Ratio and proportion — SSC CGL Quant Notes

Concepts and rules

A ratio a : b compares 2 quantities of the same kind in the same units and equals the fraction a/b (b ≠ 0). Multiplying or dividing both terms by the same non-zero number leaves it unchanged: 12 : 18 = 2 : 3.

Convert units first: 45 minutes : 2 hours = 45 : 120 = 3 : 8, not 45 : 2.

Comparing ratios: a : b > c : d exactly when a × d > b × c (all positive). With many ratios, convert each to a decimal.

Compound ratio of a : b and c : d is (a × c) : (b × d). Duplicate ratio of a : b is a² : b², triplicate is a³ : b³, sub-duplicate is √a : √b, and the inverse ratio is b : a.

Joining chains: A : B = 2 : 3 and B : C = 4 : 5. Make B the same (LCM of 3 and 4 is 12): A : B = 8 : 12, B : C = 12 : 15, so A : B : C = 8 : 12 : 15.

Dividing a total S in the ratio a : b : c: the shares are S × a/(a + b + c), S × b/(a + b + c) and S × c/(a + b + c).

Equal-multiple conditions: if A/2 = B/3 = C/4, then A : B : C = 2 : 3 : 4. If 2A = 3B = 4C, then A : B : C = 1/2 : 1/3 : 1/4 = 6 : 4 : 3 (multiply by the LCM 12).

Adding the same positive number to both terms moves a ratio towards 1 : 1. A ratio of a smaller to a larger number increases; a ratio of a larger to a smaller number decreases.

Coins and notes: the ratio of the number of coins is not the ratio of their values. Value ratio = count ratio × value of each coin.

Method and shortcuts

Write each quantity as a multiple of one unknown: if A : B = 3 : 5, take A = 3x and B = 5x. One more condition then fixes x.

When the same number k is added to or subtracted from both terms, set up (3x + k)/(5x + k) = new ratio and cross-multiply.

Income and expenditure: income a : b and expenditure c : d with savings given lets you write a × x − c × y = savings for each person and solve the pair.

Long chains A : B, B : C, C : D: multiply the first terms and the second terms, A : D = (a1 × b1 × c1) : (a2 × b2 × c2).

Shortcut check: in options, the right answer must keep the ratio exactly. Test a value like 3x by asking whether the matching 5x is also a whole number that fits the data.

Example 1: comparing 2 ratios

Question: Which is greater, 3 : 7 or 5 : 11?

Solution: Cross-multiply: 3 × 11 = 33 and 7 × 5 = 35. Since 33 < 35, 3/7 < 5/11.

Answer: 5 : 11

Example 2: joining ratios

Question: If A : B = 2 : 3 and B : C = 4 : 5, find A : B : C.

Solution: LCM of the 2 values of B (3 and 4) is 12. A : B = 8 : 12 and B : C = 12 : 15.

Answer: 8 : 12 : 15

Example 3: dividing in a ratio of fractions

Question: Divide ₹7,800 among Asha, Bela and Chitra in the ratio 1/2 : 1/3 : 1/4.

Solution: Multiply by the LCM 12: 6 : 4 : 3, total 13 parts. 1 part = 7,800 ÷ 13 = ₹600. Shares: 6 × 600 = ₹3,600, 4 × 600 = ₹2,400, 3 × 600 = ₹1,800.

Answer: ₹3,600, ₹2,400 and ₹1,800

Example 4: equal products

Question: If 2A = 3B = 4C, then A : B : C is: (a) 2 : 3 : 4 (b) 4 : 3 : 2 (c) 6 : 4 : 3 (d) 3 : 4 : 6

Solution: Let 2A = 3B = 4C = 12k. Then A = 6k, B = 4k, C = 3k. Check: 2 × 6 = 3 × 4 = 4 × 3 = 12.

Option (a) copies the coefficients; the quantities are in the inverse ratio of the coefficients.

Answer: (c) 6 : 4 : 3

Example 5: incomes, expenditures and equal savings

Question: The incomes of Mohan and Sohan are in the ratio 5 : 4 and their expenditures in the ratio 3 : 2. If each saves ₹1,600, Mohan's income is: (a) ₹4,000 (b) ₹3,200 (c) ₹4,800 (d) ₹5,000

Solution: Incomes 5x and 4x, expenditures 3y and 2y. Then 5x − 3y = 1,600 and 4x − 2y = 1,600.

Multiply the second equation by 3 and the first by 2: 12x − 6y = 4,800 and 10x − 6y = 3,200. Subtracting, 2x = 1,600, so x = 800 and y = 800.

Mohan's income = 5 × 800 = ₹4,000. Check: Mohan spends ₹2,400 and Sohan earns ₹3,200 and spends ₹1,600; both save ₹1,600.

Answer: (a) ₹4,000

Example 6: ratio after subtracting from both

Question: Two numbers are in the ratio 3 : 5. If 9 is subtracted from each, the ratio becomes 12 : 23. The smaller number is: (a) 27 (b) 55 (c) 30 (d) 33

Solution: (3x − 9)/(5x − 9) = 12/23. Cross-multiplying: 69x − 207 = 60x − 108, so 9x = 99 and x = 11.

The numbers are 33 and 55. Check: 24 : 46 = 12 : 23. Option (b) 55 is the larger number.

Answer: (d) 33

Common mistakes

Joining A : B = 2 : 3 and B : C = 4 : 5 as 2 : 3 : 5 without equalising B. Right: 8 : 12 : 15.

Reading 2A = 3B = 4C as A : B : C = 2 : 3 : 4. Right: take the reciprocals, giving 6 : 4 : 3.

Comparing quantities in different units, like 45 minutes and 2 hours, without converting.

Treating a coin-count ratio as a value ratio. Right: multiply each count by the coin value before adding values.

Adding the change to the ratio terms instead of the quantities: if 9 is subtracted from 3x and 5x, write 3x − 9 and 5x − 9, not 3 − 9 and 5 − 9.

Practice questions

Q1. A bag has ₹1, 50 paise and 25 paise coins in the ratio 5 : 6 : 8 by number, worth ₹210 in all. The number of 50 paise coins is: (a) 105 (b) 126 (c) 168 (d) 120

Q2. If A : B = 3 : 4, B : C = 6 : 7 and C : D = 14 : 15, then A : D is: (a) 3 : 5 (b) 9 : 14 (c) 1 : 2 (d) 4 : 5

Q3. Two numbers are in the ratio 4 : 7. If 6 is added to each, the ratio becomes 5 : 8. The sum of the numbers is: (a) 55 (b) 77 (c) 66 (d) 88

Answers

Q1. (b) Value ratio = 5 × 1 : 6 × 0.5 : 8 × 0.25 = 5 : 3 : 2, so 1 set of 19 coins is worth ₹10; 210 ÷ 10 = 21 sets; 50 paise coins = 6 × 21 = 126.

Q2. (a) A : D = (3 × 6 × 14) : (4 × 7 × 15) = 252 : 420 = 3 : 5.

Q3. (c) 8(4x + 6) = 5(7x + 6) gives 32x + 48 = 35x + 30, x = 6; numbers 24 and 42, sum 66.

Quick revision

Join ratios by equalising the common term with the LCM.

pA = qB = rC ⇒ A : B : C = 1/p : 1/q : 1/r.

Share = total × part ÷ sum of parts.

Take A = 3x, B = 5x and use the extra condition to find x.

Concepts and rules

A proportion a : b :: c : d says a/b = c/d. Then a × d = b × c (product of the extremes = product of the means).

Fourth proportional to a, b, c is d = b × c/a. Third proportional to a and b is c = b²/a. Mean proportional between a and b is √(a × b).

Direct proportion: y = k × x, so y1/x1 = y2/x2. Doubling x doubles y (cost and quantity, distance and petrol at fixed mileage).

Inverse proportion: x × y = k, so x1 × y1 = x2 × y2. Doubling x halves y (workers and days for a fixed job, speed and time for a fixed distance).

Compound proportion (chain rule): (M1 × D1 × H1)/W1 = (M2 × D2 × H2)/W2, where M = men, D = days, H = hours per day, W = work. Efficiency, if given, multiplies M.

Joint variation: if x varies directly as y and inversely as z, then x = k × y/z. Find k from the given set of values, then substitute the new values.

Componendo and dividendo: if a/b = c/d (with a ≠ b), then (a + b)/(a − b) = (c + d)/(c − d). It also works in reverse.

If a/b = c/d = e/f = k, then (a + c + e)/(b + d + f) = k as well, provided the denominator sum is not 0.

Method and shortcuts

Decide the type first: ask 'if one quantity doubles, does the other double (direct) or halve (inverse)?'

Chain rule layout: put the unknown on one side; for every other factor, multiply by the ratio that makes the answer larger or smaller as common sense says.

Food-and-garrison questions: count the food in man-days. Food left after t days = men × (total days − t); divide by the new number of men.

Number added to make 4 numbers proportional: solve (a + x)(d + x) = (b + x)(c + x); the x² terms cancel, leaving a linear equation.

Shortcut limit: componendo-dividendo fails when a = b (division by 0); check that the original ratio is not 1.

Example 1: fourth proportional

Question: Find the fourth proportional to 4, 9 and 12.

Solution: 4 : 9 :: 12 : d gives 4 × d = 9 × 12 = 108, so d = 27.

Answer: 27

Example 2: mean and third proportional

Question: Find the mean proportional between 9 and 16, and the third proportional to 9 and 12.

Solution: Mean proportional = √(9 × 16) = √144 = 12. Third proportional = 12²/9 = 144/9 = 16. Check: 9 : 12 :: 12 : 16.

Answer: 12 and 16

Example 3: inverse proportion

Question: 12 masons build a wall in 10 days. How many days will 15 masons take to build the same wall?

Solution: More masons, fewer days, so inverse: 12 × 10 = 15 × D, D = 120 ÷ 15 = 8.

Answer: 8 days

Example 4: making numbers proportional

Question: What number must be added to each of 6, 14, 18 and 38 so that they become proportional? (a) 4 (b) 2 (c) 3 (d) 1

Solution: (6 + x)(38 + x) = (14 + x)(18 + x) gives 228 + 44x + x² = 252 + 32x + x², so 12x = 24 and x = 2.

Check: 8 : 16 :: 20 : 40, since 8 × 40 = 16 × 20 = 320.

Answer: (b) 2

Example 5: chain rule with a bigger job

Question: 16 men working 7 hours a day finish a job in 15 days. In how many days will 21 men working 8 hours a day finish a job twice as large? (a) 18 (b) 24 (c) 20 (d) 16

Solution: (16 × 15 × 7)/1 = (21 × D × 8)/2, so 1,680 = 84 × D and D = 20.

Common-sense check: more men and more hours cut the days, the double job raises them; 15 × 16/21 × 7/8 × 2 = 20.

Answer: (c) 20

Example 6: joint variation

Question: x varies directly as y² and inversely as z. When y = 4 and z = 8, x = 6. Find x when y = 6 and z = 9. (a) 9 (b) 13.5 (c) 16 (d) 12

Solution: x = k × y²/z. From the data, 6 = k × 16/8 = 2k, so k = 3.

New x = 3 × 36/9 = 12. Option (b) 13.5 comes from ignoring z (6 × 36/16), a common slip.

Answer: (d) 12

Common mistakes

Using direct proportion for workers and days. Right: more workers means fewer days, so the product stays fixed.

Mixing up the third and fourth proportional. Third proportional to a, b uses a : b :: b : c; fourth proportional to a, b, c uses a : b :: c : d.

In garrison questions, dividing the full food by the new strength. Right: subtract the days already used first.

Forgetting the work term in the chain rule when the job size changes. A job twice as large doubles W2.

Squaring or square-rooting at the wrong step in variation: if x varies as y², use y² in both the k step and the new-value step.

Practice questions

Q1. If (√(a + 1) + √(a − 1))/(√(a + 1) − √(a − 1)) = 3, then a is: (a) 4/3 (b) 5/3 (c) 3/2 (d) 2

Q2. A car uses 12 litres of petrol for 150 km. At the same rate, petrol needed for 400 km is: (a) 30 litres (b) 36 litres (c) 32 litres (d) 28 litres

Q3. A fort of 1,200 soldiers has food for 30 days. After 6 days, 400 more soldiers join. The remaining food will last: (a) 20 days (b) 16 days (c) 24 days (d) 18 days

Answers

Q1. (b) Componendo-dividendo: √(a + 1)/√(a − 1) = (3 + 1)/(3 − 1) = 2, so (a + 1)/(a − 1) = 4 and 3a = 5, a = 5/3.

Q2. (c) Direct: 12 × 400/150 = 32 litres.

Q3. (d) Food left = 1,200 × 24 = 28,800 soldier-days; 28,800 ÷ 1,600 = 18 days.

Quick revision

a : b :: c : d ⇒ a × d = b × c.

Fourth = b × c/a; third = b²/a; mean = √(a × b).

Direct: ratio fixed; inverse: product fixed; chain rule: M × D × H/W fixed.

Variation: find k from the first data set, then substitute.