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Page No 144:Question 1:Find the amount and the compound interest on Rs 2500 for 2 years at 10% per annum, compounded annually. Answer:Principal for the first year = Rs. 2500Interest for the first year = Rs. 2500×10×1100 = Rs. 250Amount at the end of the first year = Rs. (2500 +250 ) = Rs. 2750Principal for the second year = Rs. 2750Interest for the second year = Rs. 2750×10×1100= Rs. 275Amount at the end of the second year = Rs. (2750 + 275) = Rs. 3025∴ Compound interest = Rs. 3025 - 2500 = Rs. 525 Page No 144:Question 2:Find the amount and the compound interest on Rs 15625 for 3 years at 12% per annum, compounded annually. Answer:Principal for the first year = Rs. 15625Interest for the first year = Rs. 15625×12 ×1100= Rs. 1875Amount at the end of the first year = Rs. ( 15625 + 1875) = Rs. 17500Principal for the second year = Rs. 17500Interest for the second year = Rs. 17500×12×1100= Rs. 2100Amount at the end of the second year = Rs. (17500 + 2100 )= Rs. 19600Principal for the third year = Rs. 19600Interest for the third year = Rs. 19600×12×1100= Rs. 2352 Amount at the end of the second year = Rs (19600 + 2352) = Rs. 21952 ∴ Compound interest = Rs. 21952 - 15625 = Rs. 6327 Page No 144:Question 3:Find the difference between the simple interest and the compound interest on Rs 5000 for 2 years at 9% per annum. Answer:Principal amount = Rs. 5000Simple interest = Rs. 5000×2×9100= Rs. 900The compound interest can be calculated as follows:Principal for the first year = Rs. 5000Interest for the first year = Rs. 5000×9 ×1100= Rs. 450Amount at the end of the first year = Rs. (5000 +450 ) = Rs. 5450Principal for the second year = Rs. 5450Interest for the second year = Rs. 5450×9×1100= Rs. 490.5Amount at the end of the second year = Rs. (5450 +490.5) = Rs. 5940.5∴ Compound interest = Rs. 5940.5 - 5000 = Rs. 940.5Now, difference between the simple interest and the compound interest = CI - SI =Rs. 940.5 - 900=Rs. 40.5 Page No 144:Question 4:Ratna obtained a loan of Rs 25000 from the Syndicate Bank to renovate her house. If the rate of interest is 8% per annum, what amount will she have to pay to the bank after 2 years to discharge her debt? Answer:Principal for the first year = Rs . 25000Interest for the first year = Rs. 25000×8×1100= Rs. 2000 Amount at the end of the first year = Rs. (25000 +2000) = Rs. 27000Principal for the second year = Rs. 27000Interest for the second year = Rs. 27000×8 ×1100= Rs. 2160Amount at the end of the second year = Rs. (27000 + 2160) = Rs. 29160Therefore, Ratna has to pay Rs. 29160 after 2 years to discharge her debt. Page No 144:Question 5:Harpreet borrowed Rs 20000 from her friend at 12% per annum simple interest. She lent it to Alam at the same rate but compounded annually. Find her gain after 2 years. Answer:Principal amount = Rs. 20000Simple interest = Rs. 20000×2×12100= Rs. 4800The compound interest can be calculated as follows: Principal for the first year = Rs. 20000Interest for the first year = Rs. 20000×12×1100= Rs. 2400Now, amount at the end of the first year = Rs. (20000 + 2400)= Rs. 22400Principal for the second year = Rs. 22400Interest for the second year = Rs. 22400×12×1100= Rs. 2688Now, amount at the end of the second year = Rs. (22400 + 2688) = Rs. 25088Hence, compound interest = Rs. 25088 - 20000 = Rs. 5088 Now, CI - SI =Rs. 5088-4800= Rs. 288∴ The amount of money Harpreet will gain after two years is Rs 288. Page No 144:Question 6:Manoj deposited a sum of Rs 64000 in a post office for 3 years, compounded annually at 712 % per annum. What amount will he get on maturity? Answer:Principal for the first year = Rs. 64000Interest for the first year = Rs. 64000×15×1100×2= Rs. 4800Now, amount at the end of the first year = Rs.( 64000 +4800) = Rs. 68800Principal for the second year = Rs. 68800Interest for the second year = Rs. 68800×15×1100×2 = Rs. 5160Now, amount at the end of the second year = Rs. ( 68800 + 5160) = Rs. 73960Principal for the third year = Rs. 73960Interest for the third year = Rs. 73960×15×1100×2= Rs. 5547Now, amount at the end of the third year = Rs. (73960 +5547) = Rs. 79507∴ Manoj will get an amount of Rs.79507 after 3 years. Page No 144:Question 7:Divakaran deposited a sum of Rs 6250 in the Allahabad Bank for 1 year, compounded half-yearly at 8% per annum. Find the compound interest he gets. Answer:Principal amount = Rs. 6250Rate of interest = 8% per annum = 4% for half yearTime = 1 year = 2 half yearsPrincipal for the first half year = Rs. 6250Interest for the first half year = Rs. 6250×4×1100= Rs. 250Now, amount at the end of the first half year = Rs. (6250 +250) = Rs. 6500Principal for the second half year = Rs. 6500Interest for the second half year = Rs. 6500×4×1100= Rs. 260Now, amount at the end of the second half year = Rs (6500 + 260) = Rs. 6760∴ Compound interest = Rs 6760 - 6250 = Rs 510Hence, Divakaran gets a compound interest of Rs 510. Page No 145:Question 8:Michael borrowed Rs 16000 from a finance company at 10% per annum, compounded half-yearly. What amount of money will discharge his debt after 112 years? Answer:Principal amount = Rs. 16000Rate of interest = 10% per annum = 5% for half yearTime = 112 years = 3 half yearsPrincipal for the first half year = Rs. 16000Interest for the first half year = Rs. 16000×5×1100= Rs. 800Now, amount at the end of the first half year = Rs. (16000 + 800) = Rs. 16800 Principal for the second half year = Rs. 16800Interest for the second half year = Rs. 16800×5×1100= Rs. 840Now, amount at the end of the second half year = Rs. (16800 + 840) = Rs. 17640Principal for the third half year = Rs. 17640Interest for the third half year = Rs. 17640×5×1 100= Rs. 882Now, amount at the end of the third half year = Rs. (17640 + 882) = Rs. 18522∴ The amount of money Michael has to pay the finance company after 112years is Rs 18522. Page No 151:Question 1:By using the formula, find the amount and compound interest on: Answer:Principal amount, P = Rs 6000Rate of interest, R= 9% per annumTime, n =2 years.The formula for the amount including the compound interest is given below:A = Rs. P1+R100n ⇒A = Rs. 6000 1 +91002⇒A = Rs. 6000 100+91002⇒A = Rs. 6000 1091002⇒A = Rs. 6000 1.09×1.092⇒A = Rs. 7128.6i.e., the amount including the compound interest is Rs 7128.6. ∴ Compound interest=Rs 7128.6 -6000=Rs 1128.6 Page No 151:Question 2:By using the formula, find the amount and compound interest on: Answer:Principal amount, P=Rs. 10000Rate of interest, R=11% per annum.Time, n=2 years.The formula for the amount including the compound interest is given below:A = Rs. P1+R100n ⇒A = Rs. 10000 1+111002⇒A = Rs. 10000 100+111002⇒A = Rs.10000 1111002⇒A = Rs.10000 1.11×1.112⇒A = Rs. 12321i.e., the amount including the compound interest is Rs 12321.∴ Compound interest =Rs. 12321 -10000 = Rs. 2321 Page No 151:Question 3:By using the formula, find the amount and compound interest on: Answer:Principal amount, P = Rs. 31250Rate of interest, R= 8 % per annum.Time, n =3 years.The formula for the amount including the compound interest is given below:A = Rs. P 1+R100n ⇒A = Rs. 31250 1+81003⇒A = Rs. 31250 100+81003⇒ A = Rs.31250 1081003⇒A = Rs.31250 1.08×1.08×1.083 ⇒A = Rs. 39366i.e., the amount including the compound interest is Rs 39366.∴ Compound interest=Rs. 39366 -31250=Rs. 8116 Page No 151:Question 4:By using the formula, find the amount and compound interest
on: Answer:Principal amount, P = Rs. 10240Rate of interest, R= 1212% p.a. Time, n =3 yearsThe formula for the amount including the compound interest is given below:A = Rs. P1+ R100n ⇒A = Rs. 10240 1+25100×23⇒A = Rs. 10240 1+252003⇒A = Rs. 10240 1+183⇒A = Rs. 10240 8+183⇒A = Rs. 10240 98 3⇒A = Rs. 10240 1.125×1.125×1.1253⇒A = Rs. 14580i.e., the amount including the compound interest is Rs 14580.∴ Compound interest=Rs 14580 -10240 = Rs. 4340 Page No 151:Question 5:By using the formula, find the amount and compound interest on: Answer:Principal amount, P = Rs 62500Rate of interest , R= 12% p.a.Time, n =2 years 6 months=52=21 2 yearsThe formula for the amount including the compound interest is given below:A = Rs. P1+R100n ⇒A = Rs. 62500 1+121002× 1+12×12100⇒A = Rs. 62500 1+121002×1+6 100⇒A = Rs. 62500 ×1.12×1.12×1.06⇒A = Rs. 83104i.e., the amount including the compound interest is Rs 83104.∴ Compound interest =Rs. 83104-62500 = Rs. 20604 Page No 151:Question 6:By using the formula, find the amount and compound interest on: Answer:Principal amount, P=Rs. 9000Rate of interest, R= 10% p.a. Time, n =2 years 4 months = 213years = 73 yearsThe formula for the amount including the compound interest is given below:A = Rs. P ×1+R100 n = Rs. 9000×1+101002×1+13×10 100= Rs. 9000×1.10×1.10×1.033= Rs. 11252.9≈11253i.e., the amount including the compound interest is Rs 11253.∴ Compound interest =Rs. 11253 -9000 = Rs. 2253 Page No 151:Question 7:Find the amount of Rs 8000 for 2 years compounded annually and the rates being 9% per annum during the first year and 10% per annum during the second year. Answer:Principal amount, P = Rs. 8000Rate of interest for the first year, p= 9 % p.a.Rate of interest for the second year, q= 10% p.a. Time, n =2 years.Formula for the amount including the compound interest for the first year:A = Rs. P×1+p100×1+q100 = Rs. 8000×1+9100×1+10100 = Rs . 8000×109100×110100 = Rs. 8000 × 1.09×1.1= Rs. 9592i.e., the amount including the compound interest for first year is Rs 9592. Page No 151:Question 8:Anand obtained a loan of Rs 125000 from the Allahabad Bank for buying computers. The bank charges compound interest at 8% per annum, compounded annually. What amount wil he have to pay after 3 years to clear the debt? Answer:Principal amount, P=Rs. 125000Rate of interest, R= 8% p.a.Time, n =3 yearsThe amount including the compound interest is calculated using the formula,A = Rs. P1+R100n = Rs. 125000 1+81003 = Rs. 125000 100+8100 3 = Rs. 125000 1081003= Rs. 125000 1.083= Rs. 125000 1.08×1.08×1.08= Rs. 157464∴ Anand has to pay Rs 157464 after 3 years to clear the debt. Page No 151:Question 9:Three years ago, Beeru purchased a buffalo from Surjeet for Rs 11000. What payment will discharge his debt now, the rate of interest being 10% per annum, compounded annually? Answer:Principal amount, P = Rs. 11000Rate of interest, R= 10% p.a.Time, n =3 yearsThe amount including the compound interest is calculated using the formula,A = Rs. P 1+R100n = Rs. 11000 1+101003 = Rs. 11000 100+101003 = Rs. 11000 1101003 = Rs.11000 1.13 = Rs. 11000 1. 1×1.1×1.1 = Rs. 14641Therefore, Beeru has to pay Rs 14641 to clear the debt. Page No 151:Question 10:Shubhalaxmi took a loan of Rs 18000 from surya Finance to purchase a TV set. If the company charges compound interest at 12% per annum during the first year and 1212% per annum during the second year, how much will she have to pay after 2 years? Answer:Principal amount, P = Rs. 18000Rate of interest for the first year , p= 12% p.a.Rate of interest for the second year, q= 1212 % p.a.Time, n =2 yearsThe formula for the amount including the compound interest for the first year is given below:A = P×1+p100×1+q100 = Rs. 18000×1 +12100×1+25100×2= Rs. 18000×100+12 100×1+25200= Rs. 18000×100+12100 ×1+18 = Rs. 18000×100+12100× 8+18 = Rs. 18000×112100× 98= Rs. 18000× 1.12×1.125 = Rs. 22680∴ Shubhalaxmi has to pay Rs 22680 to the finance company after 2 years. Page No 151:Question 11:Neha borrowed Rs 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months? Answer:Principal amount, P = Rs. 24000Rate of interest, R= 10% p.a.Time, n =2 years 3 months = 214 yearsThe formula for the amount including the compound interest is give n below:A = P ×1+R100n×1+14R 100= Rs. 24000 ×1+101002×1+14×10100= Rs. 24000× 100+101002×100+2.5100= Rs. 24000× 1101002×100+2.5100=Rs. 24000×1.1×1.1×1.025 = Rs. 24000×1.250= Rs. 29766Therefore, Neha should pay Rs 29766 to the bank after 2 years 3 months. Page No 151:Question 12:Abhay borrowed Rs 16000 at 712% per annum simple interest. On the same day, he lent it to Gurmeet at the same rate but compounded annually. What does he gain at the end of 2 years? Answer:Principal amount, P = Rs 16000Rate of interest, R = 152% p.a.Time, n =2 yearsNow, simple interest = Rs 16000×2×15100×2= Rs. 2400Amount including the simple interest = Rs 16000+2400 = Rs 18400The formula for the amount including the compound interest is given b elow:A = P 1+R100n = Rs. 16000 1 +15100×22 = Rs. 16000 1+152002 = Rs.16000 1+3402= Rs.16000 40+3402 = Rs. 16000 43402 = Rs. 16000 1.075×1.075i.e., the amount including the compound interest is Rs 18490.Now, CI - SI= Rs. 18490 -18400 = Rs. 90Therefore, Abhay gains Rs. 90 as profit at the end of 2 years. Page No 151:Question 13:The simple interest on a sum of money for 2 years at 8% per annum is Rs 2400. What will be the compound interest on that sum at the same rate and for the same period? Answer:Simple interest (SI ) = Rs. 2400Rate of interest, R = 8%Time, n = 2 yearsThe principal can be calculated using the formula:Sum = 100×SIR ×T⇒Sum = Rs. 100×24008×2= Rs. 15000i.e., the principal is Rs. 15000.The amount including the compound interest is calculated using the formula given below:A = P 1+R100n = Rs. 15000 1+81002 = Rs. 15000 100+81002 = Rs. 15000 1081002= Rs. 15000 1.08×1.08= Rs. 17496i.e., the amount including the compound interest is Rs. 17496.∴ Compound interest (CI) = Rs. 17496 - 15000=Rs. 2496 Page No 151:Question 14:The difference between the compound interest and the simple interest on a certain sum for 2 years at 6% per annum is Rs 90. Find the sum. Answer:Let Rs P be the sum. Then SI = P×2×6100 = Rs. 12P100= Rs.3P25Also, CI = P×1+6100 2-P= Rs. P×100+61002-P= Rs. P×53502-P= Rs. 2809P2500-P = Rs. 2809P-2500P2500= Rs.309P2500Now, CI - SI=Rs. 309P2500- 3P25= Rs. 309P-300P2500= Rs. 9 P2500Now, Rs. 90 = 9P2500⇒P = 90×25009= Rs . 25000Hence, the required sum is Rs. 25000. Page No 151:Question 15:The difference between the compound interest and the simple interest on a certain sum for 3 years at 10% per annum is Rs 93. Find the sum. Answer:Let P be the sum. Then SI = Rs P×3 ×10100= Rs 30P100= Rs 3P10Also, CI = Rs. P×1+101003-P = Rs. P×100+101003-P = Rs. P×11103-P = Rs. 1331P1000 -P= Rs. 1331P-1000P1000= Rs.331P1000Now, CI - SI=Rs 331P1000- 3P10= Rs 331P-300P 1000= Rs 31P1000Now, Rs. 93 = 31P1000⇒P = 93×100031= Rs. 3000Hence, the required sum is Rs. 3000. Page No 152:Question 16:A sum of money amounts to Rs 10240 in 2 years at 623% per annum, compounded annually. Find the sum. Answer:Let P be the sum. Rate of interest, R = 623% = 203%Time, n = 2 yearsNow, A= P×1+20100×32= Rs. P ×1+203002= Rs. P×300+203002= Rs. P× 3203002 = Rs. P× 1615×1615= Rs. 256P225 ⇒Rs. 10240 = Rs. 256P225⇒Rs. 10240×225256= P∴ P = Rs. 9000Hence, the required sum is Rs. 9000 Page No 152:Question 17:What sum of money will amount to Rs 21296 in 3 years at 10% per annum, compounded annually? Answer:Let P be the sum.Rate of interest, R =10%Time, n = 3 yearsNow, A= P×1+101003= Rs. P×100+101003= Rs. P× 1101003 = Rs. P× 1110×1110×1110= Rs. 1331P1000However , amount = Rs. 21296Now, Rs. 21296 = Rs. 1331P1000⇒Rs. 21296×10001331= P∴ P = Rs. 16000Hence, the required sum is Rs. 16000. Page No 152:Question 18:At what rate per cent per annum will Rs 4000 amount to Rs 4410 in 2 years when compounded annually? Answer:Let R% p.a. be the required rate. A = 4410P = 4000n = 2 yearsN ow, A = P 1+R100n⇒4410 = 4000 1+R100 2⇒44104000= 1+R1002⇒441400= 1+R1002⇒21202= 1+R1002⇒2120-1= R100 ⇒21-2020= R100⇒120=R100⇒R = 1×10020 = 5Hence, the required rate is 5% p.a. Page No 152:Question 19:At what rate per cent per annum will Rs 640 amount to Rs 774.40 in 2 years when compounded annually? Answer:Let the required rate be R% p.a.A = 774.40P = 640 n = 2 yearsNow, A = P 1+ R100n⇒774.40 = 640 1+R1002⇒774.40640 = 1+R1002⇒1.21= 1+R1002⇒1. 12= 1+R1002⇒1.1-1= R100⇒0.1= R100⇒R = 0.1×100= 10Hence, the required rate is 10% p.a. Page No 152:Question 20:In how many years will Rs 1800 amount to Rs 2178 at 10% per annum when compounded annually? Answer:Let the required time be n years.Rate of interest, R = 10%Principal amount, P = Rs. 1800Amount with compound interest, A = Rs. 2178Now, A = P×1+R100 n= Rs. 1800×1+10100n = Rs. 1800×100+10100 n=Rs. 1800×110100n = Rs. 1800×1110nHowever, amount =Rs. 2178Now, Rs. 2178 = Rs. 1800×1110n⇒21781800 = 1110n⇒121100 = 1110n⇒ 11102 = 1110n⇒n = 2∴ Time, n = 2 years Page No 152:Question 21:In how many years will Rs 6250 amount to Rs 7290 at 8% per annum, compounded annually? Answer:Let the required time be n years . Rate of interest, R = 8%Principal amount, P = Rs. 6250 Amount with compound interest, A = Rs. 7290Then, A = P×1+R 100n⇒A= Rs. 6250×1+8100n= Rs. 6250×100+ 8100n=Rs. 6250×108100n =Rs. 6250×2725n However, amount = Rs. 7290Now, Rs. 7290 = Rs. 6250×2725n ⇒72906250 = 2725n⇒729625 = 2725n⇒27252 = 2725n⇒n = 2∴ Time, n = 2 years Page No 152:Question 22:The population of a town is 125000. It is increasing at the rate of 2% per annum. What will be its population after 3 years? Answer:Population of the town, P = 125000Rate of increase, R= 2%Time, n = 3 yearsThen the population of the town after 3 years is given byPopulation = P×1 +R1003= 125000×1+21003= 125000×100+21003 = 125000×1021003 = 125000×51503= 125000 ×5150 ×5150×5150= 51×51×51 = 132651Therefore, the population of the town after three years is 132651. Page No 152:Question 23:Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 5%, 4% and 3% respectively, what is its present population? Answer:Let the population of the town be 50000.Rate of increase for the first year, p = 5%Rate of increase for the second year, q = 4%Rate of increase for the third year, r = 3%Time = 3 yearsNow, present population= P ×1+p100×1+q100×1+r100= 50000 ×1+5100×1+4100×1+3100=50000 ×100 +5100×100+4100×100+3100=50000 ×105100×104100 ×103100=50000 ×2120×2625×103100=21 ×26×103=56238Therefore, the present population of the town is 56238. Page No 152:Question 24:The population of a city was 120000 in the year 2009. During next year it increased by 6% but due to an epidemic it decreased by 5% in the following year. what is its population in the year 2011? Answer:Population of the city in 2009, P= 120000Rate of increase, R= 6%Time, n = 3 yearsThen the population of the city in the year 2010 is given byPopulation = P×1+R100n= 120000×1+61001 = 120000×100+6 100 = 120000×106100 = 120000×5350 = 2400 ×53 = 127200Therefore, the population of the city in 2010 is 127200.Ag ain, population of the city in 2010, P=127200Rate of decrease, R= 5% Then the population of the city in the year 2011 is given byPopulation = P ×1-R100n= 127200×1-51001= 127200×100-5100 = 127200×95100 = 127200×1920= 6360 ×19 = 120840Therefore, the population of the city in 2011 is 120840. Page No 152:Question 25:The count of bacteria in a certain experiment was increasing at the rate of 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 500000. Answer:Initial count of bacteria, P= 500000Rate of increase, R = 2%Time, n = 2 hoursThen the count of bacteria at the end of 2 hours is given byCount of bacteria = P×1+R100n =500000×1+21002=500000×100+21002=500000×1021002=500000×51502=500000×5150×5150=200×51×51 =520200Therefore, the count of bacteria at the end of 2 hours is 520200. Page No 152:Question 26:The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. Find the bacteria at the end of 3 hours if the count was initially 20000. Answer:Initial count of bacteria, P= 20000Rate of increase, R= 10%Time, n = 3 hoursThen the count of bacteria at the end of the first hour is given byCount of bacteria = P ×1+10100n=20000×1+101001=20000×100+10100 =20000×110100=20000×1110=2000×11=22000Therefore, the count of bacteria at the end of the first hour is 22000.The count of bacteria at the end of the second hour is given byCount of bacteria = P×1-10100n=22000×1-101001=22000×100-10100=22000×90100 =22000×910=2200×9=19800Therefore, the count of bacteria at the end of the second hour is 19800.Then the count of bacteria at the end of the third hour is is given byCount of bacteria = P×1+10100n=19800×1+101001=19800×100+10100=19800×110100 =19800×1110=1980×11=21780Therefore, the count of bacteria at the end of the first 3 hours is 21780. Page No 152:Question 27:A machine is purchased for Rs 625000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years? Answer:Initial value of the machine, P= Rs 625000Rate of depreciation, R= 8%Time, n = 2 yearsThen the value of the machine after two years is given byValue = P×1-R100n=Rs 625000×1-8100 2=Rs 625000×100-81002=Rs 625000×921002=Rs 625000 ×23252=Rs 625000×2325×2325=Rs 1000×23×23=Rs 529000Therefore, the value of the machine after two years will be Rs. 529000. Page No 152:Question 28:A scooter is bought at Rs 56000. Its value depreciates at the rate of 10% per annum. What will be its value after 3 years? Answer:Initial value of the scooter, P= Rs 56000Rate of depreciation, R= 10%Time, n = 3 years Then the value of the scooter after three years is given byValue = P×1-R 100n=Rs. 56000×1-101003=Rs. 56000×100-101003 =Rs. 56000×901003=Rs. 56000×9103=Rs. 56000×9 10×910×910=Rs. 56×9×9×9=Rs. 40824Therefore , the value of the scooter after three years will be Rs. 40824. Page No 152:Question 29:A car is purchased for Rs 348000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years? Answer:Initial value of the car, P= Rs 348000Rate of depreciation for the first year, p= 10%Rate of depreciation for the second year, q= 20%Time, n = 2 years.Then the value of the car after two years is given byValue = P×1-p100×1-q100=Rs. 348000×1-10100×1-20100=Rs. 348000×100-10100×100-20100=Rs. 348000×90100 ×80100=Rs. 348000×910×810=Rs. 3480×9×8=Rs. 250560∴ The value of the car after two years is Rs 250560. Page No 152:Question 30:The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 291600, for how much was it purchased? Answer:Let the initial value of the machine, P be Rs x.Rate of depreciation, R= 10%Time, n = 3 yearsThe present value of the machine is Rs 291600.Then the initial value of the machine is given byValue = P×1-R100n=Rs. x×1 -101003=Rs. x×100-101003=Rs. x×901003=Rs. x×9103∴ Present value of the machine = Rs 291600Now, Rs 291600 = Rs x×910×910×910⇒x = Rs 291600×10×10×109×9×9⇒x =Rs 291600000729⇒x = Rs 400000∴ The initial value of the machine is Rs 400000. Page No 154:Question 1:Find the amount and the compound interest on Rs 8000 for 1 year at 10% per annum, compounded half-yearly.
Answer:Principal, P = Rs. 8000Time, n = 1 year = 2 half yearsRate of interest per annum = 10%Rate of interest for half year, R = 10%2= 5%The amount with the compound interest is given byAmount = Rs. P×1+R1002= Rs. 8000×1+51002 = Rs. 8000×1051002 = Rs. 8000×21202 = Rs. 8000×21 20×2120 = Rs. 20×21×21= Rs. 8820∴ Compound interest = amount - principal = Rs. (8820 -8000) = Rs. 820 Page No 154:Question 2:Find the amount and the compound interest on Rs 31250 for 112 years at 8% per annum, compounded half-yearly. Answer:Principal, P = Rs. 31250Annual rate of interest, R = 8% Rate of interest for a half year = 12×8%= 4%Time, n = 112 years = 3 half yearsThen the amount with the compound interest is given by A = P ×1+R100 n = 31250×1+41003 = 31250×1+1253 = 31250×25+1253 = 31250×26253 = 31250×2625×2625×2625 = Rs 2×17576 = Rs 35152 Therefore, compound interest = amount - principal= Rs (35152- 31250) = Rs 3902 Page No 154:Question 3:Find the amount and the compound interest on Rs 12800 for 1 year at 712% per annum, compounded semi-annually. Answer:Principal, P = Rs. 12800Annual rate of interest, R = 152% Rate of interest for a half year = 12 152%= 154%Time, n = 1 year = 2 half yearsThen the amount with the compound interest is given by A = P ×1+R100n = 12800×1+1541002= 12800×1+15100×42 = 12800×400+ 154002 = 12800×4154002 = 12800×8380×8380 = 2×83×83 =Rs 13778Therefore, compound interest = amount - principal=Rs ( 13778-12800) = Rs 978 Page No 154:Question 4:Find the amount and the compound interest on Rs 160000 for 2 years at 10% per annum, compounded half-yearly. Answer:Principal, P = Rs. 160000Annual rate of interest, R = 10% Rate of interest for a half year = 102%= 5%Time, n = 2 years = 4 half yearsThen the amount with the compound interest is given by A = P ×1+R100n= 160000×1+5100 4 = 160000×100+51004 = 160000×1051004 = 160000×2120×2120×2120×2120= 21×21×21×21 = Rs 194481Therefore, compound interest = amount - principal=Rs (194481-160000) = Rs 34481 Page No 154:Question 5:Swati borrowed Rs 40960 from a bank to buy a piece of land. If the bank charges 1212% per annum, compounded half-yearly, what amount will she have to pay after 112 years? Also, find the interest paid by her. Answer:Principal, P = Rs. 40960Annual rate of interest, R = 25 2% Rate of interest for half year = 254%Time, n = 112 years = 3 half yearsThen the amount with the compound interest is given by A = P ×1+R100n= 40960×1+25100×43 = 40960×400+254003 = 40960×4254003 = 40960×1716× 1716×1716 = 10×17×17×17 = Rs 49130Therefore, compound interest = amount - principal=Rs (49130- 40960) = Rs 8170Therefore, Swati has to pay Rs. 49130, which includes an interest of Rs. 8170, to the bank after 112 years. Page No 154:Question 6:Mohd. Aslam purchased a house from Avas Vikas Parishad on credit. If the cost of the house is Rs 125000 and the Parishad charges interest at 12% per annum compounded half-yearly, find the interest paid by Aslam after a year and a half. Answer:Let the principal amount be P=Rs. 125000.Annual rate of interest, R = 12% Rate of interest for a half year = 6%Time, n = 112 years = 3 half yearsThen the amount with the compound interest is given by A = P ×1+R100n= Rs. 125000×1+61003 = Rs. 125000×100+61003 = Rs. 125000×1061003 = Rs. 125000×5350×5350×5350= Rs. 53×53×53 = Rs. 148877Now, CI = A - P=Rs. (148877-125000) = Rs. 23877 Therefore, Aslam has to pay an interest of Rs. 23877 to the bank after 112 years. Page No 154:Question 7:Sheela deposited Rs 20000 in a bank, where the interest is credited half-yearly. If the rate of interest paid by the bank is 6% per annum, what amount will she get after 1 year? Answer:Let the principal amount be P= Rs. 20000.Annual rate of interest, R = 6% Rate of interest for half year = 3%Time, n=1 year = 2 half yearsThen the amount with the compound interest is given by A = P ×1+R100n= Rs. 20000×1+31002 = Rs. 20000×100+31002 = Rs. 20000×1031002= Rs. 20000×103100×103100 = Rs. 2 ×103×103= Rs. 21218Therefore, Sheela gets Rs. 21218 after 1 year. Page No 154:Question 8:Neeraj lent Rs 65536 for 2 years at 1212% per annum, compounded annually. How much more could he earn if the interest were compounded half-yearly? Answer:Let the principal amount be P= Rs. 65536.Annual rate of interest, R = 252% Rate of interest for a half year = 254%Time, n = 2 years = 4 half yearsThen the amount with the compound interest is given by A = P × 1+R100n= Rs. 65536×1+25100×44 = Rs. 65536 ×400+254004 = Rs. 65536×4254004= Rs. 65536×17 164= Rs. 65536×1716×1716×1716×1716 = Rs. 17×17×17×17 = Rs. 83521Now, CI = A - P = Rs. (83521 -65536) = Rs. 17985Therefore, interest earned when compounded half yearly = Rs. 17985Amount when the interest is compounded yearly is given by A= P ×1+R100n=Rs. 65536×1+25100 ×22= Rs. 65536×200+252002 = Rs. 65536×225200 2= Rs. 65536×982= Rs. 65536×98×98 =Rs. 82944Therefore, CI = A- P =Rs. (82944-65536)=Rs. 17408∴ Difference between the interests compounded half yearly and yearly =Rs. (17985-17408)=Rs. 577 Page No 154:Question 9:Sudershan deposited Rs 32000 in a bank, where the interest is credited quarterly. If the rate of interest be 5% per annum, what amount will he receive after 6 months? Answer:Let the principal amount be P=Rs 32000. Annual rate of interest, R = 5% Rate of interest for a quarter year = 54 %Time, n=6 months= 2 quarter yearsThen the amount with the compound interest is given by A = Rs. P ×1+R100n = Rs. 32000×1+5100×42= Rs. 32000×400+54002 = Rs. 32000×4054002 = Rs. 32000×81802 = Rs. 32000×8180×8180 = Rs. 5×81×81 = Rs. 32805Therefore, Sudershan will receive an amount of Rs. 32805 after 6 months. Page No 154:Question 10:Arun took a loan of Rs 390625 from Kuber Finance. If the company charges interest at 16% per annum, compounded quarterly, what amount will discharge his debt after one year? Answer:Let the principal amount be P = Rs 390625.Annual rate of interest, R = 16% Rate of interest for a quarter year =164%=4%Time, n=1 year = 4 quarter yearsThen the amount with the compound interest is given by A = Rs. P ×1+R100n = Rs. 390625×1+41004 = Rs. 390625×100+41004 = Rs. 390625×1041004 = Rs. 390625×26254 = Rs. 390625×2625×2625×2625×2625= Rs. 26×26×26×26 = Rs. 456976Therefore, Arun has to pay Rs 456976 after 1 year. Page No 154:
Question 1:Tick (✓) the correct answer: Answer:(c) Rs. 832 A = P ×1+ R100n = Rs. 5000×1+81002 = Rs. 5000×108100 2 = Rs. 5000×27252 = Rs. 5000×2725×2725 = Rs. 8×27×27 = Rs. 5832∴ Interest = amount - principal = Rs (5832-5000)= Rs 832 Page No 154:Question 2:Tick (✓) the correct answer: Answer:(b) Rs. 3310 A = P ×1+R100n= Rs. 10000×1+101003 = Rs. 10000×1101003 = Rs. 10000×11103= Rs . 10000×1110×1110×1110 = Rs. 10×11×11×11 = Rs. 13310∴ Compound interest =amount - principal= Rs (13310 - 10000) = Rs 3310 Page No 154:Question 3:Tick (✓) the correct answer: Answer:(a) Rs 1872 Here, A = P ×1+R1001×1+12R100 = Rs 10000×1+12100×1+12×12100= Rs 10000×100 +12100×100+6100 = Rs 10000×112100×106100= Rs 10000×2825×5350 = Rs 8×28×53 = Rs 11872∴ Compound interest =amount -principal= Rs ( 11872-10000)= Rs 1872 Page No 155:Question 4:Tick (✓) the correct answer: Answer:(c) Rs 961 Here, A = P ×1+R100 2×1+14R100= Rs. 4000×1+101002×1+14×10100 = Rs. 4000×100+101002×40+140 = Rs. 4000×1101002×4140 = Rs. 4000×1110×11 10×4140 = Rs. 11×11×41 = Rs. 4961∴ Compound interest = amount -principal= Rs ( 4961- 4000)= Rs 961 Page No 155:Question 5:Tick (✓) the correct answer: Answer:(b) Rs. 5051 Here, A = Rs. P ×1+p100×1+q100×1+r100 = Rs. 25000×1 +5100×1+6100×1+8100 = Rs. 25000×105100×106100 ×108100= Rs. 25000×2120×5350×2725 = Rs. 21×53×27= Rs. 30051∴ Compound interest = amount -principal= Rs. (30051 - 25000)= Rs. 5051 Page No 155:Question 6:Tick (✓) the correct answer: Answer:(b) Rs. 510 Rate of interest compounded half yearly = 82%=4%Time = 1 year= 2 half years Now, A = P ×1+R100n = Rs. 6250×1+ 41002= Rs. 6250×1041002= Rs. 6250×2625×2625 = Rs. 10×26×26 = Rs. 6760∴ Compound interest =amount- principal= Rs. (6760-6250)= Rs. 510 Page No 155:Question 7:Tick (✓) the correct answer: Answer:(a) Rs.1209 Time = 6 months = 2 quater yearsRate compounded quarter yearly =6 4%=32%Now, A = P ×1+R100n = Rs. 40000× 1+3100×22= Rs. 40000×2032002 = Rs. 40000×203 200×203200 = Rs. 203×203 = Rs. 41209∴ Compound interest = amount-principal= Rs. 41209-Rs. 40000= Rs.1209 Page No 155:Question 8:Tick (✓) the correct answer: Answer:(b) 26460 Here, A = P ×1+R100n = Rs. 24000×1+51002 = Rs. 24000×1051002= Rs. 24000×2120×2120 = Rs. 60×21×21 = Rs. 26460 Page No 155:Question 9:Tick (✓) the correct answer: Answer:(c) Rs. 43740 Here, A = Rs. P ×1-R100n = Rs. 60000×1-10100 3 = Rs. 60000×901003 = Rs. 60000×910×910 ×910 = Rs. 60×9×9×9= Rs. 43740 Page No 155:Question 10:Tick
(✓) the correct answer: Answer:(b) Rs. 62500 Here, A = P ×1-R100n = P×1-201002 = P×801002 = P×45×45⇒40000 = 16P25∴ P = 40000×2516= Rs 62500 Page No 155:Question 11:Tick (✓) the correct answer: Answer:(a) 25000 Let P be the popultion 3 years ago.Now, present population =33275⇒33275= P×1+101003⇒33275 = P×1101003⇒33275 = P×1110×1110×1110⇒ 33275 = 1331P1000∴ P = 33275×10001331= 25000 Page No 155:Question 12:Tick (✓) the correct answer: Answer:(d) Rs 1261 Here, SI= P×5×3100⇒1200 =P×5×3100 ⇒P = 1200×1005×3 =Rs 8000Amount at the end of 3 years = Rs 8000×1+51003 = Rs 8000×1051003 = Rs 8000 ×2120×2120×2120 = Rs 21×21×21 = Rs 9261 ∴ CI = A-P = Rs ( 9261-8000)= Rs 1261 Page No 155:Question 13:Tick (✓) the correct answer: Answer:(d) Rs 480 We have: 510 =P×1+25100×22-P⇒510=⇒ P×8+182 -P⇒510 = P×98×98-P⇒510= 81P64-P⇒510 =81P-64P64 ⇒510 = 17 P64∴ P = 510×6417=Rs 1920Now, SI =P×R×T100 =Rs 1920×2×25100×2= Rs 480 Page No 155:Question 14:Tick (✓) the correct answer: Answer:(d) Rs 4096 We have Rs 4913 = P ×1+25100×43⇒Rs 4913 = P×16+1163⇒Rs 4913 =P×1716×1716×1716⇒Rs 4913 = 4913P4096⇒P = Rs 4913×40964913=Rs 4096 Page No 155:Question 15:Tick (✓) the correct answer: Answer:(c) 6% Here, A = P ×1+R100= Rs. 7500×1+R1002= Rs. 7500×1+R1002However, amount =Rs. 8427Now, Rs. 8427== Rs. 7500×1+R1002⇒Rs. 8427 Rs. 7500 = 1+R1002⇒53502= 1+R1002⇒1+R100 = 5350⇒R100= 5350-1⇒R100= 53-5050=350∴ R = 30050=6% Page No 157:Question 1:Find the amount and the compound interest on Rs 3000 for 2 years at 10% per annum. Answer:Here, A = P ×1+R100n = Rs. 3000 × 1+101002= Rs. 3000 ×1101002 = Rs. 3000 × 1110×1110 = Rs. 30×11×11 = Rs. 3630∴ CI = A-P = Rs. (3630-3000) = Rs. 630Hence, the amount is Rs. 3630 and the CI is Rs. 630. Page No 157:Question 2:Find the amount of Rs 10000 after 2 years compounded annually; the rate of interest being 10% per annum during the first year and 12% per annum during the second year. Also, find the compound interest. Answer:Here, A = P ×1+p100×1+r100 = Rs. 10000 ×1+10100×1+12100 = Rs. 10000 ×110100 ×112100 = Rs. 10000 ×1110×2825 = Rs. 40 ×11×28 = Rs. 12320∴ CI =A-P = Rs (12320-10000) = Rs. 2320Hence, the amount is Rs 12320 and the CI is Rs 2320. Page No 157:Question 3:Find the amount and the compound interest on Rs 6000 for 1 year at 10% per annum compounded half-yearly. Answer:Let the principal amount be P=Rs 6000.Rate of interest = 10% p.a.= 5% for half yearlyTime (n) = 1 year = 2 half yearsNow, A = P ×1+R100n= Rs 6000 ×1+ 51002 = Rs 6000 ×1051002= Rs 6000 ×2120× 2120= Rs 15×21×21= Rs 6615∴ CI = A-P = Rs (6615-6000) = Rs 615Hence, the amount is Rs 6615 and the CI is Rs 615. Page No 157:Question 4:A sum amounts to Rs 23762 in 2 years at 9% per annum, compounded annually. Find the sum. Answer:Amount (A) = Rs 23762Rate of interest (R) =9%Time (n) = 2 yearsNow, A = P ×1+R100n⇒Rs 23762 = P ×1+91002 ⇒Rs 23762 = P ×109100×109100⇒P = Rs 23762×100×100109×109 ⇒P = Rs 20000∴ The principal amount is Rs 20000. Page No 157:Question 5:A scooter is bought for Rs 32000. Its value depreciates at 10% per annum. What will be its value after 2 years? Answer:Let the principal amount be P=Rs 32000.Rate of interest (R) = 10%Time (n)= 2 yearsNow, A = Rs. P × 1-R100n = Rs. 32000×1-101002 = Rs. 32000×901002 = Rs. 32000×910×910 = Rs. 320 ×9×9 = Rs. 25920∴ The value of the scooter after 2 years is Rs 25920. Page No 157:Question 6:Mark (✓) against the correct answer: Answer:(b) Rs. 1050 P= Rs. 5000R= 10%n= 2 yearsNow, A = Rs. P ×1+R100 n = Rs. 5000×1+101002= Rs. 5000×1101002 = Rs. 5000×1110×1110 = Rs. 50×11×11= Rs. 6050∴ CI = A-P = Rs. ( 6050-5000) = Rs. 1050 Page No 157:Question 7:Mark (✓) against the correct answer: Answer:(c) 4410 Present population =4000 To find the population of the town after 2 years, we have: A = P ×1+R100n= 4000×1+51002= 4000×1051002 = 4000×2120×2120= 10×21×21 = 4410 Page No 157:Question 8:Mark (✓) against the correct answer: Answer:(d) 8% Here, A = Rs. P ×1+R100n⇒Rs. 5832= Rs. 5000×1+R1002⇒Rs. 5832 Rs. 5000= 1+R1002 ⇒27252 = 1+R1002⇒1+R100= 2725⇒R100 = 2725-1 =27-2525=225∴R = 100×225= 8% Page No 157:Question 9:Mark (✓) against the correct answer: Answer:(a) Rs. 1655 Here, SI =P×R×T100⇒Rs. 1500 = P× 10×3100⇒P = 1500×10010×3= Rs. 5000Now, A = P ×1+R100n= Rs. 5000×1+101003 = Rs. 5000× 1101003 = Rs. 5000×1110×1110×1110 = Rs. 5×11×11×11 = Rs. 6655∴ CI = A-P =Rs (6655-5000) = Rs 1655 Page No 157:Question 10:Mark (✓) against the correct answer: Answer:(c) Rs. 5000 Here, A = P×1+R100n = P×1+101002 = P×1101002 = P× 1110×1110Now, CI= A-P⇒Rs. 1050 = 121p100- P = 121P-100P100= 21P100∴ P = Rs 1050× 10021 =Rs 5000 Page No 157:Question 11:Fill in the blanks: Answer:(i) A= P1+R100n(ii) Compound interest(iii) A= P1-R1002, where A is the value of the machine after 2 years(iv) A= P1+R1005, where A is the population of the town after 5 years View NCERT Solutions for all chapters of Class 8 At what rate per cent per annum will a sum of 7500 amount to Rs 8427 in 2 years compounded annually a 4% B 5 C 6 D 8?7500Amount (A) = 8427Period = 2 yearsLet R be the rate of p.a., then∴ AP=(1+R100)n⇒ 84277500 = (1+R100)2⇒ 28092500=(1+R100)2⇒ (5350)2=(1+R100)2∴ 1+R100=5350⇒ R100=5350−1=350∴ R=350×100=6%
At what rate of compound interest compounded annually will 7500 become 9075 in 2 years?We have , Principal , P = Rs 7500 Time , n = 2 yrs Amount. A = P 1 + R 100 n ⇒ 9075 = 7500 1 + R 100 2 ⇒ 9075 7500 = 1 + R 100 2 ⇒ 121 100 = 1 + R 100 2 ⇒ 11 10 2 = 1 + R 100 2 ⇒ 1 + R 100 = 11 10 ⇒ R 100 = 11 10 - 1 ⇒ R 100 = 11 - 10 10 ⇒ R 100 = 1 10 ⇒ R = 10 So , rate of interest is 10.
At what rate per cent per annum will a sum of Rs 7500 give Rs 927 as compound interest in two years?Answer: answer is 6%p.a.
At what rate percent will the interest on Rupees 7500 be rupees 1500 in 4 years what will be the amount after 8 years?Answer. Your answer is: Rate percent will be 5%and the amount after 8 yrs will be Rs. 9000.
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