CAT 2025 Arithmetic Shortcuts: 1 Formula to Solve 5 Types of Questions

Importance of Arithmetic in CAT

As discussed above arithmetic has a significant topicwise weightage in CAT Quantitative Aptitude section. It makes almost 40% – 45% of the total questions asked in CAT QA. You need to focus on each topic of arithmetic to ace this section.

To know the syllabus of QA section CAT, click on: Quantitative aptitude syllabus for CAT – 2025.

Important arithmetic topics for CAT

As arithmetic is very important to ace the QA section of CAT, we should know which topic we need to focus more on. Therefore, we should know the important topics and their weightage in recent years.

Important arithmetic topics for CAT are

For complete syllabus of CAT, visit CAT syllabus 2025.

Have a look at the number of questions asked in CAT from arithmetic during last 5 years.

The Magic Formula 1: Percentage Change

CAT arithmetic shortcuts are important to increase accuracy and save time. Here, we will discuss 1 formula to solve 5 CAT question types with short approach.

The formula:

Final Value formula:
$Final\ Value = Base\ or\ Initial\ Value \times \left(1 \pm \frac{Percentage\ Change}{100}\right)$

Other form (Percentage Change formula):
$Percentage\ Change = \frac{Final\ Value - Initial\ Value}{Initial\ Value} \times 100%$

Applications of the Formula in 5 CAT question types

We are going to discuss application of percentage change formula in 5 different types of questions.

Type 1: To find the Percentage Increase or Decrease

This formula is used

• To determine the percentage increase or decrease in population of a town

• To solve the questions of depreciation or appreciation

• To determine the percentage while comparing two values

• To solve expenditure-based questions

Example:
A number is increased by 20%. By what percent should it be decreased to get the original value again?

Solution:

A number is increased by 20%:

$ \text{New Value} = \text{Initial Value} \times \left(1 + \frac{20}{100}\right) = \text{Initial Value} \times 1.2 $

$ \text{Required percentage decrease: } \text{Percentage Change} = \frac{\text{Change}}{\text{Base}} \times 100\% $

$ = \frac{\text{Initial Value} \times 1.2 - \text{Initial Value}}{\text{Initial Value} \times 1.2} \times 100\% $

$ = \frac{0.2 \times \text{Initial Value}}{1.2 \times \text{Initial Value}} \times 100\% $

$ = 16.66\% $

Type 2: Application in Profit and Loss

This formula is used

• To determine the selling price (Final Value) if Cost Price (Initial Value) and Profit or Loss percentage (Percentage Change) is given.

• To determine the Profit or Loss percentage (Percentage Change) if selling price (Final Value) and Cost Price (Initial Value) is given.

Example:

A shopkeeper buys an article at ₹1000 and gains 20%. Find the Selling Price.

Solution:

$ \text{Selling Price} = 1000 \times (1 + 0.2) = \text{₹1200} $
Similarly, if he makes a 20% loss:
$ \text{Selling Price} = 1000 \times 0.8 = \text{₹800} $

Type 3. Application in Simple Interest (SI)

Simple interest is equivalent to percentage change of the principal over time.

Example:

Principal = ₹2000, Rate = 10%, Time = 3 years. Calculate the amount after 3 years.

Solution:

Total percentage change = $10 \times 3 = 30%$

Apply the percentage change formula:

$\text{Amount} = 2000 \times \left(1 + \frac{30}{100}\right) = 2600$ Rs

Type 4. Application in Compound Interest (CI)

Repeated percentage change year after year gives the amount if interest is compounded annually.

Example:

Principal = ₹2000, Rate = 10% per annum compounded annually, Time = 3 years. Calculate the amount after 3 years.

Solution:

$ \text{Amount after 3 years} = 2000 \times \left(1 + \frac{10}{100}\right) \times \left(1 + \frac{10}{100}\right) \times \left(1 + \frac{10}{100}\right) = 2662 $ Rs
$ \text{Compound Interest} = (2662 - 2000) = 662 $ Rs

Tip → You can think of compound interest as successive percentage changes.

Type 5. Application in Successive Percentage Change

Questions bases two or more consecutive changes (increase/decrease) can also be solved with the same principle.

Shortcut:

Example:

Price of an article increases by 20% and then decreases by 10%. Find the net percentage change in the price of the article.

Solution:

Take Base Value or Initial Value = 100

$ \text{Final Value} = 100 \times \left(1 + \frac{20}{100}\right) \times \left(1 - \frac{10}{100}\right) $
$ \Rightarrow \text{Final Value} = 100 \times 1.2 \times 0.9 $
$ \Rightarrow \text{Final Value} = 108 $
$ \text{Net Effect} = 8% \ \text{increase} $

The Magic Formula 2: Weighted Average

CAT arithmetic shortcuts are important to increase accuracy and save time. Here, we will discuss 1 formula to solve 5 CAT question types with short approach where average needs to be calculated.

The formula:

$ \text{Weighted Average} = \frac{\text{Sum of (Weights × Values)}}{\text{Sum of Weights}} $

Applications of the Formula in 5 CAT question types

We are going to discuss the application of weighted average in 5 different types of questions.

Type 1: To find the average value when two or more ingredients at different price are mixed in a given ratio

This formula is used

• To determine the ratio of mixing two or more ingredients

• To determine the average price when two or more quantities are added in a ratio

• In milk and water problems

Example:

If two varieties of wheat costing ₹40 and ₹60 are mixed in a ratio of 3:4 respectively, find the average price of wheat.

Solution:

Using the formula:

$ \text{Weighted Average} = \frac{\text{Sum of (Weights × Values)}}{\text{Sum of Weights}} $
$ \text{Required Average Price} = \frac{40 \times 3 + 60 \times 4}{3 + 4} = 51.42 $

Type 2: Application in Time, Speed, and Distance

This formula is used

• To determine the average speed in TSD questions asked in CAT when different distances are covered at different speeds.

Example:

Amit covers 80 km at 40 km/h and 120 km at 80 km/h. Find his average speed for the journey.

Solution:

Time taken to cover 80 km at 40 km/h:
$ \text{Time}_1 = \frac{80}{40} = 2\ \text{hours} $

Time taken to cover 120 km at 80 km/h:
$ \text{Time}_2 = \frac{120}{80} = 1.5\ \text{hours} $

Average speed formula:
$ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} $

$ \text{Average Speed} = \frac{80 + 120}{2 + 1.5} = \frac{200}{3.5} = 57.14\ \text{km/h} $

Trick: If two equal distances are covered at different speeds, the average speed is the weighted harmonic mean.

$ \text{Average Speed} = \frac{2 \times (\text{Speed}_1 \times \text{Speed}_2)}{\text{Speed}_1 + \text{Speed}_2} $

Type 3. Application in Time and Work

When two or more workers having different efficiencies work together, their combined work rate is a weighted average of their individual rates.

Example:

A can complete a job in 10 days, B in 20 days. In how many days will they finish the work together?

Solution:

Work done by A in 1 day:
$ \text{Work}_A = \frac{1}{10} $

Work done by B in 1 day:
$ \text{Work}_B = \frac{1}{20} $

Work done by A and B together in 1 day:
$ \text{Work}_{\text{Together}} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20} $

Time taken to finish the work together:
$ \text{Time} = \frac{1}{\text{Work}_{\text{Together}}} = \frac{1}{3/20} = \frac{20}{3} \approx 6.67\ \text{days} $

Type 4. Application in Marks and Averages

The concept of weighted average is used when adding or removing, a student’s marks differ from average. Also, this concept is used to quick calculation of weighted average.

Example:

The average marks of 10 students is 40. A new student with 51 marks joins. Find the new average.

Solution:

Using the concept of weighted average:

$ \text{New Average} = \frac{10 \times 40 + 51}{10 + 1} = \frac{451}{11} = 41 $

Type 5. Application in Profit & Loss

When two or more items purchased at same rate and sold at different profit/loss percentages, the net effect can be quickly found with the help of this formula.

Shortcut:

$ \text{Final Profit or Loss %} = \frac{\text{Sum of all Profit and Loss %}}{\text{Number of Items}} $

(Here, loss percentage is taken as negative.)

Example:

A trader makes a profit of 25% on one item and a loss of 15% on another, both of equal cost price. Find the overall profit or loss percentage.

Solution:

Using the formula:

$ \text{Final Profit %} = \frac{25 + (-15)}{2} = \frac{10}{2} = 5% $

Why These Shortcuts Work

These arithmetic shortcuts for CAT are necessary to ace Arithmetic section of CAT. If we look at the topics like Profit and loss, mixtures, SI and CI, all are looked different but in reality they can be reduced to a similar logic.

As we have seen, Profit and loss, SI and CI are extension of percentages. Mixtures is an extension of average. Time and Work is extension of average and ratio.

Percentage formula unifies all the topics like profit and loss, SI and CI. Average unifies all the five topics discussed above.

Key Advantages

• It reduces time

• No need to learn specific formulas for different concepts

• It reduces confusion

• It increases accuracy

Previous Years Questions

Q1. During the first year, the population of the town increased by 12%. The next year, due to some contagious disease, it decreased by 8%. At the end of the second year, the population was 64,400. Find the population of the town at the beginning of the first year.

1. 50,000

2. 54,750

3. 62,500

4. 50,500

Solution:

Let the initial population be $x$.

Population after 12% increase: $1.12x$

Population after 8% decrease: $0.92 \times 1.12x = 64400$

$x = \frac{64400}{0.92 \times 1.12} = \frac{64400}{1.0304} = 62500$

Answer: 62,500

Q2. A notebook was sold for Rs. 28, at a profit of 12%. If it had been sold for Rs. 26.25, what would have been the profit percentage?

1. 5%

2. 7%

3. 4%

4. 6%

Solution:

Given $SP = 28$, Profit % = 12%

$CP = \frac{SP \times 100}{100 + \text{Profit %}} = \frac{28 \times 100}{112} = 25$

If $SP = 26.25$, Profit % = $\frac{26.25 - 25}{25} \times 100 = 5%$

Answer: 5%

Q3. The interest on a certain deposit at 4.5% p.a. is INR 135 in one year. How much will the additional interest in one year be on the same deposit at 5% p.a.?

1. INR 16

2. INR 15

3. INR 14

4. INR 18

Solution:

Simple Interest formula: $SI = \frac{P \times R \times T}{100}$

Given $SI = 135$, $R = 4.5%$, $T = 1$ year

$135 = \frac{P \times 4.5 \times 1}{100} \Rightarrow P = \frac{135 \times 100}{4.5} = 3000$

SI at 5%: $SI = \frac{3000 \times 5 \times 1}{100} = 150$

Additional interest = $150 - 135 = 15$

Answer: INR 15

Q4. A is 50% more efficient than B. B finished the same work in 20 days. If A and B worked together, how much time would they take to finish the same work?

1. 10 days

2. 7 days

3. 8 days

4. 9 days

Solution:

Let efficiency of B = 100 units

Efficiency of A = $100 + 50 = 150$ units

Efficiency ratio: $A:B = 150:100 = 3:2$

Total efficiency = $3 + 2 = 5$ units

Total work = Efficiency of B × Time taken by B = $2 \times 20 = 40$ units

Time taken by A + B = $\frac{\text{Total Work}}{\text{Total Efficiency}} = \frac{40}{5} = 8$ days

Answer: 8 days

Q5. Prasad goes 96 km on a bike at 16 km/hr, 124 km at 31 km/hr in a car, and 105 km at 7 km/hr in a horse cart. Find his average speed for the entire distance travelled.

1. 16 km/hr

2. 13 km/hr

3. 7 km/hr

4. 11 km/hr

Solution:

Total distance: $96 + 124 + 105 = 325$ km

Time taken: $t = \frac{96}{16} + \frac{124}{31} + \frac{105}{7} = 6 + 4 + 15 = 25$ hours

Average speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{325}{25} = 13\ \text{km/hr}$

Answer: 13 km/hr

Best resources & practice material for CAT Number System

1. Arun Sharma: A Quantitative Approach for CAT (7th Edition)
2. Quantitative Aptitude for CAT by Nishit K Sinha

CAT 2025 Preparation Resources by Careers360

The candidates can download the various CAT preparation resources designed by Careers360 using the links given below.