SSC MTS Population Problems — Study Material, 15 PYQs & Practice MCQs | ZestExam
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SSC MTS Population Problems
Study Material — 15 PYQs (2020–2020) · Concept Notes · Shortcuts
SSC MTS Population Problems is a frequently tested subtopic — 15 previous year questions from 2020–2020 papers are included below with concept notes, key rules and shortcut tricks.
The population of a city was 500,000 in 2020. If it increased by 20% in 2021, what was the population in 2021?
Exam Q 42020Previous Year Pattern
A village population decreased from 80,000 to 72,000 over one year. What was the percentage decrease?
Exam Q 52020Previous Year Pattern
If a town's population increases by 25% to reach 50,000, what was the original population?
Exam Q 62020Previous Year Pattern
A district's population was 200,000 in 2019. It increased by 15% in 2020 and then decreased by 10% in 2021. What was the population in 2021?
Exam Q 72020Previous Year Pattern
The population of a city was 500,000 in 2020. It increased by 20% in 2021 and then decreased by 10% in 2022. What is the population at the end of 2022?
Exam Q 82020Previous Year Pattern
A town's population in 2019 was 80,000. If the population increased by 15% each year for 2 years, what is the population in 2021?
Exam Q 92020Previous Year Pattern
In a village, 45% of the population are males and the rest are females. If there are 22,000 females, what is the total population of the village?
Exam Q 102020Previous Year Pattern
The population of two towns A and B are in the ratio 3:5. If town A's population increased by 20% and town B's population increased by 10%, what is the new ratio of their populations?
Exam Q 112020Previous Year Pattern
A district's population was 600,000 in 2015. By 2020, it had grown to 750,000. What is the percentage increase in population over this 5-year period?
Exam Q 122020Previous Year Pattern
The population of a city was 5,00,000 in 2015. It increased by 20% in the next 4 years and then decreased by 10% in the following 2 years. What is the population in 2021?
Exam Q 132020Previous Year Pattern
A district's population was 8,00,000 in 2010. It increased by 25% by 2015 and then by 10% by 2020. If the population in 2020 was equally distributed among 4 zones, how many people live in 3 zones?
Exam Q 142020Previous Year Pattern
In a town, the male population is 45% of the total. If the female population increased by 25% and the male population increased by 20%, the new female population is 33,000. What was the original total population?
Exam Q 152020Previous Year Pattern
In a city, the ratio of male to female population is 7:8. If the male population increases by 30% and female population increases by 20%, the new male population is 91,000. What is the new total population?
Concept Notes
Population Problems— Rules & Concept
💡
Core Concept
Read this first — the foundation of the topic
→CORE CONCEPT
Population problems follow the compound growth formula. If a population increases or decreases by a certain percentage each year, you apply that percentage repeatedly, not just once. This is different from simple interest — it's like compound interest
💡KEY RULES
Population grows or shrinks by a fixed percentage each year
2. The percentage applies to the NEW population each year, not the original
3. Use the compound formula, not simple addition/subtraction
4. Decrease and increase work the same way mathematically
🔢
Formula Block
Memorise — at least one formula appears in every paper
Final Population = Initial Population × (1 + r/100)^n
Where:
- r = rate of increase (use negative r for decrease)
- n = number of years
- If r = 5% increase, use (1 + 5/100) = 1.05
- If r = 10% decrease, use (1 - 10/100) = 0.90
📊
Exam Patterns
What examiners ask — read before attempting PYQs
1
Find final population after n years
2
Find initial population (work backwards)
3
Find rate of growth
4
Find time period
5
Mixed increase and decrease over different years
✏️
Worked Example
Solve this step-by-step before moving on
1
Step 1
Population after Year 1
= 50,000 × (1 + 10/100)
= 50,000 × 1.10
= 55,000
2
Step 2
Population after Year 2
= 55,000 × (1 + 20/100)
= 55,000 × 1.20
= 66,000
Alternative Direct Method:
= 50,000 × 1.10 × 1.20
= 50,000 × 1.32
= 66,000
COMMON MISTAKE:
Students add percentages directly: 10% + 20% = 30%, then calculate 50,000 × 1.30 = 65,000. This is WRONG because the 20% applies to the increased population, not the original. Always multiply the factors for each year.
Key Points to Remember
Population problems use compound growth formula: Final = Initial × (1 + r/100)^n
Percentage always applies to the CURRENT population, not the original amount
For decrease, use (1 - r/100) in the formula instead of (1 + r/100)
Multiple years with different rates: multiply all factors together for direct calculation
Never add percentages directly; always use multiplication of decimal factors
If asked for initial population, rearrange formula: Initial = Final ÷ (1 + r/100)^n
Exam-Specific Tips
Population formula: Final = Initial × (1 + r/100)^n where r is annual rate and n is years
For 10% increase, multiply by 1.10; for 10% decrease, multiply by 0.90
If population increases by p% one year and q% next year, combined factor = (1 + p/100) × (1 + q/100)
Compound population growth applies the percentage to the NEW amount each year, not original
For population decrease problems, the formula remains the same but r is treated as negative
Quick check: 50,000 population growing at 10% annually for 2 years = 50,000 × 1.21 = 60,500
60-Second Revision — Population Problems
Formula: Final Population = Initial × (1 + r/100)^n — this is compound, not simple
Trap: Never add percentages from different years. Multiply the growth factors instead
Decrease: Use negative r or write (1 - r/100) — both methods give same answer
Multi-year: For different rates each year, write as Initial × (1.10) × (1.20) × (0.95) etc.
Reverse: If given final population, divide backwards: Initial = Final ÷ [(1 + r/100)^n]
Quick mental check: 10% increase twice ≈ 21% total (not 20%), because second 10% acts on larger base