Fundamentals of Commercial Arithmetic

Discover how ratios compare two quantities using difference and division, read ratios with colons, and relate similar items, with examples like apples to oranges.

Compare quantities using ratios by ensuring they are of the same kind and unit, and remember ratios have no units and depend on the order of antecedent and consequent.

Explain how to generate equivalent ratios by multiplying or dividing the numerator and denominator, illustrate with 6:10 becoming 12:20 or 3:5, and reduce 24:36 to 2:3 in its simplest form.

Express statements as ratios, noting order matters, with examples such as oxygen to hydrogen at 1 to 2 and squash to water at 1 to 3.

learn to reduce ratios to simplest form by converting to a fraction and dividing by the highest common factor, with examples like 40:80 and 8:16.

Convert quantities to the same units, then simplify the ratio to its lowest form. For example, 50 cm to 5 m becomes 1:10; 8 h to 1 day becomes 1:3.

Learn to set up and simplify common word problem ratios, including girls to boys and girls to total students, and apply earnings, savings, and expenditure relationships with clear, simplified fractions.

Learn to form and simplify ratios from fruit and candy problems, and compare prices by unit conversion to a common unit, yielding a 3:1 ratio.

compare two ratios by converting them to like fractions using the least common multiple of denominators, then compare the numerators to determine which is larger.

Convert ratios to like fractions and compare the numerators by multiplying extremes, A by D and B by C, to decide whether A:B is greater than C:D.

Learn how multiplying multiple ratios yields a compounded ratio by multiplying numerators and denominators. Explore duplicate ratio, sub-duplicate ratio, triplicate ratio, sub-triplicate ratio, and reciprocal ratio.

Solve for x using duplicate and triplicate ratios, cross-multiplication, and ratio simplifications. Derive x and x:y from the given equations.

Divide a total into two parts in a given ratio by dividing into a1+a2 equal parts and applying first part = a1*a/(a1+a2) and second part = a2*a/(a1+a2).

Divide 1500 rupees between Adi and Vijay in the ratio 7:5, partition into 12 parts, assign 7 parts to Adi and 5 to Vijay, yielding 875 and 625 rupees.

Solve for two numbers in the ratio 8:7 with sum 60 by using the simplest form, giving 32 and 28, which confirms the ratio 8:7.

Apply the ratio 2:4:2:1:1 to allocate a pentagon's five interior angles that sum to 540 degrees. Compute the individual angles as 108, 216, 108, 54, and 54 degrees.

Learn to simplify ratios to their simplest form by expressing them as fractions and recognizing that two numbers in a ratio are ax and bx for some common x.

Apply the income ratio 4:3 and expenditure ratio 2:3 with 2500 rupees saved to find incomes and expenditures. Ramon earns 10000 and spends 7500; Gorgon earns 7500 and spends 5000.

Solve for two numbers in the ratio 2:3 when 15 is added to each, yielding a ratio of 11:14; cross-multiplication gives 18 and 27, with 27 the greater.

Solve a ratio problem where two numbers in 3:5 are reduced by 9 to 12:23; set (3x-9)/(5x-9)=12/23, find x=11, so the greater number is 55.

Apply ratio reasoning to a class problem. From a 4:3 boys-to-girls ratio, 20 more boys and 1 less girl yield a 2:1 ratio, giving a total of 154 students.

Set passes to failures as 4x to x, so 5x appeared; with 30 fewer appeared and 20 fewer passed, the ratio becomes 5:1 and 150 appeared.

Solve a ratio problem by setting two numbers as 4x and x, adding 5 to both, and using exact division to find x = 10, giving 40 and 10.

Determine the division of 300 rupees among p, q, and r where q is 30 more than p and r is 60 more than q, yielding p:q:r as 2:3:5.

Explore how equivalent fractions form equivalent ratios, and how a proportion uses two ratios with colon notation, such as 2:3 proportional to 4:6.

Master the cross product rule for proportions, where the product of extremes equals the product of means, enabling you to find the fourth term from any three known terms.

Explore the cross product rule, where the product of extremes equals the product of means, to solve proportions for x. See examples with 6 and 13, 2 and 50.

Explore how to determine the fourth proportional using proportional reasoning and the cross product rule, with explicit examples like 5, 4, 25 and 3, 1, 15.

Apply ratio and cross-product reasoning to compute the cost of thirteen copies given eight copies cost 488 rupees, using proportionality of copies to price.

Apply the proportion method to find how many workers earned a given wage, using the cross product rule and the product of extremes equals product of means.

Add a value x to each number so that 6+x, 15+x, 20+x, 43+x form a proportion, solve by cross-multiplication to find x=3, giving 9, 18, 23, 46.

Identify continued proportion where a by b equals b by c; the middle term is the mean proportional, and b equals the square root of ac, as shown by 2,4,8.

Compute the mean proportional between two numbers by taking the square root of their product, since B^2 = AC; for 5 and 80, it is 20.

Delivers a practical method for continued proportion and ratios, solving for x and y from x/3 = 3/12 and 3/1 = 12/y.

Explore how adding x to 16, 26, and 40 creates a continued proportion, using cross-multiplication to equate ratios and solve for x, yielding x = 9.

Explore the mean proportional in a three-number continued proportion, where B^2 = AC; apply this to x, 5, and 9 to find x, which equals 7.

Apply mean proportional and continued proportion to two numbers, derive equations for x and y, and solve to find x = 14 and y = 56.

Understand that percent expresses a fraction with hundred in the denominator, where the numerator is the percent value, as in 10 percent equals 10/100 and 25 percent equals 25/100.

Explain how percentages express fractions with a denominator of hundred and enable comparison by division, illustrated by comparing educated women in two villages and converting ratios to percent.

Convert fractions to percentages using two methods: obtain a 100 denominator by multiplying numerator and denominator, as in 1/2 to 50/100 (50%), and 3/4 to 75/100 (75%).

Convert decimals to percentages by multiplying by 100 percent, moving the decimal point to the right. See examples like 0.34 to 34%, 1.152 to 115.2%, and 18.6 to 1860%.

Convert a whole ratio to percent by writing it as a fraction and multiplying by 100 percent. For example, 3:4 becomes 3/4, or 75 percent.

Learn to convert percentages to fractions, decimals, and ratios using the percent-to-fraction approach, simplification, and decimal placement. Examples include 9%, 25%, 84%, and 150% with step-by-step conversions.

Learn to find the whole from a given percentage using simple algebra, showing 30 percent of 270 rupees equals 900 and 25 percent of 750 kilometers equals 3,000.

Express one quantity as a percentage of another by converting to the same units and applying the percentage formula. See examples in centimeters and grams.

Learn to calculate percentage increase and decrease using the change over original amount formula, with examples like fee hikes and price reductions.

Compute the third test score needed for a 60 percent overall, given three tests with max 100 marks; George must score 65 to reach the target.

Compute the original population from a 15 percent increase to 19,320 using the percentage increase formula, solving for x to arrive at 16,800.

Set 40 percent of the total marks equal to 60 and solve for the total, yielding 150 marks.

Compute a 15 percent increase on a factory worker's salary of 5700 rupees, resulting in an increase of 855 rupees and a new salary of 6555 rupees.

Calculate the two-day room rent at 3500 per day for 7000 rupees, then add a 7 percent luxury tax on the room rate to total 7490 rupees.

Calculate the nickel content in a 20-kilogram alloy by subtracting copper and zinc from the total, yielding the remaining 30 percent.

solve a percent problem by equating 25 percent of a number to 20, deducing the number is 80, then find 40 percent of the same number by proportion.

Calculate how a 5% raise on 100 rupees leads to 105, then a 5% decline on 105 yields 99.75 rupees, a net 0.25-rupee decrease.

determine the actual price per kilogram of sugar given a 10% price drop allows buying five kilograms more with 100 rupees.

Explore the concept of interest, principal, and amount, and learn how the rate of interest determines extra payments for borrowed money over time.

Explore how simple interest is calculated on the principal amount, contrasting it with compound interest calculated on the previous year's amount, using examples at 10 percent per annum.

Calculate simple interest on the principal using the formula, illustrated with 3000 rupees at 7% for three years, yielding 630 rupees interest and a total of 3630.

Apply the simple interest formula, SI = principal × rate × time / 100, and find the final amount as principal plus simple interest, illustrated with example calculations.

Master simple interest using the formula si = p r t /100 and amount = p + si through examples that find rate, principal, or time from given values.

Apply the simple interest formula with principal, rate, and time, converting months and days into years to compute interest and final amount in rupees scenarios.

Compute the principal from the amount 1368 rupees using si = prt/100 for 3.5 years at 4 percent, yielding 1200 rupees.

Find how many years it takes for a sum to double at 10 percent simple interest, using two methods that both yield ten years.

Derive the rate from a principal doubling in eight years under simple interest, then calculate when the same rate triples the principal, finding sixteen years for a hundred rupees principal.

Solve a word problem on simple interest by deriving the principal and rate from amounts at the end of two and three years, yielding a 4000 principal at 5%.

Solve a word problem on simple interest by comparing amounts after seven and forty years to find the principal and rate, then compute the rate and principal step by step.

In this simple interest word problem, increasing the rate from r to r+1 percent for three years yields 5,100 more interest, solving for principal as 170,000 rupees.

solve a word problem on simple interest with changing rates—6% for three years, 9% for five, 13% for three—sum the interest to find the principal of 8000 rupees.

Analyze a two-part loan of 7700 rupees, with the first part at 20 percent for five years and the second at 9 percent for six years, equal simple interest.

Solve a word problem on simple interest with a two-part loan of 35000 rupees, split between 10% for 4 years and 11% for 4.5 years to total 15900 interest.

Compare simple interest and compound interest using a 15,000 rupees principal at 8% for two years, showing how compound interest calculates on the previous year's amount as the new principal.

Apply a 4 percent annual rate to an 8000 rupees principal for two years with annual compounding, yielding 8652.80 rupees as the final amount and 652.80 rupees as compound interest.

Calculate the amount and interest on 25000 rupees for three years at 12 percent, compounded annually. Determine the final amount as 35123.20 rupees and the compound interest as 10123.20.

Explain the difference between simple and compound interest and how compounding frequency—annually, half-yearly, or quarterly—adjusts the rate per annum on the growing principal.

Compute compound interest on a 5,000 rupee principal over 2 years at 4 percent per annum, compounded semiannually. Determine the final amount and interest earned.

Define conversion period as the duration for calculating interest in compound interest, including annual, half-yearly, and quarterly examples. Note how frequency changes the principal and rate.

apply the formula a = p(1 + r/100)^t to compound interest. compute the final amount and the compound interest using an example of 4000 at 15% for 3 years.

Compute the compound interest and amount for a 16,000 rupee principal over two years at 15% annual interest using A = P(1+R/100)^t, yielding 21,160 rupees total and 5,160 rupees interest.

Calculate compound interest on five thousand rupees at 8 percent for three years using A = P(1 + r/100)^t; the amount is 6298.56 rupees, and the interest is 1298.56 rupees.

learn how to compute compound interest for fractional time with the formula A = P(1 + R/100)^t, handling whole and fractional years and distinguishing final amount from principal.

Explore compound interest with mixed time periods by applying the principal and rate to whole years first, then handle the fractional part; compare methods in a 10% per annum example.

Compute the amount and compound interest on 10,000 rupees at 4% per annum, compounded annually for 2.5 years, yielding 11,032.32 as amount and 1,032.32 as interest.

Explore how changing interest rates affect compound interest, calculating year-by-year growth from a principal using the prior amount as the new principal, and deriving the general formula for variable rates.

Calculate the amount after two years with changing rates of 4% and 5%, using compound interest: 25000*(1.04)*(1.05) = 27300 rupees. Then compute the compound interest as 2300 rupees.

Calculate the compound interest on a 3125 rupee principal over three years with annual rates of 4%, 5%, and 6%, using successive compounding to determine the final amount.

Compute compound interest using A = P(1 + R/100)^N, where N is number of conversion periods, and adjust the rate and time for half yearly or quarterly compounding.

Calculate compound interest with semiannual compounding on a 15000 rupees principal over 1.5 years at 8% per annum, applying a = p(1+r/100)^n to find amount and interest.

Compute the compound interest on 5000 rupees at 12% per annum with semiannual compounding over 1.5 years, using conversion periods and a halved rate.

Learn to compute compound interest using the annual and semiannual formulas, substituting principal, rate, and time into P(1 + R/100)^D or P(1 + R/200)^(2D).

Calculate the amount and compound interest on one lakh rupees at eight percent per annum, compounded semi annually for one and a half years, using the standard formula.

Explain quarterly compounding of interest on a principal amount B rupees, with four conversion periods per year, using the quarterly rate r/4 and N = 4d to compute A.

Compute the compound interest on eight thousand rupees at twenty percent per annum for nine months with quarterly compounding, yielding nine thousand two hundred sixty-one rupees.

Calculate compound interest on a 31,250 rupees at 16 percent per annum, compounded quarterly for nine months, using the formula A = P(1 + r/4)^n to obtain 35,152 and 3,902.

Generalizing compound interest formula, the amount is A = P(1 + R/(100n))^(nT), where n is the number of compounding periods per year, with n = 1, 2, 4, 12.

Learn to compute amount and interest using the compound interest formula, and deduce any missing variable when three of principal, rate, time, amount, or interest are given.

Calculate the principal borrowed when interest is 15 percent per annum for two years using the compound interest formula, and learn how annual, semiannual, and quarterly compounding affect the result.

Compute the principal in a semiannual compound interest problem. Use the semiannual formula A = P(1 + r/200)^(2t) with r = 4%, t = 1 year, giving P = 7500.

Determine rate per annum for compound interest using A = P(1 + R/100)^t, with P = 2304, A = 2500, t = 2, yielding R = 4 1/6 percent.

Determine the annual rate of interest when forty thousand rupees grows to forty eight thousand six hundred twenty point two five in two years under semi-annual compounding.

Compute the time for 4000 rupees to grow to 5324 rupees at 10 percent compound interest using a = p(1 + r)^t, by equating bases to find t.

Compute the time for a 16,000 rupee investment at 10 percent per year compounded semiannually to reach 18,522, giving t as 1.5 years.

Examine simple interest and compound interest on a 2000 rupee principal at 10% annually, showing that for the first year both methods yield 200 rupees.

Explore how compound interest grows over time, with the second conversion period yielding more interest than the first. A 100 rupee principal at 10% for two years illustrates the process.

Explain that the compound interest for a conversion period equals the amount after that period minus the amount after the previous period, illustrated with principal P and three-year annual compounding.

Explain how the compound interest of the second conversion period minus the first equals the simple interest on the first period's interest, with a 4000 rupees example.

From two-year simple interest 400 and compound interest 410, derive principal and rate by comparing yearly interest components, yielding a principal of 4000 and a rate of 5 percent.

Compute the principal from a 264.80 rupees compound interest at 10% for three years with annual compounding, yielding P = 800, then simple interest is 240 for the same sum.

Compute the rate (8%) and principal (Rs 5,000) from the given simple and compound interest data, then determine the three-year compound interest, simple interest, and their difference.

Compute compound interest on 8000 rupees for 1.5 years at 10% under annual and semi-annual compounding, yielding interests of 1240 and 1261 rupees with a 21 rupee difference.

Explore growth and compound interest using the formula P(1 + R/100)^t. See how a 10 percent per annum rate over two years grows 100 to 121, mirroring population growth.

Apply the compound interest formula to growth scenarios by equating final value to initial value times (1 + R/100)^time period, illustrated with population growth examples.

Demonstrate calculating population growth with compound growth: apply the formula P = P0(1 + r/100)^t to 16,000 at 5 percent per annum, yielding 17,640 after two years.

Apply the multi-rate compound interest formula to compute final amounts across periods. Model growth by multiplying successive rate factors, as shown with 3%, 4%, and 6% per year.

Apply the compound growth formula V = V0 (1 + R/100)^n to motorbike production, from 20,000 to 23,328 over two years, yielding an 8 percent per annum growth rate.

Apply the compound growth formula v = v0(1 + r/100)^n to back-calculate the prison population two years ago from a present population of 67,000 at 4 percent growth, yielding 62,500.

Use the compound growth formula V = V0 × (1+R/100)^N to model a population rising from 48,000 at 5 percent to 55,566. Conclude that N = 3 years.

Explore compound growth of property value at 25 percent per annum over three years using V = V0(1+0.25)^3, and calculate a final value with a 95 5/16 percent increase.

Learn how depreciation reduces value over time using the formula v0 times one minus r by hundred raised to the power n, with changing rates r1, r2, and r3.

Calculate car depreciation with a 30000 rupees cost at 20 percent annual rate over three years, using V = V0(1 - r/100)^n to find the final value and depreciation.

Apply the depreciation formula V = V0(1 - r/100)^n with r = 12% and n = 2 to find the tv's original purchase price, the butcher's price, which is 6,250 rupees.