Word problems in the quantitative section are sentences hiding equations. "Three more than twice a number is 17" is not an algebra question; it is a translation question, and the algebra that follows is two lines. Candidates who freeze here usually cannot say which of two things they are being asked to find, or they translate the sentence into the wrong equation and then solve it perfectly.
The other reason algebra deserves a lesson of its own is speed. A well-formed equation often beats arithmetic reasoning, because it does not require you to hold three quantities in your head while manipulating them.
Treat every word problem as three steps, and never skip the first.
The translation dictionary is small and worth memorising: "is" becomes equals, "more than" becomes plus, "less than" becomes minus, "of" becomes multiply, "per" becomes divide, "twice" becomes two times. The one trap: "less than" reverses order. "Five less than a number" is x - 5, not 5 - x.
Linear equations in one unknown. Collect the unknown on one side and the numbers on the other.
5(x - 3) = 2x + 6 5x - 15 = 2x + 6 3x = 21 x = 7
Check by substitution: the left side is 5 x (7 - 3) = 20, and the right side is 2 x 7 + 6 = 20. Both sides agree, so x = 7 is correct.
Quadratics. The form that appears is almost always factorable with small integers. Work the factorisation in three numbered steps rather than guessing:
Step 1 x^2 - 5x + 6 = 0
numbers that multiply to +6: (1,6) (2,3) (-1,-6) (-2,-3)
Step 2 keep the pair that ADDS to -5: (-2) + (-3) = -5
Step 3 write the factors and solve:
(x - 2)(x - 3) = 0 => x = 2 or x = 3
Check 4 - 10 + 6 = 0 and 9 - 15 + 6 = 0Both roots are positive because the constant is +6 and the middle coefficient is −5: with a positive product the pair shares a sign, and the negative sum forces both to be negative before you flip them into (x − 2) and (x − 3).
Two equations, three shapes. (1) Coefficients line up — add or subtract to eliminate: x + y = 10 with x − y = 4 gives 2x = 14, so x = 7, y = 3. (2) One variable is already isolated — substitute: y = 2x into x + y = 9 gives 3x = 9, so x = 3, y = 6. (3) Neither case — scale one equation so a coefficient matches, then subtract: 2x + 3y = 12 and x + y = 5; double the second to 2x + 2y = 10 and subtract to get y = 2, then x = 3. Always substitute the pair back into both original equations before accepting them.
Inequalities. Solve exactly as an equation, with one exception: when you multiply or divide by a negative number, flip the direction of the inequality. From 2x + 3 < 11, subtract 3 to get 2x < 8, then divide by 2 (positive, so the sign stays) for x < 4.
Expanding and simplifying. 2(3x - 4) is 6x - 8. Do this before solving whenever brackets are present, because it removes a common source of sign errors.
Degree of a polynomial. The degree is the largest total exponent on any single term after expanding — for several variables, add the exponents within that term. In x&sup5;yz² + x&sup4;y³z² + xyz + 10 the term degrees are 5+1+2 = 8, 4+3+2 = 9, 1+1+1 = 3 and 0, so the degree is 9. Constants have degree 0; you never add degrees across different terms.
Domain and range. Domain is the set of inputs the function accepts; range is the set of outputs it can produce. For a fraction, domain excludes values that make the denominator zero: f(x) = (3x − 5) / (2x + 6) is undefined when 2x + 6 = 0, so the domain is all real numbers except −3. Range is read after simplifying and tracking what the simplified form still cannot hit: (x² − 25)/(x + 5) cancels to x − 5 for x ≠ −5, so outputs are all real numbers except −5 − 5 = −10. Domain watches the denominator; range watches the cancelled-out input.
A father is currently three times as old as his son. In 12 years the father will be twice as old as his son. How old is the son now?
Name the unknowns: let s be the son's age now, and let f be the father's age now. Translate the two sentences:
f = 3s f + 12 = 2(s + 12)
The second sentence is where errors appear: it is son's age in 12 years on the right, so both ages must be advanced. Substitute the first into the second: 3s + 12 = 2s + 24, so s = 12, and f = 3 x 12 = 36.
Check against the original sentence, not the equation: the father is 36 and the son is 12, so the father is three times as old — correct. In 12 years the son is 24 and the father is 48, and 48 is twice 24 — correct. The answer is 12.
Solve for x: 5(x - 3) = 2x + 6.
If x + y = 10 and x - y = 4, what are x and y?
What are the roots of x squared - 5x + 6 = 0?
What is the degree of f(x, y, z) = x&sup5;yz² + x&sup4;y³z² + xyz + 10?
What is the domain of f(x) = (3x − 5) / (2x + 6)?
f(x) = (x² − 25)/(x + 5). What is its range?
Without looking back: what is the first step for any word problem, and how do you handle two stated relationships?
The answer is: name the unknown explicitly first; two stated relationships become two equations, which you solve together by adding, or by substituting one into the other.
Fill in the blanks from memory:
Solving 3x + 7 = 22 gives x = . Factorising x squared - 5x + 6 gives the roots and 3. If two numbers sum to 45 and differ by 13, the larger is . The degree of x&sup4;y³z² alone is , and the degree of a constant such as 10 is . The domain of (3x − 5)/(2x + 6) excludes , and after cancelling (x² − 25)/(x + 5) the range excludes .
A shop sells pens at 55 rupees each. Pens bought in a pack of 8 cost 400 rupees. You need 30 pens. What is the cheapest total, and how much do you save compared with buying single pens?
Thirty pens bought singly cost 30 x 55 = 1650 rupees. Buying packs of 8 is cheaper per pen (400 / 8 = 50 rupees), so use as many packs as possible: three packs give 24 pens for 3 x 400 = 1200 rupees, leaving 6 pens to buy singly at 6 x 55 = 330. Total 1200 + 330 = 1530 rupees, which is a saving of 1650 - 1530 = 120 rupees.
Now the step most candidates skip: could four packs be cheaper? Four packs give 32 pens for 1600 rupees — more pens, but 1600 is more than 1530, so it is worse. The algebra here is the constraint: whole packs only, and the cheapest solution is the largest number of packs that does not require buying more than needed. Writing that constraint down is what turns a word problem into a two-line computation.
Algebra handles relationships between numbers when the highest power is one. The next lesson raises the power: quadratic equations, where the unknown is squared, and where the roots can be read from factorisation, the discriminant or the sum-and-product identities — all of which rest on the factorising you have just practised.