Arithmetic is an important topic in the Arithmetic section of the SNAP Exam. You can also download this Free Arithmetic Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Arithmetic questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practising.<\/p>\n Download Arithmetic Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>A group of 20 girls has average age of 12 years. Average of first 12 from the same group is 13 years and what is the average age of other 8 girls in the group?<\/p>\n a)\u00a010<\/p>\n b)\u00a011<\/p>\n c)\u00a011.5<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the age of each of the girl in the group be $x_1,x_2,x_3,…..,x_{20}$ years<\/p>\n Average age of 20 girls = 12<\/p>\n => $\\frac{(x_1+x_2+x_3+…..+x_{20})}{20}=12$<\/p>\n =>\u00a0$(x_1+x_2+x_3+…..+x_{20})=12 \\times 20=240$ ————(i)<\/p>\n Average of first 12 girls = 13<\/p>\n => $\\frac{(x_1+x_2+x_2+….+x_{12})}{12}=13$<\/p>\n =>\u00a0$(x_1+x_2+x_3+…..+x_{12})=13 \\times 12=156$ ———–(ii)<\/p>\n Subtracting equation (ii) from (i)<\/p>\n =>\u00a0$(x_1+x_2+x_3+…..+x_{20})$ $-$ $(x_1+x_2+x_3+…..+x_{12}) = (240-156)$<\/p>\n => $(x_{13}+x_{14}+…..+x_{20})=84$<\/p>\n Dividing above equation by 8<\/p>\n => $\\frac{(x_{13}+x_{14}+……+x_{20})}{8}=\\frac{84}{8} = 10.5$<\/p>\n => Ans – (E)<\/p>\n Question 2:\u00a0<\/b>The total marks obtained by Jaya in the maths and Physics together are 50 more than the marks obtained by her in chemistry, if she got 20 marks in the physics .What are her marks in maths?<\/p>\n a)\u00a020<\/p>\n b)\u00a040<\/p>\n c)\u00a080<\/p>\n d)\u00a0Cannot Determined<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let marks obtained by Jaya in\u00a0Physics, maths and chemistry respectively be $p,m,c$<\/p>\n Marks in physics, $p=20$<\/p>\n According to ques, => $m+p=c+50$<\/p>\n => $m+20=c+50$<\/p>\n => $m-c=50-20=30$<\/p>\n There is only 1 equation and 2 variables, thus we cannot determine the marks scored by her in maths or chemistry.<\/p>\n => Ans – (D)<\/p>\n Question 3:\u00a0<\/b>Two chairs and three tables cost Rs 1025\/- and 3 chairs and 2 tables cost is Rs 1,100\/- .What is the difference between the cost one table and chair?<\/p>\n a)\u00a075<\/p>\n b)\u00a035<\/p>\n c)\u00a0125<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let cost price of 1 chair = $Rs.$ $x$ and cost price of 1 table = $Rs.$ $y$<\/p>\n => $2x+3y=1025$ ———–(i)<\/p>\n and $3x+2y=1100$ ——–(ii)<\/p>\n Subtracting equation (i) from (ii)<\/p>\n => $(3x-2x)+(2y-3y)=(1100-1025)$<\/p>\n => $x-y=75$<\/p>\n $\\therefore$ Difference between the cost one table and chair = Rs. 75<\/p>\n => Ans – (A)<\/p>\n Question 4:\u00a0<\/b>The cost of 4 cell phones and 7 digital cameras is Rs 1,25,627.what is the cost of 8 cellphones and 14 digital cameras.<\/p>\n a)\u00a0251254<\/p>\n b)\u00a0252627<\/p>\n c)\u00a0225524<\/p>\n d)\u00a0Cannot determined<\/p>\n e)\u00a0None of these<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let cost of 1 cell phone = $Rs. x$ and 1 digital camera = $Rs. y$<\/p>\n => $4x + 7y = 1,25,627$<\/p>\n Multiplying both sides by 2, we get\u00a0:<\/p>\n => $8x + 14y = 125627 \\times 2 = 251254$<\/p>\n $\\therefore$ Cost of 8 cellphones and 14 digital cameras = Rs. 2,51,254<\/p>\n Question 5:\u00a0<\/b>If a amount of Rs 41910 is distributed amongst 22 persons equally,how much each person get<\/p>\n a)\u00a01905<\/p>\n b)\u00a02000<\/p>\n c)\u00a01885<\/p>\n d)\u00a02015<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Amount distributed = Rs. 41910<\/p>\n Number of person = 22<\/p>\n Each person will get = $\\frac{41910}{22}$<\/p>\n = Rs. 1,905<\/p>\n Question 6:\u00a0<\/b>A canteen requires 112 kgs of wheat for week .how many kgs of wheat requires for 69 days<\/p>\n a)\u00a01204<\/p>\n b)\u00a01401<\/p>\n c)\u00a01104<\/p>\n d)\u00a01014<\/p>\n e)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Wheat requirement for 7 days = 112 kg<\/p>\n => Wheat requirement for 1 day = $\\frac{112}{7} = 16$ kg<\/p>\n $\\therefore$ Wheat requirement for 69 days = $16 \\times 69$<\/p>\n = 1104 kg<\/p>\n Question 7:\u00a0<\/b>The difference between 42% of a number and 28% of the same number is 210.what is the 59% of that number<\/p>\n a)\u00a0630<\/p>\n b)\u00a0885<\/p>\n c)\u00a0420<\/p>\n d)\u00a0900<\/p>\n e)\u00a0none of these<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number = $100x$<\/p>\n Acc. to ques,<\/p>\n => $(\\frac{42}{100} \\times 100x) – (\\frac{28}{100} \\times 100x) = 210$<\/p>\n => $42x – 28x = 14x = 210$<\/p>\n => $x = \\frac{210}{14} = 15$<\/p>\n => Number = $100 \\times 15 = 1500$<\/p>\n $\\therefore$ 59 % of the number = $\\frac{59}{100} \\times 1500$<\/p>\n = 885<\/p>\n Question 8:\u00a0<\/b>The average of the 5 consecutive even numbers A,B,C,D ,E is 52.what is the product of B & E<\/p>\n a)\u00a02912<\/p>\n b)\u00a02688<\/p>\n c)\u00a03024<\/p>\n d)\u00a02800<\/p>\n e)\u00a0NONE OF THESE<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the five consecutive even numbers\u00a0A,B,C,D ,E = $(x-4) , (x-2) , (x) , (x+2) , (x+4)$ respectively.<\/p>\n Average = $\\frac{A+B+C+D+E}{5} = 52$<\/p>\n => $(x-4) + (x-2) + (x) + (x+2) + (x+4) = 52 \\times 5$<\/p>\n => $5x = 52 \\times 5$<\/p>\n => $x = \\frac{52 \\times 5}{5} = 52$<\/p>\n => $B = 52 – 2 = 50$ and $E = 52 + 4 = 56$<\/p>\n $\\therefore$ Product of B & E = $50 \\times 56 = 2800$<\/p>\n Question 9:\u00a0<\/b>Pinku, Rinku and Tinku divide an amount of 4200 amongst themselves in the ratio of 7:8:6 respectively. If an amount 200 is added to their shares. What will be the new ratio?<\/p>\n a)\u00a08:9:6<\/p>\n b)\u00a07:9:5<\/p>\n c)\u00a07:8:6<\/p>\n d)\u00a08:9:7<\/p>\n e)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the amount received by\u00a0Pinku, Rinku and Tinku respectively = $7x , 8x , 6x$<\/p>\n Total amount = $7x + 8x + 6x = 4200$<\/p>\n => $21x = 4200$<\/p>\n => $x = \\frac{4200}{21} = 200$<\/p>\n => Amount with\u00a0Pinku = 7*200 = 1400<\/p>\n Rinku = 8*200 = 1600<\/p>\n Tinku = 6*200 = 1200<\/p>\n If Rs. 200 is added to their shares, respective ratio<\/p>\n = $(1400 + 200) : (1600 + 200) : (1200 + 200)$<\/p>\n = $1600:1800:1400 = 8:9:7$<\/p>\n Question 10:\u00a0<\/b>find the average of set scores? 221,231,441,359,665,525<\/p>\n a)\u00a0399<\/p>\n b)\u00a0428<\/p>\n c)\u00a0407<\/p>\n d)\u00a0415<\/p>\n e)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Set\u00a0:\u00a0221,231,441,359,665,525<\/p>\n Sum =\u00a0221 + 231 + 441 + 359 + 665 + 525 = 2442<\/p>\n => Required average = $\\frac{2442}{6}$<\/p>\n = 407<\/p>\n Question 11:\u00a0<\/b>An article when sold for 960 fetches 20% profit.What would be the percent profit \/loss if such 5 article are sold for Rs. 825\/-each?<\/p>\n a)\u00a03.125 % profit<\/p>\n b)\u00a03.125 % loss<\/p>\n c)\u00a0Neither profit nor loss<\/p>\n d)\u00a016.5 % profit<\/p>\n e)\u00a0None of these<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let cost price of an article = $Rs.$ $100x$<\/p>\n If Selling price = Rs 960<\/p>\n => Profit % = $\\frac{960-100x}{100x} \\times 100=20$<\/p>\n => $960-100x=20x$<\/p>\n => $20x+100x=120x=960$<\/p>\n => $x=\\frac{960}{120}=8$<\/p>\n Thus, cost price of 1 article = $100 \\times 8 = Rs.$ $800$<\/p>\n If selling price = Rs. 825<\/p>\n $\\therefore$\u00a0Profit %<\/span><\/strong> = $\\frac{825-800}{800} \\times 100$<\/p>\n = $\\frac{25}{8} = 3.125\\%$<\/p>\n => Ans – (A)<\/p>\n Question 12:\u00a0<\/b>The owner of an electronics shop charges his customers 22% more than the cost price .If the customer paid Rs 10,980 for DVD player then the what was the cost price of that DVD?<\/p>\n a)\u00a08000<\/p>\n b)\u00a08800<\/p>\n c)\u00a09500<\/p>\n d)\u00a09200<\/p>\n e)\u00a0none of these<\/p>\n 12)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let cost price = $Rs. 100x$<\/p>\n Selling price = Rs. 10,980<\/p>\n =>\u00a0Profit % = $\\frac{10980 – 100x}{100x} \\times 100 = 22$<\/p>\n => $10980-100x=22x$<\/p>\n => $22x+100x=122x=10980$<\/p>\n => $x=\\frac{10980}{122} = 90$<\/p>\n $\\therefore$ Cost price = $100 \\times 90 = Rs. 9000$<\/p>\n => Ans – (E)<\/p>\n Question 13:\u00a0<\/b>An item was bought at Rs. X and sold at Rs. Y, there by earning a profit of 20%. Had the value of X been 15% less and the value of Y been Rs. 76 less, a profit of 30% would have been earned. What was the value of \u2018X\u2019<\/p>\n a)\u00a0Rs. 640<\/p>\n b)\u00a0Rs.400<\/p>\n c)\u00a0Rs.600<\/p>\n d)\u00a0Rs.800<\/p>\n e)\u00a0Rs.840<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n C.P. = $Rs. x$<\/p>\n S.P. = $Rs. y$<\/p>\n Profit % = $\\frac{y – x}{x} \\times 100 = 20$<\/p>\n => $\\frac{y – x}{x} = \\frac{20}{100} = \\frac{1}{5}$<\/p>\n => $5y – 5x = x$ => $6x = 5y$<\/p>\n => $y = \\frac{6 x}{5}$ ———–(i)<\/p>\n If,\u00a0value of X been 15% less and the value of Y been Rs. 76 less<\/p>\n => $x’ = \\frac{85}{100} \\times x = \\frac{17 x}{20}$<\/p>\n => $y’ = y – 76$<\/p>\n Profit % = $\\frac{y’ – x’}{x’} \\times 100 = 30$<\/p>\n => $\\frac{(y – 76) – (\\frac{17 x}{20})}{\\frac{17 x}{20}} = \\frac{30}{100} = \\frac{3}{10}$<\/p>\n => $10 \\times [(y – 76) – (\\frac{17 x}{20}] = 3 \\times \\frac{17 x}{20}$<\/p>\n => $10y – 760 – \\frac{170 x}{20} = \\frac{51 x}{20}$<\/p>\n => $10y – \\frac{221 x}{20} = 760$<\/p>\n Using, equaiton (i), we get :<\/p>\n => $(10 \\times \\frac{6 x}{5}) – \\frac{221 x}{20} = 760$<\/p>\n => $12x – \\frac{221 x}{20} = 760$<\/p>\n => $\\frac{19 x}{20} = 760$<\/p>\n => $x = 760 \\times \\frac{20}{19}$<\/p>\n => $x = 40 \\times 20 = Rs. 800$<\/p>\n Question 14:\u00a0<\/b>Shri Ramlal purchased a TV set for Rs. 12,500 and spent Rs. 300 on transportation and Rs. 800 on installation. At what price should he sell it so as to earn an overall profit of 15% ?<\/p>\n a)\u00a0Rs. 14,560<\/p>\n b)\u00a0Rs. 14,375<\/p>\n c)\u00a0Rs. 15,460<\/p>\n d)\u00a0Rs. 15,375<\/p>\n e)\u00a0None of these<\/p>\n 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Cost price of TV = Rs. 12,500<\/p>\n Amount spent on transportation = Rs. 300 and installation = Rs. 800<\/p>\n Net spent = Rs. (12500 + 300 + 800) = Rs. 13,600<\/p>\n Let selling price = $Rs.x$<\/p>\n Profit % = $\\frac{x-13600}{13600} \\times 100=15$<\/p>\n => $x-13600=15 \\times 136$<\/p>\n => $x=2040+13600$<\/p>\n => $x=Rs.$ $15,640$<\/p>\n => Ans – (E)<\/p>\n Question 15:\u00a0<\/b>A trader sells an item to a retailer at 20% discount, but charges 10% on the discounted price, for delivery and packaging. The retailer sells it for Rs. 2046 more, thereby earning a profit of 25%. At what price had the trader marked the item?<\/p>\n a)\u00a0Rs. 9400<\/p>\n b)\u00a0Rs. 9000<\/p>\n c)\u00a0Rs. 8000<\/p>\n d)\u00a0Rs. 12000<\/p>\n e)\u00a0Rs. 9300<\/p>\n 15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let Marked price of item = $Rs. 100x$<\/p>\n => Selling price of trader = Cost price of retailer = $100x \\times \\frac{80}{100} \\times \\frac{110}{100}$<\/p>\n = $Rs. 88x$<\/p>\n Selling price of retailer = $Rs. (88x + 2046)$<\/p>\n Profit % = $\\frac{(88x + 2046) – 88x}{88x} \\times 100 = 25$<\/p>\n => $\\frac{2046}{88x} = \\frac{25}{100} = \\frac{1}{4}$<\/p>\n => $x = \\frac{2046 \\times 4}{88} = 93$<\/p>\n $\\therefore$ Marked price = $100 \\times 93 = Rs. 9,300$<\/p>\n Enroll to SNAP & NMAT 2022 Crash Course<\/a><\/p>\n Question 16:\u00a0<\/b>A starts a business with Rs. 2500. After one month from the start of the business, B joined with Rs. 4500 and A withdrew completely after eleven months from the start of the business. If the difference between A\u2019s and B\u2019s respective shares in the annual profit was Rs. 4800, what was the annual profit earned?<\/p>\n a)\u00a0Rs. 14800<\/p>\n b)\u00a0Rs. 16800<\/p>\n c)\u00a0Rs. 14400<\/p>\n d)\u00a0Rs. 11400<\/p>\n e)\u00a0Rs. 15600<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Amount invested by A = Rs. 2500 and by B = Rs. 4500<\/p>\n Both invested for 11 months.<\/p>\n Ratio of profit shared by A and B<\/p>\n = $(2500 \\times 11) : (4500 \\times 11)$<\/p>\n = $5 : 9$<\/p>\n Let total profit earned by A and B respectively = $Rs. 5x$ and $Rs. 9x$<\/p>\n => $9x – 5x = 4800$<\/p>\n => $x = \\frac{4800}{4} = 1200$<\/p>\n $\\therefore$ Total profit = $9x + 5x = 14x$<\/p>\n = $14 \\times 1200 = Rs. 16,800$<\/p>\n Question 17:\u00a0<\/b>17 articles were bought for Rs. 3,910 and sold for Rs. 4,590. How much was the approximate profit percentage per article ?<\/p>\n a)\u00a017%<\/p>\n b)\u00a012%<\/p>\n c)\u00a09%<\/p>\n d)\u00a021%<\/p>\n e)\u00a025%<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Cost price of 1 article = Rs. $\\frac{3910}{17}$ = Rs. 230<\/p>\n Selling price of 1 article = Rs. $\\frac{4590}{17}$ = Rs. 270<\/p>\n Profit % = $\\frac{270 – 230}{230} * 100$<\/p>\n = 17.39% $\\approx$ 17%<\/p>\n Question 18:\u00a0<\/b>The cost price of an article is Rs. 390. If it is to be sold at a profit of 3.12 per cent, how much would be its approximate selling price ?<\/p>\n a)\u00a0Rs. 410<\/p>\n b)\u00a0Rs. 402<\/p>\n c)\u00a0Rs. 417<\/p>\n d)\u00a0Rs. 420<\/p>\n e)\u00a0Rs. 442<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Profit obtained in selling the article at 3.12%<\/p>\n = $\\frac{3.12}{100} * 390 \\approx$ Rs. 12<\/p>\n => Selling price = 390 + 12 = Rs. 402<\/p>\n Question 19:\u00a0<\/b>The cost price of an article is Rs.1700. If it was sold at a price of Rs.2006, what was the percentage profit on the transaction?<\/p>\n a)\u00a018<\/p>\n b)\u00a012<\/p>\n c)\u00a010<\/p>\n d)\u00a015<\/p>\n e)\u00a020<\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Profit = S.P. – C.P. = 2006 – 1700<\/p>\n = Rs. 306<\/p>\n => Profit % = $\\frac{306}{1700} * 100$<\/p>\n = 18%<\/p>\n Question 20:\u00a0<\/b>21 articles were bought for 6531 and sold for Rs.9954. How much was the approximate profit percentage per article?<\/p>\n a)\u00a056%<\/p>\n b)\u00a043%<\/p>\n c)\u00a052%<\/p>\n d)\u00a049%<\/p>\n e)\u00a061%<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n C.P. of 1 article = $\\frac{6531}{21}$ = Rs. 311<\/p>\n S.P. of 1 article = $\\frac{9954}{21}$ = Rs. 474<\/p>\n => Profit on 1 article = S.P. – C.P. = 474 – 311<\/p>\n = Rs. 163<\/p>\n $\\therefore$ Profit % = $\\frac{163}{311} * 100$<\/p>\n = 52.4% $\\approx$ 52%<\/p>\n Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 21:\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 21)\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 22:\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 22)\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 23:\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 23)\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 24:\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 24)\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 25:\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 25)\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