Applications of Ratio: Income, Coins, Ages and Changing Ratios

Ratios as Units: How Many of Each

A ratio does not tell you how many there are. It tells you how a total is cut into equal parts, which we will call units. If boys : girls = 7 : 5 in a class, the class is 12 units: 7 units of boys and 5 units of girls. Almost every question in this section comes down to one job: find the value of one unit, then multiply.

The steps:

1. Write each quantity as its ratio term times k (7k boys, 5k girls).
2. Use the one fact the question gives (the total, a difference, one of the parts, a value in rupees) to find k.
3. Multiply k by the term the question asks about.

From a total. "A school has 1,350 students in Classes 9, 10 and 11 in the ratio 4 : 5 : 6." That is 15 units, so 1 unit = 1350 ÷ 15 = 90. Class 9 has 360, Class 10 has 450 and Class 11 has 540.

From a difference. "Red, green and blue pens in a box are in the ratio 3 : 5 : 8, and there are 27 more blue pens than green ones." Blue − green = 3 units = 27, so 1 unit = 9. There are 27 red, 45 green and 72 blue pens, 144 in all.

From one part. "Marks in Science, Maths and English are in the ratio 6 : 7 : 5, and the Maths score is 84." 7 units = 84, so 1 unit = 12. Science = 72, English = 60 and the total is 216.

Prices and weights of items. "The prices of a pen, a notebook and a bag are in the ratio 2 : 5 : 18, and a notebook costs ₹45 more than a pen." 3 units = ₹45, so 1 unit = ₹15 and the bag costs 18 × 15 = ₹270. If the question then asks for the cost of 3 pens, 2 notebooks and 1 bag, count units first: 3 × 2 + 2 × 5 + 1 × 18 = 34 units = ₹510. Weights of parcels or bags of grain work the same way.

Turning words into a ratio. SSC often hides the ratio in a sentence:

Part of the whole. With boys : girls = 7 : 5, boys are $\frac{7}{12}$ of the class and girls are $\frac{5}{12}$. A class of 96 with this ratio has 56 boys and 40 girls.

After you find one unit, read the question again. It often asks for something you have not worked out yet: the total, the third part or the gap between two parts.

Income, Expenditure and Savings with Ratios

Every income question rests on one line, and it holds for units just as well as for rupees:

$$\text{Income} = \text{Expenditure} + \text{Savings}$$

One person. "The ratio of Arun's income to his expenditure is 9 : 7, and he saves ₹6,400 a month." Savings are 9 − 7 = 2 units, so 1 unit = ₹3,200. His income is ₹28,800 and his expenditure ₹22,400. He saves $\frac{2}{9}$ of what he earns.

If the question gives income : savings = 5 : 1 instead, expenditure is 4 units and income : expenditure = 5 : 4. Any two of the three ratios give the third.

Two people who save the same amount. "The incomes of A and B are in the ratio 5 : 3 and their expenditures in the ratio 9 : 4. Each saves ₹2,800 a month. Find A's income."

The long way is two equations. With incomes 5x and 3x and expenditures 9y and 4y:

$$5x - 9y = 2800, \quad 3x - 4y = 2800$$

The quick way uses one fact: if both save the same amount, the gap between their incomes equals the gap between their expenditures. So make the two gaps equal.

• Incomes 5 : 3 have a gap of 2. Expenditures 9 : 4 have a gap of 5.
• Make both gaps 10: incomes 25 : 15 and expenditures 18 : 8.
• Now the units match. A saves 25 − 18 = 7 units and B saves 15 − 8 = 7 units.

7 units = ₹2,800, so 1 unit = ₹400. A's income = 25 × 400 = ₹10,000, B's income = ₹6,000, and they spend ₹7,200 and ₹3,200.

If you like formulas: with incomes a : b, expenditures c : d and equal savings S,

$$\text{A's income} = \frac{aS(d - c)}{ad - bc}$$

Here that is $\frac{5 \times 2800 \times (4 - 9)}{20 - 27} = \frac{-70000}{-7} = 10000$. The gap method is faster and harder to get wrong.

Two people who spend the same amount. Now the gap between incomes equals the gap between savings. "P and Q earn in the ratio 8 : 7 and spend the same amount. Their savings are in the ratio 3 : 2, and Q saves ₹6,000." Both gaps are 1, so the units already match: incomes 8k and 7k, savings 3k and 2k, and each spends 5k. Q's 2k = ₹6,000 gives k = ₹3,000. P earns ₹24,000, Q earns ₹21,000 and each spends ₹15,000.

[SSC Tip] Checking the options works well here: an income is right only if it gives the stated savings for both people. Also see whose income is asked. A's and B's figures, and their expenditures, are usually all among the options.

Unequal Savings, Savings Ratios and Pay Rises

Unequal savings in rupees. When the two people save different amounts, the gap trick does not apply. Write two equations and solve them.

"The incomes of A and B are in the ratio 7 : 4 and their expenditures in the ratio 5 : 2. A saves ₹5,000 and B saves ₹8,000. Find their incomes."

$$7x - 5y = 5000, \quad 4x - 2y = 8000$$

The second gives y = 2x − 4000. Putting it into the first: 7x − 10x + 20000 = 5000, so x = 5000 and y = 6000. The incomes are ₹35,000 and ₹20,000, and the expenditures ₹30,000 and ₹12,000. Check B: 20,000 − 12,000 = ₹8,000.

[Tier 2] Incomes, expenditures and savings all in ratios. "Incomes are in the ratio 3 : 5, expenditures 5 : 8 and savings 1 : 2. A saves ₹4,500. Find B's income."

$$\frac{3x - 5y}{5x - 8y} = \frac{1}{2}$$

Cross-multiplying: 6x − 10y = 5x − 8y, so x = 2y. With y = k: incomes 6k and 10k, expenditures 5k and 8k, savings k and 2k. A's savings k = ₹4,500, so B's income = 10 × 4500 = ₹45,000.

A raise of the same amount for both. The gap between two salaries does not change when each gets the same raise. So this is the units method again: make the gaps equal.

"The salaries of P and Q are in the ratio 3 : 5. Each gets a raise of ₹5,000 and the ratio becomes 2 : 3."

• Old gap: 5 − 3 = 2. New gap: 3 − 2 = 1. Write the new ratio as 4 : 6 so both gaps are 2.
• Old 3 : 5, new 4 : 6. Each salary went up 1 unit, and that is ₹5,000.
• The salaries were ₹15,000 and ₹25,000 and are now ₹20,000 and ₹30,000.

Income and expenditure both rise by rupee amounts. "A man's income and expenditure are in the ratio 4 : 3. His income goes up by ₹6,000 and his expenditure by ₹3,000, and the ratio becomes 7 : 5." No part is fixed, so use x:

$$\frac{4x + 6000}{3x + 3000} = \frac{7}{5}$$

This gives 20x + 30000 = 21x + 21000, so x = 9000. He earned ₹36,000 and spent ₹27,000; now he earns ₹42,000, spends ₹30,000 and saves ₹12,000 instead of ₹9,000.

Raises by different percentages. Multiply each term by its own multiplier. Salaries in the ratio 4 : 5 with raises of 25% and 20% become 4 × 1.25 : 5 × 1.2 = 5 : 6. There is more on this under Ratio with Percentage Changes below.

A raise of the same rupee amount keeps the gap; a raise of the same percentage keeps the ratio. If both salaries go up 10%, the ratio does not change at all, and an option claiming it does is a trap.

Coins and Notes: Number Ratio and Value Ratio

Coin questions have two different ratios, and mixing them up is the main trap.

• Number ratio: how many coins of each kind there are.
• Value ratio: how many rupees each kind is worth.

They are linked by one rule: value = number × face value. To go from number to value, multiply each term by its coin's value in rupees. To go from value to number, divide. Work in rupees throughout: 50 paise = ₹0.50, 25 paise = ₹0.25 and 10 paise = ₹0.10.

Number ratio and total value given. "A bag has ₹1, 50 paise and 25 paise coins in the ratio 3 : 4 : 8, and their total value is ₹210. How many of each are there?"

The value is 7 units = ₹210, so 1 unit = ₹30. The ₹1 coins are worth ₹90, so there are 90 of them. The 50p coins are worth ₹60, so there are 120. The 25p coins are worth ₹60, so there are 240. Check: 90 : 120 : 240 = 3 : 4 : 8.

Quicker: with 3k, 4k and 8k coins, the value is 3k + 2k + 2k = 7k = 210, so k = 30.

Value ratio given, number asked. "The values of ₹2, ₹1 and 50p coins in a box are in the ratio 6 : 5 : 4, and there are 240 coins in all." Divide each value term by its face value: $\frac{6}{2} : \frac{5}{1} : \frac{4}{0.5}$ = 3 : 5 : 8. That is 16 units = 240 coins, so 1 unit = 15. There are 45 coins of ₹2, 75 of ₹1 and 120 of 50p, worth ₹90, ₹75 and ₹60, a total of ₹225.

Four kinds of coin. ₹1, 50p, 25p and 10p coins in the number ratio 2 : 3 : 4 : 5, total value ₹150. The value ratio is 2 : 1.5 : 1 : 0.5 = 4 : 3 : 2 : 1, which is 10 units, so 1 unit = ₹15. The values are ₹60, ₹45, ₹30 and ₹15, so the counts are 60, 90, 120 and 150.

Notes. Same method. ₹10, ₹20 and ₹50 notes in the number ratio 6 : 3 : 2 have values 60 : 60 : 100 = 3 : 3 : 5. If the total is ₹4,400, then 11 units = ₹4,400 and 1 unit = ₹400. The ₹10 notes and the ₹20 notes are each worth 3 × 400 = ₹1,200, and the ₹50 notes are worth 5 × 400 = ₹2,000. So there are 1200 ÷ 10 = 120 notes of ₹10, 1200 ÷ 20 = 60 of ₹20 and 2000 ÷ 50 = 40 of ₹50.

Two kinds and no ratio. "A box has 60 coins of 50p and ₹2, worth ₹75 in all." Pretend all 60 are 50p coins: that is only ₹30. Each ₹2 coin adds ₹1.50 more, and ₹45 is missing, so there are 45 ÷ 1.5 = 30 coins of ₹2 and 30 of 50p.

[SSC Tip] The usual wrong option comes from sharing the total value straight in the number ratio. For the first bag, 210 × $\frac{8}{15}$ = 112 looks like a count of 25p coins, but it is not. Turn the number ratio into a value ratio first.

Changing Ratios: When Only One Part Changes

This type came up often in CGL 2025: students join a class, people leave a club, balls go into a bag, water is poured into milk. The ratio changes and you have to find the numbers.

The units method. When only one part changes, the other part is fixed. Make the fixed part's term the same in both ratios. Then the change in the other term, counted in units, is the number added or removed.

"Boys : girls in a class = 7 : 4. After 30 more girls join, the ratio is 7 : 5." Boys are fixed and are already 7 units in both. Girls went from 4 to 5 units, so 1 unit = 30. There were 210 boys and 120 girls; now there are 150 girls.

When the fixed term is different. "Boys : girls = 5 : 3. After 36 girls join, it becomes 10 : 9." Boys are fixed, so make the boys' term 10 in both: 5 : 3 = 10 : 6. Girls went from 6 to 9 units, so 3 units = 36 and 1 unit = 12. Boys = 120, girls = 72 before and 108 after. Check: 120 : 108 = 10 : 9.

People leaving. "Men : women in an office = 9 : 5. After 18 men leave, it becomes 3 : 2." Women are fixed. Make their term 10: 9 : 5 = 18 : 10 and 3 : 2 = 15 : 10. Men fell by 3 units = 18, so 1 unit = 6. There were 108 men and 60 women; now there are 90 men.

Balls in a bag. "Red : blue = 11 : 8. After 30 red balls are taken out, red : blue = 7 : 6." Blue is fixed, so make it 24 in both: 11 : 8 = 33 : 24 and 7 : 6 = 28 : 24. Red fell by 5 units = 30, so 1 unit = 6. There were 198 red and 144 blue balls.

Milk and water, one part added. "60 litres of a mixture has milk and water in the ratio 7 : 3. How much water must be added to make the ratio 3 : 2?" Milk is fixed. Make its term 21: 7 : 3 = 21 : 9 and 3 : 2 = 21 : 14. Water goes up by 5 units. The mixture is 30 units = 60 L, so 1 unit = 2 L and you add 10 L of water. Check: milk 42 L, water 18 + 10 = 28 L, and 42 : 28 = 3 : 2.

The same idea finds the milk to add. To turn the same 60 L into 4 : 1, the water (18 L) is fixed, so the milk must be 4 × 18 = 72 L, and you add 30 L of milk.

Taking out some of the mixture removes milk and water in the same ratio, so the ratio does not change. Replacing what was taken out with water needs a different method (see Chapter 6).

The equation way. Each of these also works with x. For the first question, $\frac{7x}{4x + 30} = \frac{7}{5}$ gives 35x = 28x + 210, so x = 30. Use it if the units feel slow, but the units method is usually quicker.

Find what did not change first. A common slip is to add the 30 new girls to the ratio term (4 + 30) instead of to the number of girls.

Changing Ratios: When Both Parts Change

The same number added to both. When the same number of boys and girls join, the gap between them stays the same. So make the gaps equal, just as with equal pay rises.

"Boys : girls = 9 : 5. The same number of boys and girls join, and the ratio becomes 5 : 3. There were 270 boys at first. How many boys joined?"

• The gap in 9 : 5 is 4. The gap in 5 : 3 is 2. Write the new ratio as 10 : 6 so both gaps are 4.
• From 9 : 5 to 10 : 6, each side went up 1 unit.
• Boys were 9 units = 270, so 1 unit = 30. 30 boys (and 30 girls) joined. Check: 300 : 180 = 5 : 3.

This is the people version of "what number added to both terms gives a new ratio" from Ratio Basics and Operations.

The total stays the same. If 12 boys leave and 12 girls join, the class size does not change. Then make the totals of the two ratios equal and compare the terms. You will see this in the solved questions.

Different numbers added or removed. Nothing is fixed, so write one equation in x.

"Red : green balls = 5 : 3. If 10 red balls are taken out and 6 green balls are put in, the ratio becomes 5 : 4."

$$\frac{5x - 10}{3x + 6} = \frac{5}{4}$$

Cross-multiplying: 20x − 40 = 15x + 30, so x = 14. There were 70 red and 42 green balls; now there are 60 red and 48 green, and 60 : 48 = 5 : 4.

[Tier 2] Two conditions and no starting ratio. "If 8 boys leave a class, boys : girls becomes 2 : 3. If instead 12 girls leave, the numbers of boys and girls become equal." The second line gives B = G − 12. Putting this into the first: G − 12 − 8 = $\frac{2}{3}$G, so $\frac{1}{3}$G = 20 and G = 60. There are 60 girls and 48 boys. Check: 40 : 60 = 2 : 3, and 48 = 60 − 12.

A ratio that changes twice. "Red : blue = 2 : 3. After 10 red balls are added, the numbers are equal. Then 6 blue balls are taken out. What is the ratio now?" The first change: blue is fixed, and red goes from 2 to 3 units, so 1 unit = 10. That means 20 red and 30 blue at first, and 30 of each after the first change. After the second change: 30 red and 24 blue, which is 5 : 4. Once you know the real numbers, follow each step with numbers, not ratios.

[SSC Tip] Spot the case first. One part changes: the other part is fixed. The same number is added to both: the gap is fixed. The same number moves from one side to the other: the total is fixed. Anything else: write an equation, or try the options, since the starting numbers must fit the first ratio.

Ratio with Percentage Changes

The rule is short: multiply each term by its own multiplier, then simplify.

Seats in subjects. "Seats for Physics, Chemistry and Maths in a college are in the ratio 5 : 7 : 8. They are increased by 40%, 50% and 75%. Find the new ratio."

5 × 1.4 : 7 × 1.5 : 8 × 1.75 = 7 : 10.5 : 14 = 14 : 21 : 28 = 2 : 3 : 4.

If the college had 1,000 seats before, the old ratio was 20 units, so 1 unit = 50 and the new total is 50 × (7 + 10.5 + 14) = 1,575 seats.

Population growing at different rates. "Men : women in a town = 6 : 5. Men increase by 5% and women by 8%." The new ratio is 6 × 1.05 : 5 × 1.08 = 6.3 : 5.4 = 7 : 6.

The total rises by a weighted average, not by (5 + 8) ÷ 2 = 6.5%:

$$\frac{6 \times 5 + 5 \times 8}{11} = \frac{70}{11} = 6\frac{4}{11}\%$$

Going the other way: "After these changes the town has 23,400 people. How many women were there before?" The new population is 6.3 + 5.4 = 11.7 units, so 1 unit = 23,400 ÷ 11.7 = 2,000. There were 10,000 women (and 12,000 men).

Decreases work the same way. "Male : female staff = 5 : 3. Then 20% of the men and 50% of the women leave." The new ratio is 5 × 0.8 : 3 × 0.5 = 4 : 1.5 = 8 : 3.

Finding the old ratio. Divide by the multipliers. "After raises of 20% and 25%, two salaries are in the ratio 9 : 10. What was the ratio before?" $\frac{9}{1.2} : \frac{10}{1.25}$ = 7.5 : 8 = 15 : 16.

Finding the percentage needed. "Boys : girls = 5 : 6. Boys increase by 20%. By what percentage must girls increase for the ratio to become 4 : 5?" Boys become 6 units, so girls must become $6 \times \frac{5}{4}$ = 7.5 units. A rise from 6 to 7.5 is 25%.

A percentage for one, rupees for the other. "The prices of two articles are in the ratio 4 : 5. The first goes up 25% and the second goes up by ₹500, and the new ratio is 10 : 11." Take the prices as 4p and 5p. The first becomes 5p, so

$$\frac{5p}{5p + 500} = \frac{10}{11}$$

which gives 55p = 50p + 5000 and p = 1000. The prices were ₹4,000 and ₹5,000.

Never add the percentages to the ratio terms (5 + 40 : 7 + 50 means nothing), and never use one average percentage for both groups.

Ages in Ratio: Two People

Age questions are friendly once you hold on to one fact: everyone gets older by the same number of years, so the difference between two ages never changes. That makes every age-ratio question a units question with the gap fixed.

Present and future. "The present ages of A and B are in the ratio 5 : 7. After 6 years the ratio will be 3 : 4. Find their present ages."

• The gap in 5 : 7 is 2. The gap in 3 : 4 is 1. Write the future ratio as 6 : 8 so both gaps are 2.
• From 5 : 7 to 6 : 8, each age goes up 1 unit, and that is 6 years.
• Present ages: 30 and 42. In 6 years: 36 and 48, which is 3 : 4.

Past and present. "8 years ago the ratio of their ages was 3 : 5, and now it is 5 : 7." Both gaps are 2, so the units already match. From 3 : 5 to 5 : 7 is 2 units = 8 years, so 1 unit = 4. Present ages: 20 and 28.

Past and future. "6 years ago the ratio was 2 : 3; 9 years from now it will be 7 : 9." The gaps are 1 and 2, so write the past ratio as 4 : 6. From 4 : 6 to 7 : 9 is 3 units, and the time between is 6 + 9 = 15 years. So 1 unit = 5 years, the ages 6 years ago were 20 and 30, and the present ages are 26 and 36.

The equation way. For the first question, $\frac{5x + 6}{7x + 6} = \frac{3}{4}$ gives 20x + 24 = 21x + 18, so x = 6. Same answer, one more step.

When will the ratio become …? "Present ages are in the ratio 3 : 5 and add up to 48." They are 18 and 30, with a gap of 12. In the ratio 2 : 3 the gap is 1 unit, so 1 unit = 12 and the ages must be 24 and 36. That happens after 6 years.

Sum given at another time. "The sum of A's and B's present ages is 60. After 5 years their ages will be in the ratio 4 : 3." In 5 years the sum is 60 + 10 = 70, so the ages will be 40 and 30, and they are now 35 and 25. Each person adds 5 years, so the sum adds 10; adding only 5 is a common slip.

Product given. "Two ages are in the ratio 4 : 5 and their product is 720." Then $4k \times 5k = 20k^2 = 720$, so $k^2 = 36$ and k = 6. The ages are 24 and 30.

[SSC Tip] The years go onto each person's age, not onto the ratio terms. Writing (5 + 6) : (7 + 6) for the ratio after 6 years is the classic wrong move.

Ages: Three People, Birth and Mixed Times

Three people, sum given. "The present ages of A, B and C are in the ratio 2 : 3 : 4 and add up to 72." 9 units = 72, so 1 unit = 8 and the ages are 16, 24 and 32. In 8 years they will be 24, 32 and 40, which is 3 : 4 : 5.

Three people, sum at another time. "Present ages are in the ratio 3 : 4 : 5. Six years ago the three ages added up to 54." Each of the three was 6 years younger, so the present sum is 54 + 3 × 6 = 72. Then 12 units = 72, and the ages are 18, 24 and 30. Adding just 6 instead of 18 is the trap.

Three people from two ratios. "In age, A : B = 2 : 3 and B : C = 4 : 5, and the three ages add up to 105." Make B's term the same: A : B : C = 8 : 12 : 15 (joining ratios is in Ratio Basics and Operations). 35 units = 105, so 1 unit = 3 and the ages are 24, 36 and 45.

At the time of birth. "When the son was born his father was 30. Now their ages are in the ratio 8 : 3." The father's age at the son's birth is simply their age gap. The gap of 30 years is 5 units, so 1 unit = 6. The father is 48 and the son is 18.

"Times as old" is a ratio too. "Ten years ago A was half as old as B. Now their ages are in the ratio 3 : 4." Ten years ago the ratio was 1 : 2 (gap 1); now it is 3 : 4 (gap 1). From 1 : 2 to 3 : 4 is 2 units = 10 years, so 1 unit = 5. They are now 15 and 20.

[Tier 2] Mixed times. Sometimes the two ages in one ratio belong to different times. "The ratio of A's age 5 years ago to B's age 5 years from now is 3 : 2. The ratio of A's age 5 years from now to B's age 5 years ago is 4 : 1." The gap trick does not work directly, so write two equations:

$$2(A - 5) = 3(B + 5), \quad A + 5 = 4(B - 5)$$

The second gives A = 4B − 25. Putting it into the first: 8B − 60 = 3B + 15, so B = 15 and A = 35. Check: 30 : 20 = 3 : 2 and 40 : 10 = 4 : 1.

Trying the options. For age questions this is often the fastest check of all. Take an option, find both ages, move both by the same years and see whether both ratios come out right.

For three people, every "n years ago" or "after n years" changes the sum by 3n, not n. For a family of four it changes by 4n.

Ratio Within Groups: Passing, Failing and Totals from Parts

Many questions give a ratio for a whole group and then a fraction or percentage for each part: boys who pass, women who are graduates, students who learn music.

Take each part in units, then take its fraction. "Boys : girls = 4 : 5. $\frac{3}{4}$ of the boys and $\frac{3}{5}$ of the girls pass." Boys are 4 units and $\frac{3}{4}$ of 4 = 3 units pass. Girls are 5 units and $\frac{3}{5}$ of 5 = 3 units pass. So 6 of the 9 units pass, which is $\frac{2}{3}$ of the class, or $66\frac{2}{3}\%$. Passed boys : passed girls = 3 : 3 = 1 : 1.

A total from one part. "In a class, boys : girls = 7 : 5. 36 girls passed, and that is $\frac{3}{5}$ of the girls. How many students are in the class?" Girls = 36 ÷ $\frac{3}{5}$ = 60 = 5 units, so 1 unit = 12. Boys = 84, and the class has 144 students.

With percentages. "Men : women in a village = 5 : 3. 60% of the men and 40% of the women can read." Readers among men : readers among women = 5 × 0.6 : 3 × 0.4 = 3 : 1.2 = 5 : 2. The part of the village that can read is (3 + 1.2) ÷ 8 = 52.5%.

The ratio of a sub-group. "Boys : girls = 3 : 2. $\frac{1}{3}$ of the boys and $\frac{1}{4}$ of the girls learn music." Music boys : music girls = $3 \times \frac{1}{3} : 2 \times \frac{1}{4}$ = 1 : $\frac{1}{2}$ = 2 : 1. Music learners are 1.5 of the 5 units, so $\frac{3}{10}$ of the school learns music.

Those who fail. "Boys : girls = 6 : 5. $\frac{1}{6}$ of the boys and $\frac{1}{5}$ of the girls fail, and 60 students fail in all." Failing boys are 1 unit and failing girls 1 unit, so 2 units = 60 and 1 unit = 30. The class has 11 × 30 = 330 students.

Two groups put together. "Class A has 40 students with boys : girls = 3 : 2, and class B has 48 with 5 : 3." Class A has 24 boys and 16 girls; class B has 30 boys and 18 girls. Together: 54 boys and 34 girls, so 27 : 17. Always find the real numbers and add them. Adding the ratios (3 + 5 : 2 + 3 = 8 : 5) is wrong.

Choosing good units. If the fractions have denominators 3 and 4, take the boys as a multiple of 3 and the girls as a multiple of 4 that still fits the ratio. Whole numbers keep the arithmetic clean and the answer easy to check.

[SSC Tip] The fraction applies to its own group, not to the whole class. "$\frac{3}{5}$ of the girls" with 5 units of girls is 3 units, not $\frac{3}{5}$ of all 9 units.

Solved Examples

Question 1

Ritu's marks in Hindi, English and Maths are in the ratio 6 : 7 : 9. She scored 27 more marks in Maths than in Hindi. Each paper is out of 100. What percentage of the total marks did she get?

Answer. Maths − Hindi = 9 − 6 = 3 units, and that is 27 marks, so 1 unit = 9. Her marks are 54, 63 and 81, which add up to 198. Out of 300, that is 198 ÷ 3 = 66%.

Question 2

The prices of a notebook, a pen and a geometry box are in the ratio 4 : 1 : 9. Three notebooks, five pens and one geometry box together cost ₹520. Find the price of the geometry box.

Answer. I count units first: 3 notebooks = 3 × 4 = 12 units, 5 pens = 5 units and 1 box = 9 units. That is 26 units = ₹520, so 1 unit = ₹20. The geometry box costs 9 × 20 = ₹180 (a notebook is ₹80 and a pen ₹20).

Question 3

The ratio of Mohan's monthly income to his expenditure is 11 : 8, and he saves ₹7,500 a month. (i) Find his expenditure. (ii) If his income goes up by ₹2,500 and his expenditure stays the same, what is the new ratio of income to expenditure?

Answer. (i) Savings = 11 − 8 = 3 units = ₹7,500, so 1 unit = ₹2,500. His expenditure is 8 × 2500 = ₹20,000 and his income ₹27,500. (ii) His income becomes ₹30,000 while he still spends ₹20,000, so the ratio is 30,000 : 20,000 = 3 : 2.

Question 4

The incomes of P and Q are in the ratio 4 : 3 and their expenditures are in the ratio 7 : 5. Each of them saves ₹4,000 a month. Find P's monthly income.

Answer. They save the same, so the income gap must equal the expenditure gap. The incomes 4 : 3 have a gap of 1 and the expenditures 7 : 5 a gap of 2, so I write the incomes as 8 : 6. Now P saves 8 − 7 = 1 unit and Q saves 6 − 5 = 1 unit. 1 unit = ₹4,000, so P earns 8 × 4000 = ₹32,000. (Q earns ₹24,000; they spend ₹28,000 and ₹20,000.)

Question 5

P and Q earn in the ratio 6 : 5 and spend the same amount every month. Their savings are in the ratio 2 : 1. If P saves ₹9,000, find Q's income.

Answer. Same spending means the income gap equals the savings gap. Both 6 : 5 and 2 : 1 have a gap of 1, so the units match: incomes 6k and 5k, savings 2k and k, and each spends 4k. P's 2k = ₹9,000, so k = ₹4,500 and Q's income is 5 × 4500 = ₹22,500. Check: both spend ₹18,000.

Question 6

[Tier 2] The incomes of A and B are in the ratio 5 : 6, their expenditures in the ratio 6 : 7 and their savings in the ratio 3 : 4. If B saves ₹8,000, find A's income.

Answer. Incomes 5x and 6x, expenditures 6y and 7y. The savings ratio gives

$$\frac{5x - 6y}{6x - 7y} = \frac{3}{4}$$

so 20x − 24y = 18x − 21y, which means 2x = 3y. Taking x = 3k and y = 2k: incomes 15k and 18k, expenditures 12k and 14k, savings 3k and 4k. B's 4k = ₹8,000, so k = ₹2,000 and A's income is 15 × 2000 = ₹30,000.

Question 7

The salaries of Ravi and Sohan are in the ratio 5 : 8. Each of them gets a raise of ₹6,000, and the ratio becomes 7 : 10. Find their salaries before the raise.

Answer. The same raise keeps the gap the same. The gap in 5 : 8 is 3 and the gap in 7 : 10 is also 3, so the units already match. Going from 5 : 8 to 7 : 10, each salary rises 2 units, and that is ₹6,000. So 1 unit = ₹3,000, and the salaries were ₹15,000 and ₹24,000. Check: 21,000 : 30,000 = 7 : 10.

Question 8

The salaries of X and Y go up by 10% and 20% respectively, and the new salaries are in the ratio 11 : 15. X's new salary is ₹33,000. Find Y's salary before the raise.

Answer. To go back, I divide each term by its multiplier: $\frac{11}{1.1} : \frac{15}{1.2}$ = 10 : 12.5 = 4 : 5. X's old salary was 33,000 ÷ 1.1 = ₹30,000, which is 4 units, so 1 unit = ₹7,500 and Y's old salary was 5 × 7500 = ₹37,500. Check: Y's new salary is ₹45,000, and 33,000 : 45,000 = 11 : 15.

Question 9

A box has ₹5, ₹2 and ₹1 coins in the ratio 2 : 5 : 9. The total value of the coins is ₹1,160. How many ₹2 coins are there?

Answer. Say there are 2k, 5k and 9k coins. Their values are 5 × 2k = 10k, 2 × 5k = 10k and 1 × 9k = 9k rupees. So 29k = 1,160 and k = 40. The number of ₹2 coins is 5 × 40 = 200. (There are 80 coins of ₹5 and 360 of ₹1.)

Question 10

A purse has ₹2, ₹1, 50 paise and 25 paise coins whose values are in the ratio 4 : 3 : 2 : 1. There are 260 coins in all. How many 25 paise coins are there, and what is the total amount?

Answer. To get the number ratio, I divide each value by the face value: $\frac{4}{2} : \frac{3}{1} : \frac{2}{0.5} : \frac{1}{0.25}$ = 2 : 3 : 4 : 4. That is 13 units = 260 coins, so 1 unit = 20. There are 40, 60, 80 and 80 coins of 25 paise. Their values are ₹80, ₹60, ₹40 and ₹20, so the total is ₹200.

Question 11

A shop's cash box has ₹10, ₹20 and ₹50 notes in the ratio 5 : 3 : 2. The total amount is ₹4,200. How many notes are there in all?

Answer. With 5k, 3k and 2k notes, the value is 50k + 60k + 100k = 210k rupees. So 210k = 4,200 and k = 20. There are 100, 60 and 40 notes, 200 notes in all. Check: 1,000 + 1,200 + 2,000 = ₹4,200.

Question 12

In a school, boys and girls are in the ratio 8 : 5. After 45 more girls are admitted, the ratio becomes 6 : 5. How many boys are there?

Answer. Only the girls changed, so the boys are fixed. I make the boys' term equal: 8 : 5 = 24 : 15 and 6 : 5 = 24 : 20. The girls went from 15 to 20 units, so 5 units = 45 and 1 unit = 9. Boys = 24 × 9 = 216. (Girls went from 135 to 180, and 216 : 180 = 6 : 5.)

Question 13

A bag has white and black balls in the ratio 7 : 4. If 12 white balls are taken out and 12 black balls are put in, the ratio becomes 6 : 5. How many white balls were there at first?

Answer. Twelve go out and twelve come in, so the total number of balls does not change. Both ratios add up to 11, so the units already match. White went from 7 units to 6 units, so 1 unit = 12, and there were 7 × 12 = 84 white balls (and 48 black). Check: 72 : 60 = 6 : 5.

Question 14

[Tier 2] If 15 boys leave a class, the ratio of boys to girls becomes 3 : 4. If instead 20 girls leave, the ratio becomes 9 : 4. How many students are in the class?

Answer. Let there be B boys and G girls. The first line gives B − 15 = $\frac{3}{4}$G, so B = $\frac{3}{4}$G + 15. The second gives 4B = 9(G − 20), so B = $\frac{9}{4}$G − 45. Setting them equal: $\frac{6}{4}$G = 60, so G = 40 and B = 45. The class has 85 students. Check: 30 : 40 = 3 : 4 and 45 : 20 = 9 : 4.

Question 15

A 45-litre mixture has milk and water in the ratio 4 : 1. (i) How much water must be added to make the ratio 3 : 2? (ii) If milk is added instead, how much is needed to make the ratio 6 : 1?

Answer. There are 36 L of milk and 9 L of water. (i) The milk stays at 36 L. For 3 : 2 the water must be 36 × $\frac{2}{3}$ = 24 L, so I add 15 L of water. (ii) Now the water stays at 9 L, and for 6 : 1 the milk must be 54 L, so I add 18 L of milk.

Question 16

The seats for Arts, Commerce and Science in a college are in the ratio 4 : 5 : 6. The seats are increased by 25%, 40% and 50% respectively. Find the new ratio. If there were 1,200 seats before, how many are there now?

Answer. Each term gets its own multiplier: 4 × 1.25 : 5 × 1.4 : 6 × 1.5 = 5 : 7 : 9. The old ratio adds to 15 units, so 1 unit = 1200 ÷ 15 = 80 seats. Now there are 80 × (5 + 7 + 9) = 1,680 seats.

Question 17

The ratio of men to women in a town is 9 : 7. The number of men goes up by 10% and the number of women by 20%, and the population becomes 45,750. How many women were there before the increase, and what is the new ratio?

Answer. With 9k men and 7k women, after the rise there are 9.9k men and 8.4k women, 18.3k people in all. So 18.3k = 45,750 and k = 2,500. There were 7 × 2500 = 17,500 women at first. The new ratio is 9.9 : 8.4 = 99 : 84 = 33 : 28.

Question 18

The prices of a fridge and a TV are in the ratio 5 : 4. The fridge's price goes up by 12% and the TV's price goes up by ₹2,000, and the new ratio is 7 : 6. Find the original price of the TV.

Answer. Take the prices as 5p and 4p. The fridge now costs 5p × 1.12 = 5.6p and the TV 4p + 2000. So

$$\frac{5.6p}{4p + 2000} = \frac{7}{6}$$

which gives 33.6p = 28p + 14000, so 5.6p = 14000 and p = 2500. The TV cost 4 × 2500 = ₹10,000 (and the fridge ₹12,500). Check: 14,000 : 12,000 = 7 : 6.

Question 19

The present ages of Anu and her uncle are in the ratio 3 : 5. Six years ago the ratio was 1 : 2. Find their present ages.

Answer. The age gap stays the same. The gap in 3 : 5 is 2 and the gap in 1 : 2 is 1, so I write the old ratio as 2 : 4. From 2 : 4 to 3 : 5, each age went up 1 unit, which is 6 years. So 1 unit = 6, and they are 18 and 30 now. Check: six years ago they were 12 and 24, which is 1 : 2.

Question 20

Eight years ago the ages of P and Q were in the ratio 2 : 5. Eight years from now they will be in the ratio 2 : 3. Find the sum of their present ages.

Answer. The gap in 2 : 5 is 3 and in 2 : 3 it is 1, so I write the future ratio as 6 : 9. From 2 : 5 to 6 : 9, each age goes up 4 units, and the time between is 8 + 8 = 16 years. So 1 unit = 4. Eight years ago they were 8 and 20, so now they are 16 and 28, and the sum is 44 years. Check: in 8 years they will be 24 and 36, which is 2 : 3.

Question 21

The present ages of A, B and C are in the ratio 3 : 5 : 7. Four years ago the sum of their ages was 63. What will be the ratio of their ages after 5 years?

Answer. Each of the three was 4 years younger, so the present sum is 63 + 3 × 4 = 75. That is 15 units, so 1 unit = 5 and their ages are 15, 25 and 35. After 5 years they will be 20, 30 and 40, so the ratio will be 2 : 3 : 4.

Question 22

The present ages of a father and his son are in the ratio 7 : 2, and the product of their ages is 504. After how many years will the ratio of their ages be 2 : 1?

Answer. With ages 7k and 2k, the product is $14k^2 = 504$, so $k^2 = 36$ and k = 6. The father is 42 and the son 12, a gap of 30 years. In the ratio 2 : 1 the gap is 1 unit, so the son must be 30 and the father 60. That happens after 18 years.

Question 23

In a class, boys and girls are in the ratio 3 : 2. In an exam, $\frac{3}{4}$ of the boys and $\frac{5}{8}$ of the girls passed. What percentage of the class passed, and what is the ratio of boys who passed to girls who passed?

Answer. To avoid fractions I take 24 boys and 16 girls (that is 3 : 2, and 24 works with quarters, 16 with eighths). Passed: 18 boys and 10 girls, 28 out of 40. So 70% of the class passed, and passed boys : passed girls = 18 : 10 = 9 : 5.

Question 24

In an office, men and women are in the ratio 7 : 4. $\frac{2}{7}$ of the men and $\frac{3}{4}$ of the women are graduates. There are 200 men who are not graduates. How many people work in the office?

Answer. Take the men as 7k: then 2k are graduates and 5k are not. So 5k = 200 and k = 40. There are 280 men and 4 × 40 = 160 women, so the office has 440 people. (120 of the women are graduates.)