Download Simple Compound Intrest Questions for XAT PDF \u2013 XAT Simple and Compound Intrest questions pdf by Cracku. Practice XAT solved Simple Compound Intrest Questions paper tests, and these are the practice question to have a firm grasp on the Simple Compound Intrest topic in the XAT exam. Top 20 very Important Simple Compound Intrest Questions for XAT based on asked questions in previous exam papers. \u00a0The XAT question papers contain actual questions asked with answers and solutions.<\/p>\n
Download Simple Compound Intrest Questions for XAT<\/a><\/p>\n Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Ms suchi deposits an amount of 24000 to obtain in a simple interest at the rate of S.I 14 p.c.p.a for 8 years .what total amont will ms suchi gets at the end of 8 years<\/p>\n a)\u00a052080<\/p>\n b)\u00a028000<\/p>\n c)\u00a050880<\/p>\n d)\u00a026880<\/p>\n e)\u00a0none of these<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Amount deposited = Rs. 24,000<\/p>\n Rate = 14 % and time = 8 years under simple interest<\/p>\n => $S.I. = \\frac{P \\times R \\times T}{100}$<\/p>\n = $\\frac{24000 \\times 14 \\times 8}{100}$<\/p>\n = $240 \\times 112 = Rs. 26,880$<\/p>\n $\\therefore$Total amount will ms Suchi gets at the end of 8 years<\/p>\n = $24000 + 26880 = Rs. 50,880$<\/p>\n Question 2:\u00a0<\/b>What would be the compound interest obtained on a amount 3000 at the rate of 8% per annum after 2 years?<\/p>\n a)\u00a0501.50<\/p>\n b)\u00a0499.20<\/p>\n c)\u00a0495<\/p>\n d)\u00a0510<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Amount = Rs. 3000<\/p>\n Time = 2 years and rate = 8% under compound interest<\/p>\n => $C.I. = P [(1 + \\frac{R}{100})^T – 1]$<\/p>\n = $3000 [(1 + \\frac{8}{100})^2 – 1]$<\/p>\n = $3000 [(\\frac{27}{25})^2 – 1] = 3000 (\\frac{729}{625} – 1)$<\/p>\n = $3000 \\times \\frac{104}{625} = 104 \\times 4.8$<\/p>\n = $Rs. 499.20$<\/p>\n Question 3:\u00a0<\/b>What would be the simple interest obtained on a amount 5670 at the rate of 6 pcpa after 3 years?<\/p>\n a)\u00a01020.60<\/p>\n b)\u00a01666.80<\/p>\n c)\u00a01336 .80<\/p>\n d)\u00a01063.80<\/p>\n e)\u00a0none of these<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Principal amount = Rs. 5,670<\/p>\n Time period = 3 years and rate = 6% under simple interest.<\/p>\n => $S.I. = \\frac{P \\times R \\times T}{100}$<\/p>\n = $\\frac{5670 \\times 6 \\times 3}{100}$<\/p>\n = $56.7 \\times 18 = Rs. 1,020.6$<\/p>\n Question 4:\u00a0<\/b>A sum of money was invested for 14 years was in Scheme A which offers simple interest at a rate of 8% p.a. The amount received from Scheme A after 14 years was then invested for two years in Scheme B which offers compound interest (compounded annually) at a rate of 10% p.a. If the interest received from Scheme B was Rs. 6,678, what was the sum invested in Scheme A?<\/p>\n a)\u00a0Rs. 15,500<\/p>\n b)\u00a0Rs. 14,500<\/p>\n c)\u00a0Rs. 16,500<\/p>\n d)\u00a0Rs. 12,500<\/p>\n e)\u00a0Rs. 15,000<\/p>\n 4)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the sum invested in scheme A = $Rs. 100x$<\/p>\n Time = 14 years and rate = 8% under simple interest<\/p>\n => $S.I. = \\frac{P \\times R \\times T}{100}$<\/p>\n = $\\frac{100x \\times 8 \\times 14}{100} = 112x$<\/p>\n => Amount invested in Scheme B = $100x + 112x = Rs. 212x$<\/p>\n Time = 2 years and rate = 10% under compound interest.<\/p>\n $C.I. = P [(1 + \\frac{R}{100})^T – 1]$<\/p>\n => $6678 = 212x [(1 + \\frac{10}{100})^2 – 1]$<\/p>\n => $6678 = 212x [(\\frac{11}{10})^2 – 1] = 212x (\\frac{21}{100})$<\/p>\n => $0.21x = \\frac{6678}{212} = 31.5$<\/p>\n => $x = \\frac{31.5}{0.21} = 150$<\/p>\n $\\therefore$ Sum invested in scheme A = $100 \\times 150 = Rs. 15,000$<\/p>\n Question 5:\u00a0<\/b>The difference between the compound interest and the simple interest for a period of 2 years at the rate\u00a0 of 10% per annum is Rs. 50. Find the principal.<\/p>\n a)\u00a0Rs. 4000<\/p>\n b)\u00a0Rs. 5000<\/p>\n c)\u00a0Rs. 5500<\/p>\n d)\u00a0Rs. 4500<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the principal amount = $Rs. 100x$<\/p>\n Time = 2 years and rate = 10%<\/p>\n => $C.I. = P [(1 + \\frac{R}{100})^T – 1]$<\/p>\n = $100x [(1 + \\frac{10}{100})^2 – 1]$<\/p>\n = $100x [(\\frac{11}{10})^2 – 1] = 100x (\\frac{121 – 100}{100})$<\/p>\n = $100x \\times \\frac{21}{100} = 21x$<\/p>\n => $S.I. = \\frac{P \\times R \\times T}{100}$<\/p>\n = $\\frac{100x \\times 10 \\times 2}{100} = 20x$<\/p>\n It is given that difference between compound and simple interest = $21x – 20x = 50$<\/p>\n => $x = 50$<\/p>\n $\\therefore$ Principal = $100 \\times 50 = Rs. 5,000$<\/p>\n Question 6:\u00a0<\/b>The simple interest accrued on an amount of Rs. 16,500 at the end of three years is Rs. 5,940. What would be the compound interest accrued on the same amount at the same rate in the same period? (rounded off to two digits after decimal)<\/p>\n a)\u00a0Rs. 6681.31<\/p>\n b)\u00a0Rs. 6218.27<\/p>\n c)\u00a0Rs. 6754.82<\/p>\n d)\u00a0Rs. 6537.47<\/p>\n e)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Principal amount = Rs. 16,500<\/p>\n Simple interest earned = Rs. 5,940 and time = 3 years<\/p>\n Let rate of interest = $R \\%$<\/p>\n => $S.I. = \\frac{P \\times R \\times T}{100}$<\/p>\n => $5940 = \\frac{16500 \\times R \\times 3}{100}$<\/p>\n => $R = \\frac{5940}{165 \\times 3}$<\/p>\n => $R = 12 \\%$<\/p>\n Now, compound interest\u00a0accrued on the same amount at the same rate in the same period = $P [(1 + \\frac{R}{100})^T – 1]$<\/p>\n = $16,500 [(1 + \\frac{12}{100})^3 – 1]$<\/p>\n = $16,500 [(\\frac{28}{25})^3 – 1] = 16,500 (\\frac{21952 – 15625}{15625})$<\/p>\n = $16,500 \\times \\frac{6327}{15625} = Rs. 6681.31$<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 7:\u00a0<\/b>What would be the compound interest accrued on an amount of Rs. 8000 at the rate of 15 p.c.p.a. in three years ?<\/p>\n a)\u00a0Rs. 4283<\/p>\n b)\u00a0Rs. 4051<\/p>\n c)\u00a0Rs. 4167<\/p>\n d)\u00a0Rs. 4325<\/p>\n e)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Principal amount = Rs. 8,000<\/p>\n Time period = 3 years and rate of interest = 15% under compound interest.<\/p>\n =>\u00a0$C.I. = P [(1 + \\frac{R}{100})^T – 1]$<\/p>\n = $8,000 [(1 + \\frac{15}{100})^3 – 1]$<\/p>\n = $8,000 [(\\frac{23}{20})^3 – 1] = 8,000 (\\frac{12167 – 8000}{8000})$<\/p>\n = $12167 – 8000 = Rs. 4,167$<\/p>\n Question 8:\u00a0<\/b>The interest received on a sum of money when invested in scheme A is equal to the interest received on the same sum of money when invested for 2 years in scheme B. Scheme A offers simple interest (p.c.p.a.) and scheme B offers compound interest (compounded annually). Both the schemes offer the same rate of interest. If the numerical value of the number of years for which the sum is invested in scheme A is same as the numerical value of the rate of interest offered by the same scheme, what is the rate of interest (p.c.p.a) offered by scheme A?<\/p>\n a)\u00a03<\/p>\n b)\u00a0$2{1 \\over {77}}$<\/p>\n c)\u00a0$3{4 \\over {99}}$<\/p>\n d)\u00a0$2{2 \\over {99}}$<\/p>\n e)\u00a02<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let sum invested in both schemes = $Rs. P$<\/p>\n Let rate of interest in both schemes = $R \\%$<\/p>\n Time period in scheme A = $R$ years<\/p>\n => Simple interest under Scheme A = $\\frac{P \\times R \\times R}{100}$<\/p>\n = $Rs. \\frac{P R^2}{100}$<\/p>\n Also, interest received from both schemes is also same, and time period under scheme B = 2 years<\/p>\n => Compound interest under scheme B = $P [(1 + \\frac{R}{100})^T – 1]$<\/p>\n => $\\frac{P R^2}{100} = P [(1 + \\frac{R}{100})^2 – 1]$<\/p>\n => $\\frac{R^2}{100} = [1 + (\\frac{R}{100})^2 + \\frac{2 R}{100}] – 1$<\/p>\n => $\\frac{R}{100} = \\frac{R}{10000} + \\frac{1}{50}$<\/p>\n => $\\frac{R}{100} (1 – \\frac{1}{100}) = \\frac{1}{50}$<\/p>\n => $\\frac{R}{100} \\times \\frac{99}{100} = \\frac{1}{50}$<\/p>\n => $R = \\frac{100 \\times 100}{99 \\times 50}$<\/p>\n => $R = \\frac{200}{99} = 2 \\frac{2}{99} \\%$<\/p>\n Question 9:\u00a0<\/b>What is the compound interest accrued on an amount of Rs. 12000 at the rate of 9 p.c.p.a. at the end of 2 years.<\/p>\n a)\u00a0Rs. 2545<\/p>\n b)\u00a0Rs. 2257.2<\/p>\n c)\u00a0Rs. 2986<\/p>\n d)\u00a0Rs. 2775.4<\/p>\n e)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total amount under compound interest<\/p>\n = $12000 (1 + \\frac{9}{100})^2$<\/p>\n = $12000 \\times 1.09 \\times 1.09$<\/p>\n = $14257.2$<\/p>\n $\\therefore$ Compound Interest = 14257.2 – 12000<\/p>\n = Rs. 2,257.2<\/p>\n Question 10:\u00a0<\/b>Ms. Anisha deposits an amount of Rs. 35000 to obtain a simple interest at the rateof 15 p.c.p.a. for 4 years. What total amount will Ms. Anisha get at the end of 4 years ?<\/p>\n a)\u00a0Rs. 56000<\/p>\n b)\u00a0Rs. 60500<\/p>\n c)\u00a0Rs. 52000<\/p>\n d)\u00a0Rs. 48500<\/p>\n e)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Principal amount = Rs. 35,000<\/p>\n Rate of interest = 15% and time period = 4 years<\/p>\n Simple interest = $\\frac{35000 \\times 15 \\times 4}{100}$<\/p>\n = Rs. $21,000$<\/p>\n Total amount Ms. Anisha will get = 35000 + 21000<\/p>\n = Rs. 56,000<\/p>\n Question 11:\u00a0<\/b>The simple interest accrued on a amount of Rs. 12,450 at the end of 6 years is Rs. 8,964. What is the rate of interest p.c.p.a. ?<\/p>\n a)\u00a08<\/p>\n b)\u00a014<\/p>\n c)\u00a010<\/p>\n d)\u00a012<\/p>\n e)\u00a0None of these<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Principal amount = Rs. 12,450<\/p>\n Let rate of interest = $r$ %<\/p>\n => Simple interest = $\\frac{12450 \\times r \\times 6}{100} = 8964$<\/p>\n => $r = \\frac{8964 \\times 100}{12450 \\times 6}$<\/p>\n => $r = 12$%<\/p>\n Question 12:\u00a0<\/b>The simple interest accrued on an amount of Rs. 19,800 at the end of three years is Rs. 7,128. What would be the compound interest accrued on the same amount at the same rate in the same period ?<\/p>\n a)\u00a0Rs. 8934.6784<\/p>\n b)\u00a0Rs. 8017.5744<\/p>\n c)\u00a0Rs. 7861.8754<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let rate of interest be $r$ %<\/p>\n => Simple interest = $\\frac{19800 \\times r \\times 3}{100} = 7128$<\/p>\n => $r = 12$%<\/p>\n Now, C.I. = $19800 [(1 + \\frac{12}{100})^3 – 1]$<\/p>\n = $19800 (1.12^3 – 1)$<\/p>\n =\u00a0$19800 * 0.405 = 8019$<\/p>\n = Rs. 8017.5744<\/p>\n Question 13:\u00a0<\/b>What will be the the compound interest acquired on sum of Rs 12,000\/- for 3 years at the rate of 10 % per annum ?<\/p>\n a)\u00a02,652<\/p>\n b)\u00a03,972<\/p>\n c)\u00a03,960<\/p>\n d)\u00a03852<\/p>\n e)\u00a0None of these<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Principal amount = Rs. 12,000<\/p>\n Time period = 3 years and rate of interest = 10% under compound interest.<\/p>\n =>\u00a0$C.I. = P [(1 + \\frac{R}{100})^T – 1]$<\/p>\n = $12,000 [(1 + \\frac{10}{100})^3 – 1]$<\/p>\n = $12,000 [(\\frac{11}{10})^3 – 1] = 12,000 (\\frac{1331 – 1000}{1000})$<\/p>\n = $12 \\times 331 = Rs. 3,972$<\/p>\n Question 14:\u00a0<\/b>The simple interest accrued on an amount of Rs. 10,530 at the end of 5 years is Rs. 6,318. What is the rate of interest p.c.p.a. ?<\/p>\n a)\u00a08<\/p>\n b)\u00a014<\/p>\n c)\u00a010<\/p>\n d)\u00a012<\/p>\n e)\u00a0None of these<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Principal amount = Rs. 10,530<\/p>\n Let rate of interest = $r$ %<\/p>\n => Simple interest = $\\frac{10530 \\times r \\times 5}{100} = 6318$<\/p>\n => $r = \\frac{6318 \\times 100}{10530 \\times 5}$<\/p>\n => $r = 12$ %<\/p>\n Question 15:\u00a0<\/b>Mr. Madhur deposits an amount of Rs. 58,750\/- to obtain a simple interest at the rate of 12 p.c:p.a:for years. What total amount will Mr. Madhur get at the end of 4 years?<\/p>\n a)\u00a0Rs. 91,230\/-<\/p>\n b)\u00a0Rs. 86,950\/-<\/p>\n c)\u00a0Rs. 74,760\/-<\/p>\n d)\u00a0Rs. 69,540\/-<\/p>\n e)\u00a0None of These<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Principal amount = Rs. 58,750<\/p>\n Rate of interest = 12% and time period = 4 years<\/p>\n Simple interest = $\\frac{58750 \\times 12 \\times 4}{100}$<\/p>\n = Rs. $28,200$<\/p>\n Total amount Mr. Madhur will get = 58750 + 28200<\/p>\n = Rs. 86,950<\/p>\n Question 16:\u00a0<\/b>The simple interest accrued on an amount of Rs. 9,530\/- at the end of 6 years is Rs. 2,859\/-. What is the rate of interest p.c:p.a:?<\/p>\n a)\u00a05<\/p>\n b)\u00a07<\/p>\n c)\u00a09<\/p>\n d)\u00a011<\/p>\n e)\u00a0None of these<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Principal amount = Rs. 9,530<\/p>\n Let rate of interest = $r$ %<\/p>\n => Simple interest = $\\frac{9530 \\times r \\times 6}{100} = 2859$<\/p>\n => $r = \\frac{2859 \\times 100}{9530 \\times 6}$<\/p>\n => $r = 5$ %<\/p>\n Question 17:\u00a0<\/b>What is the compound interest accrued on an amount of Rs. 12,000\/- at the rate of 10 p.c.p.a. at the end of 3 Years?<\/p>\n a)\u00a0Rs. 3,972\/-<\/p>\n b)\u00a0Rs. 2,567\/-<\/p>\n c)\u00a0Rs. 4,780\/-<\/p>\n d)\u00a0Rs. 5,609\/-<\/p>\n e)\u00a0None of these<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total amount under compound interest<\/p>\n = $12000 (1 + \\frac{10}{100})^3$<\/p>\n = $12000 \\times 1.1 \\times 1.1 \\times 1.1$<\/p>\n = $15972$<\/p>\n $\\therefore$ Compound Interest = 15972 – 12000<\/p>\n = Rs. 3,972<\/p>\n Question 18:\u00a0<\/b>Ms. Suchi deposits an amount of Rs. 24,000 to obtain a simple interest at the rate of 14 p.c.p.a. for 8 years. What total amount will Ms. Suchi get at the end of 8 years?<\/p>\n a)\u00a0Rs. 62080<\/p>\n b)\u00a0Rs. 28,000<\/p>\n c)\u00a0Rs. 50,880<\/p>\n d)\u00a0Rs. 26,880<\/p>\n e)\u00a0None of these<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Principal = Rs. 24,000<\/p>\n S.I. = $\\frac{24000 \\times 14 \\times 8}{100}$<\/p>\n = Rs. 26,880<\/p>\n $\\therefore$ Total amount = 24000 + 26880<\/p>\n = Rs. 50,880<\/p>\n Question 19:\u00a0<\/b>What would be the compound interest obtained on an amount of Rs. 3,000 at the rate of 8 p.c.p.a. after 2 years?<\/p>\n a)\u00a0Rs. 501.50<\/p>\n b)\u00a0Rs. 499.20<\/p>\n c)\u00a0Rs. 495<\/p>\n d)\u00a0Rs. 510<\/p>\n e)\u00a0None of these<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Amount = $P (1 + \\frac{R}{100})^T$<\/p>\n = $3000 (1 + \\frac{8}{100})^2$<\/p>\n = $3000 (1 + \\frac{2}{25})^2$<\/p>\n = $3000 \\times \\frac{27}{25} \\times \\frac{27}{25}$ = Rs. 3499.20<\/p>\n $\\therefore$ C.I. = 3499.20 – 3000<\/p>\n = Rs. 499.20<\/p>\n