Chapter 26

Ratios, Averages & Mixtures

Direct/inverse proportion, weighted averages, alligation method.

Full Chapter Notes

Source · FPSC Trap Decoder · CSS MPT Smart Notes (2026 Edition)

High-Yield Snapshot

Attribute	Value
MPT Weightage	3–4 Marks
Difficulty Level	Low–Medium
Confirmed Past Papers	2022, 2023, 2024, 2025

Trend Alert. Six confirmed questions across four past papers (2022–2025) make this a reliable source of easy marks. FPSC tests three things from this topic: simple averages (find the mean, find the missing value when the mean changes, find the youngest/oldest in an evenly spaced group), ratio problems (divide a quantity in a given ratio, find a share), and mixture problems (change the ratio of a mixture by adding or removing a component). Every question type follows a fixed template — once you know the template, the solution writes itself in under 30 seconds.

Concept Anchor

Ratio and average problems are the most mechanical in the entire mathematics section. There is no translation difficulty as in algebra, no pattern recognition as in series. The formulas are simple, the steps are fixed, and the numbers are always manageable.

Exam Strategy — The Two Failure Points to Eliminate

Students use the basic average formula when the problem requires working backwards from the average to find a missing value.
Students treat a ratio as a fraction of the whole rather than a part-to-part comparison.

This chapter eliminates both errors.

Core Formulas

Formula	Expression	Plain English
Average	Average = Sum ÷ Count	What each person gets if total is shared equally
Sum (from average)	Sum = Average × Count	Total value of all items combined
Missing value	6th obs = New sum − Old sum	Find new and old totals; subtract to find the new value
Evenly spaced (youngest)	Youngest = Avg − [(n−1)/2 × d]	Middle value = average; youngest is below by half the range
Simple ratio share	Share = (own ratio ÷ total parts) × whole	Total parts = sum of all ratio numbers
Weighted average	(n₁×a₁ + n₂×a₂) ÷ (n₁+n₂)	Multiply each average by its group size; never average the averages
Average speed (equal dist.)	2×v₁×v₂ ÷ (v₁+v₂)	Harmonic mean; always less than the arithmetic mean

Formula 1 — Average (Arithmetic Mean)

Formula Box — Average (Arithmetic Mean)

Average = Sum of all values ÷ Number of values
Sum = Average × Number of values
Missing value = New sum − Old sum

Plain English: The average is what each person would get if the total were shared equally. To find a missing value when the average changes, find both sums and subtract.

Formula 2 — Evenly Spaced Average (Children Born at Equal Intervals)

Formula Box — Evenly Spaced Values

Values: x, x+d, x+2d, …, x+(n−1)d
Average = x + (n−1)d/2 [equals the middle value]
Youngest = Average − (n−1)d/2
Oldest = Average + (n−1)d/2

Plain English: When children are born at equal time intervals, the average age equals the age of the middle child. The youngest is a fixed distance below the average.

Middle-Value Shortcut. In a group of 5 evenly spaced by 2 years with average 18: middle child (3rd) = 18. Ages: 14, 16, 18, 20, 22. Youngest = 14. Confirmed CSS MPT 2023 Special Q120.

Formula 3 — Simple Ratio: Dividing a Quantity

Formula Box — Simple Ratio

Total parts = sum of ratio numbers
Value of 1 part = Total quantity ÷ Total parts
Each share = its ratio number × value of 1 part

Plain English: A ratio of 2:3 means the total is split into 5 equal parts. One person gets 2 parts, the other gets 3 parts.

Formula 4 — Mixture: Changing the Ratio

Formula Box — Mixture Problems

Find the current amounts of each component.
Remove proportionally (if removing mixture) or add to one component only.
Set up the new ratio equation and solve for the unknown quantity.

Plain English: When you remove mixture from a container, you remove all components proportionally. When you add only one component, only that component's quantity changes.

Formula 5 — Weighted Average

Formula Box — Weighted Average

Weighted Average = (n₁×a₁ + n₂×a₂) ÷ (n₁ + n₂)

Plain English: When groups of different sizes are combined, you cannot simply average the averages. You must account for how many members each group has.

Example: Class A has 20 students averaging 70 marks. Class B has 30 students averaging 80 marks. Combined average = (20×70 + 30×80) ÷ (20+30) = (1,400 + 2,400) ÷ 50 = 3,800 ÷ 50 = 76.

CSSPrep Solved Examples

Solved Example 1 — Basic Average

Problem: If the sum of all 10 values in a data set is 10, what is the average?

Step 1: Average = Sum ÷ Count = 10 ÷ 10 = 1.

Answer: 1. (Confirmed CSS MPT 2022 Q96.)

FPSC Trap Note: FPSC placed 10 as Option B (dividing by 1 instead of the count) and 2 as Option C. Always divide the sum by the number of values.

Solved Example 2 — Evenly Spaced Ages — Finding the Youngest

Problem: The average of the ages of 5 children born at intervals of 2 years is 18. What is the age of the youngest child?

Method 1 — Middle Value Shortcut (fastest):

Middle (3rd) child = 18.
Ages: 18−4=14, 18−2=16, 18, 18+2=20, 18+4=22.
Youngest = 14.

Method 2 — Algebraic (for confirmation):

Let youngest = x. Ages: x, x+2, x+4, x+6, x+8.
Sum = 5x + 20. Average = (5x+20)/5 = x+4 = 18. x = 14. ✓

Answer: 14. (Confirmed CSS MPT 2023 Special Q120.)

Solved Example 3 — Missing Value When Average Changes

Problem: The mean of 5 observations is 10. A 6th observation is added and the new mean becomes 12. What is the 6th observation?

Step 1: Original sum = Average × Count = 10 × 5 = 50.
Step 2: New sum = New average × New count = 12 × 6 = 72.
Step 3: 6th observation = New sum − Original sum = 72 − 50 = 22.
Verify: (50 + 22) ÷ 6 = 72 ÷ 6 = 12. ✓

Answer: 22.

Solved Example 4 — Simple Ratio

Problem: Divide Rs. 900 between Ali and Bilal in the ratio 2:7. How much does each get?

Step 1: Total parts = 2 + 7 = 9.
Step 2: Value of 1 part = 900 ÷ 9 = 100.
Step 3: Ali's share = 2 × 100 = Rs. 200. Bilal's share = 7 × 100 = Rs. 700.
Verify: 200 + 700 = 900. ✓

Solved Example 5 — Ratio with Three Parts

Problem: A sum of Rs. 1,500 is divided among A, B, and C in the ratio 2:3:5. What is C's share?

Step 1: Total parts = 2 + 3 + 5 = 10.
Step 2: Value of 1 part = 1,500 ÷ 10 = 150.
Step 3: C's share = 5 × 150 = Rs. 750.
Verify: A=300, B=450, C=750. Total=1,500. ✓

Solved Example 6 — Mixture Problem

Problem: A jug contains 2 litres of a mixture of apple juice and orange juice in the ratio 1:3. We remove 400 ml from the jug. How much apple juice must be added to make the ratio 7:5?

Step 1: Total = 2,000 ml. Ratio 1:3 → 4 parts. 1 part = 500 ml. Apple = 500 ml. Orange = 1,500 ml.
Step 2: Remove 400 ml proportionally (ratio stays 1:3):
- Apple removed = (1/4) × 400 = 100 ml. Apple left = 400 ml.
- Orange removed = (3/4) × 400 = 300 ml. Orange left = 1,200 ml.
Step 3: Add x ml apple juice to reach ratio 7:5:
- (400 + x) ÷ 1,200 = 7 ÷ 5
- 5(400 + x) = 8,400 → 2,000 + 5x = 8,400 → 5x = 6,400 → x = 1,280 ml.

Answer: 1,280 ml — none of the given options (800, 1,000, 200) are correct.

FPSC Trap Note: Confirmed CSS MPT 2023 Special Q123 — Answer was D: None of these.

Solved Example 7 — Weighted Average

Problem: A class of 30 students has an average score of 70. Another class of 20 students has an average score of 80. What is the combined average of all 50 students?

Wrong method (never do this): Simple average of 70 and 80 = 75. This ignores the different class sizes.

Correct method:

Total marks, class 1 = 30 × 70 = 2,100.
Total marks, class 2 = 20 × 80 = 1,600.
Combined total = 2,100 + 1,600 = 3,700.
Combined average = 3,700 ÷ 50 = 74.

Answer: 74. The combined average (74) is closer to 70 than to 80 because there are more students in the class averaging 70. Always weight by group size.

Common Mistakes

Mistake 1 — Dividing by the Wrong Number

Wrong	Right
Sum = 10, Count = 10. Average = 10÷1 = 10.	Average = Sum÷Count = 10÷10 = 1.
Forgot to divide by the count.	Denominator = number of values.

Mistake 2 — Averaging Two Averages Without Weighting

Wrong	Right
Class A avg=70, Class B avg=80. Combined = (70+80)÷2 = 75.	Combined = (30×70 + 20×80) ÷ 50 = 74.
Simple average ignores group sizes.	Multiply each average by its group size.

Mistake 3 — Removing Only One Component in a Mixture

Wrong	Right
Remove 400 ml: subtract 400 ml only from apple juice.	Remove 400 ml proportionally from BOTH components.
Treats removal as affecting one ingredient only.	Mixture removal affects all components in the same ratio.

Wrong	Right
Ratio 2:3. Ali's share = 2÷3 of total.	Ali's share = 2÷5 of total (total parts = 2+3 = 5).
Used one part as the denominator.	Denominator = sum of ALL ratio parts.

FPSC Trap Alert

Trap 1 — The "Average the Averages" Trap. Combining two groups with different sizes and simply averaging their averages. FPSC places this wrong answer as the most attractive option. Always multiply each average by its group size, add the totals, and divide by the combined size.

Trap 2 — The "Sum = Count" Confusion Trap. When sum and count happen to be equal (sum of 10 values = 10), candidates instinctively write average = 10. Average = 10÷10 = 1. CSS MPT 2022 confirmed. Always divide.

Trap 3 — The Middle Value Trap in Even-Spaced Groups. For 5 children born 2 years apart with average age 18, FPSC places 12 as Option A. The youngest is 4 years below the average (not 2 years) because the group spans 8 years total (4 steps of 2 years). Use the middle-value shortcut: middle = average = 18, youngest = 18 − 4 = 14.

Trap 4 — The Proportional Removal Trap. In mixture problems, when you remove liquid from a mixture, you remove ALL components proportionally — not just one. If you remove 400 ml from a 1:3 apple-to-orange mixture, you remove 100 ml apple AND 300 ml orange. FPSC confirmed this in CSS MPT 2023 Special Q123.

Trap 5 — The "None of These" Mixture Trap. CSS MPT 2023 Q123's answer was 1,280 ml — none of the options (800, 1,000, 200) were correct. FPSC intentionally chose Answer D here. Always calculate your complete answer and check against all options before choosing D.

The 5-Minute Battle Card

Key Points

Average Rules

Average = Sum ÷ Count. Always divide by the count.
Sum = Average × Count. Use this to find totals.
Missing value = New total sum − Old total sum.

Evenly Spaced Values

Middle value = average. Youngest = average − [(n−1)/2 × interval].
5 children, 2-year intervals, avg=18: youngest = 18 − 4 = 14.

Ratio Rules

Total parts = sum of all ratio numbers.
Share = (own ratio number ÷ total parts) × whole.
Ratio 2:3:5 → 10 total parts. Each part = total ÷ 10.

Mixture Rules

Removing mixture: remove from ALL components proportionally.
Adding to mixture: only the added component increases; the other stays the same.

Weighted Average

Weighted avg = (n₁×a₁ + n₂×a₂) ÷ (n₁+n₂). Never average the averages.
Class 30 avg=70, Class 20 avg=80: combined = (2,100+1,600)÷50 = 74 (not 75).

Confirmed FPSC Questions

CSS MPT 2022 Q96: sum of 10 values = 10 → average = 1 (not 10).
CSS MPT 2023 Q120: 5 children, 2-year intervals, avg=18 → youngest = 14.
CSS MPT 2023 Q123: 2L mixture 1:3, remove 400 ml, add apple for 7:5 → add 1,280 ml → None of these.

Practice MCQs (FPSC Level)

Part A — Basic Recall (Q1–Q5)

Mechanical drill of ratio and average templates.

The sum of 10 values in a data set is 10. What is the average?

Show explanation

Average = Sum ÷ Count = 10 ÷ 10 = 1. Verify: 1 × 10 = 10.

Trap: FPSC plants 10 (dividing by 1) as the bait.

Repeated CSS MPT 2022 Q96

Rs. 800 is divided between X and Y in the ratio 3:5. What is Y's share?

Show explanation

Total parts = 3 + 5 = 8. Y's share = (5/8) × 800 = Rs. 500. Verify: X = 300, Y = 500, total = 800.

Trap: Treating 5/3 as the multiplier yields the wrong figure.

Repeated CSS MPT 2023, 2024

The average of the ages of 5 children born at intervals of 2 years is 18. What is the age of the youngest child?

Show explanation

Ages (2-yr intervals): x, x+2, x+4, x+6, x+8. Sum/5 = 18 → 5x + 20 = 90 → x = 14. Verify: 14+16+18+20+22 = 90.

Trap: 12 (off by one interval) is the planted near-miss.

Repeated CSS MPT 2023 Special Q120

A sum of Rs. 1,200 is divided among A, B, C in the ratio 1:2:3. What is B's share?

Show explanation

Total parts = 1 + 2 + 3 = 6. B's share = (2/6) × 1,200 = Rs. 400. Verify: A=200, B=400, C=600, sum = 1,200.

Trap: Rs. 600 is C's share, planted to bait misreaders.

Repeated CSS MPT 2024

The mean of 5 numbers is 12. What is their sum?

Show explanation

Sum = Mean × Count = 12 × 5 = 60. Verify: 60 ÷ 5 = 12.

Trap: 17 (adding instead of multiplying) is the reflex error.

Repeated CSS MPT 2022

Part B — Trap-Based (Q6–Q10)

FPSC's elite traps on weighted averages, proportional removal, and missing values.

The mean of 5 observations is 10. A 6th observation is added and the new mean becomes 12. What is the 6th observation?

Show explanation

Original sum = 5 × 10 = 50. New sum = 6 × 12 = 72. 6th observation = 72 − 50 = 22.

Trap: 12 (the new mean itself) is the reflex bait.

FPSC Elite Trap — Missing value when mean changes

Class A has 20 students with an average score of 70. Class B has 30 students with an average score of 80. What is the combined average of all 50 students?

Show explanation

Weighted avg = (20×70 + 30×80) ÷ 50 = (1,400 + 2,400) ÷ 50 = 3,800 ÷ 50 = 76.

Trap: 75 (simple average of 70 and 80) ignores group sizes — the elite trap.

FPSC Elite Trap — Weighted average, NOT simple average

A mixture contains milk and water in the ratio 4:1. If 10 litres of water is added, the new ratio becomes 2:1. What was the original quantity of the mixture?

Show explanation

Let milk = 4k, water = k. Add 10L water → 4k/(k+10) = 2/1 → 4k = 2k + 20 → k = 10. Original = 5k = 50L. Verify: 40/10 = 4:1; 40/20 = 2:1.

Trap: 40 (only the milk component) baits those who forget to add the water portion.

FPSC Elite Trap — Mixture addition

A jug contains 3 litres of a mixture of juice and water in the ratio 2:1. If 600 ml of the mixture is removed, how much juice remains?

Show explanation

3L = 3,000 ml. Juice = (2/3)×3,000 = 2,000 ml. Removing 600 ml keeps ratio 2:1, so juice removed = (2/3)×600 = 400 ml. Juice remaining = 2,000 − 400 = 1,600 ml.

Trap: 2,000 ml (the original juice amount, ignoring removal) is the planted reflex.

FPSC Elite Trap — Proportional removal

Three friends A, B, C share a profit of Rs. 2,400 in the ratio 1:2:3. B's share exceeds A's share by:

Show explanation

Total parts = 6. A = (1/6)×2,400 = 400. B = (2/6)×2,400 = 800. Difference = 800 − 400 = Rs. 400.

Trap: Rs. 800 (B's own share) baits those who skip the comparison step.

Repeated CSS MPT 2024

Part C — Elite Simulation (Q11–Q15)

Full-difficulty mixture, evenly spaced, average-speed and verification challenges.

A jug contains 2 litres of apple juice and orange juice in the ratio 1:3. 400 ml is removed. How much apple juice must be added to make the ratio 7:5?

Show explanation

2,000 ml, ratio 1:3 → Apple=500 ml, Orange=1,500 ml. Remove 400 ml (ratio kept): Apple now 400 ml, Orange 1,200 ml. Add x ml apple: (400+x)/1,200 = 7/5 → 5(400+x) = 8,400 → x = 1,280 ml. Not in A–C.

Trap: 800 / 1,000 / 1,200 all sound plausible; the true answer (1,280 ml) is absent.

Repeated CSS MPT 2023 Special Q123 — confirmed

The average of 7 consecutive even numbers is 20. What is the largest of the seven numbers?

Show explanation

For evenly spaced numbers, avg = middle (4th) value = 20. Numbers: 14, 16, 18, 20, 22, 24, 26. Largest = 26. Verify: sum = 140, 140÷7 = 20.

Trap: 28 (off by one step) is the bait.

FPSC Elite Trap — Evenly spaced even numbers

A man travels 60 km at 30 km/h and returns the same distance at 60 km/h. What is his average speed for the entire journey?

Show explanation

Time₁ = 60÷30 = 2 h, Time₂ = 60÷60 = 1 h. Total time = 3 h. Avg speed = 120 ÷ 3 = 40 km/h. Harmonic mean: 2v₁v₂/(v₁+v₂) = 2×30×60/90 = 40.

Trap: 45 km/h (arithmetic mean of 30 and 60) is the classic bait.

FPSC Elite Trap — Average speed ≠ average of speeds

Consider: (1) The average of 5 evenly spaced values equals the middle (3rd) value. (2) When removing mixture, only one component decreases. (3) Weighted average accounts for the size of each group. Which are correct?

Show explanation

Statement 1 ✓: avg of evenly spaced values = middle value. Statement 2 ✗: removing mixture reduces BOTH components proportionally. Statement 3 ✓: weighted avg accounts for group size.

Trap: Statement 2 is the planted error — sounds intuitive but contradicts the proportional-removal rule.

FPSC Elite Trap — Statement 2 is the planted error

A, B, C invest in a business in the ratio 2:3:5. At year's end, the profit is Rs. 4,000. B's share is Rs. 1,200. Is this correct?

Show explanation

Total parts = 2+3+5 = 10. B's share = (3/10) × 4,000 = Rs. 1,200. The statement is correct.

Trap: Option B uses 3/5 (only B vs C parts) instead of 3/10 of the whole.

FPSC Elite Trap — Verify the ratio share calculation

Chapter 26 — Answer Key

Chapter 26 — Ratios, Averages & Mixtures

Q	Correct	Type	Primary Trap	Why Others Fail

Quick Revision Panel

Key Points

60-Second Revision — Chapter 26

Important Formulas

Average = Sum ÷ Count
Sum = Average × Count
Missing value = New sum − Old sum
Evenly spaced youngest = Average − [(n−1)/2 × d]
Weighted average = (n₁×a₁ + n₂×a₂) ÷ (n₁+n₂)
Average speed (equal dist.) = 2×v₁×v₂ ÷ (v₁+v₂) [harmonic mean]

Key Ratio Rules

Total parts = sum of ALL ratio numbers.
Share = (own ratio number ÷ total parts) × whole.
Ratio is part-to-part, not part-to-whole.

Average Calculation Shortcuts

5 evenly spaced values: middle value = average.
5 children, 2-year intervals, avg=18: youngest = 18 − 4 = 14.
Never add just the two averages and halve them if groups differ in size.

Mixture Problem Strategy

Removing mixture: ALL components decrease proportionally.
Adding one component: only that component increases.
Set up the new ratio equation; solve for the unknown (x).

Critical Exam Reminders

Sum = Count does NOT mean Average = Sum. Average = Sum ÷ Count.
Always check every option before selecting "None of these."
Weighted average result is pulled towards the larger group.
Harmonic mean is always less than the arithmetic mean for speed problems.