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Arithmetic, Ratios, Percentages and Commercial Mathematics Notes for GATE CS

Arithmetic, Ratios, Percentages and Commercial Mathematics notes for GATE CS: 38 study cards covering concepts, formulas, shortcuts and exam traps, plus solve

arithmetic ratios percentages and commercial mathematics notes

Chapter Roadmap: Arithmetic and Commercial Mathematics

Chapter Roadmap 1 Averages, Median and Central Tendency Foundation of data representation and central values. Weightage: Moderate. 2 Ratios, Proportions and Percentages The core engine for comparative quant. Weightage: High. 3 Profit, Loss and Investment Returns Real-world financial applications. Weightage: Moderate.

Central Tendency: Finding the Center of Data

Central Tendency Provides a single value that represents the center or typical value of a dataset. Mean (Arithmetic Average) The mathematical balance point. Uses every single data point. Median The positional middle. Divides sorted data into two equal halves. Mode The most frequent value. Highlights the highest probability. Intuition Check If 8 people earn 50,000 and 1 earns 5,000,000, the mean is skewed by the millionaire. The median accurately reflects the "typical" person.

Weighted Average: Combining Different Groups

Weighted Average When merging groups of different sizes, a simple average of averages is mathematically incorrect. You must weight each average by its group size. \overset{x}{ˉ}_{co mbin e d} = \frac{n _{1} x ˉ _{1} + n _{2} x ˉ _{2} + \dots + n _{k} x ˉ _{k}}{n _{1} + n _{2} + \dots + n _{k}} Where n_{i} = number of items in group i, and \overset{x}{ˉ}_{i} = average of group i . Rule of Thumb The combined average will always be closer to the average of the larger group.

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Question 1

Level 1: Warm-up

Using the deviation shortcut, if the assumed mean is 100 and the sum of deviations for 5 observations is -25, what is the actual mean?

Question 2

Level 1: Warm-up

The ratio of two quantities is 3:5. If the sum of the quantities is strictly less than 50, what is the maximum possible integer value of the larger quantity?

Question 3

Level 1: Warm-up

Assertion (A): If the ratio of boys to girls in a class is 4:5, the total number of students can be 45. Reason (R): The total number of students must be perfectly divisible by the sum of the ratio parts.

Question 4

Level 1: Warm-up

The ratio of two positive numbers P and Q is 4 : 9 . If P = 20, what is the value of Q - P ?

Question 5

Level 1: Warm-up

The ratio of the speeds of two trains is 4 : 7 . If the sum of their speeds is at most 110 km/h, what is the maximum possible integer speed of the faster train?

Question 6

Level 1: Warm-up

Assertion (A): If the ratio of boys to girls in a school is 5 : 4, the total number of students can be 180 . Reason (R): The total number of students must be perfectly divisible by the difference of the ratio parts.

Question 7

Level 1: Warm-up

Two positive quantities X and Y are in the ratio 5 : 3 . If X - Y = 16, what is the value of X + Y ?

Question 8

Level 1: Warm-up

The ratio of the lengths of two wires is 3 : 8 . If the difference in their lengths is at most 25 cm, what is the maximum possible integer length of the longer wire?

Question 9

Level 1: Warm-up

Assertion (A): If the ratio of the number of boys to girls in a school is 3 : 4, the total number of students can be 45 . Reason (R): The total number of students must be perfectly divisible by the sum of the ratio parts.

Question 10

Level 1: Warm-up

In a dataset where a single extreme outlier is present, which measure of central tendency is robust and unaffected by the magnitude of the outlier?

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Arithmetic, Ratios, Percentages and Commercial Mathematics Notes for GATE CS

Arithmetic, Ratios, Percentages and Commercial Mathematics notes for GATE CS: 38 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Arithmetic and Commercial Mathematics

Chapter Roadmap

Averages, Median and Central Tendency

Foundation of data representation and central values. Weightage: Moderate.

Ratios, Proportions and Percentages

The core engine for comparative quant. Weightage: High.

Profit, Loss and Investment Returns

Real-world financial applications. Weightage: Moderate.

Central Tendency: Finding the Center of Data

Central Tendency

Provides a single value that represents the center or typical value of a dataset.

Mean (Arithmetic Average)

The mathematical balance point. Uses every single data point.

Median

The positional middle. Divides sorted data into two equal halves.

Mode

The most frequent value. Highlights the highest probability.

Intuition Check

If 8 people earn 50,000 and 1 earns 5,000,000, the mean is skewed by the millionaire. The median accurately reflects the "typical" person.

Weighted Average: Combining Different Groups

Weighted Average

When merging groups of different sizes, a simple average of averages is mathematically incorrect. You must weight each average by its group size.

\overset{x}{ˉ}_{co mbin e d} = \frac{n _{1} x ˉ _{1} + n _{2} x ˉ _{2} + \dots + n _{k} x ˉ _{k}}{n _{1} + n _{2} + \dots + n _{k}}

Where

n_{i}

= number of items in group

i

, and

\overset{x}{ˉ}_{i}

= average of group

i

Rule of Thumb

The combined average will always be closer to the average of the larger group.

The Median: Finding the Exact Middle

The Median

The positional center of a dataset. Step 1: Always arrange data in ascending order.

n

is ODD:
Median =

(\frac{n + 1}{2})^{th}

term.

n

is EVEN:
Median = Average of

(\frac{n}{2})^{th}

and

(\frac{n}{2} + 1)^{th}

terms.

Arithmetic, Ratios, Percentages and Commercial Mathematics: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Quantitative Aptitude MCQ

Using the deviation shortcut, if the assumed mean is 100 and the sum of deviations for 5 observations is -25, what is the actual mean?

A.
90
B.
95
C.
105
D.
125

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct application of the deviation shortcut formula, recognizable because it provides the assumed mean, the sum of deviations, and the count.

Step 1: Identify the given values: Assumed mean $A = 100$ , Sum of deviations $\sum d_{i} = - 25$ , Number of observations $n = 5$ .

Step 2: Apply the formula: $Actual Mean = A + \frac{\sum d _{i}}{n}$ .

Step 3: Calculate: $100 + \frac{- 25}{5} = 100 - 5 = 95$ .

Answer: B

Question 2 · Quantitative Aptitude MCQ

The ratio of two quantities is 3:5. If the sum of the quantities is strictly less than 50, what is the maximum possible integer value of the larger quantity?

A.
25
B.
30
C.
35
D.
40

Correct Answer:

Step-by-Step Solution

Insight: The sum of the quantities must be a multiple of the sum of the ratio parts (3+5=8).

Exam route: Sum = 8k < 50. Max integer k is 6 (since 86 = 48 < 50). Larger quantity = 5k = 5 6 = 30.

Learning route:

Let the two quantities be 3k and 5k.
The sum of the quantities is 3k + 5k = 8k.
We are given that the sum is strictly less than 50: 8k < 50.
Solve for k: k < 6.25. The maximum integer value for k is 6.
Calculate the larger quantity: 5k = 5 * 6 = 30.

Answer: 30.

Question 3 · Quantitative Aptitude MCQ

Assertion (A): If the ratio of boys to girls in a class is 4:5, the total number of students can be 45.

Reason (R): The total number of students must be perfectly divisible by the sum of the ratio parts.

A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is not the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Correct Answer:

Step-by-Step Solution

Insight: The total quantity distributed in a ratio must be perfectly divisible by the sum of the ratio parts.

Exam route: Ratio 4:5 means sum of parts = 9. Total students must be a multiple of 9. 45 is a multiple of 9, so A is true. R correctly explains this property.

Learning route:

Analyze Assertion (A): The ratio of boys to girls is 4:5. The total number of parts is 4 + 5 = 9. For the number of students to be an integer, the total must be divisible by 9. Since 45 / 9 = 5, a total of 45 is possible. Assertion A is true.
Analyze Reason (R): The reason states that the total must be divisible by the sum of the ratio parts. This is the exact mathematical rule used to verify Assertion A. Reason R is true and correctly explains A.

Answer: Both A and R are true and R is the correct explanation of A.

Question 4 · Quantitative Aptitude MCQ

The ratio of two positive numbers $P$ and $Q$ is $4 : 9$ . If $P = 20$ , what is the value of $Q - P$ ?

A.
20
B.
25
C.
45
D.
65

Correct Answer:

Step-by-Step Solution

Insight: Use the multiplier $k$ to find the actual values of $P$ and $Q$ , then calculate their difference.

Exam route: $P = 4 k = 20 ⟹ k = 5$ . $Q = 9 k = 45$ . $Q - P = 45 - 20 = 25$ .

Learning route:

Represent the numbers using the multiplier $k$ : $P = 4 k$ and $Q = 9 k$ .
Use the given value of $P$ to find $k$ : $4 k = 20 ⟹ k = 5$ .
Calculate the value of $Q$ : $Q = 9 \times 5 = 45$ .
Find the difference $Q - P$ : $45 - 20 = 25$ .

Answer: 25.

Question 5 · Quantitative Aptitude MCQ

The ratio of the speeds of two trains is $4 : 7$ . If the sum of their speeds is at most $110$ km/h, what is the maximum possible integer speed of the faster train?

A.
60
B.
65
C.
70
D.
77

Correct Answer:

Step-by-Step Solution

Insight: The sum of the speeds must be a multiple of the sum of the ratio parts ( $4 + 7 = 11$ ). Use the inequality to find the maximum multiplier $k$ .

Exam route: Sum = $11 k \leq 110 ⟹ k \leq 10$ . Max integer $k$ is 10. Faster train = $7 k = 70$ .

Learning route:

Represent the speeds as $4 k$ and $7 k$ .
The sum of their speeds is $4 k + 7 k = 11 k$ .
Apply the constraint: $11 k \leq 110$ .
Solve for $k$ : $k \leq 10$ . The maximum integer value for $k$ is 10.
Calculate the speed of the faster train: $7 k = 7 \times 10 = 70$ km/h.

Answer: 70.

Question 6 · Quantitative Aptitude MCQ

Assertion (A): If the ratio of boys to girls in a school is $5 : 4$ , the total number of students can be $180$ .

Reason (R): The total number of students must be perfectly divisible by the difference of the ratio parts.

A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is not the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Correct Answer:

Step-by-Step Solution

Insight: The total quantity distributed in a ratio must be perfectly divisible by the sum of the ratio parts, not the difference.

Exam route: Ratio $5 : 4 ⟹$ Sum of parts = 9. $180/9 = 20$ . Assertion A is true. Reason R claims divisibility by the difference ( $5 - 4 = 1$ ), which is the wrong rule. R is false.

Learning route:

Analyze Assertion (A): The ratio of boys to girls is $5 : 4$ . The total number of parts is $5 + 4 = 9$ . For the number of students to be an integer, the total must be divisible by 9. Since $180/9 = 20$ , a total of 180 is possible. Assertion A is true.
Analyze Reason (R): The reason states that the total must be divisible by the difference of the ratio parts ( $5 - 4 = 1$ ). While 180 is divisible by 1, this is not the mathematical rule that governs ratio distributions. The correct rule uses the sum of the parts. Reason R is false.

Answer: A is true but R is false.

Question 7 · Quantitative Aptitude MCQ

Two positive quantities $X$ and $Y$ are in the ratio $5 : 3$ . If $X - Y = 16$ , what is the value of $X + Y$ ?

A.
-64
B.
64
C.
32
D.
128

Correct Answer:

Step-by-Step Solution

Insight: Use the multiplier $k$ to represent the quantities, then use the given difference to find $k$ and subsequently the sum.

Exam route: $X = 5 k, Y = 3 k$ . $X - Y = 2 k = 16 ⟹ k = 8$ . $X + Y = 8 k = 64$ .

Learning route:

Represent the quantities using the multiplier $k$ : $X = 5 k$ and $Y = 3 k$ .
Use the given difference to find $k$ : $X - Y = 5 k - 3 k = 2 k$ .
Equate to the given value: $2 k = 16 ⟹ k = 8$ .
Calculate the sum $X + Y$ : $5 k + 3 k = 8 k = 8 \times 8 = 64$ .

Answer: 64.

Question 8 · Quantitative Aptitude MCQ

The ratio of the lengths of two wires is $3 : 8$ . If the difference in their lengths is at most $25$ cm, what is the maximum possible integer length of the longer wire?

A.
32
B.
36
C.
40
D.
48

Correct Answer:

Step-by-Step Solution

Insight: The difference of the lengths must be a multiple of the difference of the ratio parts ( $8 - 3 = 5$ ). Use the inequality to find the maximum multiplier $k$ .

Exam route: Difference = $5 k \leq 25 ⟹ k \leq 5$ . Max integer $k$ is 5. Longer wire = $8 k = 40$ .

Learning route:

Represent the lengths as $3 k$ and $8 k$ .
The difference in their lengths is $8 k - 3 k = 5 k$ .
Apply the constraint: $5 k \leq 25$ .
Solve for $k$ : $k \leq 5$ . The maximum integer value for $k$ is 5.
Calculate the length of the longer wire: $8 k = 8 \times 5 = 40$ cm.

Answer: 40.

Question 9 · Quantitative Aptitude MCQ

Assertion (A): If the ratio of the number of boys to girls in a school is $3 : 4$ , the total number of students can be $45$ .

Reason (R): The total number of students must be perfectly divisible by the sum of the ratio parts.

A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is not the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Correct Answer:

Step-by-Step Solution

Insight: The total quantity distributed in a ratio must be perfectly divisible by the sum of the ratio parts, not the individual parts.

Exam route: Ratio $3 : 4 ⟹$ Sum of parts = 7. $45/7$ is not an integer. Assertion A is false. Reason R correctly states the rule, so R is true.

Learning route:

Analyze Assertion (A): The ratio of boys to girls is $3 : 4$ . The total number of parts is $3 + 4 = 7$ . For the number of students to be an integer, the total must be divisible by 7. Since $45/7 \approx 6.43$ (not an integer), a total of 45 is impossible. Assertion A is false.
Analyze Reason (R): The reason states that the total must be divisible by the sum of the ratio parts ( $3 + 4 = 7$ ). This is the correct mathematical rule that governs ratio distributions. Reason R is true.

Answer: A is false but R is true.

Question 10 · Quantitative Aptitude MCQ

In a dataset where a single extreme outlier is present, which measure of central tendency is robust and unaffected by the magnitude of the outlier?

A.
Mean
B.
Median
C.
Mode
D.
Range

Correct Answer:

Step-by-Step Solution

Key idea: This is a conceptual question about the properties of central tendency measures, recognizable because it mentions an "extreme outlier" and asks for the "robust" measure.

Step 1: Recall how the mean is calculated (sum of all values). An extreme outlier will heavily pull the sum, shifting the mean.

Step 2: Recall how the median is calculated (the middle value of sorted data). The magnitude of the extreme value does not change the middle position.

Step 3: Conclude that the median is the robust measure.

Answer: B

arithmetic ratios percentages and commercial mathematics notes

Chapter Roadmap: Arithmetic and Commercial Mathematics

Chapter Roadmap

Central Tendency: Finding the Center of Data

Central Tendency

Weighted Average: Combining Different Groups

Weighted Average

Try a question

Login to view the complete notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Arithmetic, Ratios, Percentages and Commercial Mathematics Notes for GATE CS

Chapter Roadmap: Arithmetic and Commercial Mathematics

Chapter Roadmap

Central Tendency: Finding the Center of Data

Central Tendency

Weighted Average: Combining Different Groups

Weighted Average

The Median: Finding the Exact Middle

The Median

Arithmetic, Ratios, Percentages and Commercial Mathematics: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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