CSAT Percentages, Ratio and Averages CSAT

Ratio, Proportion and Averages for CSAT

CSAT ratio and averages with worked examples — combining ratios, ages, mixtures and alligation, partnership, the deviation method, replacement and weighted averages.

SyllabusCSAT · Basic numeracy (numbers and their relations, orders of magnitude, etc.) — Class X level

By 9 min read Updated Oct 2026 #csat#basic-numeracy#ratio

Prerequisites: Percentages for CSAT · Number System

The Hook: Two Grades of Rice

Connect: Everyday Puzzle → Concept

A grocer sells rice at ₹40 a kilogram and a better grade at ₹55 a kilogram. A customer wants a blend at ₹45 a kilogram. How much of each should go into the bag?

Most people start writing equations. A CSAT-trained candidate writes two subtractions — $55 - 45 = 10$ and $45 - 40 = 5$ — and answers 2 : 1 in five seconds. That shortcut is alligation, and it is one of a handful of ratio and average tools that turn CSAT word problems into mental arithmetic.

The Core Concept

1. Ratio — the language of “parts”

A ratio $a : b$ compares two quantities of the same kind: $a : b = \dfrac{a}{b}$. Think of it as parts: in $2 : 3$, the first quantity has 2 parts and the second 3 parts of the same size.

Dividing a quantity. Divide ₹7,200 among A, B and C in the ratio $2 : 3 : 4$. Total parts $= 9$; one part $= 7{,}200 \div 9 = 800$. A $= 1{,}600$, B $= 2{,}400$, C $= 3{,}200$. (Check: sum $= 7{,}200$ ✓.)

Combining two ratios. If $A : B = 3 : 4$ and $B : C = 6 : 5$, make B equal in both. $\text{LCM}(4, 6) = 12$: $A : B = 9 : 12$ and $B : C = 12 : 10$, so $A : B : C = 9 : 12 : 10$.

Comparing ratios. Which is larger, $5 : 7$ or $7 : 10$? Cross-multiply: $5 \times 10 = 50$ and $7 \times 7 = 49$. Since $50 > 49$, $5 : 7$ is larger.

2. Proportion — direct and inverse

  • $a : b = c : d$ means $ad = bc$ (product of extremes = product of means).
  • Direct proportion: if one doubles, the other doubles (cost and quantity).
  • Inverse proportion: if one doubles, the other halves (workers and days for a fixed job; speed and time for a fixed distance — see Time, Speed and Work).

3. Ages

Worked example. The ages of A and B are in the ratio $5 : 3$. After 6 years the ratio will be $7 : 5$. Find their present ages.

Let the ages be $5x$ and $3x$:

$$\frac{5x + 6}{3x + 6} = \frac{7}{5} \Rightarrow 25x + 30 = 21x + 42 \Rightarrow x = 3$$

Ages: 15 and 9. Check: after 6 years, $21 : 15 = 7 : 5$ ✓.

Option-testing shortcut: with numbers in the options, test them in the second condition — often faster than algebra.

4. Mixtures and alligation

When two ingredients of “values” $c$ (cheaper) and $d$ (dearer) are mixed to get a mean value $m$:

$$\boxed{\frac{\text{Quantity of cheaper}}{\text{Quantity of dearer}} = \frac{d - m}{m - c}}$$
graph TD
    C["Cheaper: 40"] --> M["Mean: 45"]
    D["Dearer: 55"] --> M
    M --> R1["Cheaper share = 55 - 45 = 10"]
    M --> R2["Dearer share = 45 - 40 = 5"]
    R1 --> R["Ratio 10 : 5 = 2 : 1"]
    R2 --> R

The “values” can be prices, percentages of concentration, speeds or average marks — anything averaged by weight.

Adding one component. A 60-litre mixture has milk and water in the ratio $2 : 1$ (40 L milk, 20 L water). To make it $1 : 2$, milk stays 40 L, so water must become 80 L — add 60 L of water.

5. Partnership

Profit is shared in proportion to capital × time.

Worked example. A invests ₹30,000 for 12 months and B invests ₹40,000 for 6 months. Profit ₹25,000. Ratio $= 30{,}000 \times 12 : 40{,}000 \times 6 = 3{,}60{,}000 : 2{,}40{,}000 = 3 : 2$. A gets $\dfrac{3}{5} \times 25{,}000 = ₹15{,}000$; B gets ₹10,000.

6. Averages

$$\text{Average} = \frac{\text{Sum}}{\text{Number of items}} \qquad \Longleftrightarrow \qquad \text{Sum} = \text{Average} \times \text{Number}$$

Almost every average problem is solved by converting averages to sums, doing the arithmetic on sums, and converting back.

(a) Deviation method. Marks 72, 75, 78, 81, 69. Assume 75; deviations $-3, 0, +3, +6, -6$ sum to 0, so the average is 75.

(b) Replacement. The average weight of 8 people rises by 2.5 kg when a new person replaces one who weighs 65 kg. New person $= 65 + 8 \times 2.5 = $ 85 kg (the newcomer brings the old weight plus the increase for everyone).

(c) Weighted average. Section A: 30 students averaging 60; section B: 20 students averaging 70. Combined average $= \dfrac{30 \times 60 + 20 \times 70}{50} = \dfrac{1{,}800 + 1{,}400}{50} = 64$. Note that 64 is closer to 60, because section A is larger.

(d) New entry. A batsman averages 40 runs over 10 innings. How many must he score in the 11th to raise the average to 42? Needed total $= 11 \times 42 = 462$; current total $= 400$; required score $= $ 62.

(e) Consecutive and evenly spaced numbers. The average equals the middle term, or the mean of the first and last: average of 1 to $n$ is $\dfrac{n + 1}{2}$.

Memory Anchors

  • “Parts first, values later” — find the value of one part, then multiply.
  • “Cross the X” for alligation — write the two values on top, the mean in the middle, and subtract diagonally.
  • “Average out, sum in” — convert every average to a sum before doing anything.
  • “Newcomer = old + n × change” for replacement problems (use minus if the average falls).
  • Partnership = money × months.
Confusable pairRule
Ratio vs fraction of totalIn $2 : 3$, the first is $\tfrac{2}{5}$ of the total, not $\tfrac{2}{3}$
Simple average vs weighted averageWeight by group size when groups differ
Average speedNot the simple average of two speeds for equal distances — use $\dfrac{2xy}{x + y}$

For Prelims

Facts worth a mark
  • A part of a ratio $a : b$ is a fraction $\dfrac{a}{a + b}$ of the whole.
  • Adding the same number to both terms of a ratio pulls it towards $1 : 1$.
  • Alligation works for any weighted mean — prices, concentrations, marks, speeds over equal times.
  • A weighted average always lies between the group averages, closer to the larger group.
  • If every value rises by $k$, the average rises by $k$; if every value is multiplied by $k$, so is the average.

Speed Strategy

  • Use the options. In age and number-ratio questions, test options in the condition you have not yet used.
  • Work with sums, not averages, in every multi-step average question.
  • Alligation over algebra whenever two things are mixed to get a mean.
  • Make the common term equal (LCM) before combining ratios — do not add or multiply them.
  • What to skip: three-component mixture questions with repeated replacement are rare and slow; leave them for the end.

Common Traps

Trap #1: Fraction of the whole
If boys and girls are in the ratio $3 : 2$, girls are $\tfrac{2}{5}$ (40%) of the class — not $\tfrac{2}{3}$.
Trap #2: Averaging averages
The combined average of two groups is the simple mean of their averages only if the groups are the same size. Otherwise, weight by size.
Trap #3: Changing the wrong component
When water is added to a milk–water mixture, the milk quantity stays fixed. Anchor the calculation on the component that does not change.

Practice

Prelims (CSAT)

UPSC Practice · CSAT Qra-1

Two numbers are in the ratio $3 : 5$. If 9 is added to each, the ratio becomes $3 : 4$. What are the numbers?

  1. A 9 and 15
  2. B 12 and 20
  3. C 15 and 25
  4. D 6 and 10
Solution

$\dfrac{3x + 9}{5x + 9} = \dfrac{3}{4} \Rightarrow 12x + 36 = 15x + 27 \Rightarrow x = 3$. Numbers: 9 and 15.

Check: $18 : 24 = 3 : 4$ ✓. (Option-test: $12 + 9 : 20 + 9 = 21 : 29$ ✗.)

Answer: A

UPSC-pattern practice question
UPSC Practice · CSAT Qra-2

A 60-litre mixture contains milk and water in the ratio $2 : 1$. How much water must be added to make the ratio of milk to water $1 : 2$?

  1. A 40 litres
  2. B 60 litres
  3. C 80 litres
  4. D 20 litres
Solution

Milk $= 40$ L, water $= 20$ L. Milk is unchanged, so for $1 : 2$ water must be $80$ L. Water to add $= 80 - 20 = 60$ L.

Answer: B

UPSC-pattern practice question
UPSC Practice · CSAT Qra-3

The average of 11 results is 50. The average of the first six is 49 and the average of the last six is 52. What is the sixth result?

  1. A 50
  2. B 54
  3. C 56
  4. D 58
Solution

The sixth result is counted in both groups of six.

Sixth $= (6 \times 49) + (6 \times 52) - (11 \times 50) = 294 + 312 - 550 = 56$.

Answer: C

UPSC-pattern practice question
UPSC Practice · CSAT Qra-4

In what ratio must tea costing ₹120 per kg be mixed with tea costing ₹180 per kg so that the mixture is worth ₹160 per kg?

  1. A 2 : 1
  2. B 1 : 1
  3. C 3 : 2
  4. D 1 : 2
Solution

Alligation: cheaper : dearer $= (180 - 160) : (160 - 120) = 20 : 40 = 1 : 2$.

Check: $\dfrac{1 \times 120 + 2 \times 180}{3} = \dfrac{480}{3} = 160$ ✓.

Answer: D

UPSC-pattern practice question
UPSC Practice · CSAT Qra-5

The average age of a family of five members is 24 years. The youngest member is 8 years old. What was the average age of the family at the birth of the youngest member?

  1. A 16 years
  2. B 18 years
  3. C 20 years
  4. D 22 years
Solution

Present total $= 5 \times 24 = 120$. Eight years ago, the youngest was not born and each of the other four was 8 years younger:

Total then $= 120 - 8 - 4 \times 8 = 80$, spread over 4 members → average $= 20$ years.

Answer: C

UPSC-pattern practice question

Quick Revision

SituationApproach
Divide in $a : b : c$One part $= \dfrac{\text{Total}}{a + b + c}$
Combine $A : B$ and $B : C$Make B equal using LCM
Compare ratiosCross-multiply
Mixture of twoAlligation: $(d - m) : (m - c)$
Adding one componentAnchor on the unchanged component
PartnershipCapital × time
AverageSum = average × number
ReplacementNew = old ± $n$ × change in average
Groups of unequal sizeWeighted average

Key Takeaways

Key Takeaways
  1. Ratios are parts — find one part, then everything else follows.
  2. Equalise the common term before combining ratios.
  3. Alligation solves any two-component weighted mean in one line.
  4. Convert averages to sums before doing arithmetic.
  5. Weighted averages lean towards the larger group.
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