Maths new edition

["Sales Tax and Cost of Living Index 18.13 CONCEPT APPLICATION Level 1 \t1.\t (i)\tSales tax paid by the shopkeeper = 462 \u00d7 100 . \t\t(iii)\tUsing the given information, find the total 105 amount paid. \t\t(ii)\tSelling price = `(462 - sales tax). \t9.\t (i)\tFind the total expenditure in 2000 and 2007. \t\t(iii)\tProfit % = Profit \u00d7 100. \t\t(ii)\tUse, cost of living index CP Expenditure in 2007 \t2.\t (i)\tLet the list price be `x. = Expenditure in 2000 \u00d7 100. \t\t(ii)\tx + 12% of x = 3360, find x, then find \t10.\t Cost of living index (3360 - x). \t3.\t List price = Selling price - Sales tax. = Total expenditure in the current year \u00d7 100. Total expenditure in the base year \t4.\t (i)\tLet the list price be `x. \t\t(ii)\tx + 12% of x = 3360, find x, then find \t11.\t (i)\tLet the list price be `x. (3360 - x). \t\t(ii)\tx + 8% of x = 13,500, find x. \t5.\t (i)\tLet the list price be `x. \t12.\t Cost of Living Index \t\t(ii)\tx + 12% of x = 3360, find x, then find (3360 - x). = Total expenditure in the current year \u00d7 100. Total expenditure in the base year \t6.\t (i)\tSales tax = [(616 - 50) - 500] = 66 \t\t(ii)\tPercentage of sales tax = Sales tax \u00d7 100. \t13.\t Cost of Living Index SP = Total expenditure in the current year \u00d7 100. Hints and Explanation \t7.\t (i)\tApply the formula for finding the cost of living Total expenditure in the base year index. \t14.\t (i)\tLet the total expenditure in 2000 and 2006 be \t\t(ii)\t34xx++4yy \u00d7 100 = 175, simplify and get the rela- `5x and `8x, respectively. tion between x and y. \t\t(ii)\tUse formula to find cost of the living index. \t\t(iii)\tNow, replace x or y in (3x + 4y) and verify 1\t 5.\t (i)\tCost of living index for the year 2006 from the options. = 210 \u00d7 100 , where the base year\u2019s value is 100. \t8.\t (i)\tCalculate the price after 6% discount. 120 \t\t(ii)\tFind 10% VAT on the price. \t\t(ii)\tUse formula to find total expenditures in 2006 and 2007. Level 2 \t17.\t Find the expenditure in the given 2\t 0.\t Find the sales tax in rupees. years and cost of living index \t21.\t Total tax (in `) = Total expenditure in the current year \u00d7 100. = 10 (8360) + 8 (8650) + 9 (500) Total expenditure in the base year 100 100 100 \t18.\t First, deduct the discount from the list price, and = 836 + 692 + 45 = 1573. then find the sales tax on the remaining amount. \t22.\t List price = `650. \t19.\t Let the discount be `x, then \t\tSelling price including sales tax = `728 (5675 \u2212 x)\u00d7 110 = 15675. 100","18.14 Chapter 18 \t\tLet the sales tax rate be r\u2009%. \t\tThe price of the mobile including sales tax (in `) \t\tSales tax (in `) = 728 \u2013 650 = 78 = 6075 \u2212 x + 1 (6075 \u2212 x) = 10 (6075 \u2212 x) 9 9 \t\tThe rate of sales tax = 78 \u00d7 100% = 12%. 650 10 \t\t\u2234 9 (6075 \u2212 x) = 6075 \t23.\t List price = `62,400 6075 = 10x \t\tDis=count (in `) 12=000 (62,400) 12,480 x = 607.50. \t\tPrice after discount (in `) = 49,920 \t25.\t Let the list price of the TV be `x. \t\tSal=es tax (in `) 11=000 (49,920) 4992 \t\tSales tax = ` 12 x \t\tAmount to be paid (in `) = 49,920 + 4992 = 100 54,912. 12 100 \t24.\t Let the discount be `x. \u2234 x + x = 15, 680 \t\t \t\tPrice after discount = `(6075 \u2013 x) = 15, 680 = 14, 000 x 1.12 1 11 9 1 \t\tAdditional amount to be paid (in `) = (14 \u2013 12)% 9 of x = 2% of 14,000 = 280. \t\tSales tax (in `) = 100 (6075 \u2212 x) = (6075 \u2212 x) Level 3 Hints and Explanation \t26.\t Find the list price and then proceed. (Refer to Q2) 3\t 2.\t Let its list price be `x. 2\t 7.\t (i)\tFind the total expenditures in 2006 and 2007. \t\tDiscount (=in `) 11=000 x 0.1x \t\tPrice after discount = `0.9x \t\t(ii)\tRefer to the Answer Keys for Q.13. \t30.\t Find the sales tax of each item as per their rates and \t\tSales=tax (in `) 1=080 (0.9x) 0.072x proceed. \t\tThe price including sales tax (in `) = 0.9x + 0.072x \t31.\t Let the list price be `x. = 0.972x \t\tSales tax (i=n `) 1=080 x 2 x. \t\t0.972x = 47142 \u21d2 x = 48,500. 25 3\t 3.\t Cost price = `360 \t\tSelling price including sales tax (in `) \t\tProfit = `90 = 27 x. \t\tThe price excluding sales tax = `360 + 90 = `450 25 \t\tSales tax = y (`450) = `4.5y 27 100 \t\t25 x = 351 \t\tSelling price including sales tax (in `) = 450 + 4.5y x = 13 \t\t450 + 4.5y = 495 \u21d2 y = 10. 25 \t34.\t Let the total expenditures for the year 2002 and x = 325 2007 be `4x and `7x, respectively. \t\tProfit percentage = 325 \u2212 300 (100)% = 8 1 %. \t\tCost of livin=g index 47=xx (100) 175. 300 3","Sales Tax and Cost of Living Index 18.15 \t35.\t 225 = 7x + y (100) x = 800 2x + 7y \u2234 VAT (in `) = 9 x = 144. 50 450x + 1575y = 700x + 100y \t\t1475y = 250x \t38.\t List price = `750 5.9y = x \t\tDiscou=nt (in `) 1=080 (750) 60 \t\tTotal expenditure in 2006 (in `) = 2x + 7y = \t\tPrice after discount = `690 2(5.9y) + 7y = 18.8y \t\tVAT pa=id (in `) 11=050 (690) 103.50 =\t\t 1=8180y 94y 5 \t\ty is a positive integer. \t\tAmount paid (in `) = 690 + 103.50 \t\t = 793.50. \t\tOnly option (a) satisfies this condition. 3\t 9.\t List price = `20,000 \t\t(\u2235 94y = 37600\u2234 y = 2000. In other options, y is 5 \t\tDis=count (in `) 1=040 (20,000) 800 not a positive integer). \t\tPrice after discount = `19,200 \t36.\t Let the cost of living index in 2007 be x. \t\tSal=es tax (in `) 1=050 (19, 200) 960 \t\tThe total expenditure of the family in 2007 \t\tAmount to be paid (in `) = 19,200 + 960 = 20,160 x \uf8ed\uf8eb\uf8ec1 + 25 \uf8f6 = 160 100 \uf8f8\uf8f7 \t\tx = 160 = 128 5 1.25 100 \t\tDiscount calculated (in `) = (20, 000) = 1000 Hints and Explanation y 15, 000 (100) = 128 \u21d2 y = 19, 200. \t37.\t Let the list price be `x. \t\tCalculated price after discount = `19,000 \t\tVAT (=in `) 11=080 x 9 x \t\tSales tax calculated (in `) = 4 (19, 000) 50 100 \t\tSelling price including VAT (in `) \t\t = 760 \t\tAmount actually paid (in `) = 19,000 + 760 = x + 9 x = 59 x \t\t = 19,760 50 50 59 x = 944 \t\t \u2234 The customer must have paid (20,160 \u2013 19,760), \t\t50 i.e., `400 less.","1192CChhaapptteerr aSKinimndepCmleoamtIincptseoruensdt Interest Figure 1.1 REmEmBER Before beginning this chapter, you should be able to: \u2022 Calculate percentage \u2022 Understand interest rates and time period \u2022 Compute formulae for simple interest KEY IDEaS After completing this chapter, you should be able to: \u2022 Review the concept of simple interest and solve word problems on it \u2022 Learn practical applications of simple and compound interest \u2022 Understand functions of compound interest and calculate compound interest \u2022 Know about compounding more than a year","19.2 Chapter 19 INTEREST Interest is paid to the lender by the borrower for using money for a specified period of time. Various terms and their general representations are as follows: 1.\t Principal: The original sum borrowed and denoted by P (expressed in rupees) 2.\t \u0007Time: T\u0007 he time for which money is borrowed. It is denoted by n. (n is expressed in number of periods, which is normally one year) 3.\t \u0007Rate: T\u0007 he rate at which interest is calculated on the original sum. It is denoted by r. (calculated per `100 per year) 4.\t Amount: The sum of principal and interest. It is denoted by A. Simple Interest When interest is calculated every year (or every time period) on the original principal, i.e., the sum at the beginning of the first year, such interest is called simple interest. Here, year after year, even though the interest gets accumulated and is due to the lender, this accumulated interest is not taken into account for the purpose of calculating interest for the later years. Simple interest = PNR ; where P, N and R are explained above. 100 Total amount, A=P + PNR = P \uf8eb\uf8ed\uf8ec1 + NR \uf8f6 . 100 100 \uf8f7\uf8f8 Compound Interest In compound interest, the interest is added to the principal at the end of each period to arrive at the new principal for the next period. In other words, the amount at the end of the first year (or period) will become the principal for the second year (or period); the amount at the end of the second year (or period) becomes the principal for the third year (or period), and so on. If P denotes the principal at the beginning of period 1, then P at the beginning of year 2 = P \uf8eb 1 + r\uf8f6 = PR. \uf8ec\uf8ed 100 \uf8f8\uf8f7 P at the beginning of year 3 = P \uf8eb\uf8ed\uf8ec1 + r \uf8f62 = PR 2 . 100 \uf8f8\uf8f7 P at the beginning of year (n + 1) = P \uf8eb 1 + r \uf8f6n = PRn ; where R = \uf8f1\uf8f21 + \uf8eb r \uf8f6\uf8fc . \uf8ed\uf8ec 100 \uf8f8\uf8f7 \uf8f3 \uf8ec\uf8ed 100 \uf8f8\uf8f7\uf8fd\uf8fe Hence, the amount after n years (or periods) = PRn = A. Interest = I = A \u2013 P = P[Rn \u2013 1]. Example 19.1 Find the simple interest on a sum of `1000 at an interest rate of 6% per annum, for a period of 6 years. Solution PNR 100 The formula for simple interest is I= ; where P, the principal = `1000; R the rate of interest = 6%, and N is the number of years = 6. SI = 1000 \u00d7 6 \u00d7 6 = `360. 100","Simple Interest and Compound Interest 19.3 Example 19.2 If `5000 becomes `5700 in a year\u2019s time at simple interest, what will `7000 become at the end of 5 years at the same rate of interest? Solution Amount = Principal + Interest Principal = `5000 Therefore, 5700 = 5000 + Interest Interest = `700 PNR 5000 \u00d7 R 1 100 100 I = \u21d2 700 = \u00d7 \u21d2 R = 14%. Therefore, the rate of interest = 14% per annum \u0007\\\\ The simple interest on `7000 at the rate of 14% for a period of 5 years is (7000 \u00d7 14 \u00d7 5)\/100 = `4900. Therefore, `7000 at the end of 5 years, amounts to `7000 + 4900 = `11,900. The following table gives an example of how simple interest and compound interest operate, i.e., how the principal is for various years under simple interest and compound interest. A principal at the beginning of 1st year, of `100 and a rate of 10% per annum are considered. The details are worked out for three years and shown below: (All figures pertaining to principal, interest and amount are in rupees) Under Simple Interest Under Compound Interest Principal Interest Amount Principal Interest Amount at the till the at the at the till the at the Begining Interest End End Begining Interest End End of the for the of the of the of the for the of the of the Year Year (`) Year (`) Year (`) Year (`) Year (`) Year (`) Year (`) Year (`) 1 100 10 10 110 100 10 10 110 2 100 10 20 120 110 11 21 121 3 100 10 30 130 121 12.1 33.1 133.1 As we observe in the table: In case of simple interest, \u2022\u2022 The principal remains same every year. \u2022\u2022 The interest for any year is same as that for any other year. In case of compound interest, \u2022\u2022 The amount at the end of a year is the principal for the next year. \u2022\u2022 The interest for different years is not the same. The compound interest for the first year (where compounding is done every year) is the same as simple interest for one year. Example 19.3 The cost of an electronic device reduces at the rate of 5% per annum. If its present cost is `32,490, what was its cost before two years? Choose the correct answer from the following options: (a) `37,000\t (b) `35,400\t (c) `36,000\t (d) `37,200","19.4 Chapter 19 Solution 32,490 = P \uf8ef\uf8ee\uf8f01 \u2212 R \uf8f92 100 \uf8fa\uf8fb 32,490 = P \uf8ee\uf8ef\uf8f01 \u2212 5 \uf8f92 100 \uf8fb\uf8fa \u21d2 32,490 = P \uf8ee 20 \u2212 1\uf8f92 \uf8ef\uf8f0 20 \uf8fb\uf8fa \u21d2 32,490 = P \u00d7 19 \u00d7 19 20 20 \u2234P = `36,000. Example 19.4 2 1 years under simple interest at 9% per annum earned an A certain sum lent for a period of 2 interest of `234. From the following options, find the sum that was lent. (a) `960\t (b) `1040\t (c) `1246\t (d) `1146 Solution PTR =1 100 P \u00d7 5 \u00d7 9 2 = 234 100 \u21d2 P \u00d7 45 = 234 200 \u21d2 P = 234 \u00d7 200 45 \u2234P = `1040. Compounding More Than Once a Year We just looked at calculating amount and interest when compounding is done once in a year. But, compounding can also be done for multiple times in a year. For example, interest can be added to the principal in every six months or every four months, and so on. If interest is added to the principal in every six months, we say that compounding is done twice a year. If interest is added to the principal in every four months, we say that compounding is done thrice a year. If interest is added to the principal in every three months, we say that compounding is done four times a year. The formula that we have discussed above for calculating the amount will essentially be the same, \uf8f8\uf8f6\uf8f7n, That is, Amount = P \uf8eb 1 + r \u2009but the rate (r) will not be for one year but for the time \uf8ed\uf8ec 100 period over which compounding is done and the power to which the term inside the bracket is","Simple Interest and Compound Interest 19.5 raised (n in the above case) will not be the number of years, but the number of years multiplied by the number of times compounding is done per year (this product is referred to as the total number of time periods). For example, if a sum of `10,000 is lent at the rate of 10% per annum and compounding is done in every four months (thrice a year), then the amount for two years will be equal to: 10, 000 \uf8eb 1 + 10 \u00d7 1 \uf8f62\u00d73 \uf8ed\uf8ec 3 100 \uf8f8\uf8f7 Here, the dividing factor of 3 in the rate and the multiplying factor of 3 in the exponent (multiplying the number of years)\u2014both shown by arrow marks\u2014are nothing, but the number of times compounding is done in a year. 12 k If compounding is done k times a year (i.e., once in every months), at the rate of r\u2009% per annum, then in n years the principal (P) will amount to = P \uf8ed\uf8ec\uf8eb1 + k r \uf8f6kn . \u00d7 100 \uf8f8\uf8f7 When compounding is done more than once a year, the rate of interest given in the problem is called nominal rate of interest. We can also calculate the rate of interest which will yield simple interest in one year equal to the interest obtained under the compound interest at the given nominal rate of interest. The rate of interest so calculated is called effective rate of interest. The following points helpful in solving problems should also be noted. The difference between compound interest and simple interest on a certain sum for two years is equal to the interest calculated for one year on one year\u2019s simple interest. In mathematical terms, the difference between compound interest and simple interest for two years will be equal to P(r\/100)2, which can be written as P(r\/100) (r\/100). In this, Pr\/100 is the simple interest for one year. When this is again multiplied by r\/100, it gives interest for one year on Pr\/100, i.e., interest for one year on one year\u2019s simple interest. The difference between the compound interest for k years and the compound interest for (k + 1) years is the interest for one year on the amount at the end of kth year. This can also be expressed as follows: The difference between the amount for k years and the amount for (k + 1) years under compound interest is the interest for one year on the amount at the end of the kth year. The difference between the compound interest for the kth year and the compound interest for the (k + 1)th year is equal to the interest for one year on the compound interest for the kth year. Example 19.5 At what rate of interest per annum, under compound interest, will `5120 amount to `7290 in 3 years? Solution P = `5120 A = `7290 N = 3 years","19.6 Chapter 19 A = P \uf8eb\uf8ed\uf8ec1 + r \uf8f6n 100 \uf8f7\uf8f8 7290 = 5120 \uf8ec\uf8eb\uf8ed1 + r \uf8f63 \u21d2 7290 = \uf8ed\uf8eb\uf8ec1 + r \uf8f63 100 \uf8f8\uf8f7 5120 100 \uf8f8\uf8f7 \uf8eb 9 \uf8f63 = \uf8ed\uf8eb\uf8ec1 + r \uf8f63 \u21d2 1+ r = 9 \uf8ed\uf8ec 8 \uf8f8\uf8f7 100 \uf8f8\uf8f7 100 8 r = 1 \u21d2 r = 25 = 12 1 %. 100 8 2 2 Example 19.6 A sum of money triples itself in 3 years at compound interest. In how many years will it become 9 times itself? Solution Let the sum be `p, rate be r\u2009% per annum. 3p = p \uf8ec\uf8ed\uf8eb1 + r \uf8f63 \u21d2 \uf8eb 1 + r \uf8f6 = 31\/3. 100 \uf8f8\uf8f7 \uf8ed\uf8ec 100 \uf8f7\uf8f8 Now, 9p = p \uf8ed\uf8eb\uf8ec1 + r \uf8f6n \u21d2 \uf8eb\uf8ec\uf8ed1 + r \uf8f6n = 9 100 \uf8f7\uf8f8 100 \uf8f8\uf8f7 n = 32 \u21d2 n =2\u21d2n = 6. 3 33 \u2234 In 6 years, the sum becomes 9 times itself. Example 19.7 The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is `120. Find the sum. Solution Let sum be `P. CI = P \uf8f0\uf8ee\uf8ef\uf8ef\uf8ec\uf8ed\uf8eb1 + R \uf8f6n \uf8f9 = P \uf8f0\uf8ef\uf8ee\uf8ef\uf8eb\uf8ec\uf8ed1 + 8 \uf8f62 \uf8f9 100 \uf8f7\uf8f8 \u2212 1\uf8fa 100 \uf8f8\uf8f7 \u2212 1\uf8fa \uf8fb\uf8fa \uf8fb\uf8fa CI = 104P 625 SI = P \u00d78\u00d7 2 = 4P . 100 25 Given, CI \u2013 SI = `120 = 104P \u2212 4P = 120 \u21d2 4P = 120 625 25 625 P = `18, 750 Sum = `18,750.","Simple Interest and Compound Interest 19.7 Example 19.8 The population of certain type of bacteria grows at 4%, 5% and 8% during first, second and third years, respectively. Find the population of the bacteria after 3 years, if the present population is 100,000. Solution A = P \uf8ec\uf8ed\uf8eb1 + r1 \uf8f6 \uf8eb 1 + r2 \uf8f6 \uf8eb 1 + r3 \uf8f6 100 \uf8f8\uf8f7 \uf8ec\uf8ed 100 \uf8f7\uf8f8 \uf8ec\uf8ed 100 \uf8f7\uf8f8 A = 100, 000 \uf8ed\uf8ec\uf8eb1 + 4 \uf8f6 \uf8eb\uf8ec\uf8ed1 + 5 \uf8f6 \uf8eb 1 + 8 \uf8f6 100 \uf8f7\uf8f8 100 \uf8f7\uf8f8 \uf8ec\uf8ed 100 \uf8f7\uf8f8 A = 100, 000 \uf8ee 26 \uf8f9 \uf8eb 21 \uf8f6\uf8eb 27 \uf8f6 \u21d2 A = 117, 936. \uf8f0\uf8ef 25 \uf8fa\uf8fb \uf8ec\uf8ed 20 \uf8f7\uf8f8 \uf8ec\uf8ed 25 \uf8f7\uf8f8 \\\\ The population of the bacteria after 3 years is 117,936. Example 19.9 A man borrowed `10,000 at 12% per annum, interest compounded quarterly. Find the amount that he has to pay after 9 months. Solution P = `10,000 R = 12% per annum 12 R = 4 % per quarter Time period = 9 months \u2234 n= 93= 3 \uf8ec\uf8eb\uf8ed1 3 \uf8f63 100 \uf8f8\uf8f7 A = 10, 000 + = 10, 000 \u00d7 103 \u00d7 103 \u00d7 103 = `10, 927.27. 100 100 100 Example 19.10 Find the compound interest on `50,000 for 3 years, compounded annually, and the rate of interest being 10%, 12% and 15% for the three successive years, respectively. Solution A = P \uf8eb 1 + r1 \uf8f6 \uf8eb 1 + r2 \uf8f6 \uf8eb 1 + r3 \uf8f6 \uf8ec\uf8ed 100 \uf8f8\uf8f7 \uf8ec\uf8ed 100 \uf8f8\uf8f7 \uf8ed\uf8ec 100 \uf8f7\uf8f8 A = 50, 000 \uf8ed\uf8ec\uf8eb1 + 10 \uf8f6 \uf8eb\uf8ed\uf8ec1 + 12 \uf8f6 \uf8eb 1 + 15 \uf8f6 100 \uf8f7\uf8f8 100 \uf8f8\uf8f7 \uf8ed\uf8ec 100 \uf8f7\uf8f8 A = 50, 000 \uf8eb 11 \uf8f6 \uf8eb 112 \uf8f6 \uf8eb 115 \uf8f6 \uf8ec\uf8ed 10 \uf8f8\uf8f7 \uf8ec\uf8ed 100 \uf8f8\uf8f7 \uf8ed\uf8ec 100 \uf8f7\uf8f8 A = `70,840. \u2234CI = A \u2212 P = `20,840.","19.8 Chapter 19 Example 19.11 Two persons P and Q lent certain amounts at the same rate of interest for 2 years and 3 years, respectively, under compound interest. If their final amounts are in the ratio of 3 : 5, Q\u2019s amount at the end of the first year being `8500 and P earned an interest of `510 for the first year, then find the ratio of their principals. (a) 33 : 50\t (b) 27 : 47\t (c) 35 : 61\t (d) 22 : 51 Solution P1, P2 are the principals of P and Q. \uf8eb R \uf8f6 2 \uf8ed\uf8ec 100 \uf8f8\uf8f7 P1 1 + 3 5 \uf8ec\uf8ed\uf8eb1 R \uf8f6 3 = 100 \uf8f8\uf8f7 P2 + \u21d2 P1 = 3 \u21d2 P1 = 3 5 8500 5 P2 \uf8ec\uf8ed\uf8eb1 + R \uf8f6 100 \uf8f8\uf8f7 \u21d2 P1 = 5100. P1 \u00d71\u00d7 R = 510 100 5100 \u00d7 R = 510 100 R = 10% P2 \uf8eb 1 + R \uf8f6 = 8500. \uf8ec\uf8ed 100 \uf8f7\uf8f8 \u21d2 P2 \uf8eb 1 + 10 \uf8f6 = 8500 \u21d2 P2 \u00d7 11 = 8500 \uf8ec\uf8ed 100 \uf8f7\uf8f8 10 \u21d2 P2 = 8500 \u00d7 10 . 11 \u2234 P1 : P2 = 5100 : 8500 \u00d7 10 11 = 3: 50 = 33 : 50. 11 Example 19.12 The difference between simple interest and compound interest on a sum of `20,000 for two years is `112.50. What is the annual rate of interest? Choose the correct answer from the following options: (a) 6%\t (b) 7.5%\t (c) 8.5%\t (d) 8% Solution The difference between simple interest and compound interest on a sum (x) = x \uf8eb r \uf8f62 . \uf8ed\uf8ec 100 \uf8f7\uf8f8","Simple Interest and Compound Interest 19.9 Therefore,\t\t\t 20, 000 \uf8eb r \uf8f62 = 112.5 \uf8ec\uf8ed 100 \uf8f7\uf8f8 \uf8ebr \uf8f62 = 112.5 \uf8ec\uf8ed 100 \uf8f7\uf8f8 20, 000 \uf8ebr \uf8f62 = 0.005625 \uf8ec\uf8ed 100 \uf8f7\uf8f8 \uf8eb r \uf8f6 = 0.075 \uf8ed\uf8ec 100 \uf8f8\uf8f7 \u2234r = 7.5% Example 19.13 Ravi borrowed `15,000 at the rate of 15% per annum for 2 years under simple interest. As he could not repay the loan after two years, the moneylender lender increased the rate of interest to 20% per annum for the further period. If Ravi wants to repay the entire amount at the end of a total period of 3 years and 4 months, then how much he has to pay. (in `) (a) 24,500\t (b) 24,000\t (c) 23,500\t (d) 23,000 Solution Case 1:\u2002 For the first two years P = 15,000, R = 15%, T = 2 I = 15,000 \u00d7 15 \u00d7 2 = `4500. 100 Case 2:\u2002 For the remaining time period P = 15,000, R = 20%, T = 16 months 15, 000 \u00d7 20 \u00d7 16 12 I = 100 \u21d2 I = 15,000 \u00d7 20 \u00d7 16 = `4000. 1200 \u2234 Total amount that Ravi has to pay =15,000 + 4500 + 4000 = `23,500.","19.10 Chapter 19 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t Bank P provides loan at 10% per annum simple semi-annually. Find the amount and compound interest, and bank Q provides loan at 20% per 1 annum simple interest. Vinay borrowed `1000 for interest for a period of 1 2 years. 2 years from bank P, Kumar borrowed `1000 for 2 years from bank Q. How much extra interest 1\t 4.\t Find the rate of simple interest per annum, if the Kumar have to pay? sum borrowed becomes 3 times itself in 12 years. \t2.\t The difference between the compound interest 1\t 5.\t Find the simple interest on `5600 at 20% per annum for the (p + 1)th year and compound interest for from May 22 to November 5 of the same year. (p + 2)th year is equal to the interest for one year on the compound interest for the (p + 1)th year. 1\t 6.\t If `6000 becomes `6720 in 2 years at simple (True\/False) interest, how much does a sum of `10,000 become in 5 years at the same rate of simple interest? \t3.\t Compound interest for one year on `500 calculated \t17.\t The compound interest for 2 years and simple half yearly at 20% per annum is `100. (True\/False) interest for 1 year on a certain sum at certain rate of interest are `3780 and `1800, respectively. Find \t4.\t Find the compound interest on `5000 at 10% per the principal amount and interest rate. annum for 1 year, interest compounded annually? 1\t 8.\t The compound interest on a certain sum at a cer- \t5.\t A certain sum doubles in 3 years under simple tain rate of interest for the 2nd year is `720, and for interest. In how many years will the sum become the 3rd year is `864. Find the principal and rate of 5 times itself? interest. \t6.\t Simple interest on a sum of `10,000 for 3 years at \t19.\t Farhan borrowed `25,000 at 10% per annum under compound interest. He repaid a certain amount at the rate of 20% per annum is . the end of the first year and paid `24,750 at the end of 2nd year to completely disburse the loan. What \t7.\t In what time will a sum become three times itself amount did Farhan repay at the end of the first year? at 20% per annum, at simple interest? PRACTICE QUESTIONS \t8.\t A certain sum becomes 16 times in 4 years at com- 2\t 0.\t The compound interest on `1000 is `331 for 3 pound interest, compounded annually. What is the years at certain rate of interest. What is the rate of rate of interest? interest? \t9.\t If compounding is done p times a year, at the rate \t21.\t If `6000 is lent at 10% per annum, interest being compounded annually, then what is the interest for of 5% per annum for n years, the principal x will the 3rd year? amount to . 1\t 0.\t A certain sum amounts to `73,255 in 3 years, and 2\t 2.\t What sum would amount to `17,280 in 3 years, at `84,525 in 5 years at simple interest. Find the sum. an interest of 20% per annum rate, interest com- pounded annually? \t11.\t If a certain sum becomes 3 times itself in 4 years at compound interest, in how many years will the sum \t23.\t If the rate of interest is 20% per annum com- become 81 times itself at the same rate of interest? pounded in every 6 months, then what is the effective rate of interest per annum? \t12.\t Raju borrowed `15,000 from Mahesh at the rate of 15% per annum under simple interest for 3 years. \t24.\t The difference between the compound interest Raju lent some part of money at 20% per annum and the simple interest on a certain sum for 2 years at simple interest for 3 years and the remaining part is equal to the interest calculated for 1 year on one at 12% per annum at simple interest for 3 years. If year\u2019s simple interest. (True\/False) the interest received by Raju on the money lent is equal to the interest payable, then find the sum he \t25.\t Shyam borrowed `18,000 at 15% per annum at lent at 20% simple interest? compound interest compounded annually. He repaid `10,700 at the end of the 1st year. What is 1\t 3.\t Farheen borrowed a sum of `18,000 at the the amount he should pay at the end of 2nd year rate of 20% per annum, interest compounded to completely disburse the loan?","Simple Interest and Compound Interest 19.11 \t26.\t Rajesh borrows `50,000 at simple interest, but the 2\t 8.\t The difference between compound interest at 10% rate of interest is not constant for the entire period. For the first three years it is 10% per annum, for per annum and simple interest at 8% per annum on the next two years it is 5% per annum and for next three years it is 8% per annum He repaid the entire a certain sum for 3 years is `910. Find the sum. amount after 8 years. How much he have to repay to clear the debt? 2\t 9.\t Find the simple interest on `3750 at 5 1 % per 2 \t27.\t Janardhan deposited a certain sum of money in annum for the period beginning 3rd February fixed deposit account at k% per annum inter- 2007 to 29th June 2007. est being compounded annually. If the amount of interest accrued for the 3rd and the 4th years 3\t 0.\t A sum at 12 1 % per annum amounts to `8723 in is `5000 and `6250 respectively, what is the total 2 interest accrued for the first two years? 5 years at simple interest. Find the sum. Short Answer Type Questions \t31.\t The population of a town increases at the rate of \t38.\t A moneylender found that a fall in the annual PRACTICE QUESTIONS 5% every year. Find the population of the town rate of simple interest from 7% to 6% resulted in in the year 2008, if it\u2019s population in 2005 was his amount of income being reduced by `212.50. 200,000. Find his capital. (in `) \t32.\t Suman borrowed `8000 from Mahesh at 20% \t39.\t A sum doubles itself in 4 years at simple interest. per annum at simple interest. After 3 years when How many times will it amount in 8 years at the Suman wanted to clear the debt, Mahesh insisted same rate of simple interest? to pay the amount at compound interest. How much more Suman have to pay? 4\t 0.\t Mahesh deposited `5000 in Syndicate Bank for 6 months. If the bank pays compound interest \t33.\t Jahangir borrowed `80,000 at the rate of 7% per at 12% per annum, reckoned quarterly, find the annum at compound interest, interest being com- interest received by him. pounded annually. How much should he repay at the end of the first year, so that by repaying `48,792 4\t 1.\t The difference between SI and CI (compounded at the end of second year he can clear the loan? annually) on a sum of `64,000 for 2 years is `1000. What is the rate of interest per annum? \t34.\t Sunil borrowed a certain sum from a moneylender under compound interest, interest being annually. 4\t 2.\t A sum of `150 is borrowed at simple interest at 4% If the interest for the 2nd year is 2 times the interest per annum for the first month, 8% per annum for for the first year, then what is the rate of interest? the second month, 12% per annum for the third month, and so on. What is the total amount of \t35.\t The difference between the simple interest received interest to be paid at the end of 6 months? from two different persons on `1800 for 4 years is `36. The difference between their rates of interest \t43.\t The difference between the CI and SI on a sum of is . `7200 for two years is `72. Find the rate of interest per annum. \t36.\t The compound interest is earned on a sum of money at a rate of 8% per annum for the first year and 10% 4\t 4.\t Roja invested `6000 in a bank, which paid com- per annum for the second year. Find the single equiv- pound interest, interest being compounded semi- alent rate of interest on the sum for the two years. annually. She received `10,368 after 18 months from the bank. Find the rate of interest. (per annum) \t37.\t Akhil invested `8000 in a bank, which pays com- pound interest, compounded semi-annually. He 4\t 5.\t A sum of `40 is borrowed at simple interest. It is receives `9261 after 18 months from the bank. borrowed at 2% per annum for the first month, 4% Find the rate of interest per annum. per annum for the second month, 6% per annum for the third month, and so on. What is the total interest to be paid at the end of 6 months? (in `)","19.12 Chapter 19 Essay Type Questions \t46.\t A certain sum of money at simple interest increases interests received by them is `20. Find the differ- by 50% in 5 years. What will be the compound ence between the rates of interest. interest, compounded annually on `13,000 for 3 years at the same rate? 4\t 9.\t P and Q borrowed `600 each for a period of 3 years. P paid simple interest at 25% per annum while Q \t47.\t A sum of `100,000 amounts to `171,600 in 3 years paid compound interest at 20% per annum, inter- under compound interest, interest being com- est being compounded annually. Who paid more pounded annually. It is lent at 10% per annum for interest and by how much? the first year, 20% per annum for the second year and x% per annum for the third year. Find x. \t50.\t What is the ratio of compound interest accrued on a certain sum in two years at 20% per annum 4\t 8.\t Two persons each lent `2000 at simple interest to the compound interest accrued on the same for 2 years. The difference between the simple amount in three years at 10% per annum? CONCEPT APPLICATION Level 1 \t1.\t A sum of money at simple interest amounts to \t\t(a) `2300\t\t (b) `2450 `800 in 2 years and to `1200 in 6 years. The sum \t\t(c) `2400\t\t (d) `2375 is . \t6.\t A sum of `22000 is divided into three parts such \t\t(a) `600\t\t (b) `1000 that the corresponding interests earned after 1 year \t\t(c) `400\t\t (d) `500 at 2%, per annum, 4 years at 2% per annum and 16 years at 1% per annum simple interest are equal. \t2.\t A certain sum of money becomes `2100 in 4 years Find the least of the sums which was lent. (in `) and `2550 in 7 years. Find the rate of simple inter- PRACTICE QUESTIONS est per annum. \t\t(a) 1000\t\t (b) 2000 \t\t(c) 5000\t\t (d) 4000 \t\t(a) 10%\t\t (b) 8% \t\t(c) 12%\t\t (d) 15% \t7.\t The compound interest on a sum of money for two years is `459 and the corresponding simple \t3.\t There are three amounts a, b and c, such that b is interest is `450. What is the amount under simple the simple interest on c and c is the simple interest interest on the same amount at the same rate of on a. Which of the following must always be true? interest at the end of two years? \t\t(a) a2 = bc\t\t (b) b2 = ac \t\t(a) `6325\t\t (b) `6084 \t\t(c) c2 = ab\t\t (d) a2 + b2 = c2 \t\t(c) `6075\t\t (d) `5524 \t4.\t If the compound interest on a certain sum of \t8.\t A sum of `1750 is lent out at simple interest into money for 2 years is `3280. What would the cor- two parts, the smaller part being lent at 7% per responding simple interest be, given the rate of annum and the larger part at 5% per annum. If the interest is 5% per annum? total amount of interest in one year is `98, then find the part which was lent at 5% per annum. \t\t(a) `3150\t\t (b) `3200 \t\t(c) `3100\t\t (d) `3050 \t\t(a) `525\t\t (b) `975 \t\t(c) `1225\t\t (d) `1350 \t5.\t If the compound interest on a certain sum at 8% per annum, interest compounded annually for 2 years \t9.\t A sum of money invested at compound inter- is `2496, then find the simple interest on the same est doubles itself in six years. In how many years amount at the same rate and for the same period.","Simple Interest and Compound Interest 19.13 will it become 64 times itself at the same rate of If the person\u2019s net investment is `4200, then the \u00adcompound interest? money invested at 5% is . \t\t(a) 30\t\t (b) 36 \t\t(a) `2000\t\t (b) `1000 \t\t(c) 42\t\t (d) 48 \t\t(c) `1500\t\t (d) `1200 \t10.\t Two equal sums are lent at simple interest. The first 1\t 6.\t A sum of `3000 is lent out in two parts. The smaller sum is recovered in 3 years and the second sum in part is lent at 10% per annum and the larger part is 6 years. The rate of interest per annum on the first lent at 20% per annum. If the total interest in a year is sum is 2% more than that of the second sum. Find `500, then find the sum lent at 10% per annum (in `). the total sum lent if the amount in each case is `560. \t\t(a) 800\t\t (b) 1000 \t\t(a) `530\t\t (b) `500 \t\t(c) 1200\t\t (d) 900 \t\t(c) `1480\t\t (d) `1000 1\t 7.\t The integral number of years in which a sum of \t11.\t A sum was borrowed at simple interest at R% per money at 25% per annum under compound inter- annum for 2 years. If it had been borrowed at (R + est will become more than twice itself is at least. 5)% per annum it would have become `200 more. Find the sum (in `). \t\t(a) 3\t\t (b) 2 \t\t(c) 4\t\t (d) 1 \t\t(a) 2500\t\t (b) 2000 \t\t(c) 3000\t\t (d) 1500 1\t 8.\t Find the least integral number of years in which a sum at 20% per annum compound interest will be 1\t 2.\t The ratio of the interest accrued on a sum, when more than double. invested at simple interest for 2 years and the inter- est accrued on it, if it is invested at compound \t\t(a) 4\t\t (b) 5 interest, interest being compounded annually for \t\t(c) 3\t\t (d) 6 3 years at the same rate of interest is 50 : 91. Find the rate of interest (per annum). 1\t 9.\t A sum is split into two equal parts. One of the parts is lent at simple interest at 20% per annum for \t\t(a) 10%\t\t (b) 15% 6 years. The other part is lent at 40% per annum PRACTICE QUESTIONS \t\t(c) 25%\t\t (d) 20% simple interest for 2 years. The difference in the interests is `72. Find the total sum (in `). \t13.\t Ravi took a certain amount from Raju at the rate of 8% per annum at simple interest and lent half of the \t\t(a) 180\t\t (b) 360 amount to Ramu at 8% per annum at simple inter- \t\t(c) 240\t\t (d) 270 est and the remaining amount to Raghu at 10% per annum at simple interest. If at the end of 10 years, 2\t 0.\t A sum is split into five equal parts. Each part is lent Ravi made a profit of `1250 in the deal, then find at annual rates of simple interests of 8%, 7%, 5%, 3% the amount that Ravi had taken from Raju. and 2%. They are lent for 1 year, 2 years, 3 years, 4\u00a0years and 5 years. On how many parts is the simple \t\t(a) `12,500\t\t (b) `13,125 interest at least 25% of the total interest? \t\t(c) `2245.50\t (d) Data insufficient \t\t(a) 1\t\t (b) 2 1\t 4.\t Two equal sums were lent at the same time at \t\t(c) 3\t\t (d) 4 simple interest rates of 6% and 4% per annum. The first sum was recovered 2 years earlier than 2\t 1.\t A sum of `2500 is split into two parts, one part the second sum, and the amount in each case was being lent at simple interest and the other part `930. What was the sum lent? being lent at compound interest, interest being compounded annually. At the end of two years, the \t\t(a) `820\t\t (b) `780 total amount of interest earned on the sum is `201. \t\t(c) `690\t\t (d) `750 Find the sum lent at simple interest, if both the parts are lent at an interest rate of 4% per annum. \t15.\t A person invested three different amounts at 3%, 5% and 6% per annum at simple interest. At the end of \t\t(a) `1250\t\t (b) `1625 the year, he received the same interest in each case. \t\t(c) `1875\t\t (d) `1500","19.14 Chapter 19 \t22.\t A sum is split into five equal parts. They are lent If\u00a0the rates of interest on X and Y are 8% and 4%, at annual rates of simple interests of 1%, 2%, 3%, respectively,then find the larger part between X 10% and 5%. They are lent for 6 years, 4 years, 3 and Y. years, 1 year and 1 year, respectively. The simple interest on how many parts is at least 20% of the \t\t(a) `5000\t\t (b) `6000 total interest? \t\t(c) `6500\t\t (d) `7000 \t\t(a) 1\t\t (b) 2 2\t 7.\t A borrowed a certain sum of money from B at \t\t(c) 3\t\t (d) 4 the rate of 10% per annum under simple interest and lent one-fourth of the amount to C at 8% per 2\t 3.\t A sum is split into two equal parts. One of the annum under simple interest and the remaining parts is lent at simple interest at 4% per annum for amount to D at 15% per annum under simple inter- one year. The other part is further split into three est. If at the end of 15 years, A made profit of `5850 equal parts. These are lent at compound interests, in the deal, then find the sum that A had lent to D. interests being compounded annually. These are lent at 10% per annum for 4 years, 20% per annum \t\t(a) `24,500\t\t (b) `12,000 for 2 years and 40% per annum for 1 year. The dif- \t\t(c) `9000\t\t (d) `18,600 ference between the total compound interest and simple interest is `1184.10. Find the sum. (in `) \t28.\t A sum of `62,000 is divided into three parts such that the corresponding interests earned for 3 years, \t\t(a) 3000\t\t (b) 4000 5 years and 6 years are equal. If the rates of simple \t\t(c) 5000\t\t (d) 6000 interest are 5% per annum, 4% per annum and 3% per annum, then what is the greatest of the sums \t24.\t Sreedhar borrowed `3500 at 6% per annum for that were lent? 3 years under simple interest. But, after one year he was asked to pay compound interest for the \t\t(a) `18,000\t\t (b) `22,000 remaining two years on the sum borrowed initially. \t\t(c) `24,000\t\t (d) `26,000 How much additional interest Sreedhar has to pay? 2\t 9.\t The difference between the compound interest on \t\t(a) `10\t\t (b) `14.50 a sum of `4000, interest being compounded annu- \t\t(c) `12.60\t\t (d) `12 ally and the simple interest on it for two years is `250. Find the rate of interest. (per annum) PRACTICE QUESTIONS \t25.\t Ajay borrowed `4000 at 10% per annum for 4 years under simple interest. But, after 2 years he was \t\t(a) 15%\t\t (b) 10% asked to pay compound interest for the remaining \t\t(c) 25%\t\t (d) 20% 2 years on the initially borrowed amount. How much Ajay have to pay additionally if interest was 3\t 0.\t A sum of `4000 is split into two parts. One part is compounded annually? (in`) lent at simple interest and the other at compound interest, interest being compounded annually. At \t\t(a) 40\t\t (b) 20 the end of two years, the total amount of inter- \t\t(c) 80\t\t (d) 160 est earned is `1720. Find the sum lent at simple interest, if each part is lent at 20% per annum (in `). 2\t 6.\t A sum of `10,500 is divided into two parts X and Y, such that the interest calculated on X for \t\t(a) 1200\t\t (b) 1000 4 years is equal to the interest on Y for 6 years. \t\t(c) 800\t\t (d) 1500 Level 2 \t31.\t A sum of `1500 amounts to `1680 in 3 years at 3\t 2.\t A sum of money doubles itself in 3 years. At same rate of simple interest, for a period of 9 years, how simple interest. If the interest rate is increased by many times will the sum become? 2%, it would amount to . \t\t(a) `1770\t\t (b) `1815 \t\t(a) 4\t\t (b) 6 \t\t(c) `1590\t\t (d) `1850 \t\t(c) 8\t\t (d) 9","Simple Interest and Compound Interest 19.15 \t33.\t A and B borrowed `600 and `500 respectively for a compounded annually. The difference between the period of 3 years. A paid simple interest at the rate of amounts paid by him is `15. Find each equal sum. 10% per annum, while B paid compound interest at the rate of 10% per annum compounded a\u00adnnually. \t\t(a) `1800\t\t (b) `2000 Who paid more interest and by how much? \t\t(c) `1500\t\t (d) `2500 \t\t(a) A paid more interest by `14.50. \t40.\t A certain sum of money amounts to `4200 in 3 \t\t(b) B paid more interest by `14.50. \t\t(c) A and B both paid same amount of interest. years and to `6000 in 6 years at simple interest. \t\t(d) None of the above. Find the rate of interest. \t\t(a) 12 1 % \t\t (b) 20% 2 \t\t(c) 25%\t\t (d) 30% 3\t 4.\t Due to a fall in the annual rate of interest from 4\t 1.\t A sum of money becomes four times itself in 5 6% to 5%, a person\u2019s yearly income reduces by years at a certain rate of interest, compounded annually. In how many years will it become 16 `245.25. His capital is . times itself at the same rate of interest? \t\t(a) `24,525\t\t (b) `24,600 \t\t(a) 20\t\t (b) 16 \t\t(c) `23,675\t\t (d) `24,000 \t\t(c) 12\t\t (d) 10 \t35.\t The cost of a television is `15625. Its value depre- \t42.\t The difference between simple interest and com- ciates at the rate of 8% per annum. Calculate the pound interest on a sum of `40,000 for two years total depreciation in its value at the end of 3 years. is `900. What is the annual rate of interest? \t\t(a) `3458\t\t (b) `3748 \t\t(a) 20%\t\t (b) 10% \t\t(c) `3548\t\t (d) `3845 \t\t(c) 12%\t\t (d) 15% \t36.\t The value of an ornament decreases every year at 4\t 3.\t The difference between the compound interest the rate of 5% over that of the previous year. If (compounded annually) and simple interest, for its value at the end of 2 years is `9025, then what two years on the same sum at the amount rate was its original value at the beginning of these two of interest is `370. Find the rate of interest if the years? simple interest on the amount at the same rate of interest for 1 year is `3700. \t\t(a) `12,000\t\t (b) `11,000 PRACTICE QUESTIONS \t\t(c) `10,000\t\t (d) `13,000 \t37.\t A certain sum of money triples itself in 6 years \t\t(a) 10%\t\t (b) 12% at compound interest. In how many years will it \t\t(c) 16%\t\t (d) 15% become 27 times at the same rate of compound interest? 4\t 4.\t A certain sum amounts to `13,310 after 3 years and to `16,105.10 after 5 years under compound \t\t(a) 27\t\t (b) 30 interest. Find the sum borrowed, if the interest is \t\t(c) 24\t\t (d) 18 compounded annually. (in `) \t38.\t A sum of `24,000 is divided into two parts P1 \t\t(a) 12,000\t\t (b) 8,000 and P2, such that the simple interest calculated on P1 for 3 years and on P2 for 4 years are equal. If \t\t(c) 10,000\t\t (d) 16,000 the rates of interest on P1 and P2 are 4% and 5% respectively, then find the smaller of the parts P1 \t45.\t A certain sum amounts to `77,000 in 5 years and and P2. to `68,200 in 3 years, under simple interest. If the rate of interest is increased by 2%, then in how many years will it double itself? \t\t(a) `10,000\t\t (b) `12,000 \t\t(a) 8\t\t (b) 9 \t\t(c) `8000\t\t (d) `9000 \t\t(c) 10\t\t (d) 12 3\t 9.\t A person borrowed two equal sums for two \t46.\t Find the compound interest on `40,000 at 12% years at the rate of 10% per annum, from two persons. He borrowed the first sum at simple per annum for a period of 2 years. (in `) interest and the second sum at compound interest, \t\t(a) 10,176\t\t (b) 8000 \t\t(c) 9176\t\t (d) 10,000","19.16 Chapter 19 \t47.\t A person borrowed a certain sum at 25% per annum 4\t 8.\t Find the amount on the sum of `15,625 for 18 compound interest (compounded annually) and paid `10,000 at the end of 4 years. Find the sum borrowed. months under compound interest, compounded \t\t(a) `4096\t\t (b) `5000 half yearly at the rate of 16% per annum. \t\t(c) `5016\t\t (d) `4960 \t\t(a) `19,683\t\t (b) `19,625 \t\t(c) `20,504\t\t (d) `19,625 Level 3 \t49.\t Rakesh borrowed `42,000 from Rajnikant at 6% 5\t 5.\t A certain amount of money doubles in 4 years per annum simple interest. He lent the same sum under compound interest. In how many addi- to Kishore at 10% per annum compound inter- tional years will it become 4 times of the principal est, compounded annually, for 2 years. Find the amount under the same conditions. amount earned by Rakesh in the transaction. \t\t(a) 8\t\t (b) 4 \t\t(a) `3614\t\t (b) `3550 \t\t(c) 12\t\t (d) 16 \t\t(c) `3610\t\t (d) `3780 \t50.\t A certain sum amounts to `7935 in 2 years and \t56.\t A sum of money was lent in two parts which were in the ratio of 2 : 3 for 2 years and 3 years `9125.25 in 3 years, under compound interest. respectively, both at the rate of 10% per annum at simple interest. If the difference between the Find the sum borrowed if interest is compounded interest earned is `6000, then find the total sum that was lent. annually. (in`) \t\t(a) 6000\t\t (b) 7500 \t\t(c) 8000\t\t (d) 10,000 \t\t(a) `24,000\t\t (b) `36,000 \t51.\t Ravi lent Ramu a certain sum of money at the rate \t\t(c) `60,000\t\t (d) `84,000 of 2 1 % per annum, interest compounded annu- \t57.\t The total simple interest at Rs% per annum and the 2 total compound interest at Rc% per annum for 2 years on `10,000 are equal. If Rs, Rc are integers, then find ally. After two years, Ramu paid a sum of `2560 to the minimum difference between Rs and Rc\u2009. Ravi. What amount of money did Ramu borrow PRACTICE QUESTIONS from Ravi? \t\t(a) `2036.64\t(b) `2236.64 \t\t(a) 2\t\t (b) 4 \t\t(c) `2436.64\t(d) `2636.64 \t\t(c) 8\t\t (d) 10 5\t 2.\t In how many years will a sum of `25,600, inter- \t58.\t Find the amount on `9900 at 20% per annum for 2 years at compound interest (compounded est compounded quarterly, at the rate of 25% per annually). annum, amount to `28,900? \t\t(a) 1 year\t\t (b) 1 year \t\t(a) `12,946\t\t (b) `13,548 2 \t\t(c) 2 years\t\t (d) 4 years \t\t(c) `14,256\t\t (d) `15,678 \t53.\t The value of an old bike decreases every year at \t59.\t Find the amount when `9999 is lent at sim- ple interest for 3 years and 4 months at 10% per the rate of 4% over that of the previous year. If its annum. (in `) value at the end of three years is `13824, then find its present value. \t\t(a) 13,332\t\t (b) 12,332 \t\t(c) 12,232\t\t (d) 13,333 \t\t(a) `15,625\t\t (b) `14,525 \t\t(c) `16,625\t\t (d) `15,425 \t54.\t At compound interest of 8%, 10% and 12% for \t60.\t The compound interest on a certain sum for 2 years is `882, whereas the simple interest on it is three consecutive years, the interest earned in the `840. Find the rate of interest. 3rd year is `891. Find the principal amount. (in `) \t\t(a) 6250\t\t (b) 6050 \t\t(a) 10%\t\t (b) 12% \t\t(c) 6200\t\t (d) 6225 \t\t(c) 8%\t\t (d) 15%","Simple Interest and Compound Interest 19.17 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t `200 1\t 6.\t `13,000 \t17.\t r = 10% per annum \t2.\t True \t\tPrincipal = `18,000 \t18.\t r = 20% \t3.\t False \t\tPrincipal = `3000 \t19.\t `5000 \t4.\t `500 \t20.\t 10% per annum 2\t 1.\t `726 \t5.\t 12 2\t 2.\t `10,000 \t23.\t 21% per annum \t6.\t `6000 \t24.\t True \t25.\t `11,500 \t7.\t 10 years 2\t 6.\t `82,000 \t27.\t `7200 \t8.\t 100% per annum \t28.\t `10,000 \uf8f6 pn \t29.\t `82.50 \t9.\t x \uf8eb\uf8ec1 + 5 \uf8f7 3\t 0.\t `5,368 \uf8ed 100 p \uf8f8 \t10.\t `56,350 1\t 1.\t 16 years 1\t 2.\t `5625 \t13.\t `23,958, `5958 1\t 4.\t 16 2 % per annum 3 \t15.\t `512.40 Shot Answer Type Questions \t39.\t 3 \t31.\t 231,525 4\t 0.\t `304.50 3\t 2.\t `1024 3\t 3.\t `40,000 \t41.\t 12 1 % ANSWER KEYS 3\t 4.\t 100% 2 3\t 5.\t 0.5% 3\t 6.\t 18.8% \t42.\t `10.50 3\t 7.\t 10% \t38.\t 21250 \t43.\t 10% 4\t 4.\t 40% 4\t 5.\t 1.40 Essay Type Questions 4\t 9.\t P paid an additional amount of interest of `13.20 \t46.\t `4303 than Q 4\t 7.\t `30 4\t 8.\t 0.5% \t50.\t 440 331","19.18 Chapter 19 CONCEPT APPLICATION Level 1 \t1.\u2002(a)\t 2.\u2002 (a)\t 3.\u2002 (c)\t 4.\u2002 (b)\t 5.\u2002 (c)\t 6.\u2002 (b)\t 7.\u2002 (c)\t 8.\u2002 (c)\t 9.\u2002 (b)\t 10.\u2002 (d) \t11.\u2002 (b)\t 12.\u2002 (d)\t 13.\u2002 (a)\t 14.\u2002 (d)\t 15.\u2002 (d)\t 16.\u2002 (b)\t 17.\u2002 (c)\t 18.\u2002 (a)\t 19.\u2002 (b)\t 20.\u2002 (a) \t21.\u2002 (c)\t 22.\u2002 (c)\t 23.\u2002(d)\t 24.\u2002(c)\t 25.\u2002 (a)\t 26.\u2002 (b)\t 27.\u2002 (c)\t 28.\u2002 (c)\t 29.\u2002 (c)\t 30.\u2002 (b) Level 2 33.\u2002 (a)\t 34.\u2002 (a)\t 35.\u2002 (a)\t 36.\u2002 (c)\t 37.\u2002 (d)\u2002\t 38.\u2002 (d)\t 39.\u2002 (c)\t 40.\u2002 (c) 43.\u2002 (a)\t 44.\u2002 (c)\t 45.\u2002 (c)\t 46.\u2002 (a)\t 47.\u2002 (a)\t 48.\u2002 (a) \t31.\u2002 (a)\t 32.\u2002 (a)\t \t41.\u2002 (d)\t 42.\u2002 (d)\t Level 3 51.\u2002 (c)\t 52.\u2002 (b)\t 53.\u2002 (a)\t 54.\u2002 (a)\t 55.\u2002 (a)\t 56.\u2002 (c)\t 57.\u2002 (a)\t 58.\u2002 (c) \t49.\u2002 (d)\t 50.\u2002 (a)\t \t59.\u2002 (a)\t 60.\u2002 (a) ANSWER KEYS","Simple Interest and Compound Interest 19.19 CONCEPT APPLICATION Level 1 \t1.\t (i) Interest for 4 years \t7.\t \u2009(i) \u0007Find the difference between the two interests \t\t = `(1200 \u2212 800) = `400. \t\t(ii) Interest for 2 years = `200. and proceed. \uf8f6n \t\t(iii) Sum = 800 \u2212 Interest for 2 years. \uf8f8\uf8f7 \t\t\u2009(ii) Use CI = P \uf8f0\uf8ee\uf8ef\uf8ef\uf8ec\uf8ed\uf8eb1 + R \uf8f9 and SI = PRT . 100 \u2212 1\uf8fa 100 \t2.\t (i) Interest for 3 years = `2550 \u2013 `2100 = 450. \uf8fa\uf8fb \t\t(ii) Interest for 4 years = `600. \t\t(iii) Sum = 2100 \u2212 Interest for 4 years. \uf8ee\uf8f0\uf8ef\uf8ef\uf8ed\uf8eb\uf8ec1 R \uf8f6 2 \uf8f9 P \u00d7R\u00d7 2 100 \uf8f8\uf8f7 \u2212 1\uf8fa 100 \t\t(iii) 459 = P + and 450 = , \uf8fb\uf8fa \t\t\t\u0007solve these two equations and find R, then \t3.\t (i) L\u0007 et rate of interest be r% and time period be t\u00a0years. find P. c \u00d7r \u00d7t a\u00d7r \u00d7t \t8.\t (i) L\u0007 et, `P be lent at 7% and `(1750 \u2212 P) be lent 100 100 \t\t(ii) b = and c = . at 5%. \t\t(ii) Take P \u00d77\u00d7 1 + (1750 \u2212 P) \u00d7 5 \u00d7 1 = 98. 100 100 \t\t(iii) \u0007Using the above, find the relation among a, b \t\t(iii)\u2009U\u2009\u0007 se the above information to find P and and c. \t4.\t (i) Find the principal from the given data. (1750\u00a0\u2212 P). CI P \uf8ef\uf8f0\uf8ee\uf8ef\uf8ec\uf8ed\uf8eb1 + R \uf8f6n \uf8f9 \t9.\t (i)\t`P becomes `2P in 6 years. 100 \uf8f7\uf8f8 \u2212 1\uf8fa , \t\t(ii) \u0007Use = \uf8eb R \uf8f6 6 \uf8eb R \uf8f6 \uf8fb\uf8fa \uf8ed\uf8ec 100 \uf8f7\uf8f8 \uf8ec\uf8ed 100 \uf8f8\uf8f7 \t\t(ii)\t2P = P 1 + , find 1 + value. i.e., 33228800 == P \uf8ee\uf8ef\uf8f0\uf8ef\uf8eb\uf8ed\uf8ec1 + 5 \uf8f62 \uf8f9 \t\t(iii) \u0007Now, 64P = P \uf8ec\uf8ed\uf8eb1 + R \uf8f6n and substitute Hints and Explanation 100 \uf8f8\uf8f7 \u2212 1\uf8fa . 100 \uf8f7\uf8f8 \uf8fa\uf8fb \t\t(iii) Find P from the above information. \uf8ed\uf8eb\uf8ec1 + R\uf8f6 value to find n. 100 \uf8f7\uf8f8 \t\t(iv) Now, use SI = P \u00d7R \u00d7T . 100 \t10.\t (i) \u0007Let the rate of interest on the second sum \t5.\t (i) \u0007Find the principal from the given data and be\u00a0R%. proceed. \t\t(ii) L\u0007 et the sum be `P in each case and rate of \t\t(ii) Let the sum be `P. interest on second sum be r\u2009% per annum. \uf8f0\uf8ef\uf8ef\uf8ee\uf8eb\uf8ec\uf8ed1 + R \uf8f6n \uf8f9 \t\t(iii) Take P \uf8ee\uf8f0\uf8ef1 + (r + 2)3 \uf8f9 = P \uf8ef\uf8ee\uf8f01 + r \u00d76 \uf8f9 = 560. 100 \uf8f8\uf8f7 \u2212 1\uf8fa , 100 \uf8fb\uf8fa 100 \uf8fb\uf8fa \t\t(iii) \u0007CP = P find P by substitut- \uf8fa\uf8fb \t\t\tFind P and then 2P. ing R, n and CI values. 1\t 1.\t \u2009(i) \u0007Find the difference between the simple inter- \t\t(iv) Now, find SI on P at same rate for same period. ests and proceed. \t6.\t (i) Find the interests in each case and equate them. \t\t(ii) Let the sum be `P. \t\t(ii) L\u0007et the sums lent be `x, `y and `22,000 \u2212 \t\t(iii) T\u0007 ake P \u00d7 (R + 5) \u00d7 2 \u2212 P \u00d7R\u00d72 = 200, solve (x + y). 100 100 for P. \t\t(iii) x\u00d72 = y\u00d72\u00d74 1\t 2.\t \u2009(i) \u0007Find the ratio of SI for 2 years and CI for 3 100 100 years and equate it to the given ratio. \t\t\t = [22, 000 \u2212 (x + y )] \u00d7 1 \u00d7 6 . \uf8ebP \u00d7r \u00d7 2\uf8f6 \uf8ee\uf8ef\uf8ef\uf8f0\uf8ec\uf8ed\uf8eb1 + r \uf8f63 \uf8f9 100 \uf8ed\uf8ec 100 \uf8f7\uf8f8 100 \uf8f7\uf8f8 \u2212 1\uf8fa \t\t(ii) \u00f7 P = 50 : 91. \uf8fb\uf8fa \t\t(iv) \u0007Solve the above equation and find the least of the sums. \t\t(iii) Solve the above and find r.","19.20 Chapter 19 \t13.\t (i) \u0007Let the money lent to Ramu and Raghu be ` P 2\t 1.\t \u2009\u2009\u2009\u2009(i) \u0007Let the sum lent at CI be x and the sum lent at 2 SI be (2500 \u2013 x). each. (2500 \u2212 x)4 \u00d7 2 \uf8ee\uf8ef\uf8ef\uf8f0\uf8eb\uf8ec\uf8ed1 4 \uf8f6 2 \uf8f9 100 100 \uf8f7\uf8f8 \u2212 1\uf8fa \t\t\u2009\u2009(ii) + x + = 201. \uf8fb\uf8fa \t\t(ii) P \u00d7 10 \u00d7 10 + P \u00d7 8 \u00d7 10 \u2212 \uf8eb P \u00d7 8 \u00d7 10\uf8f6 = 1250. 2 \u00d7 100 2 \u00d7 100 \uf8ed\uf8ec 100 \uf8f8\uf8f7 \t\t(iii) Solve for x. \t14.\t \u2009(i) \u0007Let the time period for the sum lent at 6% be \t22.\t \u2009(i) Let the total sum be `5x. T\u00a0years. \t\t(ii) Find the interest on each part. \t\t(ii) L\u0007 et the sum be `P in each case and second sum recovered in x years. \t\t(iii) \u0007Find the sum of the interest of 5 parts. Check which is greater than 20% of the sum of interests. \t\t(iii) P \uf8ee\uf8f0\uf8ef1 + 6(x \u2212 2)\uf8f9 = P \uf8ef\uf8f0\uf8ee1 + 4(x ) \uf8f9 = 930. Find x. 2\t 3.\t \u2009(i) Let the total sum be `2x. 100 \uf8fa\uf8fb 100 \uf8fa\uf8fb x \u00d7 4x \t\t(ii) SI = 100 . \t\t(iv) Using the above information, find P. 1\t 5.\t (i) T\u0007 ake the sum invested at 3% rate of interest to \t\t(iii) \u0007Total CI = x \uf8f0\uf8ef\uf8ef\uf8ee\uf8eb\uf8ec\uf8ed1 + 10 \uf8f64 \u2212 \uf8f9 + x \uf8ee\uf8ef\uf8ef\uf8f0\uf8eb\uf8ed\uf8ec1 + 20 \uf8f62 \uf8f9 + x \uf8f0\uf8ef\uf8ee\uf8ef\uf8eb\uf8ec\uf8ed1 + be x and that at 5% rate of interest to be y and 3 100 \uf8f7\uf8f8 1\uf8fa 3 100 \uf8f8\uf8f7 \u2212 1\uf8fa 3 that at 6% be `4200 \u2013 (x + y). \uf8fa\uf8fb \uf8fb\uf8fa \t\t(ii) x\u00d73 = y\u00d75 = [4200 \u2212x31(\uf8ef\uf8ee\uf8f0\uf8efx0\uf8eb\uf8ec\uf8ed01++y1)1]00\u00d70 6\uf8f7\uf8f6\uf8f84.\u2212 \uf8f9 + x \uf8ef\uf8ee\uf8f0\uf8ef\uf8eb\uf8ed\uf8ec1 + 20 \uf8f62 \u2212 \uf8f9 + x \uf8ef\uf8f0\uf8ee\uf8ef\uf8eb\uf8ec\uf8ed1 + 40 \uf8f6\uf8f8\uf8f71\u2212 \uf8f9 . 100 100 = 1\uf8fa 3 100 \uf8f7\uf8f8 1\uf8fa 3 100 1\uf8fa \uf8fb\uf8fa \uf8fb\uf8fa \uf8fa\uf8fb \t\t(iii) Solve the above and find y. \t\t(iv) Use, CI \u2212 SI = 1184.10. \t16.\t \u2009\u2009\u2009\u2009(i) \u0007Let the two parts be `x and `(300 \u2013 x), respectively. \t24.\t \u2009(i) F\u0007 ind the SI for 1 year and CI for the last two x \u00d7 10 (300 \u2212 x)20 Hints and Explanation \t\t\u2009\u2009(ii) In 100 + 100 = 500, find x. years. \t\t\u2009\u2009(ii) Find SI and CI as per the given data. 1\t 7.\t \u2009\u2009\u2009\u2009(i) Let the sum be `P. \t\t(iii) Find the difference between CI and SI. \t\t\u2009\u2009(ii) P \uf8eb\uf8ec\uf8ed1 + 20 \uf8f6n > 2P, solve the inequality for n. \t25.\t \u2009(i) Find SI on `4000 at 10% per annum for 4 years. 100 \uf8f7\uf8f8 \t\t\u2009\u2009(ii) F\u0007 ind sum of SI on `4000 at 10% per annum \t18.\t \u2009\u2009\u2009\u2009(i) Let the sum be `P. for 2 years and CI on 4000 at 10% per annum for 2 years. \t\t\u2009\u2009(ii) P \uf8eb\uf8ed\uf8ec1 + 20 \uf8f6n > 2P, solve the inequality for n. \t\t(iii) Find the difference of the above two. 100 \uf8f8\uf8f7 \t27.\t \u2009\u2009\u2009\u2009(i) L\u0007et A borrowed `x from B. Therefore, the money lent to C is ` x and the money lent to \t19.\t \u2009\u2009\u2009\u2009(i) \u0007Find the interests in each case and then equate 4 their difference to the given value. 3x 4 \t\t\u2009\u2009(ii) Let the total sum be `2x. D is ` . \t\t(iii) x \u00d7 20 \u00d7 6 \u2212 x \u00d7 40 \u00d7 2 = 72. \t\t\u2009 (ii) \uf8eex \u00d7 8 \u00d7 15 + 3x \u00d7 15 \u00d7 15\uf8f9 \u2212 x \u00d7 10 \u00d7 15 = 5850. 100 100 \uf8f0\uf8ef 4 100 4 \u00d7 100 \uf8fa\uf8fb 100 \t\t(iv) Solve for x and find 2x. \t\t(iii) \u0007Solve the above equation and find x, and then \t20.\t \u2009\u2009\u2009\u2009(i) Let the total sum be `5x. 3x . 4 5x \t\t\u2009\u2009(ii) 25% of 5x = ` 4 . \t28.\t (i) L\u0007et the sums lent be `x, `y and `22,000 \u2212 (x + y). \t\t(iii) F\u0007ind the interest on each sum of `x using \t\t(ii) x\u00d73\u00d75 = y\u00d75\u00d74 = [22, 000 \u2212 (x + y )]6 \u00d7 3 . 100 100 100 given data, and check which is greater than 5x . \t\t(iii) \u0007Solve the above equation and find the least of 4 the sums.","Simple Interest and Compound Interest 19.21 \t29.\t \u2009(i) Let the rate of interest be r\u2009%. \t30.\t \u2009(i) \u0007Let `x be lent at simple interest and `(4000 \u2013 x) 4000 \uf8ee\uf8ef\uf8ef\uf8f0\uf8eb\uf8ed\uf8ec1 + r \uf8f62 \uf8f9 4000 \u00d7 r \u00d72 at compound interest. 100 \uf8f8\uf8f7 \u2212 1\uf8fa 100 \t\t(ii) \u2212 = 250. \t\t(ii) \u0007x \u00d7 20 \u00d7 2+ (4000 \u2212 x) \uf8ef\uf8ef\uf8f0\uf8ee\uf8ec\uf8ed\uf8eb1 + 20 \uf8f6 2 \u2212 1\uf8f9\uf8fa = 1720, \uf8fa\uf8fb 100 x. 100 \uf8f7\uf8f8 \uf8fa\uf8fb \t\t(iii) Solve the above to find r. solve for Level 2 \t31.\t Find the interest for 1 year and the rate of interest. \t\t A P \uf8eb\uf8ed\uf8ec1 R \uf8f6n 100 \uf8f7\uf8f8 \t32.\t Assume that `P becomes `2P in 3 years and = + proceed. 25 \uf8f64 3\t 3.\t Find the interests in each case and the difference. 10, 000 = P \uf8ed\uf8ec\uf8eb1 + 100 \uf8f7\uf8f8 \t34.\t Apply the formula for finding the simple interest. 3\t 5.\t Use depreciation concept or let A = P \uf8eb 1 \u2212 r \uf8f6n , \u21d2 10, 000 = P \uf8eb\uf8ed\uf8ec1 + 1 \uf8f64 when n = 3. \uf8ed\uf8ec 100 \uf8f7\uf8f8 4 \uf8f8\uf8f7 4\t 6.\t P = `40,000, R = 12%, n = 2. \t\t\u21d2 10, 000 = P \uf8eb 5 \uf8f64 \uf8ed\uf8ec 4 \uf8f7\uf8f8 \uf8f0\uf8ef\uf8ee\uf8ef\uf8ed\uf8eb\uf8ec1 + \uf8f6n \uf8f9 CI = P R \uf8f7\uf8f8 \u2212 1\uf8fa \u21d2 P = 4\u00d74\u00d74\u00d74 \u00d7 10,000 100 5\u00d75\u00d75\u00d75 \uf8fb\uf8fa \uf8f0\uf8ef\uf8ee\uf8ef\uf8ec\uf8ed\uf8eb1 12 \uf8f62 \uf8f9 \u21d2 P = `4096. 100 \uf8f7\uf8f8 1\uf8fa = P + \u2212 \uf8fb\uf8fa R \uf8f6n Hints and Explanation 100 \uf8f7\uf8f8 \uf8f62 \t48.\t A = P \uf8eb\uf8ec\uf8ed1 + \uf8f8\uf8f7 \t\t = 40, 000 \uf8ef\uf8ef\uf8ee\uf8f0\uf8eb\uf8ec\uf8ed 28 \u2212 \uf8f9 R = 16 = 8% 25 1\uf8fa 2 \uf8fa\uf8fb = 40, 000 \uf8ee 282 \u2212 252 \uf8f9 \u21d2 A = 15, 625 \uf8ef\uf8f0\uf8ee1 + 8 \uf8f93 \uf8ef (25)2 \uf8fa 100 \uf8fa\uf8fb \uf8f0 \uf8fb = 40, 000 (28 + 25)(28 \u2212 25) 15, 625 \uf8eb\uf8ec\uf8ed 1 2 \uf8f63 25 \u00d7 25 25 \uf8f7\uf8f8 \u21d2 A = + = 64 \u00d7 53 \u00d7 3 = `10,176. \uf8f63 \uf8f7\uf8f8 \t47.\t Let the sum borrowed be `P. \u21d2 A = 15, 625 \uf8eb 27 \uf8ed\uf8ec 25 \t\tR = 25% per annum \t\tn = 4 \u21d2 A = 15, 625 \uf8eb 27 \u00d7 27 \u00d7 27 \uf8f6 \t\tA = 10,000 \t\t \uf8ec\uf8ed 25 \u00d7 25 \u00d7 25 \uf8f8\uf8f7 A = `19,683. Level 3 \uf8ec\uf8ed\uf8eb1 R\uf8f6 N 5\t 2.\t Use formula for amount and then use laws of 100 \uf8f7\uf8f8 \t51.\t A = P + indices. \uf8f6n \uf8f8\uf8f7 \t\tA = `2560, R = 2 1 %, n = 2 years, find P. \t\t28,900 = 25, 600 \uf8eb 1 + 25 , n is the number of 2 quarters. \uf8ec\uf8ed 400","19.22 Chapter 19 \t53.\t Let P be the original value. \uf8ed\uf8ec\uf8eb1 R \uf8f64 100 \uf8f8\uf8f7 \uf8ed\uf8ec\uf8eb1 \u2212 4 \uf8f63 2P = P + 100 \uf8f7\uf8f8 P = 13, 824 \uf8ed\uf8eb\uf8ec1 \u2212 1 \uf8f63 \t\t2 = \uf8eb 1 + R \uf8f64 25 \uf8f8\uf8f7 \uf8ec\uf8ed 100 \uf8f8\uf8f7 \u21d2 P = 13, 824 \uf8eb 25 \u2212 1 \uf8f6 \u21d2 21\/4 = \uf8ec\uf8ed\uf8eb1 + R\uf8f6 \uf8ed\uf8ec 25 \uf8f8\uf8f7 100 \uf8f8\uf8f7 \t\t\u21d2 P = 13, 824 \uf8eb 24 \u00d7 24 \u00d7 24 \uf8f6 \t\tLet the amount becomes 4P in \u2018n\u2019 years. \uf8ed\uf8ec 25 \u00d7 25 \u00d7 25 \uf8f7\uf8f8 \u21d2 P = 13, 824 \uf8ec\uf8ed\uf8eb1 R \uf8f6n 100 \uf8f7\uf8f8 \u21d2 P = 25 \u00d7 25 \u00d7 25 = `15,625. 4P = P + \t54.\t 1st year\u2019s interest = 8% \uf8eb 1 \uf8f6n \t\tPrincipal = P 4P = \uf8ec\uf8ed\uf8ec 24 \uf8f8\uf8f7\uf8f7 \t\tI1 = P \u00d71\u00d7 8 = 2P . \t\t4 = 2n\/4 100 25 22 = 2n\/4 \t\t2nd year: n \t\tR = 10%, Principal =P + 2P = 27P \u21d2 4 = 2 25 25 n = 8 years. 27P \t\tI 2 25 \u00d7 10 \t56.\t Let the two parts be 2P and 3P = 100 Hints and Explanation I2 = 27P . (3P ) \u00d7 3 \u00d7 10 \u2212 (2P ) \u00d7 2 \u00d7 10 = 6000 250 100 100 \t\t3rd year: \t\t90P10\u2212040P = 6000 \t\tR = 12% 50P 100 \t\tPrincipal = 27P + 27P = 297P = 6000 \u21d2 5P = `60, 000. 25 250 250 297P \u00d7 12 \t57.\t PTR = P \uf8f0\uf8ee\uf8ef\uf8ef\uf8ec\uf8ed\uf8eb1 + R \uf8f62 \u2212 \uf8f9 10, 000 \u00d7 2 \u00d7 Rs 250 100 100 \uf8f7\uf8f8 1\uf8fa 100 I3 = \uf8fb\uf8fa 100 \uf8ef\uf8ee\uf8ef\uf8f0\uf8ec\uf8eb\uf8ed1 Rc \uf8f62 \uf8f9 = 297P \u00d7 3 = 10, 000 + 100 \uf8f8\uf8f7 \u2212 1\uf8fa 6250 \t\t \uf8fb\uf8fa \t\t I3 = 891P . 200Rs = 10, 000 \uf8ee Rc2 + 2Rc \uf8f9 6250 \uf8ef\uf8f01002 100 \uf8fa \t\tGiven that interest earned in the 3rd year = `891. \uf8fb 891P 200Rs = Rc2 + 200Rc 6250 \t\t\u2234 I3 = = 891 200Rs \u2212 200Rc = Rc2 \t\t200(Rs \u2212 Rc ) = Rc2 \u21d2 P = `6250. Rc2 5\t 5.\t Time period = 4 years Rs \u2212 Rc = 200 \t\tPrincipal = P \t\tAmount = 2P Rs = Rc2 + Rc . 200","Simple Interest and Compound Interest 19.23 \t\tSince Rs and Rc are integers and Rc2 must be the \t\t= 3 years + 4 years smallest integer, it should be a multiple of 200 and 12 =\t\t 3=31 years 10 a perfect square. 3 years. \u2234 Rc2 = 400 PTR \u21d2 Rc = 20% A = P + 100 \t\t Rs = 400 + 20 200 9999 \u00d7 10 \u00d7 10 Rs = 22% 3 \t\tA = 9999 + \t\t \u2234 Minimum difference between Rs and Rc is 2. 100 \t58.\t Amount after two years = P \uf8ec\uf8eb\uf8ed1 + R \uf8f62 = 9999 + 3333 = 13,332. 100 \uf8f8\uf8f7 \t60.\t S=I1 S=I2 840 = 420. \uf8f62 2 9900 \uf8eb 1 20 \uf8f7\uf8f8 = \uf8ed\uf8ec + 100 \t\tDifference between compound interest and simple \t\t interest for 2 years is the interest on the first year\u2019s = 9900 \uf8eb 36 \uf8f6 = 396 \u00d7 36 = 14, 256. simple interest. \uf8ed\uf8ec 25 \uf8f7\uf8f8 \t\t882 \u2013 840 = R%(420) \t59.\t P = `9999 42 = R (420) \t\tR = 10% \t\t 100 \t\tT = 3 years and 4 months \u21d2 R = 10%. Hints and Explanation","1202CChhaapptteerr PRKraiontipeomo, rattiiocsn and\u00a0Variation REmEmBER Before beginning this chapter, you should be able to: \u2022 Calculate the ratio of quantities given \u2022 Understand the concepts of proportion and variation \u2022 Apply unitary method in solving problems KEy iDEaS After completing this chapter, you should be able to: \u2022 Learn about ratios and terms related \u2022 Know the properties and types of ratios \u2022 Study about proportion and variations \u2022 Know properties and types of a proportion Figure 1.1","20.2 Chapter 20 Ratio The value obtained when two similar quantities are compared by dividing one quantity with the other is called ratio. The ratio of two quantities is that value which gives us how many times one quantity is of the other. Only quantities of the same kind, i.e., the quantities with the same units can be compared. \u2009\u2009Notes\u2002 a b 1.\t The ratio of a and b is written as a : b and is measured by the fraction . \t Example: The ratio of 5 and 8 is written as 5 : 8 and measured by the fraction 5 . 8 a 2.\t I\u0007f two quantities are in the ratio a : b, then the first and the second quantities will be a+b times and b times the sum of the two quantities respectively. a+b 3 5 \t Example: If x :y =3 : 5, then x = 8 (x + y) and y = 8 (x + y ). 3.\t T\u0007 he ratio of two quantities can be found, only when both the quantities are of the same kind. \t E\u0007 xample: Ratio between 1 meter and 5 seconds cannot be found, as the quantities given are not of the same kind. 4.\t A ratio is an abstract quantity and a ratio does not have any units. \t Example: The ratio of 30 seconds and one minute is 30 seconds : 60 seconds or 1 : 2. Terms of a Ratio For a given ratio a : b, we say that a is the first term or antecedent and b is the second term or consequent. In the ratio 3 : 4, 3 is the antecedent while 4 is the consequent. Properties of a Ratio The value of a ratio remains the same, if both the terms of the ratio are multiplied or divided by the same non-zero quantity. If a, b and m are non-zero real numbers. 1.\t a = ma \u21d2 a :b = am : bm b mb a 2.\t a = m \u21d2 a:b = a : b b b m m m Simplest Form of a Ratio The ratio of two or more quantities is said to be in the simplest form, if the highest common factor (HCF) of the quantities is 1. If the HCF of the quantities is not 1, then each quantity of the ratio is divided by the HCF to convert the ratio into its simplest form. For Example, suppose there are three numbers 6, 9 and 12. The HCF of the numbers is 3. Dividing each of 6, 9 and 12 by 3, the results obtained are 2, 3 and 4. The ratio of 6, 9 and 12, in the simplest form is 2 : 3 : 4.","Ratio, Proportion and\u00a0Variation 20.3 Example 20.1 Express 81 : 93 in its simplest form. Solution The HCF of 81 and 93 is 3. We divide each term by 3. Then, =81 : 93 83=1 : 933 27 : 31 \u2234 The ratio 81 : 93 in its simplest form is 27 : 31. Comparison of Ratios a c b d Two ratios a : b and c:d can be compared in the following way: If > then a :b> c : d. Example 20.2 Compare the ratios 4 : 5 and 18 : 25. Solution 4 : 5= 45= 0.8 18 : 2=5 12=85 0.72 As, 0.8 > 0.72, \u2234 4 : 5 > 18 : 25. Alternate method: Ratios can also be compared by reducing them to equivalent fractions of a common denominator. =4 : 5 45=and 18 : 25 18 25 4 = 4 \u00d7 5 = 20 5 5 \u00d7 5 25 As, 20 > 18 ; 4 > 18 25 25 5 25 \\\\ 4 : 5 > 18 : 25. Types of Ratios 1.\t \u0007A ratio a : b, where a > b, is called a ratio of greater inequality. For a ratio of greater inequality, if a positive quantity is added to both terms, then the obtained ratio is lesser than a : b. In other words if a positive quantity x is added to the two terms in the ratio a : b (where a > b), then a + x : b + x <a : b. \t E\u0007 xample: The ratio 4 : 3 is a ratio of greater inequality, because the antecedent (4) is greater 4+1 4 than the consequent (3). It can be seen that for the ratio 4 : 3, 3+1 < 3 .","20.4 Chapter 20 2.\t \u0007A ratio a : b, where a < b, is called a ratio of lesser inequality. For a ratio of lesser inequality, if a positive quantity is added to both terms, the obtained ratio is greater than a : b. In other words, if a positive quantity x is added to the two terms of the ratio a : b (where a < b), then (a + x) : (b + x) > a : b. \t \u0007Example: The ratio 3 : 4 is a ratio of lesser inequality, because the antecedent 3 is less than the consequent (4). It can be seen that for the ratio 3 : 4, 3+1 > 3 . 4+1 4 3. \t \u0007A ratio a : b, where a = b, is called a ratio of equality. If a positive quantity is added to both the terms, the value of the ratio remains unchanged. In other words, if a positive quantity x is added to two terms in the ratio a : b, then a + x : b + x = a : b. 4.\t The duplicate ratio of a : b is a2 : b2. 5.\t The sub-duplicate ratio of a : b is a : b. 6.\t The triplicate ratio of a : b is a3 : b3. 7.\t The sub-triplicate ratio of a : b is 3 a : 3 b. 8.\t The inverse ratio of a : b is b : a. 9.\t The compound ratio of two ratios a : b and c : d is ac : bd. \u2002Note\u2002\u2002 If three quantities a, b and c are in the ratio 1 : 2 : 3, we write a : b : c = 1 : 2 : 3. It follows that a : b = 1 : 2, a : c = 1 : 3 and b : c = 2 : 3. The ratio of two quantities can be found similarly when there are more than 3 quantities given and the ratio of all quantities is known. Example 20.3 Divide `560 in the ratio 3 : 4. Solution The sum of the terms of the ratio is 3 + 4 = 7. \\\\ The first part = `\uf8eb\uf8ec\uf8ed 560 \u00d7 3\uf8f6 = `240 7 \uf8f7\uf8f8 and, the second part = `\uf8ec\uf8eb\uf8ed 560 \u00d7 4 \uf8f6 = `320. 7 \uf8f8\uf8f7 Example 20.4 Divide `780 among three friends P, Q and R in the ratio 1 : 1 : 1 . 4 3 2 Solution 1 1 1 4 3 2 Given ratio is : : . The LCM of the denominators is 12 \u21d2 1 : 1 : 1 = 1 \u00d7 12 : 1 \u00d7 12 : 1 \u00d7 12 = 3 : 4 : 6 4 3 2 4 3 2 The sum of the terms of the ratio = (3 + 4 + 6) = 13. P\u2019s share = `\uf8eb\uf8ec\uf8ed 780 \u00d7 3\uf8f6 = `180. 13 \uf8f8\uf8f7","Ratio, Proportion and\u00a0Variation 20.5 \\\\Q\u2019s share = `\uf8eb\uf8ec\uf8ed 780 \u00d7 4\uf8f6 = `240 13 \uf8f7\uf8f8 and, R\u2019s share = `\uf8eb\uf8ed\uf8ec 780 \u00d7 6 \uf8f6 = `360. 13 \uf8f7\uf8f8 Example 20.5 If x : y = 3 : 4 and y : z = 6 : 11, then find x : y : z. Solution Given x : y = 3 : 4 and y : z = 6 : 11. In both the ratios y is common. Try to make the value of y equal in both the ratios. 2 Multiply each term of the ratio 6 : 11 by 3 to make its first term (y) equal to 4. \u21d2 y : z = 6 \u00d7 2 : 11 \u00d7 2 = 4 : 22 3 3 3 \u2234x : y : z = 3 : 4 : 22 = 9 : 12 : 22 3 Example 20.6 If B = 3A and A = 4C, then find B : C. Solution B = 3A = 3(4C) = 12C \u21d2 B = 12C \u21d2 B = 12 C 1 \\\\ B : C = 12 : 1. Example 20.7 In a bag there are coins in the denominations `1, `2 and `5 in the ratio 3 : 5 : 7, respectively. If the total value of the coins in the bag is `144, find the number of coins in each denomination and also the total value of `2 coins. Solution Let the number of coins of denominations of `1, `2 and `5 be 3x, 5x and 7x, respectively. The total value of the coins in the bag = `[3x + 5x(2) + 7x(5)] = `(3x + 10x+ 35x) = `48x. Given, the total amount in the bag = `144 \u21d2 48x = 144 \u21d2 x = 3. The number of `1 coins in the bag = 3x = 9.","20.6 Chapter 20 \u2234 The number of `2 coins in the bag = 5x = 5(3) = 15 The number of `5 coins in the bag = 7x = 21. \u2234 The total value of `2 coins in the bag = `15 \u00d7 2 = `30. Example 20.8 The ratio of marks obtained by Amal, Bimal and Komal in an examination is 12 : 8 : 15. Find the marks obtained by Bimal and Komal, if Amal scored 15 marks less than that of Komal. Solution Let the marks scored by Amal, Bimal and Komal be 12x, 8x and 15x, respectively. Given that Amal scored 15 marks less than that of Komal. \u21d2 12x = 15x \u2212 15 \u21d2 3x = 15 \u21d2 x = 5. \u2234 Marks scored by Bimal = 8x = 8(5) = 40. Marks scored by Komal = 15x = 15(5) = 75. Example 20.9 If x+y = y + z = z + x , then prove that x + y + z = 0. p3 \u2212 q3 q3 \u2212 r3 r3 \u2212 p3 Solution Let x+y = y + z = z + x =k p3 \u2212 q3 q3 \u2212 r3 r3 \u2212 p3 \u21d2 x + y = k(p3 \u2013 q3); y + z = k(q3 \u2013 r3); z + x = k(r3 \u2013 p3) \u21d2 x + y + y + z + z + x = k(p3 \u2013 q3 + q3 \u2013 r3 + r3 \u2013 p3) \u21d2 2(x + y + z) = k(0) \\\\ x + y + z = 0. Example 20.10 If ba= dc= e ; then prove that each of these ratio\u2019s is equal to \uf8eb 4a2 + 3c2 \u2212 7e2 \uf8f61\/2 . f \uf8ec \uf8f7 \uf8ed 4b2 + 3d 2 \u2212 7f 2 \uf8f8 Solution Let a c e = k. b= d= f \u21d2 a = kb; c = kd; e = kf. Now, \uf8eb 4a2 + 3c2 \u2212 7e2 \uf8f61\/2 = \uf8eb (4(kb)2 + 3(kd)2 \u2212 7(kf )2 ) \uf8f61\/2 \uf8ec \uf8f7 \uf8ec \uf8f7 \uf8ed (4b2 + 3d 2 \u2212 7 f 2 ) \uf8f8 \uf8ed 4b2 + 3d2 \u2212 7 f 2 \uf8f8 = \uf8eek2(4b2 + 3d2 \u2212 7 f 2 )\uf8f91\/2 = k. \uf8ef \uf8fa \uf8f0 (4b2 + 3d2 \u2212 7 f 2) \uf8fb Hence, proved.","Ratio, Proportion and\u00a0Variation 20.7 Proportion The equality of two ratios is called proportion. \u2002Note\u2002\u2002If a : b = c : d, then a, b, c and d are said to be in proportion and the same can be represented as a : b :: c : d. It is read as \u2018a is to b is as c is to d.\u2019 Properties of Proportion 1.\t a, b, c and d are respectively known as the first, second, third and the fourth proportional. 2.\t \u0007The first and the fourth terms are called extremes while the second and the third terms are called means. 3.\t Product of extremes = Product of means 4.\t The fourth term can also be referred to as the fourth proportional of a, b and c. a c 5.\t I\u0007 f a:b=c : d, i.e., b = d , then the given proportion can be written as b : a :: d : c, i.e., b = d , by taking the reciprocals of terms on both sides. This relationship is known as a c invertendo. 6.\t b , we get a : c = b : d. This \u0007If a : b :: c : d, then multiplying both sides of the proportion by c relationship is known as alternendo. a c 7.\t Adding 1 to both sides of the proportion a : b :: c: d, we get b +1= d +1 \t \t\t\t\t\u2003\u2003\u21d2 a +b b = c d + \b(1) d \t That is, (a + b) : b = (c + d\u2009) : d \t This relationship is known as componendo. 8.\t Subtracting 1 from both sides of the proportion a : b :: c : d, we get \t \t\t\t\t a \u22121= c \u22121 \u21d2 a \u2212b = c \u2212d \b(2) b d b d \t\t\t\t(a \u2013 b) : b = (c \u2013 d) : d \t This relationship is known as dividendo 9.\t Dividing the Eq. (1) by Eq. (2), we get, a+b c+d b = d a\u2212b c \u2212d bd \u21d2 (a + b) : (a \u2013 b) = (c + d) : (c \u2013 d) \t This relationship is known as componendo and dividendo. a c e la + mc + ne b d f lb + md + nf 10.\t If ba= dc= e and l, m and n are any three non-zero numbers, then = = = . f Continued Proportion Three quantities a, b and c are said to be in continued proportion if a : b :: b : c. If a : b :: b : c, then c is called the third proportional of a and b. Mean Proportional of a and c If a : b :: b : c, then b is called the mean proportional of a and c.","20.8 Chapter 20 We have already learnt that, Product of means = Product of extremes. \u21d2b\u00d7b=a\u00d7c \u21d2 b2 = ac \u21d2 b = \u00b1 ac \u2234 The mean proportional of a and c is ac . Types of Variation Direct Variation If two quantities are related to each other such that an increase (or decrease) in the first quantity results in a corresponding proportionate increase (or decrease) in the second quantity, then the two quantities are said to vary directly with each other. Example: At constant speed, distance covered varies directly as time. This is expressed as, distance travelled \u221d time. Indirect Variation If two quantities are related to each other such that an increase (or decrease) in the first quantity results in a corresponding proportionate decrease (or increase) in the second quantity, then the two quantities are said to vary inversely with each other. Example: Number of men working together to complete a job is inversely proportional to the time taken by them to finish the job. When the number of men increases, the time taken to finish the same job decreases. \u2234 The number of men (n) working together to complete a job is inversely proportional to the time taken (t) by them to finish it. This is expressed as, n \u221d 1 . t Joint Variation When a change in a quantity depends on the changes in two or more quantities, it is said to vary jointly with those quantities. When a quantity A varies directly as B, when C is constant and varies directly as C, when B is constant, then A varies directly as the product of B and C. It is represented as A \u221d B (C is constant ) and A \u221d C (B is constant) \u21d2 A \u221d BC or A = k\u2009\u22c5\u2009BC, where k is a constant. When a quantity A varies directly as B, when C is constant and varies inversely as C, when B is constant, then A varies as B . C 1 It is represented as A \u221d B (C is constant) and A \u221d C (B is constant). \u21d2 A \u221d 1 or A = k \u22c5 1 , where k is a constant. BC BC When a quantity A varies inversely as B, when C is constant and varies inversely as C, when B is constant, then A varies inversely as BC. It is represented as A \u221d 1 (C is constant) and A \u221d 1 (B is constant). B C \u21d2 A \u221d 1 or A =k \u22c5 1 , where k is a constant. BC BC","Ratio, Proportion and\u00a0Variation 20.9 Example 20.11 If a :b= 4 : 5, find 5a \u2212 b . 10a + 3b Solution Given, a = 4 b 5 4b \u21d2 a = 5 5a \u2212 b 5 \uf8eb 4b \uf8f6 \u2212 5 3b 3 10a + 3b \uf8ed\uf8ec 5 \uf8f7\uf8f8 + 3b 11b 11 = 4b \uf8f6 = = . \uf8eb 5 \uf8f8\uf8f7 10 \uf8ec\uf8ed Example 20.12 Verify whether 6, 7, 12 and 14 are in proportion or not. Solution Ratio between 6 and 7 = 6 : 7. Ratio between 12 an=d 14 1=2 : 14 1142= 67= 6 : 7. \u21d2 6 : 7 = 12 : 14 or 6 : 7 :: 12 : 14. Hence, the numbers 6, 7, 12 and 14 are in proportion. Example 20.13 Find the fourth proportional to the numbers 8, 10 and 12. Solution Let x be the fourth proportional to 8, 10 and 12. \u21d2 8 : 10 :: 12 : x. Product of means = Product of extremes 10 \u00d7 12 = 8 \u00d7 x x = 10 \u00d7 12 = 15. 8 \u2234 The fourth proportional to 8, 10 and 12 is 15. Example 20.14 Find the mean proportional between 13 and 52. Solution Mean proportional between 13 and 52 = 13 \u00d7 52 = 676 = 26.","20.10 Chapter 20 Example 20.15 If 4 taps can fill a tank in 10 hours, then in how many hours can 6 taps fill the same tank? Solution Assume that 6 taps can do the same work in x hours. As more taps require less hours to do the same work, number of taps vary inversely as the number of hours \u2234 (Inverse ratio of number of taps) :: (Ratio of number of hours) \u21d2 6 : 4 :: 10 : x \u21d2 6 = 10 4 x \u21d2 x = 40 = 6 2 hours. 6 3 \u2234 Six taps can fill the tank in 6 2 hours. 3 Example 20.16 In a family, the consumption of power is 120 units for 18 days. Find how many units of power is consumed in 30 days. Solution Let the consumption of power in 30 days be x units The more the period of time, the more is the consumption of power. Hence the consumption of power varies directly as the number of days. \u2234 (Ratio of number of days) :: (Ratio of consumption of power) \u21d2 18 : 30 :: 120 : x Product of means = Product of extremes \u21d2 30 \u00d7 120 = 18 \u00d7 x \u21d2 x = 30 \u00d7 120 = 200 units. 18 \u2234 The consumption of power for 30 days is 200 units. Example 20.17 If (7x + 4y) : (7x \u2212 4y) :: (7p+ 4q) : (7p \u2212 4q), then show that x, y, p and q are in proportion. Solution Given, 7x + 4y = 7p + 4q 7x \u2212 4y 7p \u2212 4q Using componendo-dividendo rule, (7x + 4y) + (7x \u2212 4y) = (7p + 4q) + (7p \u2212 4q) \u21d2 14x = 14 p \u21d2 x = p (7x + 4y) \u2212 (7x \u2212 4y) (7p + 4q) \u2212 (7p \u2212 4q) 8y 8q y q \u2234 x, y, p and q are in proportion.","Ratio, Proportion and\u00a0Variation 20.11 Example 20.18 If 3k + 4l + 6m + 7n = 3k \u2212 4l + 6m \u2212 7n , then show that k, 2m, 4l and 7n are in proportion. 3k + 4l \u2212 6m \u2212 7n 3k \u2212 4l \u2212 6m + 7n Solution Given, 3k + 4l + 6m + 7n = 3k \u2212 4l + 6m \u2212 7n 3k + 4l \u2212 6m \u2212 7n 3k \u2212 4l \u2212 6m + 7n \u21d2 (3k + 4l ) + (6m + 7n ) = (3k \u2212 4l ) + (6m \u2212 7n ) (3k + 4l ) \u2212 (6m + 7n ) (3k \u2212 4l ) \u2212 (6m \u2212 7n ) \u21d2 3k + 4l = 3k \u2212 4l \u21d2 3k + 4l = 6m + 7n 6m + 7n 6m \u2212 7n 3k \u2212 4l 6m \u2212 7n \u21d2 3k = 6m \u21d2 k = 2m 4l 7n 4l 7n \u21d2 k = 4l 2m 7n \u2234 k, 2m, 4l and 7n are in proportion. Example 20.19 If a, b, c and d are in proportion, then show that a3 + 3ab2 = c3 + 3cd 2 . 3a2b + b3 3c 2d + d3 Solution \u21d2 a = c Given a, b, c and d are in proportion b d Applying componendo and dividendo, a + b = c +d a \u2212 b c \u2212d Cubing on both sides, we have (a + b)3 = \uf8eec + d \uf8f93 (a \u2212 b)3 \uf8ef\uf8f0 c \u2212 d \uf8fa\uf8fb \u21d2 a3 + 3a2b + 3ab2 + b3 = c3 + 3c2d + 3cd2 + d3 a3 \u2212 3a2b + 3ab2 \u2212 b3 c3 \u2212 3c2d + 3cd2 \u2212 d3 (a3 + 3ab2 ) + (3a2b + b3 ) = (c3 + 3cd 2 ) + (3c 2d + d3 ) (a3 + 3ab2 ) \u2212 (3a2b + b3 ) (c3 + 3cd 2 ) \u2212 (3c 2d + d3 ) \u21d2 a3 + 3ab2 = c3 + 3cd2 3a2b + b3 3c2d + d3 Hence, proved.","20.12 Chapter 20 Example 20.20 4 men, each working 6 hours per day can build a wall in 9 days. How long will 6 men, each working 3 hours per day take to finish the same work? Solution Clearly, M \u221d 1 and M \u221d 1 D H \u21d2 MDH = Constant. \u21d2 M1D1H1 = M2D2H2 \u21d2 4 \u00d7 9 \u00d7 6 = 6 \u00d7 D2 \u00d7 3 \u21d2 D2 = 12. \u2234 Six men each working 3 hours per day can finish the job in 12 days. Example 20.21 The area of a circle varies with the square of its radius. The area of a circle having radius 3 cm is C cm2. Find the area of the circle having radius 9 cm. Solution Let the area of the circle be denoted by C and its radius be denoted by r. C \u221d r2 \u21d2 C1 = r12 . C2 r22 Taking r1 = 3 cm, r2 = 9 cm and C1 = C cm2, C2 = C1 \uf8eb r22 \uf8f6 = C \uf8eb 9 \uf8f62 = 9C cm 2 . \uf8ec r12 \uf8f7 \uf8ed\uf8ec 3 \uf8f7\uf8f8 \uf8ed \uf8f8 Example 20.22 The pressure of a gas varies directly as the temperature when volume is kept constant and varies inversely as the volume when temperature is kept constant. The gas occupies volume of 400 ml when the temperature is 160 K and the pressure is 640 Pa. What is the temperature of gas whose volume and pressure are 200 ml and 320 Pa, respectively? Solution 1 V Given P\u221dT and P \u221d \u21d2P \u221d T \u21d2 PV \u221dT V \u21d2 PV = Constant T \u21d2 P1V1 = P2V2 T1 T2 \u21d2 640 \u00d7 400 = 320 \u00d7 200 \u21d2 T2 = 160 = 40 K. 160 T2 4","Ratio, Proportion and\u00a0Variation 20.13 Example 20.23 1 2 A garrison of 700 men is provisioned for 30 days at the rate of 2 kg per day per man. If 100 x days men had left the garrison, then the provisions would have lasted for at the rate of 3 1 kg 2 per day per man. Find the value of x. Choose the correct answer from the following options: (a) 18\t (b) 20\t (c) 25\t (d) 15 Solution In both cases, the provisions are the same. 700 \u00d7 30 \u00d7 2 1 2 1 = (700 \u2212 100) \u00d7 x \u00d7 3 2 . \u21d2 700 \u00d7 30 \u00d7 5 = 600 \u00d7 x \u00d7 7 2 2 \u2234 x = 25. Example 20.24 Three positive numbers are in the ratio 1 : 3 : 5. The sum of their squares is 875. Find the sum of the numbers. Choose the correct answer from the following options: (a) 45\t (b) 90\t (c) 75\t (d) 150 Solution Given that the ratio of the three numbers is 1 : 3 : 5. Let the numbers be x, 3x and 5x And also given, x2 + (3x)2 + (5x)2 = 875 \u21d2 35x2 = 875 \u21d2 x2 = 25 \u21d2 x = 5 (the numbers are positive) \u2234 The required sum = x + 3x + 5x = 9x = 9(5) = 45. Example 20.25 Two positive numbers x and y satisfy the condition 4x2 + 25y2 = 20xy. Find the value of x : y. (a) 5 : 2\t (b) 2 : 5\t (c) 3 : 2\t (d) 2 : 3 Solution Given, 4x2 + 25y2 = 20xy \u21d2 (2x \u2013 5y)2 = 0 \u21d2 2x \u2013 5y = 0 \u21d2 x = 5 \u21d2 x : y = 5: 2. y 2","20.14 Chapter 20 Example 20.26 There are three numbers such that twice the first number is equal to thrice the second number and four times the third number. Find the ratio of the first, the second and the third numbers. Choose the correct answer from the following options: (a) 2 : 3 : 4\t (b) 6 : 4 : 3\t (c) 4 : 3 : 4\t (d) 3 : 4 : 3 Solution Let the three numbers be x, y and z. Given, 2x = 3y = 4z Let 2x = 3y = 4z = k \u21d2 x = k , y = k and z = k 2 3 4 \u21d2 x : y : z = k : k : k 2 3 4 = 6 : 4 : 3. Example 20.27 `585 is to be divided among A, B and C in the ratio 3 : 4 : 6. By mistake, it is divided in the ratio 1 : 1 : 1 . Find the loss incurred to C due to this mistake (in `). Choose the correct 6 4 3 answer from the following options: (a) 10\t (b) 15\t (c) 20\t (d) 25 Solution Total money = `585 (a)\t Ratio of actual shares is 3 : 4 : 6 \t Actual share of C (in `) = 6 (585) 13 \t \t\t\t\u2003\u2003\u2009= 6(45) = 270. (b)\t Ratio of shares given = 1 : 1 : 1 6 4 3 \t \t\t\t\u2003\u2009= 2 : 3 : 4 \t Now, share of C=(in `) 49=(585) 4(65) = 260 \t Loss incurred by C (in `) = 270 \u2013 260 = 10.","Ratio, Proportion and\u00a0Variation 20.15 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t 200 metres is _____ of a kilometre. \t17.\t If 7 3 : 11 2 = 93 : (2x + 10), then what is the value 4 3 \t2.\t The result obtained after applying componendo to of x? b : (a \u2212 b) :: d : (c \u2212 d) is _____. 1\t 8.\t If x : y = 5 : 6 and y : z = 3 : 7, then find x : z. \t3.\t The simplest form of 57 : 72 is _____. \t19.\t If a : b = 2 : 3, then find (3a \u2212 b) : (2a + 3b). \t4.\t If X = 3Y = 4Z, then X :Y : Z = _____. \t20.\t Two numbers are in the ratio 3 : 4. When 7 is \t5.\t Which of the two ratios 4 : 9 and 5 : 8 is greater? added to each, the ratio becomes 4 : 5. Find the numbers. 1 \t6.\t The inverse ratio of 8 : 4 is _____. \t21.\t What is the compounded ratio of 6 1 : 8 2 and 2 3 \t7.\t The result obtained after applying dividendo to 3 1 3 4 : 5 3 ? (a + b) : b :: (c + d) : d is _____. \t8.\t The compound ratio of p : r and r : q is _____. 2\t 2.\t If (x + 2), 3, (3x \u2212 4) and 6 are in proportion, Find\u00a0x. \t9.\t The numbers 4, 7, 8 and 14 are in proportion. \t23.\t Find the fourth proportional of 4, 9 and 12. (True\/False) \t24.\t Write a ratio equal to 7 : 9 in which antecedent is 63. \t10.\t Express the ratio 1.5 litres : 250 ml in the lowest terms. \t25.\t If a = c = e = 2 , then 3a + 4c + 9e = ______ . b d f 5 3b + 4d +9f 1\t 1.\t If P and Q2 are in direct proportion. If Q doubles then P becomes _____. \t26.\t If (3a + 2b) : (5a + 7b) = 1 : 2, then find a : b. 1\t 2.\t If 10 men can do a certain amount of work in \t27.\t Arrange the ratios 4 : 5, 6 : 7, 2 : 3 and 10 : 11 in 6 days, 15 men can do the same amount of work in _____ days. the increasing order of their magnitude. 2\t 8.\t A certain sum of money is distributed among A, 1 1 1 1\t 3.\t x varies directly as the cube of y. x is 32 when y is B and C in the ratio of 2 : 3 : 4 and B gets `120. 4, then what is the value of y when x is 108? PRACTICE QUESTIONS 1\t 4.\t A varies inversely as the square of B and A is 1, Find the shares of A and C. when B is 3. What is the value of B, when A is 3? \t29.\t When a = 3, 5, 8, 10, \u2026, b = 9, 15, 24, 30, \u2026, then a and b are in _____. \t15.\t Express the ratio 3 : 7 in the lowest terms. 5 9 \t30.\t What number must be subtracted from each of \t16.\t The mean proportional between 4 and 8 is 6. the numbers 32, 38, 17 and 20, so that the terms formed will be in proportion? (True\/False) Short Answer Type Questions \t31.\t The number of boys and girls in a class are in the 3\t 5.\t The ratio of the present ages of a father and his son ratio of 3 : 4. Find the strength of the class if there is 3 : 1. The sum of their ages after five years is 58. are 10 more girls than there are boys. Find the ratio of their ages 3 years ago. \t32.\t The numbers A, A + 2, A + 12 and A + 22 are in 3\t 6.\t The ratio of the present ages of Raju and his wife proportion. Find A. is 5 : 4. Which of the following cannot be the ratio of their ages 20 years hence? \t33.\t If a3 + 2ab2 = 123 , then find the value of a : b. 2a2b + b3 136 \t\t(a) 11 : 10\t\t (b) 6 : 5 3\t 4.\t P, Q and R have to share 52 apples among them- \t\t(c) 23 : 20\t\t (d) 13 : 10 selves such that P gets thrice as many apples as Q gets and Q gets thrice as many as R gets. Find the 3\t 7.\t The mass of a liquid (in grams) varies directly with number of apples R must get. its volume (in cm3). A liquid would have a mass of","20.16 Chapter 20 20 grams if its volume is 10 cm3. Find the mass of 4\t 2.\t The volume of a sphere (V\u2009) varies directly as the the liquid in grams, if its volume is 8 cm3. cube of its radius. The volume of the sphere of \t38.\t The ratio of the present ages of Swarupa and her radius 3 cm is 36\u03c0 cm3. What is the volume of a daughter is 7 : 3. When Swarupa was 26 years old, sphere of radius 15 cm? her daughter was 6 years old. Find the present age of Swarupa. (in years) 4\t 3.\t The ratio of the tens digit and the units digit of a two digit number is 1 : 2. How many two digit 3\t 9.\t Solve: 10 + 3x + 10 \u2212 3x = 3. numbers satisfy this condition? 10 + 3x \u2212 10 \u2212 3x 4\t 4.\t The ratio of the monthly earnings of A and B is \t40.\t Find x + 2p + x + 2q , when x = 4 pq . 3 : 2. The ratio of the monthly expenditures of A x \u2212 2p x \u2212 2q p+q and B is 4 : 3. A saves `300 each month. Which of the following cannot be the monthly savings of B? 4\t 1.\t The volume (V\u2009) of a mass of gas varies directly as (in `) its absolute temperature (T\u2009) and inversely as the pressure (P) applied to it. The gas occupies a vol- \t\t(a) 180\t\t (b) 195 ume of 300 ml, when the temperature is 210 K \t\t(c) 165\t\t (d) 205 and pressure is 150 Pa. What is the temperature of gas whose volume and pressure are 200 ml and \t45.\t If 2p + 5q = 4p \u2212 3q , then find the relation\u00a0bet\u00ad 250 Pa respectively? 2r + 5s 4r \u2212 3s ween p, q, r and s. Essay Type Questions \t46.\t There are two boxes, red and white in colour. The amount more than his expected share. Find the ratio of the number of chocolates in the white box share of Y, when X got an amount of `480. to the number of biscuits in the red box is 3 : 2 and the ratio of the number of biscuits in the white 4\t 9.\t If y x = z y x = x z y , then find the value of box to the number of chocolates in the red box is +z + + 3 : 4. If the ratio of the total number of chocolates and biscuits in the white box to the total number each fraction. of chocolates and biscuits in the red box is 15 : 16. PRACTICE QUESTIONS Find the ratio of the total number of chocolates to \t50.\t The amount collected per month by the Hyderabad the total number of biscuits in the two boxes. Metro Water Works consists of two parts, a fixed charge for providing the service and a variable \t47.\t A father distributed `12250 among his four sons charge which is directly proportional to the num- A, B, C and D such that ber of kilolitres of water consumed. An amount \t\tAB=\ufffd\ufffd\u2019\u2019ss sshhaarree CB=\ufffd\ufffd\u2019\u2019ss sshhaarree C\ufffd\u2019s share 3 of `300 is charged for consuming 100 kilolitres in D\u2019\ufffd s share 4 = . Then find a particular month. It is also noticed that when the share of C. the consumption increases from 300 kilolitres per month to 400 kilolitres per month, the bill 6 4\t 8.\t Some amount is to be divided between X and Y amount increases to 5 times that of the former. in the ratio of 3 : 4. But due to wrong calculation, it was found that X got one-seventh of the total How much is the fixed charge per month? CONCEPT APPLICATION Level 1 \t1.\t If a : b = 5 : 4 and b : c = 16 : 25, then find a : b : c. \t2.\t P and Q have some coins with them. The ratio of the numbers of coins with P and Q is 7 : 8. Q has \t\t(a) 20 : 25 : 16\t\t (b) 25 : 20 : 16 5 more coins than P has. Find the number of coins \t\t(c) 25 : 16 : 20\t\t (d) 20 : 16 : 25 with Q.","Ratio, Proportion and\u00a0Variation 20.17 \t\t(a) 72\t\t (b) 40 \t\t(a) x2 \u2013 y2 + z2\t (b) x2 \u2013 y2 \u2013 z2 \t\t(c) 48\t\t (d) 80 \t\t(c) x2 + y2 + z2\t (d) None of these \t3.\t The numbers x \u2212 4, x \u2212 2, x + 2 and x + 10 are in \t12.\t Seven friends planned a tea party. The expenses proportion. Find x. per boy in rupees is numerically 1 less than the number of girls and the expenses per girl in rupees \t\t(a) 8\t\t (b) 10 is numerically 1 less than the number of boys. If the ratio of the total expenses of the boys and the \t\t(c) 6\t\t (d) 12 girls is 8 : 9, then what is the expenditure of each boy? \t4.\t If qr : pr : pq =1 :4 : 7, then find p : q . qr pr \t\t(a) 4 : 1\t\t (b) 1 : 4 \t\t(a) `2\t\t (b) `3 \t\t(c) 1 : 16\t\t (d) 16 : 1 \t\t(c) `1\t\t (d) `4 \t5.\t What number must be added to each term of the \t13.\t If a, b and c are in continued proportion, then ratio 9 : 11, so that it becomes 4 : 5. a2 : b2 is \t\t(a) 1\t\t (b) 2 \t\t(a) a2 : c2\t\t (b) a : b \t\t(c) \u20132\t\t (d) \u20131 \t\t(c) a : c\t\t (d) b : c \t6.\t A purse contains `50, `10 and `5 notes in the \t14.\t A metal X is 16 times as dense as metal Z ratio 6 : 3 : 7 and the total amount in the purse and metal Y is 7 times as dense as metal Z. In what is `730. Find the number of `5 notes in the ratio should X and Y be mixed to get an alloy 12 purse. times as dense as metal Z? \t\t(a) 7\t\t (b) 14 \t\t(a) 2 : 3\t\t (b) 5 : 2 \t\t(c) 21\t\t (d) 28 \t\t(c) 5 : 4\t\t (d) 6 : 5 \t7.\t If p : q : r : s = 3 : 4 : 7 : 8 and p + s = 55, then find \t15.\t The velocity of a freely falling body when it strikes q + r. the ground is directly proportional to the square of the time taken by it to strike the ground. It takes \t\t(a) 33\t\t (b) 55 5 seconds to strike the ground, with a velocity of PRACTICE QUESTIONS \t\t(c) 44\t\t (d) 66 10 m\/sec. If it strikes the ground with the veloc- ity of 40 m\/sec, then for how long would it have \t8.\t A metal X is 15 times as dense as metal Z and a travelled? metal Y is 8 times as dense as metal Z. In what ratio should these two metals be mixed to get an \t\t(a) 7.5 sec\t\t (b) 2.5 sec alloy which is 13 times as dense as metal Z? \t\t(c) 10 sec\t\t (d) 20 sec \t\t(a) 2 : 5\t\t (b) 2 : 3 1\t 6.\t The ratio of the number of students in two class- \t\t(c) 3 : 2\t\t (d) 5 : 2 rooms, C1 and C2, is 2 : 3. It is observed that after shifting ten students from C1 to C2, the ratio is 3 : 7. \t9.\t What must be added to each of the numbers 3, Further, how many students have to be shifted 7, 8 and 16 so that the resulting numbers are in from C2 to C1 for the new ratio to become 9 : 11? proportion? \t\t(a) 4\t\t (b) 3 \t\t(a) 10\t\t (b) 15 \t\t(c) 2\t\t (d) 1 \t\t(c) 20\t\t (d) 8 \t10.\t The cost of 570 bags of rice is `17,100. Find the \t17.\t Rakesh attended a party, where a lady asked him number of bags of rice which can be bought for his age. He said that ten years ago the sum of his `19,200. and his daughter\u2019s ages was 40 years, and the ratio of his present age to that of his daughter is 7 : 3. \t\t(a) 720\t\t (b) 800 What is the present age of Rakesh in years? \t\t(c) 640\t\t (d) 600 1\t 1.\t If x, y and z are in continued proportion, then \t\t(a) 42\t\t\t (b) 28 (x + y + z) (x \u2013 y + z) = _____. \t\t(c) 25\t\t\t (d) 30","20.18 Chapter 20 \t18.\t The mean proportional between two numbers is \t\t(a) 36,480\t\t (b) 59,520 56. If the third proportional of the same two num- \t\t(c) 72,960\t\t (d) None of these bers is 448, then find the sum of the two numbers. 2\t 5.\t In a monthly unit test, the marks scored by Bunny \t\t(a) 180\t\t (b) 440 and Sunny are in the ratio of 10 : 7 and those \t\t(c) 240\t\t (d) 140 scored by Bharat and Sunny are in the ratio of 8 : 9. If Bharat scores 112 marks, then find the marks \t19.\t The speed at which a person runs is directly pro- scored by Bunny. portional to the blood pressure in his body. If the speed of a person is 2 m\/sec his blood pressure \t\t(a) 170\t\t (b) 190 would be 120 units. Find the blood pressure in \t\t(c) \t200\t\t (d) 180 units of a person running at 8.5 m\/sec. x3 + 4xy2 15 \t\t(a) 510\t\t (b) 425 \t26.\t If 4x2y + y3 = 16 , then find x : y. \t\t(c) 340\t\t (d) 5950 \t20.\t A father wants to divide `22,515 among \t\t(a) 2 : 3\t\t (b) 3 : 2 \t\t(c) 4 : 3\t\t (d) 3 : 4 his four sons P, Q, R and S such that QP=\ufffd\u2019\u2019\ufffd ss sshhaarree QR=\u2019\ufffd\ufffd\u2019ss sshhaarree RS=\ufffd \u2019\u2019\ufffd sssshhaarree 1 . Find the \t27.\t The ratio of the tens digit and the units digit of 2 a two digits number is 2 : 3. How many possible values can it assume? share of Q. \t\t(a) `1507\t\t (b) `3002 \t\t(a) 3\t\t (b) 4 \t\t(c) 2\t\t (d) 5 \t\t(c) `6004\t\t (d) `12,008 2\t 1.\t The mean proportion of two numbers is 24 and \t28.\t The least integer which when subtracted from the their third proportion is 72. Find the sum of the antecedent and added to the consequent of the two numbers. ratio 9 : 8 gives a ratio less than the ratio 15 : 26 is \t\t(a) 11\t\t (b) 24 \t\t(a) 2\t\t (b) 3 \t\t(c) 32\t\t (d) 80 \t\t(c) 4\t\t (d) 1 PRACTICE QUESTIONS \t22.\t If 40 men can complete a job in 6 days, then find \t29.\t Ram wanted to distribute a certain amount the number of days taken by 60 men to complete between his two children Lava and Kusha in the it. ratio 5 : 7. But it was found that due to incorrect calculations Lava got one-sixth of the total amount \t\t(a) 3\t\t (b) 10 more than what he should get. Find the share of \t\t(c) 12\t\t (d) 4 Kusha, in rupees, if Lava got `560 in all. \t23.\t The cost of a precious stone varies as the cube of \t\t(a) 300\t\t (b) 350 its weight. The stone broke into 3 pieces whose \t\t(c) 400\t\t (d) 450 weights are in the ratio 1 : 2 : 3. As a result, its cost reduces. If the cost of the unbroken stone is `96,336, \t30.\t A bag contains one rupee, 50 paise and 25 paise then find the loss incurred due to breakage. coins. The ratio of the number of 1 rupee coins to that of 50 paise coins is 5 : 9 and the ratio of \t\t(a) `80,280\t\t (b) `16,056 the number of 50 paise coins to that of 25 paise coins is 2 : 1. Find the value of the 50 paise coins \t\t(c)\t `40,140\t\t (d) `8028 in the bag if the total value of the coins in the bag is `425. 2\t 4.\t A diamond falls and breaks into pieces whose weights are in the ratio 2 : 3 : 5. The value of the \t\t(a) `254\t\t (b) `180 diamond is directly proportional to the square of its weight. Find the loss incurred, if the actual cost \t\t(c) \u2009`78\t\t (d)\u2009Cannot be determined of the diamond is `96,000. (in `)","Ratio, Proportion and\u00a0Variation 20.19 Level 2 \t31.\t Seven years ago the ratio of the ages of P and Q (in \t\t(a) 48\t\t (b) 120 years) was 7 : 6. Which of the following cannot be \t\t(c) 64\t\t (d) 320 the ratio of their ages 6 years from now? 3\t 9.\t The force applied on a stationary body varies \t\t(a) 13 : 11\t\t (b) 15 : 14 \t\t(c) 13 : 12\t\t (d) 16 : 15 directly with the acceleration with which it starts \t32.\t The ratio of the monthly incomes of Ram and to move. If a force of 10 N is applied on a stationary Shyam is 3 : 4 and the ratio of their monthly expenditures is 4 : 5. If Shyam saves `400 per body, it starts to move with an acceleration of month, which of the following cannot be the 2 m\/sec2. Find the force (in N) to be applied savings of Ram? (In `\/month) on the body at which it starts to move with an acceleration of 4 m\/sec2. \t\t(a) 290\t\t (b) 280 \t\t(a) 30\t\t (b) 20 \t\t(c) 270\t\t (d) 310 \t\t(c) 40\t\t (d) 50 \t33.\t The ratio of the present ages of Anand and Bala 4\t 0.\t The ratio of the present ages of Ram and Shyam is is 8 : 3. When Anand was 30 years old, Bala was 5 3 : 2. Which of the following cannot be the ratio years old. Find the present age of Bala. (In years) of their ages 20 years ago? \t\t(a) 10\t\t (b) 12 \t\t(a) 8 : 5\t\t (b) 17 : 10 \t\t(c) 15\t\t (d) 20 \t\t(c) 9 : 5\t\t (d) 7 : 5 3\t 4.\t There are three sections A, B and C in class VIII 4\t 1.\t Ninety coins are to be distributed among P, Q and of a school. The ratio of the number of students R such that P gets twice as many coins as Q gets in A, B and C is 2 : 3 : 4. The section which has and Q gets thrice as many coins as R gets. Find the neither the maximum number of students nor the number of coins R gets. minimum number of students has a strength of 30. Find the total strength of the three sections. \t\t(a) 6\t\t (b) 3 \t\t(c) 10\t\t (d) 9 \t\t(a) 63\t\t (b) 81 \t42.\t P varies inversely with y. If y = 2, then P = 40. If PRACTICE QUESTIONS \t\t(c) 72\t\t (d) 90 P = 20 then find y. \t35.\t In a solution of 45 litres of milk and water, 40% is \t\t(a) 8 \t\t (b) 8 water. How many litres of milk must be added to \t\t(c) 4\t\t (d) 2 become the ratio of milk and water 5 : 3? 4\t 3.\t The monthly electricity bill raised by the municipal \t\t(a) 8\t\t (b) 3 corporation consists of two parts-a fixed charge for \t\t(c) 7\t\t (d) 6 providing the service and a variable charge which is directly proportional to the number of watts of 3\t 6.\t Find the triplicate ratio of (2y \u2013 x) : (2x \u2013 y), if power consumed. An amount of `500 is charged x : y = 4 : 3. for consuming 125 watts in a particular month. The ratio of the amount charged for 400 watts to \t\t(a) 64 : 125\t\t (b) 27 : 64 that of 500 watts is 21 : 25. How much is the fixed charge per month? \t\t(c) 8 : 125\t\t (d) 27 : 125 \t\t(a) `125\t\t (b) `200 \t\t(c) `250\t\t (d) `500 3\t 7.\t Nine years ago A\u2019s age and B\u2019s age were in the ratio 5 : 7. Which of the following cannot be the ratio 4\t 4.\t A solution of 30 litres of milk and water, has 70% of their ages 5 years from now? milk. How many litres of water must be added so that the volumes of milk and water will be in the \t\t(a) 11 : 13\t\t (b) 13 : 19 ratio 3 : 2? \t\t(c) 21 : 25\t\t (d) 15 : 16 \t38.\t A varies directly with x2. If x = 2 or 4, then A = \t\t(a) 6\t\t (b) 3 20. Find A if x = 8. \t\t(c) 5\t\t (d) 4","20.20 Chapter 20 \t45.\t If a +b = b+c = c +a where x + y \u2260 0 \t\t(a) 15,400\t\t (b) 16,800 xa + yb xb + yc xc + ya \t\t(c) 18,200\t\t (d) 19,600 and a + b + c \u2260 0, then each of these ratios is equal to \t48.\t The ratio of the monthly incomes of Mr Anand and Mr Milind is 9 : 10 and the ratio of their monthly \t\t(a) 1\t\t (b) 1 savings is 9 : 10. If the monthly expenditure of x+y Mr Milind is `15,000, then find the monthly expenditure of Mr Anand (in `). \t\t(c) 2 \t\t (d) 2 x + y a+b \t\t(a) `12,000\t\t (b) `9000 \t46.\t The ratio of sugar and other ingredients in the \t\t(c) `8500\t\t (d) `13,500 biscuits of three bakeries\u2014Mongunies, Karachi and Baker\u2019s Inn are 5 : 4, 13 : 12 and 29 : 24 \t49.\t The ratio of the present ages of Mr Dhuryodhana respectively. The biscuits of which bakery are the and Mr Dhushyasana is 5 : 4. Which of the fol- sweetest? lowing cannot be the ratio of their ages 5 years ago? \t\t(a) Mongunies \t\t(b) Karachi \t\t(a) 4 : 3\t\t (b) 3 : 2 \t\t(c) Baker\u2019s inn \t\t(c) 9 : 7\t\t (d) 8 : 7 \t\t(d) All the three types of biscuits are equally sweet. \t50.\t What must be added to both the terms of x : y so \t47.\t `x is divided among three persons Mr Vaayu, that the resultant ratio is inverse of the given ratio Mr Jala and Mr Agni in the ratio 3 : 4 : 7. If the (x \u2260 y)? share of Mr Jala is `5600, then find the value of x. \t\t(a) y \u2013 x\t\t (b) x \u2013 y \t\t(c) \u2013(x + y)\t\t (d) x + y Level 3 PRACTICE QUESTIONS \t51.\t Nine friends had a tea party. All boys took only 5\t 4.\t There are some students in an auditorium, some coffee and all girls took only tea. The cost per of them are dressed in white and the others are cup of coffee in rupees is numerically 2 less than dressed in blue. The ratio of the number of boys the number of girls and the cost per cup of tea in dressed in white to the number of girls dressed rupees is numerically 2 less than the number of in blue is 4 : 3 and the ratio of the total number boys. If the ratio of the total expenses of the boys of girls dressed in white to the number of boys and the girls is 5 : 6, then what is the cost of each dressed in blue is 4 : 5. The ratio of the total num- coffee? (In `) ber of boys and girls dressed in white to the total number of boys and girls dressed in blue is 12 : 13. \t\t(a) 2\t\t (b) 5 Find the ratio of the total number of boys to that \t\t(c) 7\t\t (d) 3 of girls in the auditorium. \t52.\t If x : y = 3 : 5, find the duplicate ratio of (3x + y) : \t\t(a) 13 : 15\t\t (b) 19 : 17 (5x \u2013 y). \t\t(c) 14 : 11\t\t (d) Cannot be determined \t\t(a) 49 : 25\t\t (b) 25 : 9 \t55.\t Eleven friends had a tea party. All boys took only \t\t(c) 36 : 25\t\t (d) 49 : 36 coffee and all girls took only tea. The cost per cup of coffee in rupees is numerically one more than the 5\t 3.\t The ratio of the number of students in two class- number of girls and the cost per cup of tea in rupees rooms A and B is 3 : 2. If ten students shift from is numerically equal to the number of boys. If the A to B, the ratio becomes 7 : 8. Now how many ratio of the total expenses of the boys and the girls is students must shift from A to B in order for the 7 : 6, then what is the cost of each coffee? (In `) ratio to become 8 : 7? \t\t(a) 5\t\t (b) 10 \t\t(a) 4\t\t (b) 5 \t\t(c) 15\t\t (d) 20 \t\t(c) 6\t\t (d) 7","Ratio, Proportion and\u00a0Variation 20.21 \t56.\t The ratio of the present ages of Mrs. Anoukika and \t\t(c)\t both (a) and (b) Mrs. Mythili is 5 : 6. After 6 years, Mrs. Anoukika \t\t(d) None of these would reach the present age of Mrs. Mythili. Find the sum of their present ages (in years). \t59.\t If 20 men take 15 days to complete a certain amount of work working at 10 hours per day, then \t\t(a) 66\t\t (b) 44 how many more men are required to complete \t\t(c) 33\t\t (d) 15 twice the previous work in 10 days working at 12 hours a day? \t57.\t The incomes of A and B are in the ratio 4 : 3. The expenditures of A and B are in the ratio5 : 2. If \t\t(a) 10\t\t (b) 20 B saves `3000, then which of the following cannot \t\t(c) 30\t\t (d) 40 be the savings of A? 6\t 0. \t The ratio of the amounts with Mr Umar and \t\t(a) `1500\t\t (b) `2500 Mr Gumar is 3 : 4. If Mr Gumar gives `5 to \t\t(c) `3500\t\t (d) `4500 Mr Umar, then the ratio of the amounts with Umar and Gumar is 4 : 3. Mr Umar gives `5 to 5\t 8.\t If a + b varies directly with a \u2013 b, then which of the Mr Gumar. Find the ratio of the amounts with them. following vary directly? \t\t(a) a and b \t\t(a) 3 : 5\t\t (b) 2 : 5\t \t\t(b) a2 + b2 and a2 \u2013 b2 \t\t(c) 4 : 5\t\t (d) 1 : 5 PRACTICE QUESTIONS","20.22 Chapter 20 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t 1 = 0.2 1\t 5.\t 27 : 35 5 \t16.\t False \t2.\t a = c 1\t 7.\t 65 \u2212 \u2212 a b c d 1\t 8.\t 5 : 14 \t3.\t 19 : 24 1\t 9.\t 3 : 13 \t4.\t 12 : 4 : 3 2\t 0.\t 21 and 28 \t5.\t 5 : 8 \t21.\t 135 : 256 \t6.\t 32 : 1 \t22.\t 8 \t7.\t a = c 2\t 3.\t 27 b d \t8.\t p : q 2\t 4.\t 63 : 81 \t9.\t True 2\t 5.\t 2 5 \t26.\t 3 : 1 \t10.\t 6 : 1 2 4 6 10 \t11.\t 4 times itself 27.\t 3 , 5 , 7 and 11 \t12.\t 4 \t28.\t `90, `180 1\t 3.\t 6 \t29.\t direct proportion 1\t 4.\t 3 3\t 0.\t 2 Shot Answer Type Questions 3\t 9.\t 2 \t31.\t 70 4\t 0.\t 2 3\t 2.\t 3 \t33.\t 3 : 4 4\t 1.\t 700 K \t34.\t 4 3 ANSWER KEYS 3\t 5.\t 11 : 3 3\t 6.\t 13 : 10 4\t 2.\t 4500\u03c0 cm3 3\t 7.\t 16 \t38.\t 35 \t43.\t 4 Essay Type Questions 4\t 4.\t 205 \t46.\t 18 : 13 \t45.\t ps = qr 4\t 7.\t `3360 4\t 8.\t `360 \t49.\t \u20131 \t50.\t `200","Ratio, Proportion and\u00a0Variation 20.23 CONCEPT APPLICATION Level 1 \t1.\u2002(d)\t 2.\u2002 (b)\t 3.\u2002 (c)\t 4.\u2002 (d)\t 5.\u2002 (d)\t\u20026.\u2002 (b)\t 7.\u2002 (b)\t 8.\u2002 (d)\t 9.\u2002 (c)\t 10.\u2002 (c) \t11.\u2002 (c)\t 12.\u2002 (a)\t 13.\u2002 (c)\t 14.\u2002 (c)\t 15.\u2002 (c)\t 16.\u2002 (b)\t 17.\u2002 (a)\t 18.\u2002 (d)\t 19.\u2002 (a)\t 20.\u2002 (b) \t21.\u2002 (d)\t 22.\u2002 (d)\t 23.\u2002 (a)\t 24.\u2002 (b)\t 25.\u2002 (d)\t 26.\u2002 (b)\t 27.\u2002 (a)\t 28.\u2002 (b)\t 29.\u2002 (c)\t 30.\u2002 (b) Level 2 31.\u2002 (a)\t 32.\u2002 (d)\t 33.\u2002 (c)\t 34.\u2002 (d)\t 35.\u2002 (b)\t 36.\u2002 (c)\t 37.\u2002 (b)\t 38.\u2002 (d)\t 39.\u2002 (b)\t 40.\u2002 (d) 41.\u2002 (d)\t 42.\u2002 (c)\t 43.\u2002 (c)\t 44.\u2002 (c)\t 45.\u2002 (c)\t 46.\u2002 (a)\t 47.\u2002 (d)\t 48.\u2002 (d)\t 49.\u2002 (d)\t 50.\u2002 (c) Level 3 53.\u2002 (a)\t 54.\u2002 (c)\t 55.\u2002 (d)\t 56.\u2002 (a)\t 57.\u2002 (d)\t 58.\u2002 (c)\t 59.\u2002 (c)\t 60.\u2002 (b) 51.\u2002 (a)\t 52.\u2002 (a)\t ANSWER KEYS","20.24 Chapter 20 CONCEPT APPLICATION Level 1 \t1.\t Compare both the ratios by making b equal in \t13.\t \u2009(i) \u0007If a, b and c are in continued proportion, then both of them. b2 = ac. \t2.\t Let P = 7x and Q = 8x. The difference in the \t\t(ii) Substitute b2 = ac in a2 : b2. number of coins with P and Q is 5. \t14.\t (i) Let the required ratio be x : 1. \t3.\t Product of extremes = Product of means. \t\t(ii) Let the density of metal z be k. \t4.\t Simplify the required ratio. \t\t\u2234 Density of metal x = 16k and that of y = 7k. \t5.\t Assume \u2018x\u2019 as the number to be added and apply \t\t(iii) Let the required ratio be m : 1. Product of extremes = Product of means. \t\t(iv) (16k)m + (7k)1 = (m + 1)12k, solve this for m. \t6.\t Let the number of notes be 6x, 3x and 7x. V1 t12 1\t 5.\t \u2009(i) V2 = t22 . (Apply the concept of variation). \t7.\t Let p, q, r and s be 3x, 4x, 7x and 8x. \t\t\u2009\u2009(ii) Given that V =k (where k is constant). t2 \t8.\t \u2009\u2009(i) Let the required ratio be x : 1. v1 v2 t12 t22 \t\t(ii) Let the density of metal z be k. \t\t(iii) = . \t\t\u2234 Density of metal x = 15k \t16.\t \u2009(i) C\u0007 onsider the number of students in C1 and C2 \t\tDensity of metal y = 8k. as 2x and 3x respectively. \t\t(iii) Let required ratio be m : 1. \t\t(ii) L\u0007 et the number of students in the two classes Hints and Explanation \t\t(iv) (15k)m + (8k)1 = (m + 1)(13k). C1 and C2 be 2x and 3x respectively. \t\tSolve this equation to get m. \t\t(iii) Given (2x \u2212 10) : (3x + 10) = 3 : 7. \t9.\t Let the number added be x and use, Product of \t\t\tFind x from this equation. extremes = Product of means. \t\t(iv) L\u0007 et y be the number of students to be shifted 1\t 0.\t Use the unitary method. from C2 to C1. \t11.\t \u2009(i) I\u0007 f x, y, z are in continued proportion, then \t\t(v) \u0007Substitute value of x in (2x \u2212 10 + y) : (3x + 10 \u2212 y) = y2 = xz. 9 : 11 and find y. \t\t\u2009\u2009(ii) \u0007First of all multiply the given expressions and \t17.\t \u2009(i) Frame the linear equations and solve them. simplify. \t\t\u2009(\u2009ii) T\u0007 he sum of present ages of Rakesh and his \t\t(iii) \u0007Substitute xz = y2, since x, y and z are in con- daughter is (40 + 20) years. tinued proportion. \t\t(iii) Divide 60 in the ratio 7 : 3. \t12.\t \u2009\u2009(i) C\u0007 onsider the expenses of boys and girls as x and y respectively and proceed. 1\t 8.\t \u2009\u2009\u2009(i) I\u0007 f b2 = ac, then b is the mean proportional and c is the third proportional. \t\t(ii)\t Number Number \t\t\u2009\u2009(ii) Let the two numbers be a and b. of Boys of Girls \t\t(iii) ab = (56)2. Expense b g \t\t(iv) b2 = 448a. per student g\u22121 b\u22121 S1 BP1 1\t 9.\t (i) S2 = BP2 . (Apply the concept of variation). \t\t(iii) bg \u2212b = 8 . 2 120 gb \u2212g 9 8.5 BP2 \u21d2 = \t\t(iv) S\u0007 olve b + g = 7 and the above equation and get \t\t BP2 = 510 (g \u2212 1) in rupees. \u2234"]