CSAT – Quantitative Aptitude
UPSC Civil Services – Foundation Phase (CSAT)
Quantitative Aptitude is the second section of the CSAT paper. It tests basic numeracy, logical reasoning, and problem‑solving skills at an entry‑level. The section is 25 multiple‑choice questions (MCQs) worth 100 marks, with a negative marking of –0.25 for each wrong answer.
1. Simple Explanation
| What | Why it matters |
|---|---|
| Basic Arithmetic (addition, subtraction, multiplication, division, percentages) | Most questions are straight‑forward numeracy problems that appear in daily life. |
| Ratio & Proportion | Used in “part‑whole” problems and in many probability questions. |
| Statistics (mean, median, mode, standard deviation) | Occasionally appears in data interpretation. |
| Time–Speed–Distance | Classic problem type; many practice questions. |
| Simple & Compound Interest | Optional topic; usually appears in the “optional” part of the CSAT. |
| Permutations & Combinations | Basic counting problems. |
| Probability | Includes both simple and conditional probability. |
| Word Problems | Real‑life scenarios that test reasoning and calculation. |
The key to scoring well is speed and accuracy. You should be able to solve a typical problem in 10–15 seconds.
2. Key Points (Concepts & Shortcuts)
| Concept | Formula / Shortcut | Typical Question Example |
|---|---|---|
| Percentages | (P% = \frac{P}{100}) of a number | “If 30% of 200 is X, what is X?” |
| Ratio & Proportion | (a:b = c:d \Rightarrow a \times d = b \times c) | “If 2 kg of apples cost ₹10, how much 5 kg cost?” |
| Time–Speed–Distance | (S = V \times T) | “A train covers 120 km in 2 h. What is its speed?” |
| Simple Interest | (SI = \frac{P \times R \times T}{100}) | “₹5000 at 12% p.a. for 3 years.” |
| Compound Interest (annual) | (A = P \left(1 + \frac{R}{100}\right)^T) | “₹10,000 at 5% p.a. for 4 years.” |
| Permutations | ({}_nP_r = \frac{n!}{(n-r)!}) | “In how many ways can 3 letters be chosen from ABCDE?” |
| Combinations | ({}_nC_r = \frac{n!}{r!(n-r)!}) | “Choose 2 students from 5.” |
| Probability (simple) | (\Pr(E) = \frac{\text{favourable outcomes}}{\text{total outcomes}}) | “P(rolling even on a die) = 3/6.” |
| Conditional Probability | (\Pr(A | B) = \frac{\Pr(A \cap B)}{\Pr(B)}) |
| Average / Mean | (\bar{x} = \frac{\sum x_i}{n}) | “Mean of 4, 7, 9, 12?” |
| Median | Middle value when sorted | “Median of 3, 5, 8, 10, 12?” |
| Mode | Most frequent number(s) | “Mode of 2, 3, 3, 5, 5, 5?” |
Quick Mental Tricks
| Trick | When to Use |
|---|---|
| Multiplication by 5 = multiply by 10 then divide by 2 | 5 × 7 → 70/2 = 35 |
| Multiplication by 9 = multiply by 10 then subtract the number | 9 × 8 → 80–8 = 72 |
| Square of (a+b) = (a^2 + 2ab + b^2) | 12² = (10+2)² = 100 + 40 + 4 |
| Square of (a−b) = (a^2 – 2ab + b^2) | 8² = (10–2)² = 100 – 40 + 4 |
| Divide by 25 = divide by 100 then multiply by 4 | 500 ÷ 25 = (500 ÷ 100) × 4 = 5 × 4 = 20 |
| Divide by 125 = divide by 1000 then multiply by 8 | 1250 ÷ 125 = (1250 ÷ 1000) × 8 = 1.25 × 8 = 10 |
| Add 10% quickly = add 1/10 of the number | 200 + 10% = 200 + 20 = 220 |
| Subtract 10% quickly = subtract 1/10 of the number | 200 – 10% = 200 – 20 = 180 |
3. Worked Example
Question (Time–Speed–Distance)
A person travels from City A to City B by train. The train travels at a constant speed of 60 km/h. The distance between the cities is 180 km. How long will the journey take?
Solution
-
Identify the given quantities:
- Speed (V = 60 \text{ km/h})
- Distance (S = 180 \text{ km})
-
Use the basic formula (S = V \times T).
Rearrange for time (T):
[ T = \frac{S}{V} ] -
Substitute the numbers:
[ T = \frac{180}{60} = 3 \text{ hours} ] -
Answer: 3 hours.
Why it is typical of CSAT
- Only one