Ratio and Proportion

The first job in any ratio question is to write the ratio correctly. Take the two quantities in the order the question asks, bring them to the same unit, and divide both by their HCF.

Units first. 3 hours : 100 minutes = 180 : 100 = 9 : 5. Likewise, 6 hours 40 minutes : 7 hours 30 minutes = 400 minutes : 450 minutes = 8 : 9. Never cancel numbers that are in different units.

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Lowest form. 12 : 18 is the same ratio as 2 : 3, and 8 : 4 is the same as 2 : 1. When a question hands you an unsimplified ratio like 12 : 18, read it as 2 parts to 3 parts before you do anything else.

One term given, find the other. If A : B = 4 : 7 and A = 52, then 4 parts = 52, so 1 part = 13 and B = 7 × 13 = 91. You can also go directly: B = 52 × 74 = 91. Multiply the known value by required partknown part.

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Radhika makes tea with milk and water in the ratio 5 : 1, and she has 750 mL of milk. 5 parts = 750 mL, so 1 part = 150 mL, and she needs 150 mL of water. We never needed the total here.

The census uses a "per 1000" form called the sex ratio: females per 1000 males = femalesmales × 1000. It is an ordinary ratio, scaled so that the second term becomes 1000.

Many questions do not hand you the ratio. They give a relation, and you have to pull the ratio out of it.

From an equation. If 9A = 4B, then AB = 49, so A : B = 4 : 9. The coefficients cross over: the number with A goes under B, and the number with B goes under A. In general, if pA = qB, then A : B = q : p. Check it by putting A = 4 and B = 9: you get 36 = 36.

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From fractions or percentages of two quantities. If 23 of A = 45 of B, then A : B = 45 : 23 = 12 : 10 = 6 : 5. The same crossing rule works. And if A scores 75% of the maximum marks and B scores 90%, then A : B = 75 : 90 = 5 : 6.

From "more than" or "less than" a third number. One number is 10% less than N and another is 40% more than N. Take N = 100: the numbers are 90 and 140, so the ratio is 9 : 14. Taking the common base as 100 is the quickest route here.

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Value of an expression. If a : b = 4 : 3 and you need (5a + 2b) : (5a − 2b), put a = 4 and b = 3: (20 + 6) : (20 − 6) = 26 : 14 = 13 : 7. This is allowed whenever every term of the expression has the same degree (all terms in a and b, or all in a², ab and b²), because then k cancels out.

Componendo and dividendo. If ab = cd, then a + ba − b = c + dc − d, and it works backwards too. So if (x + y) : (x − y) = 8 : 3, then xy = 8 + 38 − 3 = 115. Learn it as "sum on top, difference below, on both sides".

Differences given in a ratio. If (p − q) : (q − r) : (r − s) = 2 : 3 : 5, take p − q = 2k, q − r = 3k and r − s = 5k. Fix the last variable: put s = 0 and k = 1, which gives r = 5, q = 8 and p = 10. Any expression in p, q, r, s can now be read off directly.

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Suppose A : B = 3 : 5 and B : C = 2 : 7. Can we write A : B : C = 3 : 5 : 7? No. In the first ratio B is 5 parts and in the second B is 2 parts, so the parts are of different sizes. Before joining, make the common term (B) equal in both ratios.

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Method 1: LCM of the common term. B is 5 in one ratio and 2 in the other, and LCM(5, 2) = 10. Multiply the first ratio by 2 and the second by 5:
A : B = 6 : 10 and B : C = 10 : 35
So A : B : C = 6 : 10 : 35.

Method 2: the N-method. Write the two ratios one below the other, shifted by one place, and multiply along the letter N:
A = 3 × 2 = 6, B = 5 × 2 = 10, C = 5 × 7 = 35.
A gets the product of the two antecedents (3 × 2), C gets the product of the two consequents (5 × 7), and B gets the first consequent × the second antecedent (5 × 2). For a : b and b' : c the result is ab' : bb' : bc.

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Only A : C needed? Then leave B alone. Multiply the ratios like fractions: AC = AB × BC = 35 × 27 = 635. For a longer chain A : B, B : C, C : D, AD = AB × BC × CD.

If the common term sits in the other position, rewrite that ratio first. When a : b = 3 : 7 and c : b = 2 : 5, flip the second to b : c = 5 : 2, and the chain is ready.

Radhika and Meena have books in the ratio 3 : 2, and Meena and Sita have books in the ratio 5 : 4. Meena is common, and the LCM of 2 and 5 is 10. Radhika : Meena = 15 : 10 and Meena : Sita = 10 : 8, so Radhika : Meena : Sita = 15 : 10 : 8.

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An amount of money, a length, a group of students: something is divided in a ratio and the total is given. You have to find the shares.

If ₹T is divided between A and B in the ratio a : b, the total is a + b parts. So
A's share = aa + b × T and B's share = ba + b × T.

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For three people in the ratio a : b : c, the total is a + b + c parts, and each share is (its own parts ÷ total parts) × T.

Radhika and her two cousins share ₹900 in the ratio 1 : 3 : 5. Total parts = 9, so 1 part = ₹100. Radhika gets ₹100 and the cousins get ₹300 and ₹500. Check: 100 + 300 + 500 = 900.

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Three small variations turn up again and again:

• A difference is asked instead of a share. In 3 : 5 : 7 with a total of ₹7500, 1 part = ₹500, so "highest minus lowest" = (7 − 3) parts = 4 parts = ₹2000. The individual shares are never needed.
• The ratio comes in two pieces. A : B = 3 : 5 and B : C = 2 : 7 with a sum. Join them first (Type 3) to get 6 : 10 : 35, then divide.
• The total is hidden in another form. An average of 80 for three numbers means a sum of 240. The perimeter of a triangle is the sum of its sides. The sum of two diameters is twice the sum of the radii.

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One note on business profit. When money is invested for the same period, the profit is divided in the ratio of the investments, which is plain division in a ratio, so it belongs here. Investments for different periods are handled in the Partnership chapter.

Whole numbers only. Ratios often describe things that cannot be cut into pieces: students, pens, coins. If boys : girls = 7 : 5, the boys are 7k and the girls 5k for a whole number k. So the total, 12k, is always a multiple of 12. A total of 70 is impossible, while 72 or 168 is fine, and the smallest possible total is 7 + 5 = 12 itself. Reduce the ratio first (2 : 4 is 1 : 2, so a total of 3 is possible), and use this test only for counts that must be whole. For money, ages or lengths, k need not be a whole number: ages in the ratio 7 : 2 with total 60 years can be 4623 and 1313.

When a question asks "which of these can be the total", test every option for divisibility by the sum of the ratio terms. If a range is also given, list the possible values of k. Radhika's box has red and blue pens in the ratio 4 : 5 and the total lies between 50 and 80. The total is a multiple of 9, so it can be 54, 63 or 72.

In Type 4 the total was given. Here the question gives some other piece of information, and from it you reach the total or another share. The method stays exactly the same: get the value of 1 part from whatever is given.

• One share given. A ribbon is cut in the ratio 1 : 3 : 4 and the longest piece (4 parts) is 24 cm. Then 1 part = 6 cm and the whole ribbon (8 parts) = 48 cm.
• Difference given. Two numbers are in the ratio 7 : 4 and differ by 15. The difference is 3 parts, so 1 part = 5 and the numbers are 35 and 20.
• "So many more" given. Red : black marbles = 2 : 7 and there are 35 more black ones. 5 parts = 35, so 1 part = 7 and the total (9 parts) = 63.
• The remaining part given as a fraction. If the first piece is 58 of a rope, the second piece is 38. If 38 of the rope = 15 m, the rope is 40 m long.

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Two special forms come from the number system:

• HCF given. Two numbers are in the ratio 3 : 7 (already in lowest form) and their HCF is 6. The numbers are 3 × 6 = 18 and 7 × 6 = 42. The reason: 3 and 7 have no common factor, so the HCF of 3k and 7k is exactly k. In other words, 1 part = HCF.
• Product given. If the numbers are 4k and 3k and their product is 300, then 12k² = 300, k² = 25 and k = 5.

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Radhika's school has boys and girls in the ratio 5 : 4, with 30 more boys than girls. The gap of 1 part is 30, so there are 150 boys, 120 girls and 270 students in all.

When squares or cubes are given. Sometimes the question does not give the sum or the difference of the numbers but of their squares (or cubes). The method is the same: write the numbers as ak and bk and let the power of k carry the information. Then
sum of squares = (a² + b²)k², difference of squares = (a² − b²)k², sum of cubes = (a³ + b³)k³.
Find k² (or k³), then k, then the numbers.

Radhika thinks of two numbers in the ratio 3 : 2 whose squares add up to 325. The numbers are 3k and 2k, so 9k² + 4k² = 13k² = 325, which gives k² = 25 and k = 5. The numbers are 15 and 10. Check: 225 + 100 = 325.

Sometimes the ratio itself comes as fractions, like 13 : 15 : 16. Money cannot be handed out in fractional parts like these, so first turn the ratio into whole numbers. Multiply every term by the LCM of the denominators. The LCM of 3, 5 and 6 is 30:
13 × 30 : 15 × 30 : 16 × 30 = 10 : 6 : 5.
Now divide as usual.

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The rule does not change when the numerators are different. For 13 : 34 : 56, the LCM of 3, 4 and 6 is 12, so the ratio is 4 : 9 : 10.

Conditions in words. "A gets one-third as much as B, and B gets one-third as much as C." Start from the smallest share, or from the person who appears in both conditions. Let A = 1. Then B = 3 and C = 9, so A : B : C = 1 : 3 : 9.

"A gets half of B, and B gets one-third of C." Let A = 1, so B = 2 and C = 6, and A : B : C = 1 : 2 : 6.

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Equal multiples. "2 times X's share = 3 times Y's share = 5 times Z's share." If 2X = 3Y = 5Z = k, then X = k2, Y = k3 and Z = k5. So X : Y : Z = 12 : 13 : 15, the inverse ratio. Multiplying by the LCM 30 gives 15 : 10 : 6. There is a pattern: each share is the product of the other two multipliers (3 × 5 = 15, 2 × 5 = 10, 2 × 3 = 6).

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Fractions of two amounts work the same way. If 34 of X = 58 of Y, then X : Y = 58 : 34 = 5 : 6.

Suppose Radhika and Meena share chocolates so that 2 times Radhika's share = 3 times Meena's share. Then Radhika : Meena = 3 : 2. Radhika gets more, and that makes sense: her share is the one multiplied by the smaller number.

When two ratios are equal, the four numbers are said to be in proportion. If a : b = c : d, we write a : b :: c : d and read it as "a is to b as c is to d".

In a : b :: c : d, the terms a and d are called the extremes (the outer terms) and b and c are called the means (the middle terms). One rule solves almost everything here:
Product of extremes = product of means, that is, ad = bc.

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It comes straight from ab = cd: cross multiply and you get ad = bc.

Fourth proportional. If a, b, c, x are in proportion, x is the fourth proportional to a, b and c. From ax = bc, x = bca. The fourth proportional to 6, 9 and 14 is 9 × 146 = 21.

Continued proportion and the third proportional. If a : b :: b : c, the numbers a, b, c are in continued proportion, and b² = ac. Here c is the third proportional to a and b, so c = b²a. The third proportional to 9 and 15 is 15 × 159 = 25.

Mean proportional. In the same continued proportion, b is the mean proportional between a and c, so b = √ac. The mean proportional between 3 and 27 is √81 = 9. This is the geometric mean of the two numbers.

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The numbers 3, 6, 12 are in continued proportion because 3 : 6 = 6 : 12 = 1 : 2. So 6 is the mean proportional between 3 and 12 (√36 = 6), and 12 is the third proportional to 3 and 6 (363 = 12).

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Adding or subtracting a number to make four numbers proportional. If x is subtracted from a, b, c, d and the results are in proportion, then (a − x)(d − x) = (b − x)(c − x). The x² terms cancel and a simple linear equation is left. Putting in the options is often quicker still.

Duplicate and sub-duplicate ratios in use. If the areas of two squares are in the ratio 16 : 25, their sides are in the sub-duplicate ratio √16 : √25 = 4 : 5. If two numbers are in the ratio 4 : 5, their squares are in the duplicate ratio 16 : 25, whatever their sum may be.

Two quantities are in direct proportion when they increase or decrease together in the same ratio. If 4 pens cost ₹36, then 8 pens cost ₹72: double the pens, double the cost. We write y ∝ x, which means y = kx, so yx stays constant.

For direct proportion, then, y₁x₁ = y₂x₂. This is the unitary method you have used since school: find the value for one unit, then multiply.

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Two quantities are in inverse proportion when one increases in the same ratio as the other decreases. If 6 workers finish a job in 10 days, 12 workers take 5 days. We write y ∝ 1x, which means xy stays constant: x₁y₁ = x₂y₂.

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Radhika reads 30 pages in 2 hours. At the same pace she reads 30 × 52 = 75 pages in 5 hours (direct). To finish a 120-page book in 4 days she must read 30 pages a day, and to finish it in 3 days, 40 pages a day (inverse, since days × pages per day = 120).

Joint variation. Many physics-style questions mix the two. "Illumination is directly proportional to the power of the bulb and inversely proportional to the square of the distance" means I = k × Pd². Form this once and find k from the given data, or use the ratio form directly:
I₂I₁ = P₂P₁ × d₁²d₂².

Power relations. "Time is proportional to the cube of the radius" means t = kr³. If the radius becomes 3 times, the time becomes 3³ = 27 times. This is the triplicate ratio from the foundation, now in a real situation.

Scale models. A model in the ratio 1 : 200 means every real length is 200 times the model length. A 15 cm rod in the model is 15 × 200 = 3000 cm = 30 m in reality.

Here a ratio changes because something is added, removed or moved. The basic method never changes: take the old quantities as ak and bk, apply the change, and set the result equal to the new ratio.

ak + xbk + y = cd ⇒ d(ak + x) = c(bk + y), a simple equation in k.

In three common situations you can skip the equation altogether. Each one depends on spotting what stays the same.

1. Only one quantity changes. Boys : girls = 8 : 5, then 24 girls join and the ratio becomes 8 : 7. The boys did not change and they are 8 in both ratios, so the parts are already of the same size. The girls went from 5 parts to 7 parts, so 2 parts = 24 and 1 part = 12. Boys = 8 × 12 = 96. If the unchanged quantity has different numbers in the two ratios, make them equal with the LCM first.

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2. The same number is added to both. Adding the same number to both leaves their difference unchanged, so make the differences equal. 3 : 5 becomes 5 : 8 after adding 6. The differences are 2 and 3, and their LCM is 6. Write 3 : 5 as 9 : 15 and 5 : 8 as 10 : 16. Each term has now grown by 1 part, and that 1 part is 6. The original numbers are 9 × 6 = 54 and 15 × 6 = 90.

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3. Something moves from one to the other. The total does not change, so make the totals equal. When 6 students move from B to A, 2 : 3 (total 5) becomes 3 : 4 (total 7). The LCM is 35, so write them as 14 : 21 and 15 : 20. A has grown by 1 unit, and that unit is the 6 students who moved. So A = 14 × 6 = 84 and B = 21 × 6 = 126.

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With three quantities, find the one that did not change (or whose change is known), use it to fix the size of a part, and then check the others.

Radhika has red and blue beads in the ratio 3 : 2. She removes 10 red beads and the ratio becomes 1 : 1. Blue is unchanged and is 2 parts in the first ratio, so write 1 : 1 as 2 : 2. Red went from 3 parts to 2 parts, so 1 part = 10, and she had 30 red and 20 blue beads.

This type joins the chapter with Percentage. Two or more quantities are in a ratio, and then each of them changes by some percentage. What happens to the ratio, or to the total?

New ratio. Males : females = 4 : 3. Males increase by 5% and females by 20%. Multiply each term by its own factor:
4 × 1.05 : 3 × 1.20 = 4.2 : 3.6 = 7 : 6.
Each term gets its own multiplying factor; there is no single "average percentage" for the ratio.

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Change in the total. A budget is split 6 : 3 : 1 and the parts change by +10%, −20% and +30%. Take the total as 10 parts. New total = 6 × 1.10 + 3 × 0.80 + 1 × 1.30 = 6.6 + 2.4 + 1.3 = 10.3, so the total has gone up by 3%. The same result comes from a weighted average of the changes: 6 × 10 + 3 × (−20) + 1 × 3010 = +3%.

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Pass and fail. Boys : girls = 3 : 2, and 10% of the boys and 25% of the girls fail. Passed boys = 3 × 90% = 2.7 and passed girls = 2 × 75% = 1.5, so the ratio is 2.7 : 1.5 = 9 : 5.

Interchanged orders. Someone orders premium and standard items, the price ratio is known, and the two quantities get swapped by mistake, which changes the bill by some percentage. Let the quantities be p (premium) and s (standard), and the prices r : 1. Then
original bill = rp + s and swapped bill = rs + p.
Put "swapped bill = (1 ± change) × original bill" and solve. One useful fact: the ratio p : s does not depend on the actual quantity given, so you can test the options straight away.

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Radhika spends her pocket money on books and snacks in the ratio 3 : 1. If spending on books goes up by 20% and on snacks goes down by 20%, the new ratio is 3 × 1.2 : 1 × 0.8 = 3.6 : 0.8 = 9 : 2. Her total spending moves from 4 to 4.4, an increase of 10%.

Everything in this type stands on one relation: Income = Expenditure + Savings. You used it in Percentage; now the amounts come as ratios.

One person, income : expenditure given. If income : expenditure = 9 : 7, then savings = 9 − 7 = 2 parts, so savings are 29 of income. If the income is ₹4500, then 9 parts = 4500, 1 part = 500 and savings = ₹1000.

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Two people, incomes in a : b and expenditures in c : d, with savings given. Let the incomes be ax and bx and the expenditures cy and dy. Then
ax − cy = S₁ and bx − dy = S₂.
That is two equations in two unknowns. When both save the same amount S, the equations give ax − cy = bx − dy, and the working becomes very short.

Radhika and Meena earn in the ratio 3 : 2 and spend in the ratio 5 : 3, and each saves ₹2000. Incomes 3x and 2x, expenses 5y and 3y. 3x − 5y = 2x − 3y ⇒ x = 2y. Then 3x − 5y = 6y − 5y = y = 2000, so Radhika earns 3x = 6y = ₹12,000.

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Expenses split into several heads. If spending on four heads A, B, C, D is in the ratio 2 : 3 : 4 : 6, read every extra condition as a statement about parts. "C exceeds A by ₹10,000" means (4 − 2) parts = 10,000.

Fractions of income. If someone spends 12 of the income on essentials and 13 on rent, the fraction saved is 1 − 12 − 13 = 16.

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Mixed periods. Before comparing, bring every amount to the same period: ₹2,40,000 per annum = ₹20,000 per month.

Age questions are the most natural place for the "difference stays the same" idea. The age difference between two people never changes. If Radhika is 4 years older than her brother today, she was 4 years older ten years ago, and she will still be 4 years older thirty years from now.

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Basic method. Let the present ages be ak and bk. After n years they become ak + n and bk + n. Put ak + nbk + n = cd and solve for k. For n years ago, use ak − n and bk − n.

Quicker method: equalise the differences. A : B = 5 : 6 now and 6 : 7 after 4 years. The differences are 1 and 1, already equal. So A goes from 5 parts to 6 parts in 4 years, which makes 1 part = 4 years and A = 5 × 4 = 20 years.

If the differences are not equal, make them equal with the LCM. Son : mother = 1 : 5 now (difference 4) and 1 : 3 after 8 years (difference 2). Double the second ratio to get 2 : 6. Now the son goes from 1 part to 2 parts in 8 years, so 1 part = 8 years. Son = 1 × 8 = 8 and mother = 5 × 8 = 40.

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Radhika's age and her father's age are in the ratio 1 : 4 today, and after 10 years the ratio will be 1 : 2. The differences are 3 and 1, so write the ratios as 1 : 4 and 3 : 6. Radhika goes from 1 part to 3 parts in 10 years, so 2 parts = 10 years and 1 part = 5 years. Radhika is 5 and her father is 20. Check: after 10 years, 15 : 30 = 1 : 2.

Once you know the present ages, any past or future ratio is one addition or subtraction away.

In coin questions, keep two things apart in your mind: the number of coins and the value of the coins. Value = Number × Denomination.

If ₹10, ₹5 and ₹1 coins are in the ratio 2 : 3 : 5 by number, their values are in the ratio
2 × 10 : 3 × 5 : 5 × 1 = 20 : 15 : 5.
So if the total value is ₹480, 40 value-parts = 480 and 1 value-part = ₹12. The number multiplier is therefore 12: there are 24 coins of ₹10, 36 of ₹5 and 60 of ₹1.

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In general, if the numbers are ak, bk and ck, then total value = k × (a × d₁ + b × d₂ + c × d₃), where d₁, d₂, d₃ are the denominations. So k = Total valuea d₁ + b d₂ + c d₃.

When the value ratio is given instead. Divide each value-part by its denomination to get the number ratio. If the values of ₹1, 50 paise and 25 paise coins are in the ratio 1 : 1 : 1, the numbers are in the ratio 11 : 10.5 : 10.25 = 1 : 2 : 4. Write 50 paise as ₹0.50 and 25 paise as ₹0.25 before dividing.

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Radhika's piggy bank has ₹5 and ₹2 coins in the ratio 3 : 4 by number, and the total is ₹460. One set is worth 3 × 5 + 4 × 2 = ₹23, so k = 46023 = 20. She has 60 coins of ₹5 and 80 coins of ₹2. Check: 300 + 160 = 460.

Ratios cannot be added; quantities can. Two vessels hold milk and water in the ratios 3 : 2 and 4 : 1. If you pour them together, the new ratio is not 7 : 3. The ratio terms of different mixtures belong to different totals, so they cannot be added. Go through the actual quantities instead:

• Find the milk fraction of each mixture. A ratio of milk : water = a : b means aa + b of the mixture is milk.
• Find the quantity of milk (and of water) taken from each vessel.
• Add the milk quantities together and the water quantities together.
• Write the new ratio.

Radhika pours 35 litres from vessel X, where milk : water = 4 : 3, into a bucket. Then she pours in 45 litres from vessel Y, where milk : water = 7 : 2. From X: milk = 20 L, water = 15 L. From Y: milk = 35 L, water = 10 L. Total milk = 55 L and total water = 25 L, so the new ratio is 55 : 25 = 11 : 5.

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Equal quantities. If equal quantities are taken from the two vessels, the milk fraction of the mixture is the plain average of the two milk fractions. Take equal quantities of mixtures with milk : water = 5 : 3 and 1 : 3. The milk fractions are 58 and 14 = 28, and their average is 716. So the milk : water ratio is 7 : 9.

Finding how much to add. When the question asks "how much of the second mixture must be added to get the ratio …", take that quantity as x. Write the milk and water of the final mixture in terms of x and equate them as the new ratio demands. (When the question asks in what ratio two mixtures must be taken to get a given strength, that is alligation; you will meet it in the Mixture chapter.)

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The same working serves for any two groups that each carry their own ratio: two classes with boys and girls, two teams with men and women, two alloys with copper and zinc.

A hike question uses the word "ratio" in two different ways, so read it slowly.

"Changed in the ratio a : b". A quantity that is increased or decreased in the ratio a : b goes from a to b. So old : new = a : b and new = old × ba. Radhika's pocket money is ₹800 and it is increased in the ratio 4 : 5. New = 800 × 54 = ₹1,000. If the same amount is decreased in the ratio 5 : 4, new = 800 × 45 = ₹640.

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The percentage change comes straight from the ratio: change = b − aa × 100%. From 4 : 5 it is 14 = 25% up, and from 5 : 4 it is 15 = 20% down. The change is (b − a) parts, and the old value is a parts. So if the hike amount is given, divide it by (b − a) to get one part.

Two changes one after the other. Multiply the factors. A price increased in the ratio 3 : 4 and then decreased in the ratio 8 : 7 changes by 43 × 78 = 76, a net increase of 1623%.

Ratios across the years. Year-to-year ratios chain exactly like the combined ratios of Type 3. Suppose a company's profit is in the ratio 5 : 6 for 2022 : 2023 and 3 : 4 for 2023 : 2024. Make the common year equal: 3 : 4 becomes 6 : 8. So 2022 : 2023 : 2024 = 5 : 6 : 8. If the 2024 profit is ₹40 lakh, 8 parts = 40, one part = 5, and the 2022 profit is 5 × 5 = ₹25 lakh. The same chaining works for salaries, savings and investments over the years.

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