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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" → $-$
•
"of/times/product" → $\times$
•
"per/each/divided by" → $\div$
•
"consecutive integers" → $x, x+1, x+2$
•
"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
Application Strategy
Worked Examples
Problem Statement
The sum of three consecutive integers is 126. What is the largest of the three integers?
translating words to equations
Strategic Toolkit
$$x + (x+1) + (x+2) = 126$$
Step 1
Define the integers
Let the three consecutive integers be $x$, $x+1$, and $x+2$.
$x, \quad x+1, \quad x+2$
Step 2
Set up the equation
The problem states their sum is 126.
$x + (x+1) + (x+2) = 126$
Step 3
Simplify and solve
Combine like terms and isolate $x$.
$3x + 3 = 126 \implies 3x = 123 \implies x = 41$
Step 4
Find the largest integer
The largest integer is $x + 2$.
$x + 2 = 41 + 2 = 43$
Problem Statement
A father is 4 times as old as his son. In 10 years, he will be 3 times as old as his son. How old is the son now?
age problems
Strategic Toolkit
$$4x + 10 = 3(x + 10)$$
Step 1
Define present ages
Let the son's current age be $x$. Then the father is $4x$.
$\text{Son} = x, \quad \text{Father} = 4x$
Step 2
Write the future condition
In 10 years, both ages increase by 10, and the father is 3 times the son.
$4x + 10 = 3(x + 10)$
Step 3
Solve for $x$
Expand and solve the linear equation.
$4x + 10 = 3x + 30 \implies x = 20$
Step 4
Verify
Son is 20, father is 80. In 10 years: son is 30, father is 90. Check: $90 = 3 \times 30$.
$4(20) = 80, \quad 80 + 10 = 90 = 3(20 + 10)$ ✓
Problem Statement
Two cities are 480 km apart. Car A starts from the first city at 60 km/h and Car B starts from the second city at 40 km/h at the same time. After how many hours will they meet?
distance-speed-timerelative speed
Strategic Toolkit
$$T = \frac{D}{S_1 + S_2}$$
Step 1
Find relative speed
Since the cars move toward each other, add their speeds.
$S_{\text{relative}} = 60 + 40 = 100\text{ km/h}$
Step 2
Apply the time formula
Time to meet equals total distance divided by relative speed.
$T = \frac{480}{100} = 4.8\text{ hours} = 4\text{ h } 48\text{ min}$
Problem Statement
A car travels from City A to City B at 60 km/h and returns at 40 km/h. What is the average speed for the entire round trip?
average speedharmonic mean
Strategic Toolkit
$$S_{\text{avg}} = \frac{2 S_1 S_2}{S_1 + S_2}$$
Step 1
Let one-way distance be $d$
Let $d$ km be the distance between the cities.
$\text{One-way distance} = d\text{ km}$
Step 2
Compute total time
Time going plus time returning.
$T_{\text{total}} = \frac{d}{60} + \frac{d}{40} = \frac{2d + 3d}{120} = \frac{5d}{120} = \frac{d}{24}$
Step 3
Compute average speed
Total distance divided by total time.
$S_{\text{avg}} = \frac{2d}{d/24} = 48\text{ km/h}$
Step 4
Verify with formula
Apply the harmonic mean formula directly.
$S_{\text{avg}} = \frac{2 \times 60 \times 40}{60 + 40} = \frac{4800}{100} = 48\text{ km/h}$ ✓
Problem Statement
Pipe A can fill a tank in 12 hours and Pipe B can fill it in 15 hours. If both pipes are opened together, how long will it take to fill the tank?
work-rate
Strategic Toolkit
$$T = \frac{T_A \cdot T_B}{T_A + T_B}$$
Step 1
Find individual rates
Pipe A fills $\frac{1}{12}$ of the tank per hour. Pipe B fills $\frac{1}{15}$ per hour.
$R_A = \frac{1}{12}, \quad R_B = \frac{1}{15}$
Step 2
Find combined rate
Add the two rates since both pipes contribute to filling.
$R_{\text{combined}} = \frac{1}{12} + \frac{1}{15} = \frac{5 + 4}{60} = \frac{9}{60} = \frac{3}{20}$
Step 3
Find total time
The time is the reciprocal of the combined rate.
$T = \frac{1}{R_{\text{combined}}} = \frac{20}{3} = 6\frac{2}{3}\text{ hours} = 6\text{ h } 40\text{ min}$
Problem Statement
A shopkeeper mixes two types of rice — one costing Rs. 40/kg and another costing Rs. 60/kg. In what ratio should he mix them to sell the mixture at Rs. 52/kg?
mixturealligation
Strategic Toolkit
$$\frac{Q_1}{Q_2} = \frac{C_{\text{mix}} - C_2}{C_1 - C_{\text{mix}}}$$
Step 1
Set up alligation cross
Place dearer (60) and cheaper (40) on the left, mixture (52) in the center.
$C_1 = 60, \quad C_2 = 40, \quad C_{\text{mix}} = 52$
Step 2
Cross-subtract diagonally
The difference next to $C_2$ gives $Q_1$'s part, and the difference next to $C_1$ gives $Q_2$'s part.
$Q_1 : Q_2 = (52 - 40) : (60 - 52) = 12 : 8 = 3 : 2$
Step 3
Interpret the ratio
Mix 3 parts of the Rs. 60/kg rice with 2 parts of the Rs. 40/kg rice.
$\text{Ratio} = 3 : 2$
Problem Statement
Ali invests Rs. 5000 for 6 months and Bilal invests Rs. 8000 for 4 months in a business. If the total profit is Rs. 6200, how much does Bilal receive?
partnership
Strategic Toolkit
$$\frac{P_{\text{Ali}}}{P_{\text{Bilal}}} = \frac{I_{\text{Ali}} \times T_{\text{Ali}}}{I_{\text{Bilal}} \times T_{\text{Bilal}}}$$
Step 1
Compute investment-time products
Calculate the product of investment and time for each partner.
$\text{Ali: } 5000 \times 6 = 30000, \quad \text{Bilal: } 8000 \times 4 = 32000$
Step 2
Find profit ratio
Simplify the ratio of the products.
$P_{\text{Ali}} : P_{\text{Bilal}} = 30000 : 32000 = 15 : 16$
Step 3
Calculate Bilal's share
Bilal gets $\frac{16}{31}$ of the total profit.
$P_{\text{Bilal}} = \frac{16}{15 + 16} \times 6200 = \frac{16}{31} \times 6200 = 3200$
Problem Statement
A train 150 metres long is running at 72 km/h. How long does it take to cross a 250-metre platform?
distance-speed-timetrains
Strategic Toolkit
$$T = \frac{L_{\text{train}} + L_{\text{platform}}}{S}$$
Step 1
Convert speed to m/s
Multiply km/h by $\frac{5}{18}$ to get m/s.
$S = 72 \times \frac{5}{18} = 20\text{ m/s}$
Step 2
Find total distance to cover
The train must clear its own length plus the platform length.
$D = 150 + 250 = 400\text{ m}$
Step 3
Calculate time
Divide total distance by speed.
$T = \frac{400}{20} = 20\text{ seconds}$
Problem Statement
What is the angle between the hour and minute hands of a clock at 3:20?
clocksangular speed
Strategic Toolkit
$$\theta = |30H - 5.5M|$$
Step 1
Identify $H$ and $M$
The time is 3 hours and 20 minutes.
$H = 3, \quad M = 20$
Step 2
Apply the clock angle formula
Substitute into $\theta = |30H - 5.5M|$.
$\theta = |30(3) - 5.5(20)| = |90 - 110| = 20°$
Memorize formulas
Distance-Speed-Time
Foundation of all motion problems — find the missing quantity when any two are known.
Formula
$D = S \times T$
Average Speed (Equal Distances)
Use when a journey covers equal distances at two different speeds — it is the harmonic mean, not the arithmetic mean.
Formula
$S_{\text{avg}} = \frac{2S_1 S_2}{S_1 + S_2}$
Combined Work Time
Time for two workers to complete a job together when working alone they take $T_1$ and $T_2$.
Formula
$T = \frac{T_1 \cdot T_2}{T_1 + T_2}$
Alligation Ratio
Find the mixing ratio when combining two components to achieve a target value.
Formula
$\frac{Q_1}{Q_2} = \frac{C_{\text{mix}} - C_2}{C_1 - C_{\text{mix}}}$
Boat and Stream Speeds
Derive boat speed and stream speed from downstream and upstream speeds.
Formula
$S_{\text{boat}} = \frac{S_{\text{down}} + S_{\text{up}}}{2}, \quad S_{\text{stream}} = \frac{S_{\text{down}} - S_{\text{up}}}{2}$
Clock Angle
Find the angle between the hour and minute hands at any time.
Formula
$\theta = |30H - 5.5M|$
Review
1
A stationery store sells notebooks for Rs. 50 each. During a sale, the store collected Rs. 4,200 from notebook sales and also received Rs. 200 in separate cash donations. How many notebooks were sold?
Ans
84
Review
Tap to reveal
Set up the equation
The donation is separate from notebook revenue, so only the Rs. 4,200 figure matters for finding the notebook count.
$$50n = 4200$$
Solve for $n$
Divide both sides by the price per notebook.
$$n = \frac{4200}{50} = 84$$
2
The sum of three consecutive even integers is 78. What is the product of the smallest and the largest integer?
Ans
672
Review
Tap to reveal
Express the integers algebraically
Three consecutive even integers differ by 2 each, so represent them as $x$, $x+2$, and $x+4$.
$$x + (x+2) + (x+4) = 78$$
Solve for the smallest integer
Simplify and solve the equation.
$$3x + 6 = 78 \implies x = 24$$
Compute the required product
Multiply the smallest (24) by the largest ($24 + 4 = 28$).
$$24 \times 28 = 672$$
3
A father is currently 4 times as old as his son. In 12 years, the father will be exactly twice as old as his son. How old is the son right now?
Ans
6 years
Review
Tap to reveal
Define variables and set up the equation
Let son = $x$, father = $4x$. Express the future relationship.
$$4x + 12 = 2(x + 12)$$
Solve for the son's age
Expand the right side and isolate $x$.
$$4x + 12 = 2x + 24 \implies 2x = 12 \implies x = 6$$
4
A delivery truck travels at a constant speed of 60 km/h for 2 hours and 30 minutes. How far does the truck travel?
Ans
150 km
Review
Tap to reveal
Convert time to decimal hours
2 hours 30 minutes $= 2 + \frac{30}{60} = 2.5$ hours.
$$T = 2.5 \text{ hours}$$
Apply the distance formula
Multiply speed by time.
$$D = 60 \times 2.5 = 150 \text{ km}$$
5
Two cities are 420 km apart. Car A departs from city X heading east at 80 km/h, and Car B departs from city Y heading west at 60 km/h. If they leave at the same time, after how many hours will they meet?
Ans
3 hours
Review
Tap to reveal
Determine relative speed
When two objects move toward each other, add their speeds to get the closing rate.
$$S_{\text{combined}} = 80 + 60 = 140 \text{ km/h}$$
Calculate meeting time
Divide the total distance by the combined speed.
$$T = \frac{420}{140} = 3 \text{ hours}$$
6
A driver travels from Lahore to Islamabad at 90 km/h and returns along the same route at 60 km/h. If the distance each way is 360 km, what is the average speed for the round trip?
Ans
72 km/h
Review
Tap to reveal
Apply the average speed formula for equal distances
When a round trip covers the same distance at two different speeds, use the harmonic-mean-like formula.
$$S_{\text{avg}} = \frac{2 \times 90 \times 60}{90 + 60}$$
Compute the result
Simplify the fraction to get the average speed.
$$S_{\text{avg}} = \frac{10800}{150} = 72 \text{ km/h}$$
7
A boat can travel 24 km downstream in 2 hours and the same 24 km upstream in 3 hours. What is the speed of the stream?
Ans
2 km/h
Review
Tap to reveal
Find downstream and upstream speeds
Divide the distance by the time in each direction.
$$S_{\text{down}} = \frac{24}{2} = 12 \text{ km/h}, \quad S_{\text{up}} = \frac{24}{3} = 8 \text{ km/h}$$
Apply the stream speed formula
The stream speed is half the difference between downstream and upstream speeds.
$$S_{\text{stream}} = \frac{12 - 8}{2} = 2 \text{ km/h}$$
8
A train 200 meters long crosses a platform 300 meters long in 25 seconds. What is the speed of the train in km/h?
Ans
72 km/h
Review
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Find total distance to cross
The train must travel its own length plus the platform length to fully clear the platform.
$$D = 200 + 300 = 500 \text{ meters}$$
Calculate speed in m/s then convert to km/h
Divide by time to get m/s, then multiply by $18/5$ to convert.
$$\frac{500}{25} = 20 \text{ m/s} = 20 \times \frac{18}{5} = 72 \text{ km/h}$$
9
Worker A can complete a painting job in 12 days and Worker B can complete it in 18 days. If they work together, how many days will it take to finish the job?
Ans
7.2 days
Review
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Compute individual work rates
Worker A does $1/12$ of the job per day, Worker B does $1/18$ per day.
$$\text{Rate}_A = \frac{1}{12}, \quad \text{Rate}_B = \frac{1}{18}$$
Add rates and invert for total time
The combined rate is the sum of individual rates. Total time is the reciprocal.
$$\frac{1}{12} + \frac{1}{18} = \frac{5}{36} \implies T = \frac{36}{5} = 7.2 \text{ days}$$
10
A water tank can be filled by Pipe A in 10 hours and by Pipe B in 15 hours. Pipe C can empty the full tank in 12 hours. If all three pipes are opened simultaneously, how long will it take to fill the tank?
Ans
12 hours
Review
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Set up the combined rate equation
Pipes A and B contribute positive rates; Pipe C (emptying) contributes a negative rate.
$$\text{Rate} = \frac{1}{10} + \frac{1}{15} - \frac{1}{12}$$
Compute the net rate using LCM
Find a common denominator to add and subtract the fractions.
$$\frac{6}{60} + \frac{4}{60} - \frac{5}{60} = \frac{5}{60} = \frac{1}{12}$$
Invert to find total time
The reciprocal of the net rate gives the time to fill the tank.
$$T = 12 \text{ hours}$$
11
A grocer mixes two varieties of rice: one costing Rs. 40 per kg and another costing Rs. 60 per kg. In what ratio should the two varieties be mixed so that the resulting mixture costs Rs. 48 per kg?
Ans
2 : 3
Review
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Apply the alligation formula
Place the cheaper price and dearer price on opposite sides, with the mixture price in the middle.
$$\frac{Q_{\text{expensive}}}{Q_{\text{cheap}}} = \frac{48 - 40}{60 - 48} = \frac{8}{12}$$
Simplify the ratio
Reduce $8 : 12$ to its simplest form.
$$2 : 3$$
12
Ahmed invests Rs. 60,000 for the entire year in a business, while Bilal invests Rs. 40,000 for only 9 months. If the total annual profit is Rs. 39,000, what is Bilal's share of the profit?
Ans
Rs. 13,000
Review
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Calculate investment-months for each partner
Multiply each partner's investment amount by the number of months invested.
$$\text{Ahmed} = 60000 \times 12 = 720{,}000, \quad \text{Bilal} = 40000 \times 9 = 360{,}000$$
Determine the profit-sharing ratio
Simplify the ratio of investment-months.
$$720000 : 360000 = 2 : 1$$
Calculate Bilal's share
Bilal gets $\frac{1}{3}$ of the total profit.
$$\frac{1}{3} \times 39000 = \text{Rs. } 13000$$
13
What is the angle between the hour hand and the minute hand of a clock at exactly 3:20?
Ans
20°
Review
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Apply the clock angle formula
Use $\theta = |30H - 5.5M|$ with $H = 3$ and $M = 20$.
$$\theta = |30(3) - 5.5(20)| = |90 - 110|$$
Compute the absolute value
The absolute difference gives the angle between the hands.
$$\theta = 20°$$
14
At what time between 4 o'clock and 5 o'clock will the hands of a clock first form a right angle (90°)?
Ans
4 : 05 $\frac{5}{11}$ min
Review
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Set up the clock angle equation
For a right angle, set $|30H - 5.5M| = 90$ with $H = 4$.
$$|120 - 5.5M| = 90$$
Solve both cases
Case 1: $120 - 5.5M = 90 \implies M = \frac{60}{11} = 5\frac{5}{11}$. Case 2: $120 - 5.5M = -90 \implies M = \frac{420}{11} = 38\frac{2}{11}$.
$$M = 5\frac{5}{11} \text{ (first occurrence) or } M = 38\frac{2}{11} \text{ (second occurrence)}$$
15
If $n$ is a positive integer, which of the following expressions must always be odd?
Ans
$2n + 1$
Review
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Analyze each expression for parity
Consider what happens when $n$ is odd versus even for each expression.
$$2n \text{ is always even} \implies 2n + 1 \text{ is always odd}$$
Verify with counterexamples for distractors
Show that the other expressions can be even. For example, when $n = 3$: $n^2 + 1 = 10$ (even), $n(n+1) = 12$ (even), $n^2 + n = 12$ (even).
$$\text{Only } 2n + 1 \text{ is guaranteed odd for all positive integers } n$$
16
Two trains start from stations 240 km apart at the same time and travel toward each other. The faster train moves at 70 km/h and the slower at 50 km/h. How far from the faster train's starting station will the two trains meet?
Ans
140 km
Review
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Find the meeting time
Add the two speeds and divide the total distance by the combined speed.
$$T = \frac{240}{70 + 50} = \frac{240}{120} = 2 \text{ hours}$$
Calculate the faster train's distance
Multiply the faster train's speed by the meeting time.
$$D_{\text{fast}} = 70 \times 2 = 140 \text{ km}$$
17
Five years ago, the ratio of Ali's age to Sara's age was 3 : 2. Five years from now, the ratio will be 5 : 4. How old is Ali right now?
Ans
20 years
Review
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Set up the ratio equations
Express past and future ages in terms of a common variable $x$, then use the future ratio to create an equation.
$$\frac{3x + 10}{2x + 10} = \frac{5}{4}$$
Solve for $x$ and find Ali's current age
Cross-multiply and solve, then add 5 to the past age to get the current age.
$$12x + 40 = 10x + 50 \implies x = 5 \implies \text{Ali now} = 3(5) + 5 = 20$$
18
A fruit seller buys 60 oranges and 40 apples for a total of Rs. 2,400. If each orange costs Rs. 10 less than each apple, what is the price of one apple?
Ans
Rs. 30
Review
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Express the total cost as an equation
Let apple price $= a$, orange price $= a - 10$. Write the total cost equation.
$$60(a - 10) + 40a = 2400$$
Solve for the apple price
Expand and simplify to isolate $a$.
$$100a - 600 = 2400 \implies 100a = 3000 \implies a = 30$$
19
A boat covers a certain distance downstream in 4 hours and the same distance upstream in 6 hours. If the speed of the stream is 5 km/h, what is the speed of the boat in still water?
Ans
25 km/h
Review
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Find the distance using the stream speed
Stream speed $= \frac{D/4 - D/6}{2} = \frac{D}{24}$. Set this equal to 5 km/h.
$$\frac{D}{24} = 5 \implies D = 120 \text{ km}$$
Compute both directional speeds
Find downstream and upstream speeds from the known distance and times.
$$S_{\text{down}} = \frac{120}{4} = 30, \quad S_{\text{up}} = \frac{120}{6} = 20$$
Apply the boat speed formula
The boat's still-water speed is the average of downstream and upstream speeds.
$$S_{\text{boat}} = \frac{30 + 20}{2} = 25 \text{ km/h}$$
20
A project can be completed by 5 workers in 20 days. After working for 8 days, 2 workers leave the team. How many additional days are needed to finish the project?
Ans
20 days
Review
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Calculate total and remaining work
Total work in worker-days, minus the work already done in the first 8 days.
$$\text{Total} = 5 \times 20 = 100, \quad \text{Remaining} = 100 - 40 = 60 \text{ worker-days}$$
Compute days with reduced workforce
Divide remaining work by the number of workers who stayed.
$$\frac{60}{5 - 2} = \frac{60}{3} = 20 \text{ days}$$
Question 1 of 20
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1
A stationery store sells notebooks for Rs. 50 each. During a sale, the store collected Rs. 4,200 from notebook sales and also received Rs. 200 in separate cash donations. How many notebooks were sold?