One Interest is one of the key topics in the IBPS PO Prelims exam. Here you can download free Interest questions PDF with answers for IBPS PO Prelims 2022 by Cracku. These questions will help you to practice and solve the Interest questions in the IBPS PO exams. Utilize this best PDF practice set which includes the answers in detail. Click on the below link to download the Interest Usage MCQ questions PDF for IBPS PO Prelims 2022 for Free.<\/p>\n
Download Interest Questions for IBPS PO Prelims<\/a><\/p>\n Download IBPS PO Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>Atul borrowed a sum of \u20b912000 and agreed to repay it by paying \u20b94800 at the end of first year and \u20b99240 at the end of second year. Whatis the rate of compound interest compounded annually?<\/p>\n a)\u00a010%<\/p>\n b)\u00a08%<\/p>\n c)\u00a012%<\/p>\n d)\u00a0$\\frac{8}{5}\\%$<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Question 2:\u00a0<\/b>A person borrowed a sumof \u20b930800 at 10% p.a. for 3 years, interest compounded annually. At the end of twoyears, he paid a sumof \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 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Question 3:\u00a0<\/b>An amount of Rs. 38000 was deposited on the simple interest at the rate of 9% per annum for two years and another amount of Rs. 42000 was deposited on the compound interest at the rate of 12% per annum for one year compounded half-yearly. Find out the total interest obtained from both of the amounts.<\/p>\n a)\u00a0Rs. 12497.2<\/p>\n b)\u00a0Rs. 12262.2<\/p>\n c)\u00a0Rs. 12031.2<\/p>\n d)\u00a0Rs. 12656.2<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n An amount of Rs. 38000 was deposited on the simple interest at the rate of 9% per annum for two years. Question 4:\u00a0<\/b>An amount of Rs. 56000 was deposited for 1.5 years at 18% rate of interest per annum compounded half-yearly. Find out the interest obtained from it.<\/p>\n a)\u00a0Rs. 16185.236<\/p>\n b)\u00a0Rs. 16363.412<\/p>\n c)\u00a0Rs. 16521.624<\/p>\n d)\u00a0Rs. 16942.848<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Here the interest is compounded half yearly. It means a 9% rate of interest for six months. Question 5:\u00a0<\/b>Find the difference between the Compound and Simple Interest for a sum of Rs 1000 at a 10% rate of interest per annum\u00a0for 2 years.<\/p>\n a)\u00a0Rs 10<\/p>\n b)\u00a0Rs 20<\/p>\n c)\u00a0Rs 100<\/p>\n d)\u00a0None of the above<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The difference between the Compound and Simple Interest for 2 years is given as = Take Free IBPS PO Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>A sum of money doubles itself in 15 years. By how much time it will become 8 times at Compound Interest?<\/p>\n a)\u00a050 years<\/p>\n b)\u00a040 years<\/p>\n c)\u00a045 years<\/p>\n d)\u00a0None of the above<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n If money will become m times in n years. Question 7:\u00a0<\/b>What will be the approximate compound interest on a sum of Rs. 16500 invested for 3 years at the rate of 15% per annum compounded annually?<\/p>\n a)\u00a0Rs.8418<\/p>\n b)\u00a0Rs.8594<\/p>\n c)\u00a0Rs.8680<\/p>\n d)\u00a0Rs.8742<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Compound interest = $P(1+\\frac{R}{100})^{n} – P$ = $16500\\times1.15^{3} – 16500$<\/p>\n = $16500\\times 1.520875 – 16500$ = Rs.8594 (approx.) Question 8:\u00a0<\/b>The difference between the simple interest and the compound interest earned on a certain sum of money at 15% per annum for 2 years is Rs.558. Find the difference between the simple interest and the compound interest earned on the same sum of money at 20% per annum for 2 years.<\/p>\n a)\u00a0Rs.1116<\/p>\n b)\u00a0Rs.992<\/p>\n c)\u00a0Rs.948<\/p>\n d)\u00a0Rs.864<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the sum invested be Rs.10000x. Amount earned after 2 years at compound interest = $10000x\\times\\dfrac{115}{100}\\times\\dfrac{115}{100} = Rs.13225x$.<\/p>\n Compound interest = Rs.13225x – Rs.10000x = Rs.3225x.<\/p>\n Given, 3225x – 3000x = Rs.558. Hence, The sum invested = Rs.10000x = Rs.24800.<\/p>\n Simple Interest earned on Rs.24800 at 20% per annum = $\\dfrac{24800\\times20\\times2}{100} = Rs.9920$.<\/p>\n Amount earned after 2 years at 20% per annum compound interest = $24800\\times1.2\\times1.2 = Rs.35712$. Question 9:\u00a0<\/b>If the simple interest earned on a sum of Rs.x at 12.5% per annum for \u2018t\u2019 years is the same as the simple interest earned on a sum of Rs.y at 16.67% per annum for \u20182t\u2019 years, then find the ratio between the sum invested at 12.5% per annum and 16.67% per annum respectively.<\/p>\n a)\u00a07:5<\/p>\n b)\u00a06:7<\/p>\n c)\u00a08:3<\/p>\n d)\u00a04:1<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Sum = Rs.x Sum = Rs.y $\\dfrac{x\\times12.5\\times t}{100} = \\dfrac{y\\times16.67\\times2t}{100}$<\/p>\n $\\dfrac{x\\times1}{8} = \\dfrac{y\\times2}{6}$<\/p>\n $\\dfrac{x}{8} = \\dfrac{y}{3}$<\/p>\n Hence, x:y = 8:3.<\/p>\n Question 10:\u00a0<\/b>Find the compound interest on Rs 7540 at 8% per annum for 1 year, when compounded half yearly.<\/p>\n a)\u00a0Rs. 454.66<\/p>\n b)\u00a0Rs. 254.66<\/p>\n c)\u00a0Rs. 615.264<\/p>\n d)\u00a0Rs. 554.664<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to the question, Question 11:\u00a0<\/b>A man invested 50% of his capital money at 8% per annum, 25% of his capital money at 6% per annum and the remaining capital money he invested at 10% per annum at simple interest. If at the end of a year he receives the total interest of Rs 450, then find the overall capital money he invested.<\/p>\n a)\u00a0Rs 5400<\/p>\n b)\u00a0Rs 4500<\/p>\n c)\u00a0Rs 5256<\/p>\n d)\u00a0Rs 5625<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to the question, Question 12:\u00a0<\/b>If the sum of money kept in compound interest triple itself in 3 years, then find the time taken by it to be 9 times of itself.<\/p>\n a)\u00a09 years<\/p>\n b)\u00a06 years<\/p>\n c)\u00a05 years<\/p>\n d)\u00a012 years<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to the question, Question 13:\u00a0<\/b>If the compound interest for the third year at 9.09% per annum compounded annually is Rs.720. Find the principal.<\/p>\n a)\u00a0Rs.8575<\/p>\n b)\u00a0Rs.6655<\/p>\n c)\u00a0Rs.4875<\/p>\n d)\u00a0Rs.6425<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n 9.09% = $\\dfrac{1}{11}$. Question 14:\u00a0<\/b>The rate of interest at which a certain sum of money at simple interest amounts to Rs.18174 in 2 years and Rs.22035 in 5 years is:<\/p>\n a)\u00a08.25%<\/p>\n b)\u00a06.75%<\/p>\n c)\u00a06.67%<\/p>\n d)\u00a015%<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, Amount for 2 years = Rs.18174. Question 15:\u00a0<\/b>A certain sum of money amounts to Rs.14725 at 6.25% per annum simple interest\u00a0in 3 years. What approximate interest would have been obtained if the same sum is invested at the same rate of compound interest in the same time period?<\/p>\n a)\u00a0Rs.2583<\/p>\n b)\u00a0Rs.2473<\/p>\n c)\u00a0Rs.2613<\/p>\n d)\u00a0Rs.2437<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the principal be Rs.P. $0.1875P = 14725-P$ $\\approx 14873$<\/p>\n Hence, Interest = 14873 – 12400 = Rs.2473.<\/p>\n Question 16:\u00a0<\/b>A loan was repaid in two equal instalments of Rs.51577.5 each in two years. If the rate of interest is 15% compounded annually, then the difference between the amount of loan taken and the total interest paid is:<\/p>\n a)\u00a0Rs.64895<\/p>\n b)\u00a0Rs.64545<\/p>\n c)\u00a0Rs.68750<\/p>\n d)\u00a0Rs.65800<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let Principal = Rs.x. Total interest paid = 103155 – 83850 = 19305. Question 17:\u00a0<\/b>A certain sum of money amounts to Rs.101167.5 in 4 years 6 months at 7.69% per annum simple interest. Find the interest obtained by the same sum of money at 15.38% per annum simple interest for 6.8 years.<\/p>\n a)\u00a0Rs.78621.6<\/p>\n b)\u00a0Rs.72186.6<\/p>\n c)\u00a0Rs.76218.6<\/p>\n d)\u00a0Rs.71626.8<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 7.69% = $\\dfrac{1}{13}$ $22.5P+65P = 101167.5$ Question 18:\u00a0<\/b>Which of the following yields the highest interest of all when invested for 2 years? a)\u00a0Only I<\/p>\n b)\u00a0Both II and III<\/p>\n c)\u00a0Only III<\/p>\n d)\u00a0Both I and III<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n I. Principal = Rs.24000 II. Principal = Rs.19440 Interest = Rs.24000-19440 = Rs.4560.<\/p>\n III. Principal = Rs.28750. Question 19:\u00a0<\/b>A person invested a certain sum of money at 15% Simple interest per annum for three years. He then withdraws the whole amount and invested the whole amount in a scheme which offers 20% per annum for the first year and 10% per annum for the second year. He withdraws the whole amount which is equal to Rs.5742. What is the sum which he invested initially?<\/p>\n a)\u00a0Rs.2500<\/p>\n b)\u00a0Rs.3000<\/p>\n c)\u00a0Rs.4000<\/p>\n d)\u00a0Rs.4500<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the initial investment be Rs.x.<\/p>\n Amount after 3 years = $\\dfrac{x\\times15\\times3}{100}+x = Rs.1.45x$<\/p>\n Amount after next two year = $1.45x\\times1.2\\times1.1 = Rs.1.914x$.<\/p>\n Given, 1.914x = 5742<\/p>\n x = $\\dfrac{5742}{1.914} = 3000$<\/p>\n Hence, Option B is the correct answer.<\/p>\n Question 20:\u00a0<\/b>In approximately how many years will a certain sum of money become 21 times of itself at 7% per annum Compound interest compounded annually?<\/p>\n a)\u00a045 years<\/p>\n b)\u00a084 years<\/p>\n c)\u00a068 years<\/p>\n d)\u00a039 years<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the sum of money be Rs.x $x\\times\\dfrac{107}{100}^T = 21x$<\/p>\n $\\dfrac{107}{100}^T = 21$<\/p>\n $1.07^T = 21$<\/p>\n From the options, Hence, Option A is the correct answer.<\/p>\n
\nInterest = $\\frac{38000\\times9\\times2}{100}$
\n= 6840
\nAnother amount of Rs. 42000 was deposited on the compound interest at the rate of 12% per annum for one year compounded half-yearly.
\nInterest = 42000 of 106% of 106% – 42000
\n= $42000\\times1.06\\times1.06 – 42000$
\n= 47191.2 – 42000
\n= 5191.2
\nTotal interest = 6840+5191.2
\n= Rs. 12031.2
\nHence, option c is the correct answer.<\/p>\n
\nInterest = 56000 of 109% of 109% of 109% – 56000
\n= $56000\\times1.09\\times1.09\\times1.09 – 56000$
\n= 72521.624 – 56000
\n= Rs. 16521.624
\nHence, option c is the correct answer.<\/p>\n
\n$P\\times (\\dfrac{R}{100})^2$, where P is Principle amount and R is the rate of interest.
\nHere P = Rs 1000 and R = 10%
\nFrom the above formula, we get the difference between Ci and SI for two years
\n= $1000\\times (\\dfrac{10}{100})^2$
\nRs 10.<\/p>\n
\nSo, $m^n$ times in n x number of years
\n2 times will become in 15 years
\nSo, $2^3$ will become double in 3 x 15 years = 45 years.<\/p>\n
\n= $16500(1+\\frac{15}{100})^{3} – 16500$
\n= $16500(\\frac{115}{100})^{3} – 16500$<\/p>\n
\n= $16500( 1.520875 – 1)$
\n= $16500 \\times0.520875$
\n= 8594.4375<\/p>\n
\nHence, option b is the correct answer.<\/p>\n
\nSimple interest earned on Rs.10000x at 15% per annum for 2 years = $\\dfrac{10000x\\times15\\times2}{100} = Rs.3000x$.<\/p>\n
\n225x = 558
\nx = 2.48<\/p>\n
\nCompound interest = Rs.35712 – Rs.24800 = Rs.10912.
\nHence, The required difference = Rs.10912 – Rs.9920 = Rs.992.<\/p>\n
\nRate of interest = 12.5% per annum.
\nTime period = \u2018t\u2019 years.<\/p>\n
\nRate of interest = 16.67% per annum.
\nTime period = 2t years.<\/p>\n
\nRate = 8%,
\nPrincipal = Rs 7540
\nTime = 2 half years
\nCompound Interest = $P[(1+\\frac{r}{200})^2 – 1]$
\nCI = $7540[(1+\\frac{8}{200})^2 – 1]$
\nCI = $7540[(\\frac{208}{200})^2 – 1]$
\nCI = $7540 \\times (\\frac{43264}{40000}-1)$, solving this we get,
\nCI = Rs. 615.264.<\/p>\n
\nLet his capital be x
\nSo,
\n50% of x = x\/2
\nSimple Interest = $\\frac{ x . 8 . 1}{200}$
\n25% of x = x\/4
\nSimple interest = $\\frac{x . 6 . 1}{400}$
\nRemaining amount = x\/4
\nSimple interest = $\\frac{x . 10 . 1}{400}$
\nNow on adding
\n$\\frac{ x . 8 . 1}{200}$ + $\\frac{x . 6 . 1}{400}$ + $\\frac{x . 10 . 1}{400}$ = 450
\n$\\frac{ x . 8 . 1}{200}$ + $\\frac{x . 3 . 1}{200}$ + $\\frac{x . 5 . 1}{200}$ = 450
\n8x + 3x + 5x = 90000
\n16x = 90000
\nX = 5625
\nSo his investment = Rs 5625<\/p>\n
\nAmount = $Principal(1 + \\frac{rate}{100})^{time}$
\nAmount gets double in three years so,
\n3 x Amount = $Principal(1 + \\frac{rate}{100})^{3}$
\nAmount gets 9 times in,
\n3 x 3 x Amount = $Principal(1 + \\frac{rate}{100})^{n}$
\nWe know,
\n$(Principal(1 + \\frac{rate}{100})^{3})^2$ = $Principal(1 + \\frac{rate}{100})^{n}$
\nSo
\nN = 6 years<\/p>\n
\nLet the Principal be Rs.1331P.
\nAmount after the first year = $1331P\\times\\dfrac{12}{11} = Rs.1452P$.
\nPrincipal for the second year = Rs.1452P.
\nAmount after the second year = $1452P\\times\\dfrac{12}{11} = Rs.1584P$.
\nPrincipal for the third year = Rs.1584P.
\nAmount after the third year = $1584P\\times\\dfrac{12}{11} = Rs.1728P$.
\nInterest for the third year = 1728P – 1584P = Rs.144P.
\nGiven, 144P = 720 \u21d2 P = 5.
\nHence, Principal = Rs.1331P = Rs.6655.<\/p>\n
\nAmount for 5 years = Rs.22035.
\nThen, Simple Interest for 3 years = Rs.22035 – Rs.18174 = Rs.3861.
\nSimple Interest for 2 years = $\\dfrac{2\\times3861}{3} = Rs.2574$.
\nPrincipal = Amount for 2 years – Simple Interest for 2 years = Rs.18174 – Rs.2574 = Rs.15600.
\nLet the rate of interest be R%.
\n$\\dfrac{15600\\times R\\times2}{100} = 2574$
\n$312R = 2574$
\n$R = 8.25\\%$<\/p>\n
\nRate of interest = 6.25%.
\nTime period = 3 years.
\nInterest = Rs.14725 – P.
\n$\\dfrac{P\\times6.25\\times3}{100} = 14725 – P$<\/p>\n
\n$1.1875P = 14725$
\n$P = Rs.12400$
\nIf Rs.12400 is invested at compound interest,
\nAmount after three years = $12400\\times\\dfrac{106.25}{100}\\times\\dfrac{106.25}{100}\\times\\dfrac{106.25}{100}$<\/p>\n
\nTotal amount repaid = 51577.5 $\\times$2 = Rs.103155
\nLoan amount = $\\dfrac{51577.5}{1.15}+\\dfrac{51577.5}{1.15\\times1.15} = 44850+39000 = Rs.83850$.<\/p>\n
\nHence, Required difference = 83850 – 19305 = Rs.64545.<\/p>\n
\nLet the principal be Rs.65P.
\nGiven, $\\dfrac{65P\\times4.5}{13} + 65P = 101167.5$<\/p>\n
\n$87.5P = 101167.5$
\nP = 1156.2.
\nThen, Principal = 65P = Rs.75153.
\nRate of interest = 15.38% = $\\dfrac{2}{13}$
\nTime period = 6.8 years.
\nInterest = $\\dfrac{75153\\times2\\times6.8}{13} = Rs.78621.6$.<\/p>\n
\nI. Rs.24000 invested at 12.5% per annum Simple Interest.
\nII. Rs.19440 invested at 11.11% per annum Compound interest compounded annually.
\nIII. Rs.28750 invested at 10% per annum compounded half-yearly.<\/p>\n
\nRate of interest = 12.5%
\nTime period = 2 years.
\nInterest = $\\dfrac{24000\\times12.5\\times2}{100} = Rs.6000$.<\/p>\n
\nRate of interest = 11.11% = $\\dfrac{1}{9}$
\nAmount after two years = $19440\\times\\dfrac{10}{9}\\times\\dfrac{10}{9} = Rs.24000$<\/p>\n
\nRate of interest = 10% per annum compounded half-yearly.
\nAmount after two years = $28750\\times1.05\\times1.05\\times1.05\\times1.05 = Rs.34945.8$.
\nInterest = Rs.34945.8 – Rs.28750 = Rs.6195.8
\nHence, Option C is the correct answer.<\/p>\n
\nLet the time period be \u2018T\u2019 years.
\nAmount after T years = Rs.343x<\/p>\n
\nA) $1.07^\\text{45} \\approx 21$
\nB) $1.07^\\text{84} \\approx 294$
\nC) $1.07^\\text{68} \\approx 100$
\nD) $1.07^\\text{39} \\approx 14$<\/p>\n