, P1 = ` 20,000, 20000 × 8, SI1 = SI at 8% p.a. for 1st year = `, = ` 1600, 100, 2. Then find the amount which will be paid or received. This becomes principal for the, next year., Amount at the end of 1st year = P1 + SI1 = ` 20000 + ` 1600, = ` 21600 = P2 (Principal for 2nd year), 2021–22
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COMPARING QUANTITIES, , 127, , 3. Again find the interest on this sum for another year., SI2 = SI at 8% p.a.for 2nd year = `, , 21600 × 8, 100, , = ` 1728, 4. Find the amount which has to be paid or received at the end of second year., Amount at the end of 2nd year = P2 + SI2, = ` 21600 + ` 1728, = ` 23328, Total interest given = ` 1600 + ` 1728, = ` 3328, Reeta asked whether the amount would be different for simple interest. The teacher, told her to find the interest for two years and see for herself., SI for 2 years = `, , 20000 × 8 × 2, = ` 3200, 100, , Reeta said that when compound interest was used Heena would pay ` 128 more., Let us look at the difference between simple interest and compound interest. We start, with ` 100. Try completing the chart., Under, Simple Interest, First year, , Second year, , Principal, , `, , 100.00, , `, , 100.00, , Interest at 10%, , `, , 10.00, , `, , 10.00, , Year-end amount, , `, , 110.00, , `, , 110.00, , Principal, , `, , 100.00, , `, , 110.00, , Interest at 10%, , `, , 10.00, , `, , 11.00, , `, , 121.00, , Year-end amount, Third year, , Under, Compound Interest, , ` (110 + 10) = ` 120, , Principal, , `, , 100.00, , `, , 121.00, , Interest at 10%, , `, , 10.00, , `, , 12.10, , `, , 133.10, , Year-end amount, , ` (120 + 10) = ` 130, , Note that in 3 years,, Interest earned by Simple Interest = ` (130 – 100) = ` 30, whereas,, Interest earned by Compound Interest = ` (133.10 – 100) = ` 33.10, Note also that the Principal remains the same under Simple Interest, while it changes, year after year under compound interest., 2021–22, , Which, means you, pay interest, on the, interest, accumulated, till then!
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128, , MATHEMATICS, , 8.7 Deducing a Formula for Compound Interest, Zubeda asked her teacher, ‘Is there an easier way to find compound interest?’ The teacher, said ‘There is a shorter way of finding compound interest. Let us try to find it.’, Suppose P1 is the sum on which interest is compounded annually at a rate of R%, per annum., Let P1 = ` 5000 and R = 5. Then by the steps mentioned above, , 1., , SI1 = `, , 5000 × 5 × 1, 100, , so, A1 = ` 5000 +, , 5000 × 5 × 1, 100, , SI1 = `, , or, , A1 = P1 + SI1 = P1 +, , 5 , , = P2, = ` 5000 1 +, 100 , 2., , or, , SI2 =, , 5000 × 5 , 5 , 1 +, , 100, 100 , , P2 × R × 1, 100, , R , R, , = P1 1 +, ×, , 100 100, =, , 5 , 5000 × 5, , A2 = ` 5000 1 +, + `, 100, 100 , , 5 , , = ` 5000 1 +, 100 , , P1R, 100, , R , , = P2, = P1 1 +, 100 , , 5 5 ×1, , ×, SI2 = ` 5000 1 +, 100 100, =`, , P1 × R × 1, 100, , or, , 5 , , 1 +, , 100 , , P1R , R , 1 +, , 100, 100 , , A2 = P2 + SI2, , 5 , , 1 +, , 100 , , R , R , R , , + P1, 1+, = P1 1 +, , , , 100 , 100 100 , , 2, , R , , = P1 1 +, 100 , , 5 , , = ` 5000 1 +, = P3, 100 , , R , , 1 +, , 100 , 2, , R , , = P1 1 +, = P3, 100 , Proceeding in this way the amount at the end of n years will be, , R , , An = P1 1 +, 100 , , Or, we can say, , R , , A = P 1 +, 100 , 2021–22, , n, , n
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COMPARING QUANTITIES, , 129, , So, Zubeda said, but using this we get only the formula for the amount to be paid at, the end of n years, and not the formula for compound interest., Aruna at once said that we know CI = A – P, so we can easily find the compound, interest too., Example 11: Find CI on ` 12600 for 2 years at 10% per annum compounded, annually., n, R , , 1, +, Solution: We have, A = P , , where Principal (P) = ` 12600, Rate (R) = 10,, 100 , Number of years (n) = 2, 2, , 2, 10 , , 11, = ` 12600 1 +, = ` 12600 , 100 , 10 , , 11 11, ×, = ` 15246, 10 10, CI = A – P = ` 15246 – ` 12600 = ` 2646, , TRY THESE, 1. Find CI on a sum of ` 8000 for, 2 years at 5% per annum, compounded annually., , = ` 12600 ×, , 8.8 Rate Compounded Annually or Half Yearly, (Semi Annually), Time period and rate when interest not compounded, You may want to know why ‘compounded, annually’ was mentioned after ‘rate’. Does it, mean anything?, It does, because we can also have interest, rates compounded half yearly or quarterly. Let, us see what happens to ` 100 over a period of, one year if an interest is compounded annually, or half yearly., P = ` 100 at 10% per, annum compounded annually, The time period taken is 1 year, , 100 × 10 × 1, = Rs 10, 100, A = ` 100 + ` 10, = ` 110, , I=`, , annually, The time period after which the interest is added each, time to form a new principal is called the conversion, period. When the interest is compounded half yearly,, there are two conversion periods in a year each after 6, months. In such situations, the half yearly rate will be, half of the annual rate. What will happen if interest is, compounded quarterly? In this case, there are 4, conversion periods in a year and the quarterly rate will, be one-fourth of the annual rate., , P = ` 100 at 10% per annum, compounded half yearly, The time period is 6 months or, , 100 × 10 ×, , 1, year, 2, , 1, 2, , =`5, 100, A = ` 100 + ` 5 = ` 105, Now for next 6 months the P = ` 105, 1, 105 × 10 ×, 2 = ` 5.25, So, I = `, 100, I=`, , and A = ` 105 + ` 5.25 = ` 110.25, 2021–22, , Rate, becomes, half
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130, , MATHEMATICS, , Do you see that, if interest is compounded half yearly, we compute the interest two, times. So time period becomes twice and rate is taken half., , TRY THESE, Find the time period and rate for each ., 1, 1. A sum taken for 1 years at 8% per annum is compounded half yearly., 2, 2. A sum taken for 2 years at 4% per annum compounded half yearly., , THINK, DISCUSS AND WRITE, A sum is taken for one year at 16% p.a. If interest is compounded after every three, months, how many times will interest be charged in one year?, 1, Example 12: What amount is to be repaid on a loan of ` 12000 for 1 years at 10%, 2, per annum compounded half yearly., , Solution:, Principal for first 6 months = ` 12,000, , Principal for first 6 months = ` 12,000, , 1, years., 2, Therefore, compounding has to be done 3 times., , There are 3 half years in 1, , Time = 6 months =, Rate = 10%, , 12000 × 10 ×, Rate of interest = half of 10%, , I=`, , n, , 5 , , = ` 12000 1 +, 100 , , 3, , 21 21 21, ×, ×, 20 20 20, = ` 13,891.50, , = ` 12000 ×, , 1, 2 = ` 600, , 100, A = P + I = ` 12000 + ` 600, , = 5% half yearly, , R , , A = P 1 +, 100 , , 6, 1, year = year, 12, 2, , = ` 12600. It is principal for next 6 months., 1, 12600 × 10 ×, 2 = ` 630, I=`, 100, Principal for third period = ` 12600 + ` 630, = ` 13,230., 1, 13230 × 10 ×, 2 = ` 661.50, I=`, 100, A = P + I = ` 13230 + ` 661.50, = ` 13,891.50, , 2021–22
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COMPARING QUANTITIES, , TRY THESE, Find the amount to be paid, 1. At the end of 2 years on ` 2,400 at 5% per annum compounded annually., 2. At the end of 1 year on ` 1,800 at 8% per annum compounded quarterly., Example 13: Find CI paid when a sum of ` 10,000 is invested for 1 year and, 1, 3 months at 8 % per annum compounded annually., 2, , Solution: Mayuri first converted the time in years., 1 year 3 months = 1, , 3, 1, year = 1 years, 12, 4, , Mayuri tried putting the values in the known formula and came up with:, 17 , , A = ` 10000 1 +, 200 , , 1, , 1, 4, , Now she was stuck. She asked her teacher how would she find a power which is fractional?, The teacher then gave her a hint:, Find the amount for the whole part, i.e., 1 year in this case. Then use this as principal, to get simple interest for, , 1, year more. Thus,, 4, , 17 , , A = ` 10000 1 +, , 200 , , = ` 10000 ×, Now this would act as principal for the next, for, , 217, = ` 10,850, 200, , 1, year. We find the SI on ` 10,850, 4, , 1, year., 4, , 1, × 17, 4, 100 × 2, , 10850 ×, SI = `, , = `, , 10850 × 1 × 17, = ` 230.56, 800, , 2021–22, , 131
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132, , MATHEMATICS, , Interest for first year = ` 10850 – ` 10000 = ` 850, And, interest for the next, , 1, year = ` 230.56, 4, , Therefore, total compound Interest = 850 + 230.56 = ` 1080.56., , 8.9 Applications of Compound Interest Formula, There are some situations where we could use the formula for calculation of amount in CI., Here are a few., (i) Increase (or decrease) in population., (ii) The growth of a bacteria if the rate of growth is known., (iii) The value of an item, if its price increases or decreases in the intermediate years., Example 14: The population of a city was 20,000 in the year 1997. It increased at, the rate of 5% p.a. Find the population at the end of the year 2000., Solution: There is 5% increase in population every year, so every new year has new, population. Thus, we can say it is increasing in compounded form., Population in the beginning of 1998 = 20000 (we treat this as the principal for the 1st year), Increase at 5% =, , 5, × 20000 = 1000, 100, , Population in 1999 = 20000 + 1000 = 21000, Increase at 5% =, , 5, × 21000 = 1050, 100, , Population in 2000 = 21000 + 1050, = 22050, Increase at 5% =, , Treat as, the Principal, for the, 3rd year., , 5, × 22050, 100, , = 1102.5, At the end of 2000 the population = 22050 + 1102.5 = 23152.5, or,, , 5 , , Population at the end of 2000 = 20000 1 +, , 100 , = 20000 ×, = 23152.5, So, the estimated population = 23153., 2021–22, , Treat as, the Principal, for the, 2nd year., , 3, , 21 21 21, ×, ×, 20 20 20
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COMPARING QUANTITIES, , Aruna asked what is to be done if there is a decrease. The teacher then considered, the following example., Example 15: A TV was bought at a price of ` 21,000. After one year the value of the, TV was depreciated by 5% (Depreciation means reduction of value due to use and age of, the item). Find the value of the TV after one year., Solution:, Principal = ` 21,000, Reduction = 5% of ` 21000 per year, 21000 × 5 × 1, = ` 1050, 100, value at the end of 1 year = ` 21000 – ` 1050 = ` 19,950, Alternately, We may directly get this as follows:, , =`, , 5 , , , value at the end of 1 year = ` 21000 1 −, 100 , 19, = ` 21000 ×, = ` 19,950, 20, , TRY THESE, 1. A machinery worth ` 10,500 depreciated by 5%. Find its value after one year., 2. Find the population of a city after 2 years, which is at present 12 lakh, if the rate, of increase is 4%., , EXERCISE 8.3, 1. Calculate the amount and compound interest on, 1, (a) ` 10,800 for 3 years at 12 % per annum compounded annually., 2, 1, (b) ` 18,000 for 2 years at 10% per annum compounded annually., 2, 1, (c) ` 62,500 for 1 years at 8% per annum compounded half yearly., 2, (d) ` 8,000 for 1 year at 9% per annum compounded half yearly., (You could use the year by year calculation using SI formula to verify)., (e) ` 10,000 for 1 year at 8% per annum compounded half yearly., 2. Kamala borrowed ` 26,400 from a Bank to buy a scooter at a rate of 15% p.a., compounded yearly. What amount will she pay at the end of 2 years and 4 months to, clear the loan?, (Hint: Find A for 2 years with interest is compounded yearly and then find SI on the, 4, years)., 2nd year amount for, 12, 2021–22, , 133
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134, , MATHEMATICS, , 3. Fabina borrows ` 12,500 at 12% per annum for 3 years at simple interest and, Radha borrows the same amount for the same time period at 10% per annum,, compounded annually. Who pays more interest and by how much?, 4. I borrowed ` 12,000 from Jamshed at 6% per annum simple interest for 2 years., Had I borrowed this sum at 6% per annum compound interest, what extra amount, would I have to pay?, 5. Vasudevan invested ` 60,000 at an interest rate of 12% per annum compounded, half yearly. What amount would he get, (i) after 6 months?, (ii) after 1 year?, 6. Arif took a loan of ` 80,000 from a bank. If the rate of interest is 10% per annum,, 1, find the difference in amounts he would be paying after 1 years if the interest is, 2, (i) compounded annually., (ii) compounded half yearly., 7. Maria invested ` 8,000 in a business. She would be paid interest at 5% per annum, compounded annually. Find, (i) The amount credited against her name at the end of the second year., (ii) The interest for the 3rd year., 1, years at 10% per, 2, annum, compounded half yearly. Would this interest be more than the interest he, would get if it was compounded annually?, , 8. Find the amount and the compound interest on ` 10,000 for 1, , 1, 9. Find the amount which Ram will get on ` 4096, if he gave it for 18 months at 12 %, 2, per annum, interest being compounded half yearly., 10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum, (i) find the population in 2001., (ii) what would be its population in 2005?, 11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the, rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially, 5, 06,000., 12. A scooter was bought at ` 42,000. Its value, depreciated at the rate of 8% per annum., Find its value after one year., , 2021–22
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COMPARING QUANTITIES, , 135, , WHAT HAVE WE DISCUSSED?, 1. Discount is a reduction given on marked price., Discount = Marked Price – Sale Price., 2. Discount can be calculated when discount percentage is given., Discount = Discount % of Marked Price, 3. Additional expenses made after buying an article are included in the cost price and are known, as overhead expenses., CP = Buying price + Overhead expenses, 4. Sales tax is charged on the sale of an item by the government and is added to the Bill Amount., Sales tax = Tax% of Bill Amount, 5. GST stands for Goods and Services Tax and is levied on supply of goods or services or both., 6. Compound interest is the interest calculated on the previous year’s amount (A = P + I), 7. (i) Amount when interest is compounded annually, , R , , = P 1 +, ;, 100 , n, , P is principal, R is rate of interest, n is time period, , (ii) Amount when interest is compounded half yearly, , R , , = P 1 +, , , 200 , , 2n, , R, is half yearly rate and, 2, 2n = number of ’half-years’, , 2021–22
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136, , MATHEMATICS, , NOTES, , 2021–22
