Ratios, Percentages and Proportions

Ratio-based total count

A ratio tells you only the relative sizes of groups, not how many members each group actually has. If two groups are in the ratio a:b, their counts can be written as ak and bk for some positive integer k, called the multiplier or common factor. Students, animals, coins and balls are countable, so k must be a whole number, and the total (a+b)k must be a multiple of the sum of the ratio terms, a+b. This divisibility test is the core of every ratio-based total count question and is the first check to make against the options. For a 2:5 mixture, the total must be a multiple of 7, so 35 is possible but 36 is not.

The same idea can be stated with fractions. Each group equals its fraction \frac{a}{a+b} or \frac{b}{a+b} of the whole, and that number must be a whole number. Multiplying the total by a decimal fraction such as 0.7 is a convenient check, and a non-integer result immediately eliminates an option.

When people are added to or removed from each group, the new numbers are no longer proportional to the old ones, so the original ratio cannot be applied directly. Write the original counts as ak, bk, ck, write the new counts as ak+c and so on, and equate two of the new ratios. Cross-multiplying gives a linear equation in k, from which the original total (a+b+c)k follows. For example, if 4:9 becomes 2:3 after 6 is added to each group, then \frac{4k+6}{9k+6}=\frac{2}{3} gives k=1, so the original total is 13. Always substitute back into all the ratios to confirm.

When groups are transferred or merged, the count in each category of the receiving group is the sum of its old count and the transferred count. First compute the actual counts in each source using the total and the sum of the ratio terms, then combine them. The unit value of a ratio is the total divided by the sum of its terms. For instance, a group of 80 in the ratio 3:5 has a unit of 10 and holds 30 and 50. If these are added to a group holding 12 and 30, the combined counts are 42 and 80, giving 21:40, which differs from 3:5. The combined ratio must always be recomputed from the summed counts and never carried over from a source ratio.

Percentages and ratios are both relative measures, but a percentage is always taken against a stated base. Suppose 40% of residents own a car and 25% of car owners are women. The two percentages use different bases, so they cannot be added or compared without a common total. Convert each share into counts against its own base before comparing, because the share of one subgroup is not the share of the whole population. A correct inference therefore depends on which base each percentage refers to.

When an amount is divided in a given ratio, each share equals its term times a unit value x, so the total is the sum of the terms times x. A difference between two shares depends only on the difference of their terms, so an expression such as (p-q)x often gives x directly, after which every share follows. For example, if an amount is shared in the ratio 2:3:5 and the largest share exceeds the smallest by 150, then 3x=150, so x=50 and the total is 10x=500.

Some problems link two groups through a reversed ratio, such as the girls-to-boys ratio in one group equal to the boys-to-girls ratio in another. If the common number of girls is g and the two group totals are T_1 and T_2, the condition \frac{g}{T_1-g}=\frac{T_2-g}{g} gives g^2=(T_1-g)(T_2-g), which simplifies to g=\frac{T_1T_2}{T_1+T_2}. This value must be a positive whole number. For totals of 20 and 30, g=12, so the first group has 8 boys and the second has 18 boys, and both group ratios equal 3:2. This simplified formula is useful as a quick check.

A ratio can also describe the sizes of overlapping sets, such as people who like one option, the other option, or both. Find the number who like at least one set by subtracting those who like neither from the total, then apply |A\cup B|=|A|+|B|-|A\cap B|. Use the ratio to express one set in terms of the other, which reduces the relation to a single equation. For example, among 50 people, 10 like neither set and 5 like both, and A is twice B. The union is 40, so A+B-5=40 gives 3B=45, hence B=15 and A=30.

When items have different unit values, a ratio of counts is not a ratio of money or value. Multiply each count by its unit value to get the value of each group, then find the share of the total value. Counts of 3 and 1 coins of ₹5 and ₹10 are worth ₹15 and ₹10, so the ₹5 coins contribute \frac{15}{25}=60\% of the money, even though there are only three times as many of them.

Pie charts and percentage distributions express each category as a share of a total treated as 100. Solve for the counts or ratio terms first, divide each by the common total, and multiply by 100. The shares must add up to exactly 100%, so a chart whose shares do not sum to 100 is immediately wrong. When two linear relations among the categories are given, substitute one into the other until every category is expressed in a single unknown.

Several pairs cause frequent errors. A ratio 7:3 does not mean the counts are 7 and 3; it means the counts are multiples of 7 and 3. A part-to-part ratio is different from a part-to-whole fraction, so 3:5 corresponds to \frac{3}{8} and \frac{5}{8} of the total. Percentages of different bases must never be compared directly.

Key expressions:

• in the ratio a:b — the two quantities are in proportion a to b, so the counts are ak and bk. e.g. Sand and cement are mixed in the ratio 2:5.
• common factor / multiplier — the integer k that scales the ratio terms to actual counts.
• sum of the parts — a+b in a two-part ratio, which the total must be a multiple of.
• in proportion to — the same meaning as in the ratio, often used when sharing an amount.
• of the applicants / of the qualifiers — percentages with different bases, which must be converted to counts before they are compared.
• neither — the number outside both sets, which is subtracted from the total before applying inclusion–exclusion.

Chained ratio combination

Chained ratio combination is the process of joining two or more ratios that share a common term into one ratio between the first and last quantities. The key step is to make the shared term equal in both ratios before combining them, because a ratio fixes only relative size. For example, if a:b = 2:3 and b:c = 4:5, the shared term b is 3 in the first ratio and 4 in the second. Multiplying the first ratio by 4 and the second by 3 gives a:b = 8:12 and b:c = 12:15, so a:b:c = 8:12:15 and a:c = 8:15.

Never add the terms of two ratios or invert one ratio without checking the shared term. A reliable method is to fix one quantity at a convenient value, such as p = 1, and then find each following quantity by multiplying with the factor of the next ratio. The same method works in reverse when the question asks for a ratio that skips over intermediate quantities.

Percentage phrases must be converted to multipliers before chaining. A quantity that is x\% more than another equals (1 + x/100) times that quantity, so a value that is 50% more than s equals 1.5s, not 0.5s. Similarly, x\% less means multiplying by (1 - x/100). Ratios with fractional terms should be cleared by multiplying every term by the LCM of the denominators. For instance, x:y:z = \frac{1}{2}:\frac{1}{3}:\frac{1}{4} becomes 6:4:3 after multiplying by 12. An expression such as \frac{x+z-y}{y} is a ratio of homogeneous terms, so it depends only on the ratio and takes the same value for x = 6k, y = 4k, z = 3k with any k \neq 0.

Mixture problems involve two different ratios: the ratio of quantities and the ratio of unit prices. The cost of a component equals its quantity multiplied by its unit price, so the cost ratio cannot be read directly from the quantity ratio. If quantities are in the ratio m:n and unit prices in the ratio p_1:p_2, the cost ratio is mp_1 : np_2. For example, quantities in the ratio 2:3 with unit prices 5:4 give costs 10:12 = 5:6.

The share of a component in the total cost is its own cost divided by the total cost, not its share of the quantity. If a batch uses 3 parts of one material and 1 part of another, and the unit prices are in the ratio 1:2, the second material's cost share is \frac{2}{3 \cdot 1 + 1 \cdot 2} = \frac{2}{5} of the total.

In such problems, introduce a single unknown for the unit cost of the cheaper item and express the other unit cost through the given ratio. Then write the total cost as one equation. Solving that equation gives the unit cost, and the cost of the requested component follows by multiplying it by its quantity. A common error is to substitute the total cost directly into a single component's cost and forget to scale by the quantity ratio.

After solving, check the answer by recomputing both the total and the ratio of costs. If a candidate value gives a total or a cost ratio that differs from the given data, the setup has been misread, often because the quantity ratio was used as the cost ratio or because a ratio was reversed.

When the total cost is fixed, the cost of a single component can also be found as a fraction of the total. Write each component's cost as a multiple of a common unit and divide by the sum of those multiples. This avoids solving for the unit price explicitly and works whenever all costs are proportional to the same unknown.

Counting problems often give the number of items of each denomination in a ratio, but the money held is count times value. The ratio of counts must therefore be multiplied by the face values before finding the percentage share. If the counts of ₹5 and ₹20 coins are in the ratio 1:1, the value ratio is 5:20, so the ₹5 coins hold \frac{5}{25} = 20\% of the money, not 50%. The percentage share of a part equals the part divided by the total, multiplied by 100.

For profit problems, the cost price equals the selling price divided by (1 + r/100), where r\% is the profit. This cost price is the value to which a mixture of two items must average. When two items with unit costs c_1 and c_2 are mixed in quantities x and y to reach an average unit cost m, the relation m = \frac{c_1 x + c_2 y}{x + y} holds, and the ratio is x:y = (c_2 - m):(m - c_1). For example, unit costs of 40 and 70 that average 50 require x:y = 20:10 = 2:1, and indeed \frac{2(40) + 70}{3} = 50.

Proportional expressions and constant ratio

Two quantities are proportional when their ratio is fixed. If a is proportional to b, there is a constant k, called the constant of proportionality, such that a = kb, so a/b = k stays the same as the values change. The statement a : b = c : d means a/b = c/d, and cross-multiplication gives ad = bc. A useful companion fact is that if \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \lambda, then \frac{a+c+e}{b+d+f} = \lambda, because each numerator is \lambda times its denominator.

The standard method is to turn the proportional statement into an equation with one constant, eliminate the variables, and then ask what the requested quantity depends on. If (x+y) is proportional to (x-y), write \frac{x+y}{x-y} = k. This gives x(1-k) = -y(1+k), so \frac{x}{y} = \frac{k+1}{k-1} provided k \neq 1. The case k = 1 would force y = 0, which is excluded when the variables are non-zero. Since k is fixed, x/y is a constant that depends on neither x nor y separately. Substituting t = x/y instead gives a quadratic in t with fixed roots, which leads to the same conclusion.

When three variables are each equal to a common fraction, such as \frac{x}{A} = \frac{y}{B} = \frac{z}{C} = \lambda, we have x = \lambda A, y = \lambda B and z = \lambda C. Any combination px + qy + rz then becomes \lambda(pA + qB + rC). If the bracket pA + qB + rC is zero, the whole combination is zero for every value of \lambda. For example, if \frac{x}{a-b} = \frac{y}{b-c} = \frac{z}{c-a}, then x + y + z = \lambda \cdot 0 = 0, since (a-b) + (b-c) + (c-a) = 0. The key step is to expand the bracket term by term and look for cancellation.

A ratio is unchanged when both of its terms are multiplied by the same non-zero quantity. So if x : y = (a+1) : (a-1), write x = k(a+1) and y = k(a-1) with k \neq 0. Then x^2 - y^2 = 4ak^2 and x^2 + y^2 = 2(a^2+1)k^2. The common factor k^2 cancels, leaving (x^2 - y^2) : (x^2 + y^2) = 2a : (a^2+1). The identities (a+1)^2 - (a-1)^2 = 4a and (a+1)^2 + (a-1)^2 = 2(a^2+1) are the tools needed here.

Percentage shares of a whole form a proportional system as well. The shares sum to 100, and each condition such as 'one share plus twice another share equals 50' is a linear equation in the shares. Solve enough of these together with the total, then check the remaining conditions. When the options are pie charts, substituting each option's values into every condition is often the fastest route.

• proportional to — having a constant ratio with another quantity, so x proportional to y means x = ky. e.g., The distance covered at a fixed speed is proportional to the time taken.
• constant of proportionality — the fixed value k in a = kb; it is often the quantity asked for.
• in the ratio m : n — the two quantities equal mt and nt for some t. e.g., A sum divided in the ratio 3 : 4 gives parts 3t and 4t.
• cross-multiply — from a/b = c/d obtain ad = bc; valid when the denominators are non-zero.
• constant versus depends on — a quantity is constant if its value is the same for all admissible variable values; it depends on x only if changing x alone changes it.
• sum of the parts — the percentage shares of a whole must add up to 100, which gives one equation in every percentage question of this type.

Discount and profit from marked price

The marked price (MP) is the list price printed on an item. The selling price (SP) is the amount the buyer finally pays, and the cost price (CP) is what the seller paid. A discount is the reduction from MP, so Discount = MP - SP, and the discount percentage is measured on the marked price: \text{Discount}\% = \frac{MP - SP}{MP} \times 100. A profit is SP - CP, and the profit percentage is measured on the cost: \text{Profit}\% = \frac{SP - CP}{CP} \times 100. A loss is the reverse, and its percentage is also measured on cost. Confusing the base of the two percentages is the most common error in this topic.

Because profit is a percentage of CP, SP = CP\left(1 + \frac{p}{100}\right) for a profit of p\%, so CP = \frac{SP}{1 + p/100}. When the cost is given only through a ratio with another item, the chain of steps is: ratio gives CP, profit gives SP, MP - SP gives the discount, and dividing by MP gives the percentage. Each step uses a different base, so write each base down. For example, an item with cost 400 sold at 25% profit has SP = 500. If its MP is 800, the discount is 300, which is 37.5% of MP, while the profit is still 25% of cost.

If two items are sold at the same selling price, one at a gain of p\% and one at a loss of p\%, the net result is always a loss. The two cost prices are \frac{SP}{1 + q} and \frac{SP}{1 - q}, where q = p/100. Their sum is \frac{2}{1 - q^2} SP, which exceeds the total selling price 2SP. The net loss is therefore q^2 of the total cost, so a 10% gain and a 10% loss give a net loss of 1% of the total cost, not 2%.

When the price includes tax on the selling price at rate r, the amount paid equals SP(1 + r). To recover the cost, first divide the amount paid by (1 + r) to get SP, then divide SP by (1 + \text{profit}) to get CP. The tax is applied to the selling price only and is never applied to the cost, so the steps must be reversed in order.

• marked price (list price) — the price printed on the item, on which discount percentages are calculated. e.g., A 20% discount on a marked price of 500 is 100.
• discount on marked price — the reduction is always a percentage of MP, not of SP or CP.
• profit on cost — gain percentages are computed on CP, so SP = CP \times (1 + \text{profit}\%).
• inclusive of tax — the stated amount already contains tax; divide by (1 + r) to remove it.
• net gain or net loss — the overall result of two transactions, measured against the combined cost price.

Tiered percentage fee calculation

A tiered or slab-based fee applies different rules to different ranges of the base amount. Each slab has a threshold T. The phrase 'up to T' means the rule covers amounts F \le T. The phrase 'exceeds T' or 'the amount above T' means the rule applies only to the excess F - T, not to the whole amount. The word 'plus' means a fixed charge is added to the excess charge. A typical schedule therefore has the form O(F) = aF for F \le T and O(F) = b + c(F - T) for F > T, where b is the charge at the threshold. A percentage on the lower slab applies to the entire fee, whereas on the upper slab the percentage applies only to the excess.

The two formulas need not agree at F = T. For example, with T = 2{,}00{,}000, a = 8\% and b = 12{,}000, the lower formula gives 16,000 at F = T while the upper formula gives 12,000. So the first task is to determine which slab the actual fee falls into. A safe method is to assume the fee exceeds the threshold, solve, and check whether the solution really exceeds T. If it does not, use the lower formula instead.

Charges are applied in the order stated. Here the overhead is added to the fee, and the tax is then applied to the sum of fee and overhead. If the tax rate is r, the total payable is (1 + r)(F + O), so the tax multiplies the whole sum and not the fee alone. Multiplying the bracket out before solving avoids mistakes.

The total payable increases with the fee, so the largest permitted fee is found by setting the total equal to the limit L and solving. On the upper slab, (1 + r)\left(F + b + c(F - T)\right) = L is linear in F, giving F = \frac{L/(1+r) - b + cT}{1 + c}. Then check that this value exceeds T. If it does not, the answer must come from the lower slab, and the limit may not be binding at all. Indian grouping (1,00,000 is one lakh; 10,00,000 is ten lakh) must be read carefully when checking these numbers.

• up to — an inclusive upper limit, so the value at the threshold belongs to the lower slab.
• exceeds / in excess of — the rule applies only to the part beyond the threshold.
• overhead (service charge) — an amount added on top of the fee, distinct from tax.
• inclusive / exclusive of tax — whether the stated figure already contains tax; a tax on the total means the tax base includes every earlier addition.
• maximum payable — an upper limit on the total, solved by setting the total equal to the limit and checking which slab the solution lies in.

Two-set percentage overlap

When a group is split by two properties, the two subgroups usually overlap, so their sizes cannot simply be added. The principle of inclusion and exclusion gives the size of the union as |A \cup B| = |A| + |B| - |A \cap B|, where |A \cap B| is the number of members who have both properties. The overlap is subtracted once because it was counted in both |A| and |B|.

Members who have neither property form the complement of the union within the whole group, so |\overline{A \cup B}| = N - |A \cup B|, where N is the total. When a question asks for people who like neither of two things, this complement is the target quantity, and reporting |A| or |B| instead is a common error.

Because a percentage is always relative to the total, every value in the formula must be on the same base before it is added or subtracted. For example, if 30% belong to A, 50% belong to B and 20% belong to both, then the union is 30 + 50 - 20 = 60\% and the group in neither set is 40\%.

A percentage is relative to its own base, so a share of one set cannot be read as a count without converting it to absolute numbers. If a group is p\% of a total T_1 and another group is q\% of a total T_2, the counts are \frac{p}{100}T_1 and \frac{q}{100}T_2, and any conclusion about which count is larger must be drawn from these counts, not from the percentages alone.

When one group is a fixed multiple of another, turn the ratio into an equation and combine it with the union condition. Suppose a group of 200 has 72 students in neither set and 16 in both. The union is 128, so R + C - 16 = 128, where R and C are the two group sizes. If R = 3C, then 4C = 144, giving C = 36 and R = 108.

A relation such as one group being one-quarter of another is handled the same way: write the smaller group in terms of the larger one, then substitute into the union equation.

A subset condition supplies an extra equation. If every member of A also belongs to B, then |B| = |A \cap B| + |B \text{ only}|, which often gives the unknown overlap directly. If two sets are disjoint, the overlap term is zero and drops out of the formula.

• neither ... nor ... — not belonging to either group; e.g., 'neither tea nor coffee' denotes the complement of the union.
• at least one of — belonging to the union; e.g., 'at least one of the two courses' is counted with the inclusion–exclusion formula.

Mixture ratio from average price

A mixture sold at a single average price has a cost that is a weighted average of its ingredients. If x units of a material costing c_1 per unit are mixed with y units costing c_2 per unit, the average cost is \bar{c} = \dfrac{c_1 x + c_2 y}{x + y}, which rearranges to x : y = (c_2 - \bar{c}) : (\bar{c} - c_1). This is the alligation rule, and it is the fastest route once the target average is known.

The per-unit cost must be computed from the total cost and the quantity, so a price quoted for 5 kg is divided by 5 before it is used in the mixture. For example, ₹800 for 5 kg is ₹160 per kg.

Cost price and selling price must be separated before the ratio is computed. If a seller earns a profit of p\% on cost, then \text{SP} = \text{CP}\left(1 + \frac{p}{100}\right), so the cost per unit is \text{CP} = \dfrac{\text{SP}}{1 + p/100}. Using the selling price directly as the mixture cost gives the wrong ratio.