Download Compound Intrest Questions for IIFT PDF \u2013 IIFT Compound Intrest questions pdf by Cracku. Practice IIFT solved Compound Intrest Questions paper tests, and these are the practice question to have a firm grasp on the Compound Intrest topic in the IIFT exam. Top 20 very Important Compound Intrest Questions for IIFT based on asked questions in previous exam papers. \u00a0The IIFT question papers contain actual questions asked with answers and solutions.<\/p>\n
Download Compound Intrest Questions for IIFT<\/a><\/p>\n Get IIFT 2023 Mocks at Just Rs. 299<\/a><\/p>\n Question 1:\u00a0<\/b>What is the difference (in \u20b9) between the compound interest, when interest is compounded 6-monthly, and the simple interest on a sum of \u20b910,000 for $1\\frac{1}{2}$ years at 10% p.a.?<\/p>\n a)\u00a0102.25<\/p>\n b)\u00a087<\/p>\n c)\u00a076.25<\/p>\n d)\u00a091.5<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n If interest is half-yearly, Then<\/p>\n Time =\u00a0$1.5\\times\\ 2=3$<\/p>\n Rate =\u00a0$\\frac{10}{2}=\\ 5\\%$<\/p>\n $A=P\\left(1+\\frac{r}{100}\\right)^n$<\/p>\n =\u00a0$A=10,000\\left(1+\\frac{5}{100}\\right)^3$<\/p>\n $A=10,000\\ \\times\\ \\frac{21}{20}\\times\\ \\frac{21}{20}\\times\\ \\frac{21}{20}=11576.25$<\/p>\n C.I = A\u00a0– P = 11576.25 – 10000 = \u20b91576.35<\/p>\n Simple Interest =\u00a0$S.I\\ =\\frac{\\left(P\\times\\ R\\times\\ T\\right)}{100}$<\/p>\n =\u00a0$\\frac{\\left(10000\\times\\ 10\\times\\ 1.5\\right)}{100}$<\/p>\n =\u00a0\u20b91500<\/p>\n Required Difference = \u20b91576.25 – \u20b91500 =\u00a0\u20b976.25<\/p>\n Hence,\u00a0Option C<\/strong> is correct.<\/p>\n Question 2:\u00a0<\/b>A sum of \u20b925600 is invested on simple interest partly at 7% per annum and the remaining at 9% per annum. The total interest at the end of 3 years is \u20b95832. How much money(in \u20b9) was invested at 9% per annum?<\/p>\n a)\u00a018000<\/p>\n b)\u00a07600<\/p>\n c)\u00a09600<\/p>\n d)\u00a016000<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the amount invested in 9% per annum = P<\/p>\n Amount invested in 7% per annum =\u00a025600 – P<\/p>\n The total interest at the end of 3 years is \u20b95832.<\/p>\n P x 3 x $\\frac{9}{100}$ +\u00a0(25600 – P) x 3 x\u00a0$\\frac{7}{100}$ = 5832<\/p>\n $\\frac{9}{100}$P + 1792 –\u00a0$\\frac{7}{100}$P = 1944<\/p>\n $\\frac{2}{100}$P = 152<\/p>\n P = \u20b97600<\/p>\n $\\therefore$\u00a0 Amount invested at 9% per annum = P = \u20b97600<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 3:\u00a0<\/b>Two equal sums were lent on simple interest at 6% and 10% per annum respectively. The first sum was recovered two years later than the second sum and the amount in each case was \u20b91105. What was the sum (in \u20b9) lent in each scheme?<\/p>\n a)\u00a0900<\/p>\n b)\u00a0850<\/p>\n c)\u00a0936<\/p>\n d)\u00a0891<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the sum lent in each scheme = P<\/p>\n Let the sum at 6% and 10%\u00a0simple interest was recovered after ‘t+2’ years and ‘t’\u00a0years respectively.<\/p>\n Amount obtained from\u00a0simple interest at 6% = P +\u00a0$\\frac{P\\times(t+2)\\times6}{100}$<\/p>\n 1105 =\u00a0P +\u00a0$\\frac{P\\times(t+2)\\times6}{100}$<\/p>\n 1105 – P =\u00a0$\\frac{P\\times(t+2)\\times6}{100}$…….(1)<\/p>\n Amount obtained from simple interest at 10% = P + $\\frac{P\\times t\\times10}{100}$<\/p>\n 1105 =\u00a0P + $\\frac{P\\times t\\times10}{100}$<\/p>\n 1105 – P =\u00a0$\\frac{P\\times t\\times10}{100}$……..(2)<\/p>\n From (1) and (2),<\/p>\n $\\frac{P\\times(t+2)\\times6}{100}$ =\u00a0$\\frac{P\\times t\\times10}{100}$<\/p>\n 6t + 12 = 10t<\/p>\n t = 3<\/p>\n Substituting t = 3 in equation (2), we get P = 850<\/p>\n The sum lent in each scheme =\u00a0\u20b9850<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 4:\u00a0<\/b>A sum of \u20b99500 amounts to \u20b911495 in 2 years at a certain rate percent per annum, interest compounded yearly. What is the simple interest (in \u20b9) on the same sum for the same time and double the rate?<\/p>\n a)\u00a03990<\/p>\n b)\u00a03420<\/p>\n c)\u00a04560<\/p>\n d)\u00a03800<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the rate of interest = R<\/p>\n 11495 = 9500(1+$\\frac{R}{100}$)$^2$<\/p>\n (1+$\\frac{R}{100}$)$^2$ =\u00a0$\\frac{11495}{9500}$<\/p>\n (1+$\\frac{R}{100}$)$^2$ = $\\frac{121}{100}$<\/p>\n 1+$\\frac{R}{100}$ =\u00a0$\\frac{11}{10}$<\/p>\n $\\frac{R}{100}$ =\u00a0$\\frac{1}{10}$<\/p>\n R = 10%<\/p>\n Simple interest =\u00a0$\\frac{9500\\times2\\times20}{100}$<\/p>\n =\u00a0\u20b93800<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 5:\u00a0<\/b>A person borrowed a sum of \u20b930800 at 10% p.a. for 3 years, interest compounded annually. At the end of two years, he paid a sum of \u20b913268. At the end of 3rd year, he paid \u20b9 x to clear of the debt. What is the value of x ?<\/p>\n a)\u00a026400<\/p>\n b)\u00a026510<\/p>\n c)\u00a026200<\/p>\n d)\u00a026620<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Amount to be paid after 2 years = 30800(1+$\\frac{10}{100}$)$^2$<\/p>\n =\u00a030800($\\frac{11}{10}$)$^2$<\/p>\n =\u00a030800($\\frac{121}{100}$)<\/p>\n =\u00a0\u20b937268<\/p>\n Amount paid by the person =\u00a0\u20b913268<\/p>\n Remaining amount =\u00a0\u20b937268 –\u00a0\u20b913268 =\u00a0\u20b924000<\/p>\n Amount to be paid at the end of 3rd year to clear debt(i.e, compound interest on\u00a0\u20b924000 for next 1 year) = 24000(1+$\\frac{10}{100}$)$^1$<\/p>\n $\\Rightarrow$\u00a0\u00a0x = 24000($\\frac{11}{10}$)<\/p>\n $\\Rightarrow$\u00a0\u00a0x =\u00a0\u20b926400<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 6:\u00a0<\/b>At what rate percent per annum will \u20b97200 amountto \u20b97938 in one year, if interest is compounded half yearly?<\/p>\n a)\u00a05<\/p>\n b)\u00a08<\/p>\n c)\u00a012<\/p>\n d)\u00a010<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the rate of interest per annum = R<\/p>\n According to the problem,<\/p>\n 7938 = 7200(1+$\\frac{\\frac{R}{2}}{100}$)$^2$<\/p>\n 3969 =\u00a03600(1+$\\frac{\\frac{R}{2}}{100}$)$^2$<\/p>\n 441 = 400(1+$\\frac{\\frac{R}{2}}{100}$)$^2$<\/p>\n (1+$\\frac{\\frac{R}{2}}{100}$)$^2$ = $\\frac{441}{400}$<\/p>\n 1+$\\frac{\\frac{R}{2}}{100}$ =\u00a0$\\frac{21}{20}$<\/p>\n $\\frac{R}{200}$ =\u00a0$\\frac{1}{20}$<\/p>\n R = 10%<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 7:\u00a0<\/b>A sum at simple interest becomes two times in 8 years at a certain rate of interest p.a. The time in which the same sum will be 4 times at the same rate of interest at simple interest is:<\/p>\n a)\u00a025 years<\/p>\n b)\u00a020 years<\/p>\n c)\u00a030 years<\/p>\n d)\u00a024 years<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the principal amount = P<\/p>\n Rate of interest = R<\/p>\n Time = 8 years<\/p>\n Amount = 2P<\/p>\n $\\Rightarrow$\u00a0 2P = Principal amount + Simple Interest<\/p>\n $\\Rightarrow$\u00a0 2P = P + $\\frac{\\text{P}\\times8\\times \\text{R}}{100}$<\/p>\n $\\Rightarrow$\u00a0 P = \u00a0$\\frac{\\text{P}\\times8\\times \\text{R}}{100}$<\/p>\n $\\Rightarrow$ \u00a0R = 12.5%<\/p>\n Let the time in which sum will be 4 times at the same rate of interest = T<\/p>\n i.e, Amount = 4P<\/p>\n $\\Rightarrow$\u00a0 4P = P + $\\frac{\\text{P}\\times \\text{T}\\times12.5}{100}$<\/p>\n $\\Rightarrow$\u00a0 3P =\u00a0$\\frac{\\text{P}\\times \\text{T}\\times12.5}{100}$<\/p>\n $\\Rightarrow$\u00a0 3 =\u00a0$\\frac{\\text{T}}{8}$<\/p>\n $\\Rightarrow$\u00a0 T = 24 years<\/p>\n $\\therefore\\ $The time in which the same sum will be 4 times at the same rate of interest = 24 years<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 8:\u00a0<\/b>Suresh lent out a sum of money to Rakesh for 5 years at simple interest. At the end of 5 years, Rakesh paid 9\/8 of the sum to Suresh to clear out the amount. Find the rate of simple interest per annum.<\/p>\n a)\u00a03.5% p.a.<\/p>\n b)\u00a02.5% p.a.<\/p>\n c)\u00a03% p.a.<\/p>\n d)\u00a02% p.a.<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the sum of money lent out by Suresh to Rakesh = P<\/p>\n Amount paid by Rakesh to Suresh after 5 years =\u00a0$\\frac{9}{8}$P<\/p>\n Let the rate of simple interest = R%<\/p>\n $\\Rightarrow$\u00a0 P + $\\frac{\\text{P}\\times5\\times \\text{R}}{100}$ = $\\frac{9}{8}$P<\/p>\n $\\Rightarrow$\u00a0 1 + $\\frac{R}{20}$ =\u00a0$\\frac{9}{8}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{R}{20}$ =\u00a0$\\frac{9}{8}-1$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{R}{20}$ =\u00a0$\\frac{1}{8}$<\/p>\n $\\Rightarrow$\u00a0 R = 2.5%<\/p>\n $\\therefore\\ $Rate of simple interest per annum = 2.5%<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 9:\u00a0<\/b>If the difference between the compound interest and simple interest on a certain sum of money for three years at 10% p.a. is \u20b9 558, then the sum is:<\/p>\n a)\u00a0\u20b9 18,500<\/p>\n b)\u00a0\u20b9 15,000<\/p>\n c)\u00a0\u20b9 16,000<\/p>\n d)\u00a0\u20b9 18,000<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the principal sum = P<\/p>\n Rate = 10%<\/p>\n Time = 3 years<\/p>\n Compound interest on the sum = P$\\left(1+\\frac{10}{100}\\right)^3-$ P = P$\\left(\\frac{110}{100}\\right)^3-$ P = P$\\frac{1331}{1000}-$ P =\u00a0$\\frac{331}{1000}$P<\/p>\n Simple interest on the sum = $\\frac{P\\times3\\times10}{100}$ = $\\frac{3}{10}$P<\/p>\n According to the problem,<\/p>\n $\\frac{331}{1000}$P $-$\u00a0$\\frac{3}{10}$P = 558<\/p>\n $\\Rightarrow$ \u00a0$\\frac{331P-300P}{1000}$ = 558<\/p>\n $\\Rightarrow$ \u00a0$\\frac{31P}{1000}$ = 558<\/p>\n $\\Rightarrow$ \u00a0 P = 18000<\/p>\n $\\therefore\\ $The principal sum =\u00a0\u20b9 18,000<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 10:\u00a0<\/b>The sum of simple interest on a sum at 8% p.a. for 4 years and 8 years is \u20b9960. The sum is:<\/p>\n a)\u00a0\u20b9800<\/p>\n b)\u00a0\u20b91100<\/p>\n c)\u00a0\u20b91000<\/p>\n d)\u00a0\u20b9900<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the Principal amount = P<\/p>\n Rate = 8%<\/p>\n Given, the sum of simple interest on P for 4 years and 8 years is \u20b9960<\/p>\n $\\Rightarrow$ \u00a0$\\frac{P\\times4\\times8}{100}+\\frac{P\\times8\\times8}{100}=960$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{32P}{100}+\\frac{64P}{100}=960$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{96P}{100}=960$<\/p>\n $\\Rightarrow$ \u00a0 P = 1000<\/p>\n $\\therefore\\ $The required sum =\u00a0\u20b91000<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 11:\u00a0<\/b>The difference between the compound interest on a sum of \u20b9 8,000 for 1 year at the rate of 10% per annum, interest compounded yearly and half yearly is:<\/p>\n a)\u00a0\u20b940<\/p>\n b)\u00a0\u20b910<\/p>\n c)\u00a0\u20b930<\/p>\n d)\u00a0\u20b920<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, Principal amount =\u00a0\u20b9 8,000<\/p>\n Rate = 10%<\/p>\n When the interest is compounded yearly, time period = 1 year<\/p>\n Compound interest when compounded yearly =\u00a0$8000\\left(1+\\frac{10}{100}\\right)^1=8000\\left(\\frac{110}{100}\\right)=8800$<\/p>\n When the interest is compounded yearly, time period = 2 half-years<\/p>\n Rate =\u00a0$\\frac{10}{2}$ = 5%<\/p>\n Compound interest when compounded half yearly =\u00a0$8000\\left(1+\\frac{5}{100}\\right)^2=8000\\left(\\frac{105}{100}\\right)^2=8820$<\/p>\n $\\therefore$ Required difference = 8820 – 8800 = \u20b920<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 12:\u00a0<\/b>There is a 60% increase in an amount in 5 years at simple interest. What will be the compound interest on \u20b9 6,250 for two years at the same rate of interest, when the interest is compounded yearly?<\/p>\n a)\u00a0\u20b9 1,480<\/p>\n b)\u00a0\u20b9 1,560<\/p>\n c)\u00a0\u20b9 1,500<\/p>\n d)\u00a0\u20b9 1,590<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the rate of interest = R%<\/p>\n Principal amount = P<\/p>\n Time = 5 years<\/p>\n $\\Rightarrow$\u00a0 Amount = $\\frac{160}{100}P$<\/p>\n $\\Rightarrow$\u00a0 P +\u00a0$\\frac{P\\times5\\times R}{100}$ =\u00a0$\\frac{160}{100}\\text{P}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{P\\times5\\times R}{100}$ =\u00a0$\\frac{160}{100}\\text{P}$ – $\\text{P}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{P\\times5\\times R}{100}$ =\u00a0$\\frac{60}{100}\\text{P}$<\/p>\n $\\Rightarrow$ \u00a0R = 12%<\/p>\n Compound interest on \u20b9 6,250 for two years at 12% = $6250\\left(1+\\frac{12}{100}\\right)^2-6250$<\/p>\n $=6250\\left(\\frac{112}{100}\\right)^2-6250$<\/p>\n $=6250\\left(1.12\\right)^2-6250$<\/p>\n $=6250\\left(1.2544\\right)-6250$<\/p>\n $=6250\\left(0.2544\\right)$<\/p>\n $=$ \u20b9 1590<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 13:\u00a0<\/b>If the present amount is \u20b9 87,750 with 8% rate of interest in four years, then what was the principal amount?<\/p>\n a)\u00a0\u20b9 69,345.6<\/p>\n b)\u00a0\u20b9 78,456.34<\/p>\n c)\u00a0\u20b9 56,896.98<\/p>\n d)\u00a0\u20b9 66,477.2<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, Amount =\u00a0\u20b9 87,750<\/p>\n Rate = 8%<\/p>\n Time = 4 years<\/p>\n Let Principal amount = P<\/p>\n $\\Rightarrow$\u00a0 P + $\\frac{P\\times4\\times8}{100}=87750$<\/p>\n $\\Rightarrow$\u00a0 P + $\\frac{32P}{100}=87750$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{132P}{100}=87750$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{P}{100}=664.7727$<\/p>\n $\\Rightarrow$\u00a0 P = 66477.27<\/p>\n $\\therefore\\ $Principal amount = \u20b9 66477.27<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 14:\u00a0<\/b>A man has \u20b910,000. He lent a part of it at 15% simple interest and the remaining at 10% simple interest. The total interest he received after 5 years amounted to \u20b96,500. The difference between the parts of the amounts he lent is:<\/p>\n a)\u00a0\u20b91,750<\/p>\n b)\u00a0\u20b92,500<\/p>\n c)\u00a0\u20b92,000<\/p>\n d)\u00a0\u20b91,500<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Total Amount = \u20b910,000<\/p>\n Let the amount lent at 15% = $x$<\/p>\n $=$>\u00a0 Amount lent at 10% = $10000-x$<\/p>\n Total interest he received after 5 years amounted to \u20b96,500<\/p>\n $=$> \u00a0$\\frac{x\\times15\\times5}{100}+\\frac{\\left(10000-x\\right)\\times10\\times5}{100}=6500$<\/p>\n $=$> \u00a0$\\frac{3x}{4}+\\frac{1}{2}\\left(10000-x\\right)=6500$<\/p>\n $=$> \u00a0$\\frac{3x}{4}-\\frac{x}{2}+5000=6500$<\/p>\n $=$> \u00a0$\\frac{3x-2x}{4}=6500-5000$<\/p>\n $=$> \u00a0$\\frac{x}{4}=1500$<\/p>\n $=$> \u00a0$x=6000$<\/p>\n $\\therefore\\ $Difference between the parts of amounts he lent =\u00a0$x-\\left(10000-x\\right)=6000-\\left(10000-6000\\right)=$\u20b9 2000<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 15:\u00a0<\/b>Ram deposited an amount of \u20b9 8,000 in a bank\u2019s savings account with interest 6.5% compounded monthly. What amount will he get at the end of 18 months?<\/p>\n a)\u00a0$\u20b9 8816.97$<\/p>\n b)\u00a0$\u20b9 8788.98$<\/p>\n c)\u00a0$\u20b9 8907.56$<\/p>\n d)\u00a0$\u20b9 8790.54$<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Principal amount (P) = \u20b9 8,000<\/p>\n Rate of interest (R) = 6.5% per annum = $\\frac{6.5}{12}\\%$ per month<\/p>\n Time (n) = 18 months<\/p>\n $\\therefore\\ $Amount =\u00a0$P\\left(1+\\frac{R}{100}\\right)^n$<\/p>\n $=8000\\left(1+\\frac{6.5}{12\\times100}\\right)^18$<\/p>\n $=8000\\left(1+0.0054167\\right)^{18}$<\/p>\n $=8000\\left(1.0054167\\right)^{18}$<\/p>\n $=$\u20b9 8816.97<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 16:\u00a0<\/b>Latha deposited an amount of $\u20b9 35,000$ in a bank with simple interest 11% per annum. How much interest will she earn after one year?<\/p>\n a)\u00a0$\u20b9 3,370$<\/p>\n b)\u00a0$\u20b9 3,220$<\/p>\n c)\u00a0$\u20b9 3,500$<\/p>\n d)\u00a0$\u20b9 3,850$<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Principal amount (P) =\u00a0$\u20b9 35,000$<\/p>\n Rate of Simple interest (R)% = 11%<\/p>\n Time (T) = 1 year<\/p>\n $\\therefore\\ $Simple interest earned by Latha =\u00a0$\\frac{\\text{PTR}}{100}=\\frac{35000\\times1\\times11}{100}=\u20b9 3,850$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 17:\u00a0<\/b>The difference of simple interest on a sum of money for 8 years and 10 years is $\u20b9 200.$ If the rate of interest is 10 % p.a, then what is the sum of money?<\/p>\n a)\u00a0$\u20b9 1,600$<\/p>\n b)\u00a0$\u20b9 1,000$<\/p>\n c)\u00a0$\u20b9 1,200$<\/p>\n d)\u00a0$\u20b9 1,400$<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the Principal amount = P<\/p>\n Given, rate of interest = 10%<\/p>\n The difference of simple interest on the sum of money for 8 years and 10 years is $\u20b9 200$<\/p>\n $=$> \u00a0$\\frac{P\\times10\\times10\\ }{100}-\\frac{P\\times8\\times10}{100}=200$<\/p>\n $=$> \u00a0$\\frac{100P\\ }{100}-\\frac{80P}{100}=200$<\/p>\n $=$> \u00a0$\\frac{20P\\ }{100}=200$<\/p>\n $=$> \u00a0$P=\u20b9 1000$<\/p>\n $\\therefore\\ $Sum of money =\u00a0$\u20b9 1000$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 18:\u00a0<\/b>At which rate of simple interest does an amount become double in 12 years?<\/p>\n a)\u00a0$7 \\frac{4}{5}%$<\/p>\n b)\u00a0$7 \\frac{1}{2}%$<\/p>\n c)\u00a0$8%$<\/p>\n d)\u00a0$8 \\frac{1}{3}%$<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Time (T) = 12 years<\/p>\n Let the Principal amount = P<\/p>\n Rate of Simple Interest = R<\/p>\n According to the Problem,<\/p>\n Amount = 2P<\/p>\n $=$>\u00a0 Principal + Simple Interest = 2P<\/p>\n $=$>\u00a0 P +\u00a0$\\frac{\\text{P}\\times\\text{T}\\times \\text{R}}{100}$ = 2P<\/p>\n $=$>\u00a0\u00a0$\\frac{\\text{P}\\times12\\times \\text{R}}{100}$ = 2P – P<\/p>\n $=$> \u00a0$\\frac{\\text{P}\\times12\\times \\text{R}}{100}$ = P<\/p>\n $=$> \u00a0$\\text{R}=\\frac{100}{12}$<\/p>\n $=$> \u00a0$\\text{R}=8\\frac{1}{3}$%<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 19:\u00a0<\/b>A person invested a total of \u20b99,000 in three parts at 3%, 4% and 6% per annum on simple interest. At the end of a year, he received equal interest in all the three cases. The amount invested at 6% is:<\/p>\n a)\u00a0$\u20b93,000$<\/p>\n b)\u00a0$\u20b95,000$<\/p>\n c)\u00a0$\u20b92,000$<\/p>\n d)\u00a0$\u20b94,000$<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the amount invested by the person at 3%, 4% and 6% are $x$, $y$, $z$ respectively<\/p>\n $=$> \u00a0$x+y+z=9000$ ………….(1)<\/p>\n Interest on amount $x$ at 3% after 1 year = $\\frac{x\\times3\\times1}{100}=\\frac{3x}{100}$<\/p>\n Interest on amount $y$ at 4% after 1 year =\u00a0$\\frac{y\\times4\\times1}{100}=\\frac{4y}{100}$<\/p>\n Interest on amount $z$ at 6% after 1 year =\u00a0$\\frac{z\\times6\\times1}{100}=\\frac{6z}{100}$<\/p>\n Given, interest received after 1 year on $x$, $y$, $z$ amounts are equal<\/p>\n $=$> \u00a0$\\frac{3x}{100}=\\frac{4y}{100}=\\frac{6z}{100}$<\/p>\n $=$> \u00a0$3x=4y=6z$ ……………….(2)<\/p>\n Sustituting values from (2) in (1)<\/p>\n $=$> \u00a0$x+\\frac{3}{4}x+\\frac{3}{6}x=9000$<\/p>\n $=$> \u00a0$x+\\frac{3}{4}x+\\frac{1}{2}x=9000$<\/p>\n $=$> \u00a0$\\frac{4x+3x+2x}{4}=9000$<\/p>\n $=$> \u00a0$\\frac{9x}{4}=9000$<\/p>\n $=$> \u00a0$x=4000$<\/p>\n $\\therefore\\ $ $y=\\frac{3}{4}x=\\frac{3}{4}\\times4000=3000$ and<\/p>\n $z=\\frac{3}{6}x=\\frac{3}{6}\\times4000=2000\\ $<\/p>\n $\\therefore\\ $The amount invested at 6% Simple interest =\u00a0$\u20b92,000$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 20:\u00a0<\/b>In how may years will a sum of $\u20b9 320$ amount to $\u20b9 405$ if interest is compounded at 12.5% per annum?<\/p>\n a)\u00a02 years<\/p>\n b)\u00a01 year<\/p>\n c)\u00a0$1 \\frac{1}{2}$ years<\/p>\n d)\u00a0$2 \\frac{1}{2}$ years<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Principal (P) =\u00a0$\u20b9 320$<\/p>\n Rate (R)% = 12.5%<\/p>\n Amount (A) =\u00a0$\u20b9 405$<\/p>\n Let the required number of years = n<\/p>\n $=$> \u00a0$\\text{P}\\left(1+\\frac{\\text{R}}{100}\\right)^n=405$<\/p>\n $=$> \u00a0$320\\left(1+\\frac{12.5}{100}\\right)^n=405$<\/p>\n $=$> \u00a0$320\\left(\\frac{112.5}{100}\\right)^n=405$<\/p>\n $=$> \u00a0$\\left(\\frac{1125}{1000}\\right)^n=\\frac{405}{320}$<\/p>\n $=$> \u00a0$\\left(\\frac{9}{8}\\right)^n=\\frac{81}{64}$<\/p>\n $=$> \u00a0$\\left(\\frac{9}{8}\\right)^n=\\frac{9^2}{8^2}$<\/p>\n $=$> \u00a0$\\left(\\frac{9}{8}\\right)^n=\\left(\\frac{9}{8}\\right)^2$<\/p>\n $=$> \u00a0$n=2$<\/p>\n Hence, the correct answer is Option A<\/p>\n