Translate a real scenario — ages, work rates, or mixtures — into an equation, then solve it. The algebra is rarely hard; the setup is where marks disappear. Every question comes with a written breakdown of how the equation was built, not just how it was solved.
"5 years older than twice his age" and "twice 5 years older than his age" translate to different equations. Read the sentence right to left if the order isn't clicking.
Two workers finishing a job in 6 hours and 3 hours don't finish together in 9 hours, or in the average of the two — rates add, times don't. Convert to "job per hour" first.
A mixture problem hides two conditions in one sentence — the total amount and the total concentration both have to check out, not just one of them.
Straight from the bank — one per difficulty tier. Reveal the answer to see the explanation you'd get in a real session.
In 5 years, Raj will be twice as old as he was 3 years ago. How old is Raj now?
C — 11. Let x be Raj's age now. "In 5 years" is x+5; "3 years ago" is x−3. The sentence says the first equals twice the second: x+5 = 2(x−3). Distributing gives x+5 = 2x−6, so x = 11. Forgetting to distribute the 2 — writing x+5 = 2x−3 instead — gives 8 (B), the most common slip on this question type. 6 (A) drops the "+5" from the setup entirely.
Pipe A can fill a tank in 6 hours. Pipe B can fill the same tank in 3 hours. Working together, how many hours will it take to fill the tank?
A — 2. Rates add, not times. Pipe A fills 1/6 of the tank per hour and Pipe B fills 1/3 per hour; combined that's 1/6 + 1/3 = 1/2 of the tank per hour, so the tank fills in 2 hours. 4.5 (C) comes from averaging the two times instead of combining rates — a shortcut that only works when the two rates are equal. 9 (D) comes from simply adding the times together, which would only make sense if the pipes worked one after another.
A chemist has 10 liters of a solution that is 20% acid. How many liters of pure acid must be added to make the mixture 50% acid?
C — 6. Let x be the liters of pure acid added. The total acid must equal 50% of the new total volume: 0.20(10) + x = 0.50(10 + x). That gives 2 + x = 5 + 0.5x, so 0.5x = 3 and x = 6. 3 (A) comes from ignoring that adding x liters also grows the total volume — solving 0.20(10) + x = 0.50(10) instead, which leaves the denominator stuck at 10.
Word problems are an algebra skill wearing a disguise — the setup is what's being tested, and that's exactly what untimed drilling builds before the clock gets added.
For every problem, write out the full equation from the words alone before doing any algebra. Untimed practice is where you build that separation between setup and solving.
Move to timed sessions once you can set up an equation without hesitating. Word problems run longer than pure algebra since there's a sentence to translate first.
Combine word problems with linear equations and quadratic equations in a mock — all three end in the same solving step, just with different setups.
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