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Download Simple and Compound Interest Questions for IBPS PO Prelims<\/a><\/p>\n Download IBPS PO Previous Papers<\/a><\/p>\n Question 1:\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 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The difference between the Compound and Simple Interest for 2 years is given as = Question 2:\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 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n If money will become m times in n years. Question 3:\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 3)\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 4:\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 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to the question, Question 5:\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 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to the question, Question 6:\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 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, Amount for 2 years = Rs.18174. Question 7:\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 7)\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 8:\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 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let Principal = Rs.x. Total interest paid = 103155 – 83850 = 19305. Question 9:\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 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 7.69% = $\\dfrac{1}{13}$ $22.5P+65P = 101167.5$ Question 10:\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 10)\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. Take Free IBPS PO Mock Tests<\/a><\/p>\n Question 11:\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 11)\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 Question 12:\u00a0<\/b>A sum of money of Rs.15000 is lent in two parts such that one part is lent at Simple Interest at 11.11% per annum for two years and the other part is lent at Compound Interest at 12.5% per annum compounded annually for two years. If the difference between the interests earned from both the schemes is Rs.507, then find the interest earned on the sum of money which was invested at simple interest is now invested at\u00a08.33% per annum Compound interest compounded annually for 2 years.<\/p>\n a)\u00a0Rs.1677.77<\/p>\n b)\u00a0Rs.1432.67<\/p>\n c)\u00a0Rs.1756<\/p>\n d)\u00a0Rs.1237.5<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, Principal = Rs.15000.<\/p>\n Let the sum of money invested at Simple interest be Rs.9x. (Since, 11.11% = $\\dfrac{1}{9}$)<\/p>\n Interest earned in two years = $\\dfrac{9x\\times2}{9} = Rs.2x.$<\/p>\n Sum of money invested at Compound Interest = Rs.(15000-9x).<\/p>\n Amount earned in two years = $(15000-9x)\\times\\dfrac{9}{8}\\times\\dfrac{9}{8}$<\/p>\n Interest earned through Compound Interest = $(15000-9x)\\times\\dfrac{81}{64}-(15000-9x)$<\/p>\n Given, $(15000-9x)\\times\\dfrac{81}{64}-(15000-9x) – 2x = 507$<\/p>\n $\\dfrac{151875}{8}-\\dfrac{729x}{64}-15000+9x-2x=507$<\/p>\n $9x-2x-\\dfrac{729x}{64}+\\dfrac{151875}{8}-15000=507$<\/p>\n $\\dfrac{31875}{8}-\\dfrac{281x}{64}=507$<\/p>\n $\\dfrac{281x}{64}=\\dfrac{31875}{8}-507$<\/p>\n $\\dfrac{281x}{64}=\\dfrac{27819}{8}$<\/p>\n $281x=222552$<\/p>\n $x=792$.<\/p>\n Therefore, Sum of money invested at Simple Interest = 9x = Rs.7128.<\/p>\n Hence, Required Interest = $7128\\times\\dfrac{13}{12}\\times\\dfrac{13}{12} = Rs.8365.5$.<\/p>\n Hence, Interest earned = 8365.5 – 7128 = Rs.1237.5<\/p>\n Question 13:\u00a0<\/b>Simple interest on a certain sum of money is Rs 500 for two years\u00a0and the compound interest on the same sum of money is Rs 560 for two years on the same rate of interest per year. So find the rate of interest for which they invested?<\/p>\n a)\u00a020%<\/p>\n b)\u00a024%<\/p>\n c)\u00a022%<\/p>\n d)\u00a015%<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to the question, $p(r\/100)^{2}$ = 60<\/p>\n $25000\\times (r\/10000) = 60$<\/p>\n solving this we get r = 24%<\/p>\n Question 14:\u00a0<\/b>Two friends invested in the ratio 7:6 in compound interest at 20% and 10% respectively, then what will be the ratio of their interest after one year.<\/p>\n a)\u00a03:7<\/p>\n b)\u00a077:60<\/p>\n c)\u00a07:3<\/p>\n d)\u00a06:7<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n According to the question, Question 15:\u00a0<\/b>If a sum of money after 4 years amounts to Rs 6600 and after 7 years it amounts to 7800 at Simple interest, then find the principal and interest.<\/p>\n a)\u00a05000,8%<\/p>\n b)\u00a04500,12%<\/p>\n c)\u00a05000,10%<\/p>\n d)\u00a04500,8%<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n So interest for 3 years=7800-6600 Question 16:\u00a0<\/b>A person took a loan of certain amount at simple interest of 7% per annum for 2 years and paid an interest of Rs.280. If he took the same amount at compound interest at the same rate of interest and for the same time period, then the interest would be?<\/p>\n a)\u00a0Rs.289.8<\/p>\n b)\u00a0Rs.324.6<\/p>\n c)\u00a0Rs.312.8<\/p>\n d)\u00a0Rs.329.8<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the sum be Rs. x \u21d2 $x = 2000$ Compound Interest for Rs.2000 at 7% per annum for 2 years: Therefore, Compound Interest = 14.49% of 2000 = Rs.289.8<\/p>\n Question 17:\u00a0<\/b>A sum of money amounts to Rs.680 in 3 years and to Rs.750 in 5 years. Then find the principal.<\/p>\n a)\u00a0Rs.560<\/p>\n b)\u00a0Rs.575<\/p>\n c)\u00a0Rs.450<\/p>\n d)\u00a0Rs.350<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given that the amount after 3 years = Rs.680 Question 18:\u00a0<\/b>A certain sum of money doubles itself in 20 years at simple interest. Then in how many years, will it become 8 times itself?<\/p>\n a)\u00a0120 years<\/p>\n b)\u00a0140 years<\/p>\n c)\u00a0180 years<\/p>\n d)\u00a0200 years<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let Principal be Rs.100 => $R = 5$%<\/p>\n Now, Amount should be 8 times of Principal => $\\Large\\frac{100\\times5\\times T}{100}$ $= 700$<\/p>\n => T $= 140$ years<\/p>\n Hence, Required time period $= 140$ years<\/p>\n Question 19:\u00a0<\/b>A sum of money doubles itself in 20 years at simple interest. Find the rate of interest.<\/p>\n a)\u00a04 %<\/p>\n b)\u00a07 %<\/p>\n c)\u00a010 %<\/p>\n d)\u00a05 %<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the principal be Rs.100 => $20x = 100$ Question 20:\u00a0<\/b>The difference between Simple Interest and Compound Interest of a certain amount of money for 2 years at 5% compound interest is Rs.5. Then find the sum.<\/p>\n a)\u00a0Rs.400<\/p>\n b)\u00a0Rs.800<\/p>\n c)\u00a0Rs.1200<\/p>\n d)\u00a0Rs.2000<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the principal be Rs.10000
\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
\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
\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
\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
\nLet the principle be p and rate be r
\nGiven
\n$\\frac{prt}{100} = 500$ where t = 2 solving this we get pr = 25,000
\n$p(r\/100)^{2}$ = 560-500 = 60<\/p>\n
\nLet principal = 7x and 6x
\nWe know compound interest =\u00a0 P(1+r\/100) -P
\nTherefore ratio of interest =P(1+r\/100) -P : p(1+R\/100) -p
\nI.e\u00a0 7x(1+20\/100) -7x : 6x(1+10\/100) -6x solving this we get,
\nThe ratio of interest = 7:3<\/p>\n
\n=1200
\nInterest for 1 year=1200\/3
\n=Rs 400
\nSo for 4 years interest=4*400
\n=1600
\nA=P+I
\nP=6600-1600
\nP=Rs 5000
\nFor 4 years interest =1600
\nSo 5000*r*4\/100 =1600
\nr=8%<\/p>\n
\nGiven, $\\dfrac{x \\times 7 \\times 2}{100} = 280$<\/p>\n
\nTherefore, Principal $= Rs.2000$<\/p>\n
\nEffective compound interest percentage for 2 years $= 7 + 7 + \\dfrac{7\\times7}{100} = 14.49$ %<\/p>\n
\nAmount after 5 years = Rs.750
\nThen, S.I. for 2 years = Rs.750-Rs.680 = Rs.70
\nThen, S.I. for 3 years = 70$\\times\\frac{3}{2}$ = Rs.105
\nThen Principal = Rs.680-Rs.105 = Rs.575<\/p>\n
\nThen, Amount after 20 years = Rs.200
\nInterest = Rs.200-Rs.100 = Rs.100
\nLet Rate of interest = R%
\n$\\Large\\frac{100\\times20\\times R}{100}$ $= 100$<\/p>\n
\n=> Amount $= 8*100 = Rs.800$
\nInterest $= Rs.800 – Rs.100 = Rs.700$
\nLet time period = T years<\/p>\n
\nThen, Final amount = Rs.100*2 = Rs.200
\nInterest = Rs.200-Rs.100 = Rs.100
\nTime period = 20 years
\nLet rate of interest = R %
\nThen, $\\frac{100\\times20\\times R}{100} = 100$<\/p>\n
\n=> $x = 5$
\nTherefore, Rate of interest $= 5$ %<\/p>\n
\nSimple Interest $= \\Large\\frac{10000\\times2\\times5}{100}$ $= Rs.1000$
\nAmount at the end of 1st year at 5% compound interest $= 105$% of $10000 = Rs.10500$
\nAmount at the end of 2 years $= 105$% of $10500 = Rs.11025$
\nCompound interest $= 11025-10000 = Rs.1025$
\nDifference between SI and CI = Rs.1025-1000 = Rs.25
\nGiven difference = Rs.5
\n25 \u2192 5
\n10000 \u2192 ?
\n$\\Large\\frac{10000\\times5}{25}$ $= Rs.2000$
\nTherefore, Principal $= Rs.2000$<\/p>\n