Numerical Ability MCQ Quiz - Objective Question with Answer for Numerical Ability - Download Free PDF
Last updated on Jun 23, 2025
Latest Numerical Ability MCQ Objective Questions
Numerical Ability Question 1:
Comprehension:
Read the given passage and answer the questions that follow
The following table has the number of enrolled students for a particular course and year.
Total number of seats are given in brackets with respect to each course.
| Year |
Number of Enrolled Students |
|||
|
BTech |
MTech |
MPhil |
MBA (60) |
|
| 2016 | 96 | 24 | 12 | 52 |
| 2017 | 110 | 26 | 14 | 57 |
| 2018 | 108 | 25 | 13 | 54 |
| 2019 | 112 | 28 | 15 | 56 |
| 2020 | 104 | 27 | 16 | 48 |
Which year has the minimum vacant seats?
Answer (Detailed Solution Below)
Numerical Ability Question 1 Detailed Solution
Given:
| Year |
Number of Enrolled Students |
||||
|
BTech |
MTech |
MPhil |
MBA (60) |
Total (230) |
|
| 2016 | 96 | 24 | 12 | 52 | 180 |
| 2017 | 110 | 26 | 14 | 57 | 207 |
| 2018 | 108 | 25 | 13 | 54 | 200 |
| 2019 | 112 | 28 | 15 | 56 | 211 |
| 2020 | 104 | 27 | 16 | 48 | 195 |
Formula Used:
Total Seats = Enrolled Seats + Vacant Seats
Calculation:
Vacant seats in 2016 = 230 - 180 = 50
Vacant seats in 2017 = 230 - 207 = 23
Vacant seats in 2018 = 230 - 200 = 30
Vacant seats in 2019 = 230 - 211 = 19
Vacant seats in 2020 = 230 - 195 = 35
∴ Year which has the minimum vacant seats is 2019
Numerical Ability Question 2:
Comprehension:
Read the given passage and answer the questions that follow
The following table has the number of enrolled students for a particular course and year.
Total number of seats are given in brackets with respect to each course.
| Year |
Number of Enrolled Students |
|||
|
BTech |
MTech |
MPhil |
MBA (60) |
|
| 2016 | 96 | 24 | 12 | 52 |
| 2017 | 110 | 26 | 14 | 57 |
| 2018 | 108 | 25 | 13 | 54 |
| 2019 | 112 | 28 | 15 | 56 |
| 2020 | 104 | 27 | 16 | 48 |
Which course has the highest percentage increase in enrollment of students from 2016-2017?
Answer (Detailed Solution Below)
Numerical Ability Question 2 Detailed Solution
Given:
| Year |
Number of Enrolled Students |
||||
|
BTech |
MTech |
MPhil |
MBA (60) |
Total (230) |
|
| 2016 | 96 | 24 | 12 | 52 | 180 |
| 2017 | 110 | 26 | 14 | 57 | 207 |
| 2018 | 108 | 25 | 13 | 54 | 200 |
| 2019 | 112 | 28 | 15 | 56 | 211 |
| 2020 | 104 | 27 | 16 | 48 | 195 |
Percentage increase in enrollment of students from 2016-2017 is
In BTech = [(110 - 96)/96] × 100 = (14/96) × 100 = 14.58%
In MTech = [(26 - 24)/24] × 100 = (2/24) × 100 = 8.33%
In MPhil = [(14 - 12)/12] × 100 = (2/12) × 100 = 16.66%
In MBA = [(57 - 52)/52] × 100 = (5/52) × 100 = 9.61%
∴ The course which has the highest percentage increase in enrollment of students from 2016-2017 is MPhil.
Numerical Ability Question 3:
Comprehension:
Read the given passage and answer the questions that follow
The following table has the number of enrolled students for a particular course and year.
Total number of seats are given in brackets with respect to each course.
| Year |
Number of Enrolled Students |
|||
|
BTech |
MTech |
MPhil |
MBA (60) |
|
| 2016 | 96 | 24 | 12 | 52 |
| 2017 | 110 | 26 | 14 | 57 |
| 2018 | 108 | 25 | 13 | 54 |
| 2019 | 112 | 28 | 15 | 56 |
| 2020 | 104 | 27 | 16 | 48 |
What is the average number of vacant seats of B. Tech for years (2016-2020)?
Answer (Detailed Solution Below)
Numerical Ability Question 3 Detailed Solution
Given:
| Year |
Number of Enrolled Students |
||||
|
BTech |
MTech |
MPhil |
MBA (60) |
Total (230) |
|
| 2016 | 96 | 24 | 12 | 52 | 180 |
| 2017 | 110 | 26 | 14 | 57 | 207 |
| 2018 | 108 | 25 | 13 | 54 | 200 |
| 2019 | 112 | 28 | 15 | 56 | 211 |
| 2020 | 104 | 27 | 16 | 48 | 195 |
Calculation:
The number of vacant seats of B.Tech for years (2016-2020) is
In the year 2016 = 120 - 96 = 24
In the year 2017 = 120 - 110 = 10
In the year 2018 = 120 - 108 = 12
In the year 2019 = 120 - 112 = 8
In the year 2020 = 120 - 104 = 16
So, the average number of vacant seats of B. Tech for years (2016-2020) is
⇒ (24 + 10 + 12 + 8 + 16)/5
⇒ 70/5 = 14
∴ The average number of vacant seats of B. Tech for years (2016 - 2020) is 14
Numerical Ability Question 4:
Comprehension:
Read the given passage and answer the questions that follow
The following table has the number of enrolled students for a particular course and year.
Total number of seats are given in brackets with respect to each course.
| Year |
Number of Enrolled Students |
|||
|
BTech |
MTech |
MPhil |
MBA (60) |
|
| 2016 | 96 | 24 | 12 | 52 |
| 2017 | 110 | 26 | 14 | 57 |
| 2018 | 108 | 25 | 13 | 54 |
| 2019 | 112 | 28 | 15 | 56 |
| 2020 | 104 | 27 | 16 | 48 |
Find out the difference between average vacant seats of M Tech and MBA for 2016-2020
Answer (Detailed Solution Below)
Numerical Ability Question 4 Detailed Solution
Given:
| Year |
Number of Enrolled Students |
||||
|
BTech |
MTech |
MPhil |
MBA (60) |
Total (230) |
|
| 2016 | 96 | 24 | 12 | 52 | 180 |
| 2017 | 110 | 26 | 14 | 57 | 207 |
| 2018 | 108 | 25 | 13 | 54 | 200 |
| 2019 | 112 | 28 | 15 | 56 | 211 |
| 2020 | 104 | 27 | 16 | 48 | 195 |
Formula Used:
Total Seats = Enrolled Seats + Vacant Seats
Calculation:
Vacant Seats in MTech for 2016-2020 = (30 - 24) + (30 - 26) + (30 - 25) + (30 - 28) + (30 - 27)
Now, average vacant seats of M Tech = (6 + 4 + 5 + 2 + 3)/5
⇒ 20/5 = 4
Vacant Seats in MBA for 2016-2020 = (60 - 52) + (60 - 57) + (60 - 54) + (60 - 56) + (60 - 48)
Now, average vacant seats of M Tech = (8 + 3 + 6 + 4 + 12)/5
⇒ 33/5 = 6.6
So, the difference between average vacant seats of M Tech and MBA for 2016-2020 = 6.6 - 4 = 2.6
∴ The difference between average vacant seats of M Tech and MBA for 2016-2020 is 2.6
Numerical Ability Question 5:
There are 6 rooms on the first floor of a hotel, with 3 rooms on each side of the corridor, each room exactly opposite to the other room. The number of ways in which three guests are accommodated in three of the six rooms, one in each room, such that no two guests are in adjacent rooms or in the opposite rooms is
Answer (Detailed Solution Below)
Numerical Ability Question 5 Detailed Solution
Step-by-Step Solution:
Step 1: Define the problem
We need to select 3 rooms out of the 6 such that the selected rooms satisfy the conditions. Adjacent rooms are defined as rooms next to each other on the same side (e.g., Room A and Room B), and opposite rooms are defined as rooms directly facing each other across the corridor (e.g., Room A and Room D).
Step 2: Eliminate invalid combinations
We begin by eliminating combinations where any two rooms are adjacent or opposite:
- Adjacent rooms on Side 1: (A, B), (B, C)
- Adjacent rooms on Side 2: (D, E), (E, F)
- Opposite rooms: (A, D), (B, E), (C, F)
Step 3: Count valid combinations
Next, we select 3 rooms such that no two rooms are adjacent or opposite. The valid combinations can be identified as follows:
- Select one room from Side 1 and two rooms from Side 2, or vice versa.
- The selected rooms must not be adjacent or opposite.
Case 1: Select one room from Side 1 and two rooms from Side 2
- From Side 1, valid choices: A or C (since B is adjacent to both A and C).
- From Side 2, valid choices depend on the room selected from Side 1:
- If A is selected, valid rooms from Side 2 are E and F (D is opposite to A).
- If C is selected, valid rooms from Side 2 are D and E (F is opposite to C).
Total combinations for Case 1:
- If A is selected: {A, E, F}
- If C is selected: {C, D, E}
Number of combinations = 2.
Case 2: Select two rooms from Side 1 and one room from Side 2
- From Side 1, valid choices depend on avoiding adjacent rooms:
- Valid pairs: {A, C} (B is adjacent to both A and C).
- From Side 2, the room must not be opposite to any selected room from Side 1:
- If {A, C} is selected, valid room from Side 2 is E (D is opposite to A, and F is opposite to C).
Total combinations for Case 2:
- {A, C, E}
Number of combinations = 1.
Total Number of Valid Combinations:
Adding the combinations from both cases:
Case 1: 2 combinations
Case 2: 1 combination
Total = 2 + 1 = 3 combinations.
Step 4: Factorial arrangements within rooms
Since each guest can be assigned to any of the selected rooms, we calculate permutations for arranging 3 guests in 3 rooms:
Permutations = 3 × 2 × 1 = 6.
Final Calculation:
Total arrangements = Number of valid combinations × Permutations
Total arrangements = 3 × 6 = 18.
Correct Answer: Option 4 (18)
Top Numerical Ability MCQ Objective Questions
A batsman makes a score of 97 runs in 15th inning and thus increases his average by 5. Find his average after 15th inning.
Answer (Detailed Solution Below)
Numerical Ability Question 6 Detailed Solution
Download Solution PDFGiven:
The batsman scores 97 runs in the 15th inning.
This increases his average by 5 runs.
Formula used:
Average = Total runs / Number of innings
Solution:
Let x be the batsman's average after 14 innings.
We know that the total runs scored by the batsman in 14 innings is 14x.
According to the question:
14x + 97 = 15(x + 5)
14x + 97 = 15x + 75
x = 22
Average after 15th innings = x + 22 = 5 + 22 = 27
∴ Option 3 is the correct answer.
Direction: Study the given graph and answer the question that follows.
The expenditure on Interest on Loans is by what percentage more than the expenditure on Taxes ?
Answer (Detailed Solution Below)
Numerical Ability Question 7 Detailed Solution
Download Solution PDFGiven:
Percentage of Taxes = 5
Percentage of Interest on Loans = 7.5
Calculation:
Difference between the expense of Interest on Loans and Taxes = 7.5 - 5
⇒ 2.5
∴ The required percentage = 2.5/5 × 100
⇒ 50%
Study the given graph and answer the question that follows.
In which year was the percentage increase in the revenue as compared to that in its preceding year below 6%?
Answer (Detailed Solution Below)
Numerical Ability Question 8 Detailed Solution
Download Solution PDFFormula used:
Percentage Increase = [
Calculation:
Increase percent in 2015 = (350 - 320)/ 320 × 100 = 30/320 × 100 = 9.37%
Increase percent in 2016 = (380 - 350)/ 350 × 100 = 30/350 × 100 = 8.57%
Increase percent in 2017 = (400 - 380)/ 380 × 100 = 20/380 × 100 = 5.26%
Decrease percent in 2018 = (400 - 280)/ 400 × 100 = 120/400 × 100 = 30%
Increase percent in 2019 = (300 - 280)/ 280 × 100 = 20/280 × 100 = 7.14%
∴ In 2017 the percentage increase in the revenue as compared to its preceding year is below 6%.
The line graph shows the production (in tonnes) and the sales (in tonnes) of a company.
What percentage (approximately) of the total production of the company is the total sales of the company in all the years together? (correct to 2 decimal places)
Answer (Detailed Solution Below)
Numerical Ability Question 9 Detailed Solution
Download Solution PDFCalculation:
Total production = (600 + 628 + 750 + 500 + 570) tones = 3048 tons
Total sales = (450 + 525 + 500 + 400 + 516) tons = 2391 tons
∴ Percentage of the total sales = 2391/3048 × 100
⇒ 78.44%
A and B are two typists. One afternoon, they were each given 80 pages for typing. They divided the work equally, but B finished 40 minutes before A, who took 4 hours for the same. The next afternoon, they were again given 121 pages to type. However, this time they decided to divide the work such that they finished typing simultaneously. How many pages did A have to type?
Answer (Detailed Solution Below)
Numerical Ability Question 10 Detailed Solution
Download Solution PDFGiven data:
A's time for 80 pages: 4 hours
B's time for 80 pages: 3 hours 20 minutes
Concept: Work and time ratio.
Calculation:
The ratio of work rates of A and B = B's time/A's time = 4/3.333 = 1.2
Ratio of pages = 1.2 : 1 = 6 ∶ 5
A's share of 121 pages = 5/11 × 121 = 55 pages
Hence, A had to type 55 pages.
A bus travels from station P to station Q at a speed of 70 km/h, and from station Q to station P at a speed of 90 km/h. What is the average speed of the bus during the entire journey?
Answer (Detailed Solution Below)
Numerical Ability Question 11 Detailed Solution
Download Solution PDFGiven:
Speed from station P to station Q: 70 km/h
Speed from station Q to station P: 90 km/h
Formula:
Average speed = 2xy/x + y
Solution:
Average speed = (2 × 70 × 90)/70 + 90 = 78.75 km/h
Therefore, the average speed of the bus during the entire journey is 78.75 km/h
Study the given bar chart and answer the question that follows.
The bar chart shows the production of medicines by a company (in 1000 tonnes) over six years 2010 to 2015.
If in each of the following years, the percentage increase in production as compared to the previous year is taken, then the ratio for increase production of the year 2011 to increase production of year 2015 is:
Answer (Detailed Solution Below)
Numerical Ability Question 12 Detailed Solution
Download Solution PDFFormula used:
Percentage Increase = (Production of current year - Production of previous year)/ Production of Previous year × 100
Calculation:
Percentage increase in the production of medicines of the year 2011 = (60 - 40)/40 × 100 = 20/40 × 100 = 50%
Percentage increase in the production of medicines of the year 2015 = (80 - 60)/60 × 100 = 20/60 × 100 = 100/3%
∴ The required ratio = 50% : 100/3%
⇒ 1 : 2/3
⇒ 3 : 2
A company has six departments. The given bar graph shows the number of employees in each department for the years 2015 and 2018. The company added 80 new employees and fired some employees between the years 2015 and 2018. All the employees who were fired were from department C. If the employees of department C had NOT been fired, then the employee strength of department C in 2018 would have been greater by______.
Answer (Detailed Solution Below)
Numerical Ability Question 13 Detailed Solution
Download Solution PDFCalculation:
In 2015, No. of employees in all department = 20 + 40 + 50 + 70 + 80 + 30 = 290
In 2018, No. of employees in all department = 30 + 70 + 80 + 90 + 60 + 30 = 360
Difference = 360 - 290 = 70
According to question,
No of employees added = 80
Fired employees from C = 80 - 70 = 10
Strength of C in 2018 = 80 (Given)
If not fired, strength of C in 2018 = 80 + 10 = 90
Strength of department C in 2018 would have been greater by = (10/80) × 100
⇒ 12.5%
∴ Option 1 is the correct answer.
Study the histogram that shows the marks obtained by students in an examination and answer the question that follows.
By what percentage is the number of students who obtained marks between 200 and 300 more than the number of students who obtained 350 or more marks?
Answer (Detailed Solution Below)
Numerical Ability Question 14 Detailed Solution
Download Solution PDFGiven:
No of the students who obtained marks between 200 and 300
⇒ 45 + 60 = 105
No of students who obtained marks between 350 or more marks
⇒ 40 + 35 = 75
Calculation:
⇒ The required percent = (105 - 75)/75 × 100
⇒ 30/75 × 100
⇒ 40%
∴ The answer is 40%.
Study the following bar graph and answer the question that follows.
In how many years was the export more than the average for the given period?
Answer (Detailed Solution Below)
Numerical Ability Question 15 Detailed Solution
Download Solution PDFCalculation:
Average Export diamonds for 7 years = Total export / Total no of years
⇒ (7.2 + 8.5 + 9.8 + 10.9 + 11.8 + 11.5 + 12.4)/7 = 72.1/7
⇒ 10.3 crores
In the given 4 years, the export is more than the average export:
⇒ 2011 = 10.9 crores
⇒ 2012 = 11.8 crores
⇒ 2013 = 11.5 crores
⇒ 2014 = 12.4 crores
∴ The no of years where export is more than the average export is 4.