Page 1 :
warm-up!, , jous class. you have already leamt about, sath jnterest and its applications., , - is money lent or borrowed or invested is called, a Principal (P) and the extra charge paid for using, aye money is called interest (I)., , $ sple interest is the interest calculated on the, , none) borrowed or lent out for a fixed time period., , | rere is n0 change in the principal throughout the, pened set in case of simple interest (SI)., , : PxRxT .., . simple interest, ST = sien the principal, , , , (py. the rate of interest (R) and the time (7)., « Amount (A) is calculated by adding the simple, jaterest to the principal., } tatis. A= P+ SL, Exomple I: Find the simple interest (SI) and amount, (4) for the principal of %800 at the rate of 5% per, num for 3 years., Solution: Given that P = 2800; R = 5% per annum,, , T=3 years., 2 PXRxT _ 800x3x5, ipl st= PRR =, Simple Interest, ST 700 oO, 12000 _.120, 100, Exercise, , 1. Choose the correct option., , a. If principal = 7600, rate = 5% per annum, and, time = 2 years. What is the simple interest?, i.%600 ii. 750 iii. 730 iv. 760, , b. What principal amount will earn a simple interest, of 2375 at 5% p.a. rate of interest ina year?, 1.33750 ii. 25700 _ iii. 27500 iv. 37000, , c. Inhow many years will %4800 yield an interest, of 2960 at 5% per annum simple interest?, i4 ii, 2 iii. 3 iv. |, , => Simple interest, SI = 7120, , Amount, A = P + SI = %800 + 7120 = 7920, => Amount, A = %920, Therefore. simple interest, Example 2: In how many years will 72400 yield an, interest of 2720 at 10% per annum simple interest?, Solution: Principal, P = 72400; SI = 2720;, , Rate of interest, R = 10% p.aSIx100 _ 720x100 _ 72000 _3, , Wehave. T= “pg = 2400%10 24000, , => T=3 years, , Therefore, %2400 will yield an interest of 2720 in 3, , years., , Example 3: Irfan deposits 24200 in a bank for 4 years, which will earn him a simple interest of 8% p.a. What, , is the amount he gets after 4 years?, , Solution: Principal, P = 74200; Time, T=4years: oy, Rate of interest, R = 8% p.a- gpa PXRXT 2 ROOKIE 2 1344 = SI=TIH, , , , js 2120 and amount is 920., , 100 100, We have, Amount, A = P + ST, = 74200 + T1344 = W554., , Therefore, Irfan gets an amount of @55-4 after 4+ years., , 2, Find the simple interest for P = 85600 and, A=%6720., , 3, Find the Principal, when SI = 71200, R = 25% per, annum; and T = 2:years., , 4. If 2980 amounts to 1372 in | year, find the rate of, interest per annum., , 5. How many years will it take for a sum of $8250 to, earn an interest of 83960 at the rate of 8% p.a.?, , 6. Interest obtained on a sum of %2640 for 6 years is, 2792. Calculate the rate of interest,, , pee VYPIA VAL AY

Page 2 :
ComPounp INTEREST, , tus look at the fo, Jeevan deposited 2], 3 years in asim, Teceives F109, , llowing example., , 5,000 at 4% per annum for, , Ple interest based scheme. He, , as interest at the end of 3 years., i Px ExT - seed = 1800, e interest earned by Jeevan = SI = 21800., , Manu also deposited the same amount at the same, , Tate of interest. But, he earned %1872.96 as an, Interest at the end of 3 years,, , How did Manu earn more, interest than Jeevan?, Let us see how the interest, , is being calculated in this, Case., , Interest for the 1st year, = PXRXxT _15000x1x4, ~ 100 = —F99, , For the 2nd year, interest is calculated on a new, Principal given by 15,000 + %600 = 315,600, Interest for the 24 year = woxixt = %624, , = %600, , Interest for the 34 year is calculated on 215,600, +%624 = 216,224, , Interest for the 3 year = ixd = %648.96, , Total interest earned by Manu for 3 years = 7600, , + %624 + 2648.96 = %1872.96, , Observe that in this case, the principal and the, , interest earned on it are increasing over the, , Period. Also, the interest is calculated on the, , Previous year’s amount ie,, the interest earned, , is added to the Principal and the total amount, , is considered as the new Principal for the next, , year. Such an interest earned on the sum of the, , Principal and the accumulated interests of the, , Previous years is called compound interest (Cl), , We can see that,, , * For the same amount of money invested, at, the same rate for the same time period, the, compound interest is more than the simple, interest., , 1872.96 > 21800 ie. CI> SI, , * Let us now calculate the simple interest earned, by Jeevan for the first year., , fe PXRxT = 15000x 4x1, , 100 100, = %600 = Cl for one year, , , , oer ree, , str, , , , , imple interest is SAME AS the Con, Thus, the 5 first year. Py, , interest for the, , tite Oa, , ‘on of interest is based on the, , lati :, . e us year’s amount when the Mteres _, prev!, , ounded. be es, ea in the successive periods is Caley, « Intel, , as Cleat, the sum of the principal and the iMteresy d, obtained for the last period., , , , , , , , , , , , , , , f iod, me time period, compoung i, Within the given t ; Me, . imes calculated every Months, o, ah, is aie or once a year. The time inten”, she wii the interest is added to the previo, year’s principal each time to get the Principay iy, the next period, is determined based on Whethey, the interest is compounded quarterly, half yey, or annually. :, This time period when the interest is addeq tof, existing principal to obtain the new Principal i, called the conversion period., We can say that, when the interest is Compound, annually, the time lag between the Successive, conversions of interest to principal is one Year, or, there is only one conversion over a year,, Similarly, when the interest is compounded sen., annually, there are two such conversions during, year, and four such conversions happen when the, interest is compounded quarterly,, , Interest Compounded Annually, Example 4: Sofia borrowed a sum of 37500 for, , Pay at the end of 2 years., Solution: Step 1: Find the interest (simple interest) fa, one year. P= 27500, T= | year, R = 3% p.a., , Interest, J = PXRXT = 7500x3x1, , 100 TG = 2225, Amount at the end of 1* year = 7500 + 225 = 27705., Step 2: Principal = Amo,, , unt at the end of 1° year, =37725,T=] Year, R= 3% p.a,, Interest, 7 = PRRxT 7725x3x1, , Amount at the end of 24 year = 27725 + 2231.15, , = 7956.75

Page 3 :
r CO, , und Interest, CI = Total interest, Comms = 225 + 231.75 = 8456.75, , oy, ” sofia has to pay a sum of 37956.75 at the end of, ie years and the total interest paid is 2456.75, , to be paid after, , 2 sink! Will Sofia pay the same interest if she e =], | charged simple interest rate? ; |, , PXRxT 759, te st, St PXRXT _ 7500%3x2, simple Interes 100 = qq = 8450, , Note that, 3456.75 > 2450, ic. Cl> SI., , , , this method of calculation of compound interest, will get lengthier with the increase in the time period, , qd for different conversion periods. Let us deduce a, , formula to calculate the compound interest for ease of, calculations., , formula for Compound Interest, , Let P be the principal and R be the rate of interest, compounded annually,, , PXR, rest for the 1* year, J = ———, Inte! 00, Amount at the end of It year= py PR, 100, = a(t + x), 100, This amount becomes the principal for the, 24 year., af + a) xR, Interest for the 2"4 year = 100), 100, = (i +) x Rk, 100 100, , Amount at the end of 2"4 year, , aPiageX.) ap 12%). 8, 100 100) * 100, , 2, , R R R, , = ae —| =P} 1+—, (+5) (+5) ( +#), , This amount becomes the principal for the, , 34 year., 2, [+t] xR, , 100, , , , Interest for the 34 year =, , Amount at the end of 3% year, \2 \2, R ( R R, = ake 1+—] x —, rl +t) PU tTog) * 00, , =1(tog) (tea) (8), , Continuing in the similar manner, the amount at, the end of n years is given by,, R n, A= vf 1+ nl, 100, We know that, CI = Amount - Principal = A —P, R\" R\" 4, = —| -P= 1+—| , cl r(1+%) P=P ( a, Note that the conversion period for the interest, compounded annually is 1 year., Example 5: Calculate the compound interest on a, sum of 713,200 for 4 years at a rate of 6% per annum, compounded annually., Solution: We have, Principal, P = 713,200,, , Rate of interest, R = 6%, Time period = 4 years > n=4, , Amount = (1 + x), , 100, 6 \ 3\, =a=13200(1+;5) = 13200 (1+ 5], 53)" a, = 13200 | =] = 13200 (1.06), A=216,664.7, , CI=A — P=216,664.7 — 213,200 = %3464.7, Thus, the compound interest on $13,200 for 4 years at, arate of 6% p. a. compounded annually is 73464.7., , Example 6: Find the compound interest on a sum of, %2160 for 3 years at 162% p.a. compounded annually., Solution: P =%2160; Time = 3 years; R= 162.% pa., , We have, Amount = (i +5) (Here, n = 3), 3, 165 50 3, A =2160|1+4+—3| = 50), 216o(14 =], a 2160 32 * = 2160 (1.166) = 33424, 300) OO) RS, , CI=A — P=%3424 - 2160 = T1264., Hence, the compound interest on 2160 for 3 years at, , 165% p.a. compounded annually is 71264.

Page 4 :
When the Time Period is Given as a Fraction, , Example 7: Jithin invested, , |, a sum of 88560 for 25, Years at 2], , % p.a. compounded annually. How much, , amount will he get after 25 years?, , Solution: P =%g560: Time period for investment, 1, , =2, > years >n=21.p, , iR=21% pa., 2 2 ‘, We know that, Amount = >| 1+ a), , mt, , 21)\2, 14+—, 100, , 5 =, , 21) 2\2, = 8s6o( 7) = 8560 (7), 100 10, , (On applying the laws of exponents.), , 5, = 8s6o{ 2) = 8560 x (1.1)5 = 213,785.96, , Jithin will get an amount of 213,785.96 after 25, years., , A= 8560,, , 1, Example 8: Jaya took a loan of %6000 for 25 years at, , 4% p.a. compounded annually. How much should she, repay? Also, find the compound interest paid by Jaya., , 1, Solution: P = %6000, n= 23, R=4% pa., RY, Je = oe, We know that, A (i + x), , Let us first calculate the amount to be paid, after 2 years,, , 4y, A= 6000{ +5), , 616 _ 48x 676, 625 5, Now, %6489.6 serves as the principal for, , = 6000 x = %6489.6, , , , the remaining 5 year of the loan period., , 6489.6 x 4 x J, 3 = %36,528, Thus, the amount at the end of 21 years, %6489.6 + %86.528 = %6576,128,, Hence, Jaya should repay an amount of %6576.128 at, the end of 2 years,, , , , CI = %6576.128 - %6000 = 3576, 128, Thus, the compound interest paid by Jaya on 26000, , for 2 ; years at 4% pa, compounded annually is, %576.128., , , , , , , , , , a sum of €25,000 at §, , ‘ %o, 1, Vikram invested i, , in a simple interest based sche, vears ina simp: b, Li interest earned. Also, find the, Fin + received by Vikram, if the interest, a compounded annually. What is the, Wiesaite between the interests carned jp, ; ., ¢ situations? ;, 2 nant the following when the interest j,, ounded annually:, ene when P = %5000; T= 4 years;, = 6% p.a., b. ui when P = £12,500; T= 5 years;, R=4% pa., 3, Rahul took a loan of €1,02,400 for 2 years a, vig p.a. compounded annually. How much, 2, , should he repay? 1, 4, Find CI ona sum of €10,000 for 25 years at, , an interest of 3% p.a. compounded annually,, 5. Find the amount to be repaid on a loan of, , 315,000 for 2 years and 9 months at an, , interest of 1.5% p.a. compounded annually,, , , , , , Interest Not Compounded Annually, , We have calculated the compound interest, , when the interest is compounded annually. But, this is just an ideal situation, we may have to, find the compound interest when the interest is, compounded quarterly or half-yearly, i.e. the, interest is calculated every three months or every, six months respectively., , In such cases, the time Period as well as the rate, of interest need to be converted appropriately., , Interest Compounded Half-yearly, Conversion period, , A year has 12 months. When the interest is, compounded half-yearly, the interest is calculated, , twice in a year, that is, every 6 months. Here, the, Conversion period is 6 months., , Rate of interest, If the rate of interest is, then, Correspondin;, rate will change as, Rate of interest =, for each half Year,, , given as R% per annum,, € to the conversion period, the, , 9, , R, Half of the annual rate = 2 %

Page 5 :
terest Compounded Quarterly, ott v version period, when the interest is compounded, tert is calculated 4 times jn, eve 3 months. And thus, the ¢, is ey months., , quarterly, the, 4 year, that is,, Onversion period, , ate of interest, , when the interest is compounded quarterly, the, te of interest given as R% Per annum becomes, , 4 ve fourth of the annual rate R, qherefore, rate of interest, , ! R, = of R%= =, each quarter year, 4 a, , When the interest is compounded annually, we, R ,, , 00, When the interest is compounded half yearly, rate, , | of interest = 7? half-yearly and period = 2n half, years., , formula for Compound Interest, | ae dea( +k, , f, , t, , R\", , 2n, =P\14+— = R, 2A ; orA ats; ), , f R 2a, Cl=P|}1+—]| and ( = ], | When the interest is compounded quarterly,, , 5 R, rate of interest = — % quarterly and period = 4n, , | quarter-years., R, , R 4n, >A=P| 1+ or=P[ 14, 100 400, , dn, , , , , , , , , , * Incase of interest being compounded halfyearly, the time period becomes twice the full, years and the rate per annum is halved., , * Incase of interest being compounded, , quarterly, the time period becomes four times, , the full years and the rate per annum becomes, one-fourth., , , , , , Fvample 9 Vidya took a loan of &15,000 for, , 1 years at 6% p.a. compounded ee How, , ae amount should she pay after I= years?, , Solution: Here, P = 215,000; Time period = 1 years, = 5 years R=6% pa., Since the interest is compounded half-yearly, we have,, , n=2x $= 3nd R= s% = 3% half-yearly., = tsoon(1+ 3) = ‘a 22) = 216,390,905, , 1, +. Vidya should repay 716,390.905 after 5 years., Example 10: Find the compound interest on 220,000, , for 25 years at 3% p.a., interest compounded semi, annually., Solution: Since the interest is compounded semiannually i.e., 2 times in a year, we have,, , P=820,00;n=2x 25=2x 5 =5:, R= = semi-annually., 3 3., , o $, A = 20000] 14-2. = 20000{ 1+ 35), 00 200, , = zoono( 23) = 21,545.68, , CI =A — P= 221,545.68 - 720,000 = 71545.68, , «. Thus, the compound interest on 220,000 for 22, years at 3% p.a. compounded semi-annually is“, %1545.68., , Example 11: Find the compound interest on %25,000, for one year at 8% p.a. compounded quarterly., , Solution: Since the interest is compounded quarterly, we have,, , P=805,000;n= 4x 1=4; R= © % = 2% quarterly., 2) 1\, A=25000(1+-2-) = a5000( 141, ( 2] 25 (+3), , s1\*, = 25000 0) = %27,060.8, , CL. =A — P =227,060.8 - 725,000 = %2060.8, <. Thus, the compound interest is 2060.8.