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Download Average and Interest Questions for SSC MTS<\/a><\/p>\n Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>A fruit seller has a sale of \u20b910,435, \u20b99,927,\u20b910,855 \u20b910,230 and \u20b99,562 for five consecutive months.How much sale (in \u20b9) must he have in the sixth month so that he gets an average sale of \u20b910,500?<\/p>\n a)\u00a09,231<\/p>\n b)\u00a08,231<\/p>\n c)\u00a08,991<\/p>\n d)\u00a011,991<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n As we know,<\/p>\n Average = $\\frac{Sumofobservation}{total\\ number\\ of\\ observation}$<\/p>\n Let the sale of 6th month be x<\/p>\n Average =\u00a0$\\frac{\\left(10435+9927+10855+10230+9562+x\\right)}{6}=10500$<\/p>\n =\u00a0$51009\\ +\\ x=63000$<\/p>\n x = 11991<\/p>\n Hence, option D is correct.<\/p>\n Question 2:\u00a0<\/b>The average weight of 6 people increased by 2.5 kg when a new person came in place of other person weighing 55 kg. what can be the weight of new person (in kg)?<\/p>\n a)\u00a070<\/p>\n b)\u00a062.5<\/p>\n c)\u00a060<\/p>\n d)\u00a075.5<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total weight increased = number of persons\u00a0$\\times\\ $ increased weight<\/p>\n i.e;$6\\times\\ 2.5=15$<\/p>\n Weight of new person = weight of replaced person + Total increased weight<\/p>\n i.e; 55 + 15 = 70 kg<\/p>\n Hence, Option A is correct.<\/p>\n Question 3:\u00a0<\/b>The average of eleven numbers is 56. The average of first three numbers is 52 and that of next five numbers is 60. The 9th and 10th number are 3 and 1 more than the 11th number respectively. What is the average of 9th and 11th numbers?<\/p>\n a)\u00a053.5<\/p>\n b)\u00a052<\/p>\n c)\u00a052.5<\/p>\n d)\u00a054<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average of first three numbers is 52.<\/p>\n Sum of the first three numbers = 52 x 3 = 156<\/p>\n The average of next five numbers is 60.<\/p>\n Sum of the next five numbers = 60 x 5 = 300<\/p>\n Let the 11th number be ‘n’.<\/p>\n The average of eleven numbers is 56.<\/p>\n Sum of the eleven numbers = 56 x 11 = 616<\/p>\n 156 + 300 + (n + 3) + (n + 1) + n = 616<\/p>\n 3n + 460 = 616<\/p>\n 3n = 156<\/p>\n n = 52<\/p>\n Average of 9th and 11th numbers = $\\frac{(n+3)+n}{2}$<\/p>\n = $\\frac{107}{2}$<\/p>\n = 53.5<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 4:\u00a0<\/b>The average of 23 numbers is 51. The average of first 12 numbers is 49 and the average of last 12 numbers is 54. If the twelfth number is removed, then the average of the remaining numbers (correct to two decimal places) is:<\/p>\n a)\u00a053:25<\/p>\n b)\u00a050.45<\/p>\n c)\u00a051.75<\/p>\n d)\u00a052.65<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average of 23 numbers is 51.<\/p>\n Sum of the 23 numbers = 51 x 23 = 1173…….(1)<\/p>\n The average of first 12 numbers is 49.<\/p>\n Sum of the first 12 numbers = 49 x 12 = 588<\/p>\n Sum of the first 11 numbers + 12th number = 588…..(2)<\/p>\n The average of last 12 numbers is 54.<\/p>\n Sum of the last 12 numbers = 54 x 12 = 648<\/p>\n 12th number +\u00a0Sum of the last 11 numbers = 648…..(3)<\/p>\n Adding (2) and (3),<\/p>\n Sum of the first 11 numbers + 12th number +\u00a012th number + Sum of the last 11 numbers = 588 + 648<\/p>\n (Sum of the first 11 numbers + 12th number +\u00a0Sum of the last 11 numbers) +\u00a012th number = 1236<\/p>\n Sum of the 23 numbers +\u00a012th number = 1236<\/p>\n 1173 +\u00a012th number = 1236<\/p>\n 12th number = 63<\/p>\n Required average =\u00a0$\\frac{1173-63}{22}$<\/p>\n = 50.4545<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 5:\u00a0<\/b>The average weight of a certain number of students in a class is 55.5 kg. If 4 students with average weight 60 kg join the class, then the average weight of all students in the class increases by 360 g. The number of students in the class, initially, is:<\/p>\n a)\u00a046<\/p>\n b)\u00a031<\/p>\n c)\u00a041<\/p>\n d)\u00a036<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the initial number of students = n<\/p>\n Average of the weight of ‘n’ students = 55.55<\/p>\n Sum of the weight of ‘n’ students = 55.55n<\/p>\n Sum of the weight of 4 students = 60 x 4 = 240 kg<\/p>\n According to the problem,<\/p>\n $\\frac{55.55n+240}{n+4}=55.55+\\frac{360}{1000}$<\/p>\n $\\frac{55.55n+240}{n+4}=55.55+0.36$<\/p>\n $\\frac{55.55n+240}{n+4}=55.86$<\/p>\n 55.55n + 240 = 55.86n + 223.44<\/p>\n 0.36n = 16.56<\/p>\n n = 46<\/p>\n The initial number of students = n = 46<\/p>\n Hence, the correct answer is Option A<\/p>\n Take a free SSC MTS Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>The average monthly salary of 60 employees of a factory is \u20b929900. If two officers are getting \u20b990000 each and the average salary of 8 supervisors is \u20b965000, then what is the average salary (in \u20b9) of the remaining employees?<\/p>\n a)\u00a022680<\/p>\n b)\u00a029080<\/p>\n c)\u00a021080<\/p>\n d)\u00a021880<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average monthly salary of 60 employees of the factory is \u20b929900.<\/p>\n Total monthly salary of 60 employees of the factory = 29900 x 60 = \u20b91794000<\/p>\n Two officers are getting \u20b990000 each.<\/p>\n Sum of the salary of two officers = 2 x 90000 = \u20b9180000<\/p>\n The average salary of 8 supervisors is \u20b965000.<\/p>\n Total salary of 8 supervisors = 65000 x 8 =\u00a0\u20b9520000<\/p>\n Total salary of remaining 50 employees of the factory =\u00a01794000 –\u00a0180000 –\u00a0520000 =\u00a0\u20b91094000<\/p>\n Average of remaining 50 employees of the factory =\u00a0$\\frac{1094000}{50}$ =\u00a0\u20b921880<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 7:\u00a0<\/b>The average daily production of toys in a factory in the month of December is 512. If the average production during first 20 days is 515 and that of the last 13 days is 510, then what is the average of production on 19 and 20 December?<\/p>\n a)\u00a01058<\/p>\n b)\u00a0513<\/p>\n c)\u00a0529<\/p>\n d)\u00a0512<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n The average daily production of toys in a factory in the month of December is 512.<\/p>\n Total production of toys in the month of December = 512 x 31 = 15872……..(1)<\/p>\n The average production during first 20 days is 515.<\/p>\n Total production during first 20 days = 515 x 20 = 10300<\/p>\n Production during first 18 days + 19 December + 20 December = 10300……..(2)<\/p>\n The average production during last 13 days is 510.<\/p>\n Total production during last 13 days = 510 x 13 = 6630<\/p>\n 19 December + 20 December +\u00a0Production during last 11 days = 6630……..(3)<\/p>\n Solving (2) + (3) – (1), we get<\/p>\n 19 December + 20 December =\u00a010300 +\u00a0 6630 – 15872<\/p>\n 19 December + 20 December = 1058<\/p>\n Total production on 19 and 20 December = 1058<\/p>\n Average of production on 19 and 20 December =\u00a0$\\frac{1058}{2}$ = 529<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 8:\u00a0<\/b>The average of 40 numbers is 36. The average of the first 25 numbers is 31 and the average of last 16 numbers is 43. Find the 25th number.<\/p>\n a)\u00a021<\/p>\n b)\u00a023<\/p>\n c)\u00a024<\/p>\n d)\u00a022<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, average of 40 numbers = 36<\/p>\n $\\Rightarrow$ Sum of 40 numbers = 36 x 40 = 1440………………(1)<\/p>\n The average of the first 25 numbers is 31<\/p>\n $\\Rightarrow$ Sum of first 25 numbers = 31 x 25 = 775<\/p>\n $\\Rightarrow$ Sum of first 24 numbers + 25th number = 775 ……(2)<\/p>\n The average of last 16 numbers is 43<\/p>\n $\\Rightarrow$ Sum of last 16 numbers = 43 x 16 = 688 ………….(3)<\/p>\n Adding equations (2) and (3),<\/p>\n Sum of first 24 numbers + 25th number +\u00a0Sum of last 16 numbers = 775 + 688<\/p>\n $\\Rightarrow$ \u00a0Sum of first 24 numbers +\u00a0Sum of last 16 numbers +\u00a025th number = 1463<\/p>\n $\\Rightarrow$\u00a0 Sum of 40 numbers + 25th number = 1463<\/p>\n $\\Rightarrow$ \u00a01440 +\u00a025th number = 1463<\/p>\n $\\Rightarrow$ \u00a025th number = 23<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 9:\u00a0<\/b>The average of 24 numbers is 26. The average of the first 15 numbers is 23 and that of the last 8 number is 33. Find the 16$^{th}$ number.<\/p>\n a)\u00a016<\/p>\n b)\u00a017<\/p>\n c)\u00a015<\/p>\n d)\u00a018<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0the average of the first 15 numbers is 23<\/p>\n $\\Rightarrow$\u00a0 Sum of first 15 numbers = 23 x 15 = 345<\/p>\n Average of last 8 numbers is 33<\/p>\n $\\Rightarrow$\u00a0 Sum of last 8 numbers = 33 x 8 = 264<\/p>\n Average of 24 numbers is 26<\/p>\n $\\Rightarrow$\u00a0 Sum of 24 numbers = 26 x 24 = 624<\/p>\n $\\Rightarrow$\u00a0 Sum of first 15 numbers + 16th number +\u00a0Sum of last 8 numbers = 624<\/p>\n $\\Rightarrow$\u00a0 345 + 16th number + 264 = 624<\/p>\n $\\Rightarrow$ \u00a016th number = 15<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 10:\u00a0<\/b>The average of four consecutive odd natural numbers is eight less than the average of three consecutive even natural numbers. If the sum of these three even numbers is equal to the sum of above four odd numbers, then the average of four original odd numbers is:<\/p>\n a)\u00a032<\/p>\n b)\u00a024<\/p>\n c)\u00a018<\/p>\n d)\u00a036<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the four consecutive odd natural numbers are a, a+2, a+4, a+6<\/p>\n Let the three consecutive even natural numbers are b, b+2, b+6<\/p>\n Average of four consecutive odd natural numbers = $\\frac{a+a+2+a+4+a+6}{4}=\\frac{4a+12}{4}=a+3$<\/p>\n Average of three consecutive even natural numbers = $\\frac{b+b+2+b+4}{3}=\\frac{3b+6}{3}=b+2$<\/p>\n The average of four consecutive odd natural numbers is eight less than the average of three consecutive even natural numbers.<\/p>\n $\\Rightarrow$ \u00a0$a+3=b+2-8$<\/p>\n $\\Rightarrow$\u00a0\u00a0$b-a=9$ ………(1)<\/p>\n The sum of the three even numbers is equal to the sum of four odd numbers.<\/p>\n $\\Rightarrow$ \u00a0$b+b+2+b+4=a+a+2+a+4+a+6$<\/p>\n $\\Rightarrow$\u00a0 $3b=4a+6$<\/p>\n $\\Rightarrow$ \u00a0$3b-4a=6$ ……(2)<\/p>\n Solving 3(1)-(2) we get,\u00a0 $a = 21$<\/p>\n $\\therefore\\ $Average of four consecutive odd natural numbers =\u00a0$a+3=21+3=24$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 11:\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 11)\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 12:\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 12)\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 13:\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 13)\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 14:\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 14)\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 15:\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 15)\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 16:\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 16)\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 17:\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 17)\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\\ $The time in which the same sum will be 4 times at the same rate of interest = 24 years<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 18:\u00a0<\/b>Suresh lent out a sum of money to Rakesh for 5 years at simple interest. At the end of 5 years, Rakesh paid 9\/8 of the sum to Suresh to clear out the amount. Find the rate of simple interest per annum.<\/p>\n a)\u00a03.5% p.a.<\/p>\n b)\u00a02.5% p.a.<\/p>\n c)\u00a03% p.a.<\/p>\n d)\u00a02% p.a.<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the sum of money lent out by Suresh to Rakesh = P<\/p>\n Amount paid by Rakesh to Suresh after 5 years =\u00a0$\\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\\ $Rate of simple interest per annum = 2.5%<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 19:\u00a0<\/b>If the difference between the compound interest and simple interest on a certain sum of money for three years at 10% p.a. is \u20b9 558, then the sum is:<\/p>\n a)\u00a0\u20b9 18,500<\/p>\n b)\u00a0\u20b9 15,000<\/p>\n c)\u00a0\u20b9 16,000<\/p>\n d)\u00a0\u20b9 18,000<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the principal sum = P<\/p>\n Rate = 10%<\/p>\n Time = 3 years<\/p>\n Compound interest on the sum = P$\\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 20:\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 20)\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