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

Solve 8+ Arithmetic, Ratios, Percentages and Commercial Mathematics previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

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.

Arithmetic, Ratios, Percentages and Commercial Mathematics: Solved Questions with Step-by-Step Explanations (8 Problems)

Question 1 · Quantitative Aptitude · 2026_Set2 MCQ

The values of Stock A and Stock B on a particular day are Rs. 50 and Rs. 80, respectively. An investor invests Rs. 100 in Stock A and Rs. 80 in Stock B. He sells all the stocks the next day when the value of Stock A is Rs. 55 and Stock B is Rs. 70. The profit made by the investor is Rs. ________

A.
0
B.
5
C.
10
D.
20

Correct Answer:

Step-by-Step Solution

Insight: Calculate the number of shares purchased for each stock using the initial investment and price, then find the total selling value.

Exam route: Shares of A = 100 / 50 = 2. Shares of B = 80 / 80 = 1. Selling value = 2 55 + 1 70 = 110 + 70 = 180. Total cost = 180. Profit = 180 - 180 = 0.

Learning route:

Step 1: Find the number of shares bought for Stock A. Investment = Rs. 100, Price = Rs. 50. Shares of A = 100 / 50 = 2.

Step 2: Find the number of shares bought for Stock B. Investment = Rs. 80, Price = Rs. 80. Shares of B = 80 / 80 = 1.

Step 3: Calculate the total selling value the next day. Price of A = Rs. 55, Price of B = Rs. 70.

Selling value of A = 2 * 55 = 110.

Selling value of B = 1 * 70 = 70.

Total selling value = 110 + 70 = 180.

Step 4: Calculate profit. Total cost = 100 + 80 = 180. Profit = Total selling value - Total cost = 180 - 180 = 0.

Trap warning: A common mistake is to just average the percentage changes or add the price differences (55 - 50 + 70 - 80 = -5) without weighting by the number of shares.

Verification: Cost = 180. Final value = 180. Profit = 0. Matches option A.

Question 2 · Quantitative Aptitude · 2025_Set1 MCQ

The average marks obtained by a class in an examination were calculated as 30.8. However, while checking the marks entered, the teacher found that the marks of one student were entered incorrectly as 24 instead of 42. After correcting the marks, the average becomes 31.4. How many students does the class have?

A.
25
B.
28
C.
30
D.
32

Correct Answer:

Step-by-Step Solution

Insight: The increase in the average is caused entirely by the correction in the misrecorded mark, distributed equally among all students.

Exam route: Error in sum = 42 - 24 = 18. Increase in average = 31.4 - 30.8 = 0.6. Number of students = 18 / 0.6 = 30.

Learning route:

Step 1: Identify the error in the total sum. The mark 24 was entered instead of 42. The sum was short by 42 - 24 = 18.

Step 2: Identify the increase in the average. The average went from 30.8 to 31.4, an increase of 31.4 - 30.8 = 0.6.

Step 3: Relate the sum error to the average error. The extra 18 marks were distributed equally among $N$ students, increasing the average by 0.6.

So, $18/ N = 0.6$ .

Step 4: Solve for $N$ . $N = 18/0.6 = 30$ .

Trap warning: Some students might subtract 24 from 42 and get 18, then divide by the initial average 30.8, which is incorrect. The 18 is the total increase in the sum, not a single student's mark.

Verification: Initial sum = 30 * 30.8 = 924. Correct sum = 924 - 24 + 42 = 942. Correct average = 942 / 30 = 31.4. Matches perfectly.

Question 3 · Quantitative Aptitude · 2024_Set2 MCQ

A person sold two different items at the same price. He made 10% profit in one item, and 10% loss in the other item. In selling these two items, the person made a total of

A.
1% profit
B.
2% profit
C.
1% loss
D.
2% loss

Correct Answer:

Step-by-Step Solution

Insight: Same selling price with equal-percent profit and loss always produces a net loss of

(x /10)^{2} %

. Exam route: Here

x = 10

. Net loss

= (10/10)^{2} = 1%

. Learning route: Step 1: Let the common selling price be

S

. Step 2: Cost price of the profit item:

C P_{1} = \frac{100 S}{100 + 10} = \frac{100 S}{110}

. Step 3: Cost price of the loss item:

C P_{2} = \frac{100 S}{100 - 10} = \frac{100 S}{90}

. Step 4: Total cost

= 100 S (\frac{1}{110} + \frac{1}{90}) = 100 S \cdot \frac{200}{9900} = \frac{20000 S}{9900} = \frac{200 S}{99}

. Step 5: Total selling price

= 2 S

. Since

\frac{200 S}{99} > 2 S

, there is a loss. Step 6: Loss

= \frac{200 S}{99} - 2 S = \frac{2 S}{99}

. Loss \%

= \frac{2 S /99}{200 S /99} \times 100 = 1%

. Trap warning: The tempting but wrong answer is "no profit no loss" — equal percentages on the same cost price would cancel, but here they are on different cost prices that produce the same selling price. Verification: Plug

S = 110

C P_{1} = 100

C P_{2} = 110 \times 100/90 = 122.22

. Total

C P = 222.22

, total

S P = 220

. Loss

= 2.22

, loss \%

= 2.22/222.22 = 1%

Question 4 · Quantitative Aptitude · 2024_Set1 MCQ

Consider the following sample of numbers:

9, 18, 11, 14, 15, 17, 10, 69, 11, 13

The median of the sample is

A.
$13.5$
B.
$14$
C.
$11$
D.
$18.7$

Correct Answer:

Step-by-Step Solution

Insight: Median is the positional middle. With an even count, it is the average of the two central values after sorting.

Exam route: Sort the data: $9, 10, 11, 11, 13, 14, 15, 17, 18, 69$ . $n = 10$ (even). Median $= (5 th + 6 th) /2 = (13 + 14) /2 = 13.5$ .

Learning route:

Step 1: Arrange the data in ascending order: $9, 10, 11, 11, 13, 14, 15, 17, 18, 69$ .

Step 2: Count $n = 10$ , which is even. The median is the average of the $(n /2)$ -th and $(n /2 + 1)$ -th terms, i.e., the 5th and 6th terms.

Step 3: 5th term $= 13$ , 6th term $= 14$ . Median $= (13 + 14) /2 = 13.5$ .

Trap warning: Option B ( $14$ ) is just the 6th term without averaging. Option C ( $11$ ) is the mode (most frequent value). Option D ( $18.7$ ) is close to the mean, which is pulled up by the outlier $69$ .

Verification: Five values $\leq 13.5$ and five values $\geq 13.5$ — the definition of the median is satisfied.

Question 5 · Quantitative Aptitude · 2024_Set1 MCQ

The number of coins of ₹1, ₹5, and ₹10 denominations that a person has are in the ratio $5 : 3 : 13$ . Of the total amount, the percentage of money in ₹5 coins is

A.
$21%$
B.
$14 \frac{2}{7} %$
C.
$10%$
D.
$30%$

Correct Answer:

Step-by-Step Solution

Insight: The ratio given is of the number of coins, not their monetary value. You must convert the count ratio into a value ratio by multiplying each part by its denomination.

Exam route: Let the number of coins be $5 k, 3 k, 13 k$ . Their values are $5 k \times 1 = 5 k$ , $3 k \times 5 = 15 k$ , and $13 k \times 10 = 130 k$ . Total value $= 5 k + 15 k + 130 k = 150 k$ . The percentage in ₹5 coins is $\frac{15 k}{150 k} \times 100 = 10%$ .

Learning route:

Step 1: Assign the multiplier $k$ . The number of ₹1, ₹5, and ₹10 coins are $5 k, 3 k$ , and $13 k$ respectively.

Step 2: Convert the number ratio into a value ratio by multiplying each count by its denomination.

Value of ₹1 coins $= 5 k \times 1 = 5 k$
Value of ₹5 coins $= 3 k \times 5 = 15 k$
Value of ₹10 coins $= 13 k \times 10 = 130 k$

Step 3: Total amount $= 5 k + 15 k + 130 k = 150 k$ .

Step 4: Required percentage $= \frac{15 k}{150 k} \times 100 = 10%$ .

Trap warning: Option B ( $14 \frac{2}{7} %$ ) comes from taking $\frac{3}{5 + 3 + 13} = \frac{3}{21}$ , which is the ratio of the number of ₹5 coins to the total number of coins. The question asks for the percentage of the amount, not the count.

Verification: If $k = 1$ , coins are 5, 3, 13; values are ₹5, ₹15, ₹130; total ₹150; ₹15 is exactly 10%.

Question 6 · Quantitative Aptitude · 2022 MCQ

In a recently conducted national entrance test, boys constituted 65% of those who appeared for the test. Girls constituted the remaining candidates and they accounted for 60% of the qualified candidates.

Which one of the following is the correct logical inference based on the information provided in the above passage?

A.
Equal number of boys and girls qualified
B.
Equal number of boys and girls appeared for the test
C.
The number of boys who appeared for the test is less than the number of girls who appeared
D.
The number of boys who qualified the test is less than the number of girls who qualified

Correct Answer:

Step-by-Step Solution

Insight: Percentages of the qualified pool tell us the boys-vs-girls split among qualifiers directly — no need to know the pass rates.

Exam route: Girls are $60%$ of qualified $⟹$ boys are $40%$ of qualified. So fewer boys qualified than girls. Option D.

Learning route:

Step 1: Let total appeared $= 100$ . Boys appeared $= 65$ , girls appeared $= 35$ .

Step 2: Girls account for $60%$ of qualified candidates. Therefore boys account for the remaining $40%$ of qualified candidates.

Step 3: Compare the options:

A: "Equal number of boys and girls qualified" — false, since $60% \neq = 40%$ .
B: "Equal number of boys and girls appeared" — false, $65 \neq = 35$ .
C: "Boys who appeared $<$ girls who appeared" — false, $65 > 35$ .
D: "Boys who qualified $<$ girls who qualified" — true, since $40% < 60%$ of the same qualified pool.

Trap warning: Students often try to compute actual numbers of qualifiers, but the pass rates are not given. The inference must come purely from the qualified-pool split.

Verification: Whatever the total qualified count $Q$ is, boys qualified $= 0.4 Q$ and girls qualified $= 0.6 Q$ , so boys $<$ girls always holds.

Question 7 · Quantitative Aptitude · 2021_Set2 MCQ

The number of students in three classes is in the ratio 3:13:6. If 18 students are added to each class, the ratio changes to 15:35:21.

The total number of students in all the three classes in the beginning was:

A.
22
B.
66
C.
88
D.
110

Correct Answer:

Step-by-Step Solution

Insight: Adding the same constant to every term of a ratio preserves the absolute difference between terms. Use that invariant to set up a quick equation.

Exam route: Initial ratio $3 : 13 : 6 ⟹$ parts sum $= 22$ . After $+ 18$ , ratio $15 : 35 : 21$ . Using first two terms: $\frac{3 x + 18}{13 x + 18} = \frac{15}{35} = \frac{3}{7} ⟹ 21 x + 126 = 39 x + 54 ⟹ x = 4$ . Total $= 22 x = 88$ .

Learning route:

Step 1: Let initial students be $3 x, 13 x, 6 x$ . Total $= 22 x$ .

Step 2: After adding 18 to each class: $3 x + 18, 13 x + 18, 6 x + 18$ .

Step 3: The new ratio of the first two classes is $\frac{3 x + 18}{13 x + 18} = \frac{15}{35} = \frac{3}{7}$ .

Step 4: Cross-multiply: $7 (3 x + 18) = 3 (13 x + 18) ⟹ 21 x + 126 = 39 x + 54$ .

Step 5: Solve: $18 x = 72 ⟹ x = 4$ .

Step 6: Initial total $= 22 x = 22 \times 4 = 88$ .

Trap warning: Option A ( $22$ ) is just the sum of the ratio parts, forgetting the multiplier. Option B ( $66$ ) and D ( $110$ ) come from mis-solving the equation or mis-scaling.

Verification: Initial $= 12, 52, 24$ . After $+ 18$ : $30, 70, 42$ . Ratio $= 30 : 70 : 42 = 15 : 35 : 21$ . Matches.

Question 8 · Quantitative Aptitude · 2021_Set1 MCQ

The ratio of boys to girls in a class is 7 to 3.

Among the options below, an acceptable value for the total number of students in the class is:

A.
21
B.
37
C.
50
D.
73

Correct Answer:

Step-by-Step Solution

Insight: The total quantity must be a multiple of the sum of the ratio parts.

Exam route: Ratio is 7:3. Sum of parts = 10. Total students must be a multiple of 10. Among the options, only 50 is a multiple of 10.

Learning route:

Step 1: Let the number of boys be $7 k$ and girls be $3 k$ .

Step 2: Total students $= 7 k + 3 k = 10 k$ .

Step 3: Since $k$ must be an integer (you cannot have a fraction of a student), the total number of students must be a multiple of 10.

Step 4: Check the options: 21, 37, 50, 73. Only 50 is divisible by 10.

Trap warning: Students might just look for a multiple of 7 or 3 individually. The sum of the parts is the key invariant.

Verification: If total is 50, $k = 5$ . Boys = 35, Girls = 15. Ratio = 35:15 = 7:3. Matches perfectly.

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

Chapter Roadmap: Arithmetic and Commercial Mathematics

Chapter Roadmap

Central Tendency: Finding the Center of Data

Central Tendency

Arithmetic, Ratios, Percentages and Commercial Mathematics: Solved Questions with Step-by-Step Explanations (8 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

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