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 be x. 60% of x is girls. There is one girl out of students. 60 Therefore, 60% of x = 18 So, 18 girls are out of how many students? 60 100 × x = 18 OR Number of students = × 18 100 60 18 × 100 or, x = =30 = 30 60 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 = . 1223 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. 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 are 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: Amita said that she would first find Sunita used the unitary method. the increase in the price, which is 20% of 20% increase means, ` 34,000, and then find the new price. ` 100 increased to ` 120. 20% of ` 34000 = ` 20 34000 100 × = ` 6800 OR So, ` 34,000 will increase to? Increased price = ` 120 34000 100 × New price = Old price + Increase = ` 40,800 = ` 34,000 + ` 6,800 = ` 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 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. (ii) Find 10% of this, i.e., ` × 580 = ` 58 . 10 100 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) No, I will spend ` 3 on paper to wrap the gift and tape. So my expenditure is ` 8. 2 This gives me a profit of ` 2, which is, × 100 =25% only. 8 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? 30030 P Profit =×100% = % =10% P% = × 100 3000 3 CP 8.5 Sales Tax/Value Added 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 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 On ` 450, the tax paid would be = ` × 450 100 = ` 22.50 Bill amount = Cost of item + Sales tax = ` 450 + ` 22.50 = ` 472.50. 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 Hence when price including tax is ` 3300, the original price = ` ×3300 = ` 3000. 110 1. Find the buying price of each of the following when 5% ST is added on the purchase of (a) A towel at ` 50 (b) Two bars of soap at ` 35 each (c) 5 kg of flour at ` 15 per kg ST means Sales Tax, which we pay when we buy items. This sales tax 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. These days however, the prices include the tax known as Value Added Tax (VAT). COMPARING QUANTITIES 2. If 8% VAT is included in the prices, find the original price of (a) A TV bought for ` 13,500 (b) A shampoo bottle bought for ` 180 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. 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’. COMPARING QUANTITIES 3. Again find the interest on this sum for another year. 21600 ×8 SI2 = SI at 8% p.a.for 2nd year = ` 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. 20000 ××82 SI for 2 years = ` = ` 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. 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. 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% per annum. Then by the steps mentioned above 5000 ××51 P1 ×× R1 1. SI1 = ` or SI1 = ` 100 100 5000 ××51 PR so, A1 = ` 5000 + or A1 =P1 + SI1 = P1 +1 100 100 ⎛ 5 ⎞⎛ R ⎞ = ` 5000 ⎜1+ ⎟⎠ = P=P1+=P ⎝ 21 ⎜⎟ 2 100 ⎝ 100⎠ ⎛ 5 ⎞ 5×1 P2 ×× R1 2. SI2 = ` 5000 ⎜1+ ⎟× or SI2 = ⎝ 100⎠ 100 100 5000 ×5 ⎛ 5 ⎞⎛ R ⎞ R = ` ⎜1 + =1 + ⎟× ⎟ P1⎜ 100 ⎝ 100⎠⎝ 100⎠ 100 PR ⎛ R ⎞ =1⎜1+⎟ 100 ⎝ 100⎠ ⎛ 5 ⎞ 5000 ×5 ⎛ 5 ⎞ = ` 5000 1 ++` 1+ =P2 + SI2 A2⎜⎟ ⎜⎟ A2 ⎝ 100 ⎠ 100 ⎝ 100 ⎠ ⎛ 5 ⎞⎛ 5 ⎞⎛ R ⎞ R ⎛ R ⎞ = ` 5000 1 ⎜+ ⎜1 + =P1 + ⎟+P1 1 +⎟ ⎟⎟ 1 ⎜⎜ ⎝ 100⎠⎝ 100⎠⎝ 100⎠ 100 ⎝ 100⎠ ⎛ 5 ⎞2 ⎛ R ⎞⎛ R ⎞ = ` = P3 =P1 + 1 + 5000 1 ⎟ 1 ⎟⎜ ⎟ ⎜+ ⎜ ⎝ 100⎠⎝ 100⎠⎝ 100⎠ ⎛ R ⎞2 =P1⎜+ = 3 1 ⎟ P ⎝ 100⎠ Proceeding in this way the amount at the end of n years will be ⎛ R ⎞n A =P1 n1 ⎜+ ⎟ ⎝ 100⎠ ⎛ R ⎞n Or, we can say A= P1⎜+ ⎟ ⎝ 100⎠ 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. Find the time period and rate for each . 1. Asum taken for 11 years at 8% per annum is compounded half yearly. 2. Asum taken for 2 years at 4% per annum compounded half yearly. 2 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: COMPARING QUANTITIES 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. 31 1 year 3 months = 1 12 year = 1 4 years Mayuri tried putting the values in the known formula and came up with: 1 ⎛ 17 ⎞14 A = ` 10000 1 ⎜+ ⎟ ⎝ 200⎠ 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 1 to get simple interest for year more. Thus, 4 ⎛ 17 ⎞ A = ` 10000 ⎜1+⎟ ⎝ 200⎠ 217 = ` 10000 × = ` 10,850 2001 Now this would act as principal for the next year. We find the SI on ` 10,850 4 for 4 year. 1 10850 ××17 4 SI = ` 100 ×2 10850 ××1 17 = ` = ` 230.56 800 Interest for first year = ` 10850 – ` 10000 = ` 850 1 And, interest for the next 4 year = ` 230.56 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) 5 Increase at 5% = × 20000 = 1000 100 Population in 1999 = 20000 + 1000 = 21000 5 Increase at 5% = × 21000 = 1050 100 Population in 2000 = 21000 + 1050 Treat as = 22050 the Principal for the 5 Increase at 5% = × 22050 3rd year. 100 = 1102.5 At the end of 2000 the population = 22050 + 1102.5 = 23152.5 53 ⎛⎞ or, Population at the end of 2000 = 20000 ⎜1+⎟ ⎝⎠ 100 21 21 21 = 20000 ××× 20 20 20 = 23152.5 So, the estimated population = 23153. COMPARING QUANTITIES Aruna asked what is to be done if there is a decrease. The teacher then considered the following example. Example 15: ATV 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 ××51 = ` = ` 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 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 2 % per annum compounded annually. (b) ` 18,000 for 2 1 2 years at 10% per annum compounded annually. (c) ` 62,500 for 1 1 2 years at 8% per annum compounded half yearly. (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 2nd year amount for years). 12 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, find the difference in amounts he would be paying after 11 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 8. Find the amount and the compound interest on ` 10,000 for 1 2 years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually? 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. COMPARING QUANTITIES WHAT HAVE WE DISCUSSED? NOTES