Solving Real-World Arithmetic Word Problems

Welcome back.

In the previous lesson, you practised extracting exactly what a passage supports. General Aptitude now shifts from verbal evidence to numerical modelling: turning a short story about prices, workers, travel, or marks into a small set of quantities and relationships.

For GATE, these questions are rarely difficult because of advanced mathematics. They become difficult when the base quantity, unit, or relationship is chosen incorrectly. This lesson builds a common framework for ratios, proportions, percentages, averages, profit and loss, speed–time–distance, and time–work problems.

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Start with the structure, not the arithmetic

Every arithmetic word problem has three ingredients:

1. Quantities: money, distance, people, marks, hours, articles, or work.
2. A relationship: a ratio, percentage, average, rate, or change.
3. A requested quantity: the unknown you must calculate.

Before calculating, write a compact mathematical model. For example:

• “The ratio of boys to girls is $3:5$” means boys and girls can be written as $3k$ and $5k$.
• “A price rises by $20\%$” means the new price is $1.20$ times the old price.
• “A can finish work in $12$ days” means A completes $\frac{1}{12}$ of the work per day.
• “A car travels at $60$ km/h” means $d=60t$, if $t$ is measured in hours.

The calculation should follow the model. If you begin manipulating numbers before deciding what they represent, you are likely to apply the correct formula to the wrong base.

A useful GATE habit is to annotate the quantities with units:

$$\text{Speed}=\frac{\text{distance}}{\text{time}}$$ $$\text{Work rate}=\frac{\text{work}}{\text{time}}$$

If units do not agree, pause. For instance, do not combine minutes with kilometres per hour without converting minutes to hours.

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Ratios and proportions: represent quantities using common parts

A ratio compares quantities in a specified order. Thus,

$$A:B=3:5$$

means

$$\frac{A}{B}=\frac{3}{5}$$

and can be represented as

$$A=3k,B=5k$$

for some common multiplier $k$.

Dividing a total in a ratio

Suppose a fund of $96$ is divided between A and B in the ratio $3:5$.

The total number of parts is

$$3+5=8$$

One part is

$$\frac{96}{8}=12$$

Therefore,

$$A=3\times 12=36$$ $$B=5\times 12=60$$

The important idea is that a ratio is not itself an absolute quantity. It specifies only the relative number of equal parts.

Using a difference instead of a total

If two numbers are in the ratio $4:7$ and their difference is $27$, write them as $4k$ and $7k$. Their difference is:

$$7k-4k=3k$$

So,

$$3k=27$$ $$k=9$$

The numbers are $36$ and $63$.

Here, the difference in ratio parts, not their sum, is the relevant quantity.

Proportions

A proportion says that two ratios are equal:

$$\frac{a}{b}=\frac{c}{d}$$

Cross multiplication gives:

$$ad=bc$$

Use this only after preserving the order of quantities. For example, if $5$ machines make $240$ components in a fixed time, and all machines have equal productivity, then $8$ machines make $x$ components in the same time:

$$\frac{5}{240}=\frac{8}{x}$$ $$5x=8\times 240$$ $$x=384$$

The correspondence must stay consistent: machines with machines, components with components.

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Percentages: identify the base before applying the percent

The word “percent” means “per hundred.” Therefore:

$$p\%=\frac{p}{100}$$

The central relationship is:

$$\text{percentage}=\frac{\text{part}}{\text{base}}\times 100$$

The base is the quantity of which the percentage is taken.

For example, $15\%$ of $400$ is:

$$\frac{15}{100}\times 400=60$$

But if $60$ is what percent of $400$, then:

$$\frac{60}{400}\times 100=15\%$$

These are inverse forms of the same relationship.

Percentage changes use the original value as the base

If a quantity changes from $x$ to $y$, its percentage change is:

$$\frac{y-x}{x}\times 100$$

The denominator is the original quantity $x$.

If a salary rises from $40,000$ to $46,000$, the increase is $6,000$. Therefore:

$$\frac{6,000}{40,000}\times 100=15\%$$

When wording says “$A$ is $p\%$ more than $B$,” use $B$ as the base. “Than” usually identifies the comparison base.

Multipliers are faster than repeated percentage calculations

Change	Multiplier
Increase by $p\%$	$1+\frac{p}{100}$
Decrease by $p\%$	$1-\frac{p}{100}$
Increase by $20\%$	$1.20$
Decrease by $20\%$	$0.80$

If a price first increases by $20\%$ and then decreases by $20\%$, the final multiplier is:

$$1.20\times 0.80=0.96$$

So the final price is $96\%$ of the original price: an overall decrease of $4\%$.

This is a classic trap. Equal percentage increases and decreases do not cancel because they use different bases.

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Averages: work with totals, not intuition

An average is equal distribution of a total:

$$\text{Average}=\frac{\text{Total}}{\text{Number of observations}}$$

Equivalently:

$$\text{Total}=\text{Average}\times \text{Number of observations}$$

Suppose five test scores have an average of $68$. Their total is:

$$5\times 68=340$$

If a sixth score of $83$ is added, the new total becomes:

$$340+83=423$$

The new average is:

$$\frac{423}{6}=70.5$$

Adding, removing, or replacing an observation

For these questions, first recover the old total. Then modify that total.

• Adding a member: add the new value and increase the count.
• Removing a member: subtract the removed value and reduce the count.
• Replacing a value: subtract the old value, add the new one; the count stays unchanged.

Do not average two averages directly unless the two groups have equal sizes. If one group has $10$ students and another has $40$, the second group must contribute four times as much to the combined average.

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Profit, loss, discount, and the correct reference price

Use these terms precisely:

• Cost Price, $CP$: purchase price or investment.
• Selling Price, $SP$: price received on selling.
• Marked Price, $MP$: displayed or list price.
• Discount: reduction from marked price.

The core relationships are:

$$\text{Profit}=SP-CP$$ $$\text{Loss}=CP-SP$$ $$\text{Profit percent}=\frac{SP-CP}{CP}\times 100$$ $$\text{Loss percent}=\frac{CP-SP}{CP}\times 100$$

Profit or loss percentage is calculated on cost price. A discount percentage is calculated on marked price.

A layered price problem

An article has cost price $800$. It is marked $25\%$ above cost price, and then a discount of $10\%$ is offered.

First calculate marked price:

$$MP=800\times 1.25=1000$$

Now apply the discount to $MP$:

$$SP=1000\times 0.90=900$$

Profit is:

$$900-800=100$$

Therefore profit percentage is:

$$\frac{100}{800}\times 100=12.5\%$$

A frequent error is to subtract the discount percentage directly from the markup percentage and conclude a $15\%$ profit. That does not work because the two percentages have different bases.

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Rates and travel: distance, speed, and time

The fundamental identity is:

$$d=st$$

where $d$ is distance, $s$ is speed, and $t$ is time.

Its rearrangements are:

$$s=\frac{d}{t}$$ $$t=\frac{d}{s}$$

Keep units consistent

The standard conversion is:

$$1\text{ km/h}=\frac{5}{18}\text{ m/s}$$ $$1\text{ m/s}=\frac{18}{5}\text{ km/h}$$

For example:

$$72\text{ km/h}=72\times \frac{5}{18}=20\text{ m/s}$$

Average speed is not usually an arithmetic mean

Average speed always means:

$$\text{Average speed}=\frac{\text{total distance}}{\text{total time}}$$

Suppose a car covers $60$ km at $30$ km/h and another $60$ km at $60$ km/h.

Time for the first part:

$$\frac{60}{30}=2\text{ hours}$$

Time for the second part:

$$\frac{60}{60}=1\text{ hour}$$

Thus:

$$\text{Average speed}=\frac{120}{3}=40\text{ km/h}$$

It is not $45$ km/h. An arithmetic mean of speeds is valid only when the durations at those speeds are equal.

Relative speed

For objects moving in opposite directions, their separation changes at the sum of their speeds.

For objects moving in the same direction, their separation changes at the difference of their speeds.

If two trains travel towards each other at $45$ km/h and $55$ km/h, their relative speed is:

$$45+55=100\text{ km/h}$$

If they are $250$ km apart, meeting time is:

$$\frac{250}{100}=2.5\text{ hours}$$

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Time and work: convert completion times into daily rates

Time-and-work questions are rate questions. Instead of kilometres per hour, the rate is “fraction of the whole task per day.”

If A completes a task in $12$ days, A’s one-day work is:

$$\frac{1}{12}$$

If B completes the same task in $18$ days, B’s one-day work is:

$$\frac{1}{18}$$

Working together, their daily work is:

$$\frac{1}{12}+\frac{1}{18}=\frac{3+2}{36}=\frac{5}{36}$$

So the time required for one whole task is:

$$\frac{1}{5/36}=\frac{36}{5}=7.2\text{ days}$$

The central rule is:

Add workers’ rates, not their completion times.

Workers joining or leaving

When a worker leaves, split the problem into periods:

1. Calculate the work completed in the first period.
2. Subtract it from the whole work.
3. Use the rate of the remaining workers for the unfinished part.

The same model handles pipes and cisterns. A filling pipe has a positive rate; a leak or outlet has a negative rate.

If one pipe fills a tank in $6$ hours and a leak empties it in $12$ hours, their combined rate is:

$$\frac{1}{6}-\frac{1}{12}=\frac{1}{12}$$

Thus, with the leak present, the tank fills in $12$ hours.

Direct and inverse relationships

With constant worker efficiency and daily hours:

• More workers means less time for the same work.
• More work means more time for the same workforce.
• More working hours per day means fewer days for the same work.

These are useful checks, but rate equations remain safer when the wording includes workers joining, leaving, working at different efficiencies, or changing schedules.

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A compact method for GATE arithmetic questions

Under time pressure, use this sequence:

1. Name the requested quantity.
   Is the question asking for a share, a percentage, a price, a time, or a rate?

2. Write the governing relation.
   Examples include $A=3k$, $d=st$, or $\text{work}=\text{rate}\times \text{time}$.

3. Identify the base.
   For percentage, profit, loss, and discounts, explicitly write what quantity receives the percentage.

4. Check units and direction.
   If speed decreases for a fixed distance, time must increase. If a discount is applied, selling price must be lower than marked price.

5. Estimate before finalising.
   A $10\%$ discount should not produce a selling price greater than the marked price. Two workers together should not take longer than the faster worker working alone.

When reviewing a mistake, classify it accurately:

• A wrong formula or relation is a conceptual error.
• Choosing the wrong percentage base is an interpretation error.
• A fraction, sign, or arithmetic slip is a calculation error.
• Spending too long on routine arithmetic is a time-management error.

That classification will make later revision much more targeted than simply re-solving every question.

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Key takeaways

Most GATE arithmetic problems reduce to a few stable models:

• A ratio $a:b$ can be represented as $ak$ and $bk$.
• A proportion equates two consistently ordered ratios.
• Percentages require a clearly identified base; successive changes are handled with multipliers.
• Average is total divided by count, so adding or removing values should be handled through totals.
• Profit and loss percentages use $CP$; discounts use $MP$.
• Travel problems rely on $d=st$, with average speed calculated from total distance divided by total time.
• Work problems rely on rates: a task finished in $n$ days corresponds to a daily rate of $\frac{1}{n}$.

In the next lesson, you will extend this numerical foundation to equations, inequalities, exponents, logarithms, and numerical sequences—tools that make more compact GATE aptitude questions manageable.