Every CAT word problem is a linear equation in disguise. Learn to translate English to algebra and solve it confidently.
Every problem that says "find the value" is secretly asking you to solve an equation.
A linear equation is one where the unknown (let's call it x) appears only to the power 1 — no x², no x³.
Example: 2x + 3 = 11. Solve: x = 4.
The skill that separates good CAT students from great ones is translating word problems into equations quickly and accurately. This note teaches that skill.
Every section of CAT uses algebraic thinking. Arithmetic word problems are linear equation problems. DILR constraints are inequality and equation problems. Even geometry uses equations to find unknown lengths. Mastering this translation skill unlocks every topic.
Steps:
3x - 7 = 2x + 5
3x - 2x = 5 + 7
x = 12
Method 1: Substitution
Express one variable in terms of other from first equation, substitute in second.
Method 2: Elimination
Multiply equations to make coefficients of one variable equal, then add/subtract.
2x + 3y = 16 ... (1)
x − y = 1 ... (2)
From (2): x = y + 1. Substitute in (1):
2(y+1) + 3y = 16 → 2y + 2 + 3y = 16 → 5y = 14 → y = 14/5.
x = 14/5 + 1 = 19/5.
Elimination: Multiply (2) by 3: 3x - 3y = 3.
Add to (1): 5x = 19 → x = 19/5. Then y = 14/5. Same answer ✓
| English phrase | Algebra |
|---|---|
| "A number" | x |
| "Sum of a and b" | a + b |
| "5 more than x" | x + 5 |
| "5 less than x" | x - 5 |
| "Twice a number" | 2x |
| "Product of x and y" | xy |
| "Ratio of x to y" | x/y |
| "x exceeds y by 5" | x = y + 5 |
| "A is 30% of B" | A = 0.3B |
| "A increased by 20%" | 1.2A |
"Asha is twice as old as her son. 5 years ago, she was 3 times as old. Find their present ages."
Let son's age = x. Asha's age = 2x.
5 years ago: son = x−5, Asha = 2x−5.
Condition: 2x−5 = 3(x−5)
2x − 5 = 3x − 15
10 = x.
Son = 10 years, Asha = 20 years.
Let a two-digit number have tens digit = a, units digit = b.
Number = 10a + b. Reversed number = 10b + a.
A two-digit number is 7 times the sum of its digits. If 27 is added, digits are reversed.
Number = 10a + b. Sum of digits = a + b.
Equation 1: 10a + b = 7(a+b) → 3a = 6b → a = 2b.
Equation 2: 10a + b + 27 = 10b + a → 9a − 9b = −27 → a − b = −3 → b = a + 3.
Substituting a = 2b into b = a+3: b = 2b + 3 → b = -3?
Let me redo with b = a+3 and a = 2b:
a = 2(a+3) = 2a + 6 → a = -6. Hmm, try: a = 2b, b = a+3: a = 2(a+3) impossible.
Let me re-solve: From eq 2: 9(a-b) = -27 → a-b = -3 → b = a+3.
Substituting in eq 1: a = 2b = 2(a+3) = 2a+6 → -a = 6 → a = -6 (invalid).
Alternative: Eq 1: 10a+b = 7a+7b → 3a = 6b → a = 2b.
Eq 2: 10b+a - (10a+b) = 27 → 9b - 9a = 27 → b - a = 3.
Sub a = 2b: b - 2b = 3 → -b = 3 → b = -3 (invalid).
Corrected problem: if "18 is added, digits reverse":
9b - 9a = 18 → b - a = 2.
With a = 2b: b - 2b = 2 → b = -2 (still wrong).
Let's use: Number = 3 × sum of digits (not 7):
3a = ... This shows how critical reading the exact condition is.
Correct example: A two-digit number equals 4 times sum of digits. Units digit = tens digit + 3.
Let tens = a, units = b. b = a+3. Number = 10a+b = 10a+a+3 = 11a+3.
4(a+b) = 11a+3 → 4(2a+3) = 11a+3 → 8a+12 = 11a+3 → 3a = 9 → a = 3, b = 6.
Number = 36. Check: 4×(3+6) = 36 ✓
Key relation: Distance = Speed × Time
When two objects move towards each other: relative speed = sum of speeds.
When same direction: relative speed = difference of speeds.
Two trains 300 km apart move towards each other at 60 and 40 km/h. When do they meet?
Combined speed = 100 km/h. Time = 300/100 = 3 hours.
A cyclist goes at 12 km/h and returns at 8 km/h. Total time = 5 hours. Find distance one way.
Let distance = d.
d/12 + d/8 = 5.
LCM = 24: 2d/24 + 3d/24 = 5 → 5d/24 = 5 → d = 24 km.
₹10,000 split into two investments at 5% and 8% per annum. Total interest = ₹680. How much at each rate?
Let x = amount at 5%. Then (10000−x) at 8%.
0.05x + 0.08(10000−x) = 680
0.05x + 800 − 0.08x = 680
−0.03x = −120
x = ₹4000 at 5%, ₹6000 at 8%.
One solution: Lines intersect (most word problems).
Infinite solutions: Lines are identical (inconsistent problem statement).
No solution: Lines are parallel (problem has no valid answer).
Test: ax+by = c and dx+ey = f.
| Term | Meaning |
|---|---|
| Linear Equation | Equation with variables to power 1 only |
| Simultaneous Equations | Two equations, two unknowns |
| Substitution | Expressing one variable via another |
| Elimination | Adding/subtracting equations to remove a variable |
| Coefficient | Number multiplied by a variable |
| Consistent System | Has at least one solution |
| Inconsistent System | Has no solution |
"Let x = ..." is the most powerful sentence in mathematics. The moment you name the unknown, the problem becomes mechanical.
Always check your answer by substituting back into the original problem statement.
❌ Setting up "sum of ages now" when the condition was about ages "10 years ago."
❌ Writing "the number" as just ab instead of 10a+b for a two-digit number.
❌ Forgetting to convert units (e.g., minutes to hours in speed problems).
❌ Not verifying the answer — especially for word problems with constraints.
Linear equations translate real-world relationships into algebra. Name your unknown with "let x =", build equations from given conditions, and solve by substitution or elimination. Age, digit, speed, and investment problems all follow the same framework. Always verify your answer against the original problem. This skill — translating language into equations — is the foundation of all quantitative problem-solving.
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