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Word Problems and Problem Solving
Translating Words to Equations
Every word problem begins with a story. Your job is to extract the numbers, identify the unknown variable, and write an algebraic equation that captures the relationship described in words.
Define Variables First: Assign a letter (usually $x$) to the quantity the problem asks you to find. Every other unknown should be expressed in terms of $x$.
Keywords Matter: "is" means $=$, "more than" means $+$, "less than" means $-$, "of" usually means $\times$, "per" means $\div$.
Check Units: If the answer asks for time in minutes but the speed is in km/h, convert units before setting up the equation.
Re-read the Question: After solving, plug your answer back into the original word statement to verify it makes sense.
Common Word-to-Symbol Translations
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"is/was/will be" → $=$
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"more than/greater than" → $+$
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"less than/fewer than" → $-$
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"of/times/product" → $\times$
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"per/each/divided by" → $\div$
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"consecutive integers" → $x, x+1, x+2$
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"consecutive even integers" → $x, x+2, x+4$
When a problem involves two related quantities, you can use a single variable by expressing one quantity in terms of the other. This creates one equation with one unknown — the simplest form of a system of equations.
$$\text{Quantity A} + \text{Quantity B} = \text{Total}$$
Most two-quantity problems reduce to "part + part = whole" or a direct relationship between the parts.
$\text{Quantity A}$=One part expressed in terms of $x$(depends on context)
$\text{Quantity B}$=The other part expressed in terms of $x$(depends on context)
Part-Whole Pattern: If two quantities add to a total, write one as $x$ and the other as $(\text{total} - x)$. Example: two numbers add to 100 → $x$ and $(100 - x)$.
Ratio Pattern: If two quantities are in a ratio $a : b$, write them as $ax$ and $bx$.
Difference Pattern: If one quantity exceeds another by $d$, write them as $x$ and $(x + d)$.
Age Problems
Age problems test your ability to handle unknown variables that change uniformly over time. The key insight: the difference between two people's ages never changes.
$$\text{Age}_{\text{future}} = \text{Age}_{\text{present}} + n$$
Every person ages by the same amount $n$ over $n$ years, so age differences stay constant.
$\text{Age}_{\text{present}}$=Current age of the person(years)
$n$=Number of years into the future (or past, if negative)(years)
$n > 0$
→Future age — add $n$ to current age.
$n < 0$
→Past age — subtract $|n|$ from current age.
Constant Difference: If Ali is 10 years older than Sara today, Ali will always be 10 years older — past, present, and future.
"$k$ Times as Old": "$A$ is 3 times as old as $B$" means $A = 3B$. Read carefully — "3 times older than" is ambiguous and usually means $A = 4B$.
Sum of Ages Pattern: If the sum of ages at two different times is given, set up two equations and subtract them to eliminate the constant.
Reverse Problems: If told "in 5 years, I will be twice as old as you were 3 years ago", write each person's age at the referenced time before equating.
Distance, Speed, and Time
The distance-speed-time relationship is the backbone of all motion word problems. Every problem about moving objects — cars, trains, boats, runners — boils down to this triangle.
$$D = S \times T$$
Distance equals speed multiplied by time. Given any two, you can find the third.
$D$=Distance travelled(km, m, miles)
$S$=Speed (rate of travel)(km/h, m/s)
$T$=Time taken(hours, seconds, minutes)
$S = \frac{D}{T}$
→Finding speed from distance and time.
$T = \frac{D}{S}$
→Finding time from distance and speed.
Unit Consistency: If speed is in km/h, time must be in hours. Convert minutes to hours by dividing by 60: $45\text{ min} = 0.75\text{ h}$.
Average Speed: When two speeds are used for equal distances, $S_{\text{avg}} = \frac{2S_1 S_2}{S_1 + S_2}$ (harmonic mean). This is NOT the arithmetic mean.
Relative Speed: When two objects move toward each other, add their speeds. When one overtakes the other, subtract the slower speed from the faster.
Key Unit Conversions
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1 km/h = $\frac{5}{18}$ m/s
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1 m/s = $\frac{18}{5}$ km/h
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1 hour = 60 minutes = 3600 seconds
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1 km = 1000 m = 100000 cm
Train problems add one twist to the basic motion equation: the relative speed must account for the train's own length. A train "passing a pole" must cover its own length; "passing a platform" must cover its length plus the platform's length.
$$T = \frac{L_{\text{train}} + L_{\text{object}}}{S_{\text{relative}}}$$
The time for a train to completely pass an object equals total length to cover divided by relative speed.
$L_{\text{train}}$=Length of the train(metres)
$L_{\text{object}}$=Length of the object (platform, another train); zero for a pole/person(metres)
$S_{\text{relative}}$=Relative speed (add if opposite direction, subtract if same direction)(km/h or m/s)
Passing a pole (standing person)
→$L_{\text{object}} = 0$, so $T = \frac{L_{\text{train}}}{S}$.
Two trains, same direction
→$S_{\text{relative}} = |S_1 - S_2|$.
Two trains, opposite direction
→$S_{\text{relative}} = S_1 + S_2$.
Always Convert to Metres/Seconds: Train lengths are in metres and speeds in km/h. Convert speed first: multiply by $\frac{5}{18}$.
Same vs Opposite Direction: Same direction = subtract speeds (one is chasing). Opposite direction = add speeds (they are approaching).
Two Trains Crossing: Total distance = sum of both train lengths. Use relative speed based on direction.
Boat and stream problems are a special case of relative motion. The stream's speed either adds to or subtracts from the boat's own speed in still water, creating downstream speed and upstream speed.
$$S_{\text{down}} = S_{\text{boat}} + S_{\text{stream}}, \quad S_{\text{up}} = S_{\text{boat}} - S_{\text{stream}}$$
Downstream, the current helps the boat. Upstream, the current fights it.
$S_{\text{boat}}$=Speed of the boat in still water(km/h)
$S_{\text{stream}}$=Speed of the current(km/h)
$S_{\text{down}}$=Effective speed going downstream(km/h)
$S_{\text{up}}$=Effective speed going upstream(km/h)
Finding $S_{\text{boat}}$
→$S_{\text{boat}} = \frac{S_{\text{down}} + S_{\text{up}}}{2}$
Finding $S_{\text{stream}}$
→$S_{\text{stream}} = \frac{S_{\text{down}} - S_{\text{up}}}{2}$
Deriving Boat Speed: Add downstream and upstream speeds, then divide by 2 — the stream effects cancel out.
Deriving Stream Speed: Subtract upstream from downstream, then divide by 2.
Round Trip Time: Total time = $\frac{D}{S_{\text{down}}} + \frac{D}{S_{\text{up}}}$. The boat covers $2D$ total distance.
Work and Rate Problems
In work-rate problems, each worker has a constant rate of completing a job. When workers collaborate, their rates add. The central question is always: how long does it take together?
$$\text{Work} = \text{Rate} \times \text{Time}, \quad \text{Rate} = \frac{1}{\text{Time to complete alone}}$$
If a person can finish a job in $T$ days, their rate is $\frac{1}{T}$ jobs per day. When multiple people work together, their individual rates sum.
$\text{Rate}$=Fraction of the job completed per unit time(jobs/day)
$T$=Time for one person to complete the entire job alone(days, hours)
$\text{Work}$=Portion of the job completed (1 = full job)(fraction of job)
Two workers together
→Combined rate $= \frac{1}{T_1} + \frac{1}{T_2}$, so time $= \frac{T_1 \cdot T_2}{T_1 + T_2}$
Partial work
→If A works for $d$ days, work done = $\frac{d}{T_A}$. Remaining = $1 - \frac{d}{T_A}$.
Reciprocal Relationship: Faster workers have smaller $T$ but larger $\frac{1}{T}$. A worker who finishes in 3 days has rate $\frac{1}{3}$, which is greater than rate $\frac{1}{6}$ for a 6-day worker.
LCM Shortcut: Set the total work to the LCM of the individual times. Then each worker's rate becomes a whole number, making arithmetic easier.
Leaving/Joining Midway: Break the problem into phases — before and after a worker leaves or joins. Track work completed in each phase.
Pipes and Cisterns: Inlet pipes fill (positive rate), outlet pipes drain (negative rate). Net rate = sum of inlet rates minus sum of outlet rates.
Mixture and Alligation
Mixture problems involve combining two or more components with different properties (cost, concentration, percentage) to create a blend. The total quantity of each property is conserved — what goes in must come out.
$$C_1 \cdot Q_1 + C_2 \cdot Q_2 = C_{\text{mix}} \cdot (Q_1 + Q_2)$$
The total value (cost × quantity or concentration × quantity) of the mixture equals the sum of the values of its components.
$C_1, C_2$=Property values of the two components (cost per kg, concentration %)(depends on context)
$Q_1, Q_2$=Quantities of the two components(kg, litres)
$C_{\text{mix}}$=Property value of the resulting mixture(same as $C_1$)
Conservation Principle: Total cost = sum of individual costs. If you mix 3 kg at Rs. 10/kg and 2 kg at Rs. 20/kg, total cost = $30 + 40 = 70$, mixture cost = $70 \div 5 = 14$/kg.
Replacement/Removal: If $x$ litres are removed from a $V$-litre mixture and replaced, the amount of the original substance becomes $V \cdot \left(1 - \frac{x}{V}\right)^n$ after $n$ replacements.
Repeated Dilution: After each removal-and-replacement, multiply the remaining fraction of the original substance.
Alligation is a visual shortcut for finding the ratio in which two components should be mixed to achieve a desired mixture property. It works for cost, concentration, speed, and percentage problems.
$$\frac{Q_1}{Q_2} = \frac{C_{\text{mix}} - C_2}{C_1 - C_{\text{mix}}}$$
The ratio of quantities is determined by the differences between each component's value and the target mixture value. Cross-subtract and write the ratio.
$Q_1 / Q_2$=Ratio in which to mix component 1 and component 2(ratio (unitless))
$C_1 - C_{\text{mix}}$=Difference between component 1's value and the target(same as $C$)
$C_{\text{mix}} - C_2$=Difference between the target and component 2's value(same as $C$)
Alligation Cross: Write $C_1$ (dearer) and $C_2$ (cheaper) on the left, $C_{\text{mix}}$ in the center. Cross-subtract diagonally: $C_1 - C_{\text{mix}}$ goes opposite $C_2$, and $C_{\text{mix}} - C_2$ goes opposite $C_1$. Those are your ratio parts.
Mean Price: The mixture's cost is always between $C_1$ and $C_2$ — it is a weighted average.
Extension to Three Components: Use alligation pairwise, or set up the weighted average equation directly.
Partnership and Profit Sharing
In partnership problems, two or more people invest money in a business. Profit is shared in proportion to the product of each person's investment amount and the time period for which it was invested.
$$\frac{\text{Profit}_A}{\text{Profit}_B} = \frac{I_A \times T_A}{I_B \times T_B}$$
Each partner's share of profit equals their share of the total investment-time product.
$I_A, I_B$=Investment amounts by partners A and B(currency (Rs.))
$T_A, T_B$=Time periods the investments were held(months, years)
$\text{Profit}_A, \text{Profit}_B$=Individual profit shares(currency (Rs.))
Same investment time
→Profit ratio = investment ratio ($T_A = T_B$ cancels).
Same investment amount
→Profit ratio = time ratio ($I_A = I_B$ cancels).
Investment-Time Product: A person investing Rs. 5000 for 6 months has the same claim as someone investing Rs. 3000 for 10 months (both = 30000 invest-months).
Midway Changes: If a partner withdraws or adds capital mid-year, split the year into phases with different investment amounts, then sum the products.
Salary + Profit: If one partner takes a fixed salary, subtract it from the total profit first, then distribute the remainder by the ratio.
Clocks and Calendars
Clock problems test your understanding of angular speed — the hour and minute hands move at different rates, creating specific angle relationships at given times.
$$\theta = |30H - 5.5M|$$
The angle between the hour and minute hands at $H$ hours and $M$ minutes.
$\theta$=Angle between the two hands(degrees)
$H$=Hour (on a 12-hour clock)(hours)
$M$=Minutes past the hour(minutes)
$30H$=Hour hand's base position at hour $H$ (30° per hour)(degrees)
$5.5M$=Minute hand's position minus hour hand's drift in $M$ minutes(degrees)
$\theta > 180°$
→The reflex angle is $360° - \theta$.
Hands overlap
→$30H = 5.5M$, so overlaps occur at $M = \frac{60H}{11}$ minutes past $H$.
Speed of Hands: Minute hand moves at $6°$ per minute ($360° \div 60$). Hour hand moves at $0.5°$ per minute ($30° \div 60$). Relative speed = $5.5°$ per minute.
Hands Coincide: The hands overlap 11 times every 12 hours (every $\frac{720}{11} \approx 65.45$ minutes).
Hands at Right Angle: The hands form a 90° angle 22 times every 12 hours.
Hands in a Straight Line: The hands form a straight line (0° or 180°) 11 times every 12 hours.
Clock Key Facts
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Minute hand: $6°$ per minute
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Hour hand: $0.5°$ per minute
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Relative speed: $5.5°$ per minute
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Overlaps: 11 times in 12 hours
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Right angles: 22 times in 12 hours
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Straight line: 11 times in 12 hours
Calendar problems test your understanding of modular arithmetic — specifically how days cycle every 7 and how leap years shift the pattern.
$$\text{Day shift} = (\text{Number of years} + \text{Number of leap years}) \pmod{7}$$
Each normal year shifts the day of the week forward by 1. Each leap year shifts it by 2. After finding the total shift, take modulo 7 to find the new day.
$\text{Day shift}$=Total days the weekday moves forward(days (mod 7))
$\text{Number of years}$=Years between the two dates(years)
$\text{Number of leap years}$=Leap years in that interval (each adds one extra day)(count)
Leap Year Rule: A year is a leap year if divisible by 4, except century years which must be divisible by 400. So 2000 is a leap year, 1900 is not.
Century Year Shift: 100 years shift by 5 days, 200 years by 3 days, 300 years by 1 day, 400 years by 0 days.
Same Calendar Year: Two years share the same calendar if the total day shift between them is a multiple of 7.
GMAT-Style Quantitative Reasoning
GMAT-style problems emphasize logical reasoning over brute-force algebra. The best test-takers use back-solving and number picking to avoid lengthy equations entirely.
Back-Solving: Start with answer choice B or C (the middle value). If it is too small, eliminate A and B, then try D. If too large, eliminate C, D, and E, then try B. This halves the search space.
Number Picking: For problems with variables in the answer choices, substitute easy numbers. Pick $x = 2$, $y = 3$ — avoid 0 and 1 since they make too many expressions equal.
Elimination: Before computing, scan choices for obvious traps. If two choices sum to a number in the problem, one of them is likely the answer.
Estimation: When exact values are messy, round strategically. If choices are far apart (e.g., 12, 25, 48, 95), rough estimation is enough.
Strategic Answer Choice Patterns
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If two choices are reciprocals, the correct answer is often one of them
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If two choices are negatives of each other, check your signs carefully
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The most complicated-looking answer is often correct (test-makers make wrong answers look simple)
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If all choices are algebraic, try number picking instead of solving