Question bank — Word problems using linear equations

False. Doubling $x$ generally breaks the equality; a specific $x$ is what makes $ax+b=c$ true. What people mean is subtler: once you choose one $x$, every quantity built from it scales together automatically — but that is substitution, not "doubling the equation."

False. Any valid unknown reaches the answer, but a poor choice (e.g. the harder quantity) can force fractions or extra steps. The choice affects difficulty, not correctness.

False for a single linear equation: once one count is $x$ and the total count is known, the other count is $23-x$. Two unknowns without a second equation is Simultaneous Linear Equations, a different tool.

False. The algebra can be perfect; negativity only signals a problem if the quantity can't be negative (people, ages, note counts). For a temperature or a bank balance, negative is a real answer.

True — both mean multiply by $2$, giving $2x$. Contrast with "two years older," which is $x+2$: a shift, not a scaling.

True. If son is $x$ and father $3x$, after $12$ years the difference stays $2x$ but the ratio drifts from $3$ toward $1$ — which is exactly why age-ratio problems have a unique solution.

True only if they start together. Meeting means their travel times match when they set off simultaneously; if one starts late, you must subtract that delay from its time. The constraint is "same clock reading," not "same number typed in."

True and expected: rate is "tanks per hour," and two fast pipes can fill more than one tank per hour. It is the fraction of a single job per hour, not a probability, so it is not capped at $1$.

The multiplier attaches to the smaller quantity. "Father is $3$ times the son" means father $=3\times$son, so if son $=x$ then father $=3x$ — the reasoning was inverted.

"More than" is addition, and order does not matter for it: $5$ more than $x$ is $x+5$. The student both used subtraction and reversed it — two separate slips producing a wrong sign.

They summed the counts and set them equal to the value. The value equation must weight each count by its worth: $5x+10(23-x)=155$. Counting and valuing are different constraints (see Ratio and Proportion for why units must match).

The son's future age must also gain $12$: the right side is $2(x+12)=2x+24$, not $2x+12$. The "$+12$" belongs inside the doubling because you double the son's future age.

You cannot add times for parallel work — two pipes together must be faster than either alone, yet $15>9$. Add rates ($\frac{1}{6}+\frac{1}{9}$) instead; the time is the reciprocal, $\frac{18}{5}=3.6$ h.

Consecutive even numbers jump by $2$, so it is $x,x+2,x+4$. Step $1$ describes any consecutive integers, which would mix odd and even.

Both orders actually give the same final price (multiplication commutes), so setting them equal yields $0=0$ — no information. Order-independence means that "equation" is not a genuine constraint to solve.

Because one equation can only pin down one unknown. Expressing everything through a single symbol (using Algebraic Expressions) collapses many facts into one solvable statement.

The algebra faithfully solves whatever equation you wrote — even a wrong one. Substituting back into the original words tests the translation itself, not just the arithmetic.

"Moving across" is just shorthand for applying the same inverse operation to both sides. Dividing both sides by $5$ keeps the equality true; there is no separate rule, only balance preservation.

"Per" gives a rate — a fixed amount for each unit. Total $=$ rate $\times$ number of units, so "₹5 per note, $x$ notes" becomes $5x$. It is repeated addition compressed into a product.

They can cover equal ground by trading speed against time: a slow walker over long time matches a fast one over short time, since $D=R\times T$ holds the product fixed while the factors vary.

Then every other term is $x$ plus a positive step ($x+1,x+2,\dots$), so the equation has only additions and no negative offsets to track — fewer sign errors.

A single ratio "father $=3\times$ son" has infinitely many age pairs. The second condition (a ratio at another time) adds the missing constraint, turning a family of solutions into one — this is the core idea behind Applications of Algebra.

The equation is satisfied by $7.5$, but people are indivisible, so the scenario has no valid answer. Either the problem is flawed or you mistranslated a constraint — a genuine fractional-person result signals inconsistency.

The statement is always true, meaning your "constraint" carried no new information (often you used the same fact twice). The problem is under-determined; you need another independent relationship.

No value of $x$ can work — the conditions contradict each other. In word-problem terms, the story as told is impossible (e.g. asking two speeds to meet that are travelling apart).

Not for pipes filling together; it would mean help slows you down. But if one "pipe" is a drain (negative rate), the net rate can shrink and the time can legitimately exceed a single filler's time.

No — it violates the physical setup. The algebra found a root, but the interpretation step rejects it; recheck which quantity got the multiplier or which age gained the time shift.

The only consistent count is zero of each. Any positive count of a positive-value note forces a positive total, so ₹0 forces the degenerate empty case — a valid but trivial solution.

Zero travel time means the two objects are already together at the start (distance between them is $0$), or the "meeting" is the starting instant. It is a boundary answer, not an error.