Tabulation MCQ Quiz - Objective Question with Answer for Tabulation - Download Free PDF
Last updated on Jun 19, 2025
Latest Tabulation MCQ Objective Questions
Tabulation Question 1:
Comprehension:
Consider the following table that shows the percentage (%) distribution of workforce of India and ratio of male to female workforce of India in different employment sectors during a given year. Total workforce in all the sectors put together is 80 lakhs. Based on the data contained in the table, answer the question :
Sector - wise Distribution of Workforce of India
| Employment Sector | Workforce of India | |
| Percentage (%) Distribution | Ratio of Male to Female (M : F) | |
| Service (SE) | 15% | 3 ∶ 2 |
| Sales (SA) | 12% | 5 ∶ 3 |
| Construction and Maintenance (M) | 9% | 5 ∶ 4 |
| Professional (PR) | 18% | 5 ∶ 7 |
| Management (MA) | 21% | 3 ∶ 4 |
| Production and Transport (PT) | 19% | 5 ∶ 3 |
| Others (OT) | 6% | 3 ∶ 5 |
The number of female workforce in SE and PR sectors put together is what percent of the number of male workforce in CM sector?
Answer (Detailed Solution Below)
Tabulation Question 1 Detailed Solution
Given:
Total workforce in all the sectors put together is 80 lakhs.
Calculation:
Total Workforce in Service (SE) = 15% of 80 lakhs
⇒ \(\frac{15}{100}\) × 80 lakhs = 12 lakhs
Ratio (Male : Female) = 3 : 2
Female workforce in SE = 12 lakh × \(\frac{2}{5}\) = 4.8 lakhs
Total Workforce in Professional (PR) = 18% of 80 lakhs
⇒ \(\frac{18}{100}\) × 80 lakhs = 14.4 lakhs
Ratio (Male : Female) = 5 : 7
Female workforce in PR = 14.4 lakh × \(\frac{7}{12}\) = 8.4 lakhs
Total Workforce in Construction and Maintenance (M) = 9% of 80 lakhs
⇒ \(\frac{9}{100}\) × 80 lakhs = 7.2 lakhs
Ratio (Male : Female) = 5 : 4
Male workforce in CM = 7.2 lakh × \(\frac{5}{9}\) = 4 lakhs
Required percentage = [(female workforce in SE and PR sectors put together)/(the number of male workforce in CM sector)] × 100
⇒ [(4.8 + 8.4)/4] × 100 = 330%
∴ The required percentage is 330%.
Tabulation Question 2:
Comprehension:
Consider the following table that shows the percentage (%) distribution of workforce of India and ratio of male to female workforce of India in different employment sectors during a given year. Total workforce in all the sectors put together is 80 lakhs. Based on the data contained in the table, answer the question :
Sector - wise Distribution of Workforce of India
| Employment Sector | Workforce of India | |
| Percentage (%) Distribution | Ratio of Male to Female (M : F) | |
| Service (SE) | 15% | 3 ∶ 2 |
| Sales (SA) | 12% | 5 ∶ 3 |
| Construction and Maintenance (M) | 9% | 5 ∶ 4 |
| Professional (PR) | 18% | 5 ∶ 7 |
| Management (MA) | 21% | 3 ∶ 4 |
| Production and Transport (PT) | 19% | 5 ∶ 3 |
| Others (OT) | 6% | 3 ∶ 5 |
What is the average number of male workforce in all the sectors put together? (rounded - off to two decimal places)
Answer (Detailed Solution Below)
Tabulation Question 2 Detailed Solution
Given:
Total workforce in all the sectors put together is 80 lakhs.
Calculation:
Total Workforce in Service (SE) = 15% of 80 lakhs
⇒ \(\frac{15}{100}\) × 80 lakhs = 12 lakhs
Ratio (Male : Female) = 3 : 2
Male workforce = 12 lakh × \(\frac{3}{5}\) = 7.2 lakh
Total Workforce in Sales (SA) = 12% of 80 lakhs
⇒ \(\frac{12}{100}\) × 80 lakhs = 9.6 lakhs
Ratio (Male : Female) = 5 : 3
Male workforce = 9.6 lakh × \(\frac{5}{8}\) = 6 lakhs
Total Workforce in Construction and Maintenance (M) = 9% of 80 lakhs
⇒ \(\frac{9}{100}\) × 80 lakhs = 7.2 lakhs
Ratio (Male : Female) = 5 : 4
Male workforce = 7.2 lakh × \(\frac{5}{9}\) = 4 lakhs
Total Workforce in Professional (PR) = 18% of 80 lakhs
⇒ \(\frac{18}{100}\) × 80 lakhs = 14.4 lakhs
Ratio (Male : Female) = 5 : 7
Male workforce = 14.4 lakh × \(\frac{5}{12}\) = 6 lakhs
Total Workforce in Management (MA) = 21% of 80 lakhs
⇒ \(\frac{21}{100}\) × 80 lakhs = 16.8 lakhs
Ratio (Male : Female) = 3 : 4
Male workforce = 16.8 lakh × \(\frac{3}{7}\) = 7.2 lakhs
Total Workforce in Production and Transport (PT) = 19% of 80 lakhs
⇒ \(\frac{19}{100}\) × 80 lakhs = 15.2 lakhs
Ratio (Male : Female) = 5 : 3
Male workforce = 15.2 lakh × \(\frac{5}{8}\) = 9.5 lakhs
Total Workforce in Others (OT) = 6% of 80 lakhs
⇒ \(\frac{6}{100}\) × 80 lakhs = 4.8 lakhs
Ratio (Male : Female) = 3 : 5
Male workforce = 4.8 lakh × \(\frac{3}{8}\) = 1.8 lakhs
Average = (Total number of male workforce in all the sectors) / (Total sectors)
Average = (7.2 + 6 + 4 + 6 + 7.2 + 9.5 + 1.8)/7
⇒ \(\frac{41.7}{7}\) = 5.96 lakhs
∴ The average number of male workforce in all the sectors is 5.96 lakhs.
Tabulation Question 3:
Comprehension:
Consider the following table that shows the percentage (%) distribution of workforce of India and ratio of male to female workforce of India in different employment sectors during a given year. Total workforce in all the sectors put together is 80 lakhs. Based on the data contained in the table, answer the question :
Sector - wise Distribution of Workforce of India
| Employment Sector | Workforce of India | |
| Percentage (%) Distribution | Ratio of Male to Female (M : F) | |
| Service (SE) | 15% | 3 ∶ 2 |
| Sales (SA) | 12% | 5 ∶ 3 |
| Construction and Maintenance (M) | 9% | 5 ∶ 4 |
| Professional (PR) | 18% | 5 ∶ 7 |
| Management (MA) | 21% | 3 ∶ 4 |
| Production and Transport (PT) | 19% | 5 ∶ 3 |
| Others (OT) | 6% | 3 ∶ 5 |
The number of male workforce in SA and MA sectors put together is approximately what percent of the total number of workforce in PT sector?
Answer (Detailed Solution Below)
Tabulation Question 3 Detailed Solution
Given:
Total workforce in all the sectors put together is 80 lakhs.
Formula Used:
Required percentage = [(Male workforce in SA + Male workforce in MA sectors)/(Total number of workforce in PT sector)] × 100
Calculation:
Total Workforce in Sales (SA) = 12% of 80 lakhs
⇒ \(\frac{12}{100}\) × 80 lakhs = 9.6 lakhs
Ratio (Male : Female) = 5 : 3
Male workforce = 9.6 lakh × \(\frac{5}{8}\) = 6 lakhs
Total Workforce in Management (MA) = 21% of 80 lakhs
⇒ \(\frac{21}{100}\) × 80 lakhs = 16.8 lakhs
Ratio (Male : Female) = 3 : 4
Male workforce = 16.8 lakh × \(\frac{3}{7}\) = 7.2 lakhs
Total Workforce in Production and Transport (PT) = 19% of 80 lakhs
⇒ \(\frac{19}{100}\) × 80 lakhs = 15.2 lakhs
Required percentage = [(Male workforce in SA + Male workforce in MA sectors)/(Total number of workforce in PT sector)] × 100
⇒ \(\frac{6 + 7.2}{15.2}\) × 100
⇒ \(\frac{13.2}{15.2}\) × 100 = 86.84 ≈ 87%
∴ The required percentage is 87%.
Tabulation Question 4:
Comprehension:
Consider the following table that shows the percentage (%) distribution of workforce of India and ratio of male to female workforce of India in different employment sectors during a given year. Total workforce in all the sectors put together is 80 lakhs. Based on the data contained in the table, answer the question :
Sector - wise Distribution of Workforce of India
| Employment Sector | Workforce of India | |
| Percentage (%) Distribution | Ratio of Male to Female (M : F) | |
| Service (SE) | 15% | 3 ∶ 2 |
| Sales (SA) | 12% | 5 ∶ 3 |
| Construction and Maintenance (M) | 9% | 5 ∶ 4 |
| Professional (PR) | 18% | 5 ∶ 7 |
| Management (MA) | 21% | 3 ∶ 4 |
| Production and Transport (PT) | 19% | 5 ∶ 3 |
| Others (OT) | 6% | 3 ∶ 5 |
What is the ratio of the number of female workforce in CM sector compared to the number of male workforce in PR and OT sectors put together?
Answer (Detailed Solution Below)
Tabulation Question 4 Detailed Solution
Given:
Total workforce in all the sectors put together is 80 lakhs.
Calculation:
Total Workforce in Construction and Maintenance (CM) = 9% of 80 lakhs
⇒ \(\frac{9}{100}\) × 80 lakhs = 7.2 lakhs
Ratio (Male : Female) = 5 : 4
Female workforce in CM = 7.2 lakh × \(\frac{4}{9}\) = 3.2 lakhs
Total Workforce in Professional (PR) = 18% of 80 lakhs
⇒ \(\frac{18}{100}\) × 80 lakhs = 14.4 lakhs
Ratio (Male : Female) = 5 : 7
Male workforce in PR = 14.4 lakh × \(\frac{5}{12}\) = 6 lakhs
Total Workforce in Others (OT) = 6% of 80 lakhs
⇒ \(\frac{6}{100}\) × 80 lakhs = 4.8 lakh
Ratio (Male : Female) = 3 : 5
Male workforce in OT = 4.8 lakh × \(\frac{3}{8}\) = 1.8 lakhs
Required ratio = (The number of female workforce in CM sector) : (The number of male workforce in PR and OT sectors put together)
Required ratio = 3.2 : (6 + 1.8) = 3.2 : 7.8 = 16 : 39
∴ The required ratio is 16 : 39.
Tabulation Question 5:
Comprehension:
Consider the following table that shows the percentage (%) distribution of workforce of India and ratio of male to female workforce of India in different employment sectors during a given year. Total workforce in all the sectors put together is 80 lakhs. Based on the data contained in the table, answer the question :
Sector - wise Distribution of Workforce of India
| Employment Sector | Workforce of India | |
| Percentage (%) Distribution | Ratio of Male to Female (M : F) | |
| Service (SE) | 15% | 3 ∶ 2 |
| Sales (SA) | 12% | 5 ∶ 3 |
| Construction and Maintenance (M) | 9% | 5 ∶ 4 |
| Professional (PR) | 18% | 5 ∶ 7 |
| Management (MA) | 21% | 3 ∶ 4 |
| Production and Transport (PT) | 19% | 5 ∶ 3 |
| Others (OT) | 6% | 3 ∶ 5 |
The difference in the number of female workforce in SA and MA sectors put together and the number of female workforce in PT sector is :
Answer (Detailed Solution Below)
Tabulation Question 5 Detailed Solution
Given:
Total workforce in all the sectors put together is 80 lakhs.
Calculation:
Total Workforce in Sales (SA) = 12% of 80 lakhs
⇒ \(\frac{12}{100}\) × 80 lakhs = 9.6 lakhs
Ratio (Male : Female) = 5 : 3
Female workforce = 9.6 lakh × \(\frac{3}{8}\) = 3.6 lakhs
Total Workforce in Management (MA) = 21% of 80 lakhs
⇒ \(\frac{21}{100}\) × 80 lakhs = 16.8 lakhs
Ratio (Male : Female) = 3 : 4
Female workforce = 16.8 lakh × \(\frac{4}{7}\) = 9.6 lakhs
Total Workforce in Production and Transport (PT) = 19% of 80 lakhs
⇒ \(\frac{19}{100}\) × 80 lakhs = 15.2 lakhs
Ratio (Male : Female) = 5 : 3
Female workforce = 15.2 lakh × \(\frac{3}{8}\) = 5.7 lakhs
Difference = (The number of female workforce in SA and MA sectors put together) - (the number of female workforce in PT sector)
Difference = (3.6 + 9.6) - 5.7 = 7.5 lakhs
∴ The difference in the number of female workforce in SA and MA sectors put together and the number of female workforce in PT sector is 7.5 lakhs.
Top Tabulation MCQ Objective Questions
Comprehension:
Given below the data about three employees of a call center named ZINTOCA. In the given table, data about calls received by them, calls selected for further lead and calls finally received is given:
| A | B | C | |
| Initial calls(% out of total calls) | 40% | 30% | 30% |
| Lead calls (% out of initial calls) | 80% | x% | 88% |
| Final received calls(% out of lead calls) | 90% | 80% | y% |
Total number of initial calls in the company is 24000 and total number of calls not received by company C is 4464 less than the total number of calls received by company A. Find the value of y.
Answer (Detailed Solution Below)
Tabulation Question 6 Detailed Solution
Download Solution PDFCalculation:
Let the total number of calls be 1000z.
Initial calls (% out of total calls) of A = (1000z × 40/100) = 400z
Initial calls (% out of total calls) of B = (1000z × 30/100) = 300z
Initial calls (% out of total calls) of C = (1000z × 30/100) = 300z
Lead calls (% out of initial calls) of A = (400z × 80/100) = 320z
Lead calls (% out of initial calls) of B = (300z × x/100) = 3zx
Lead calls (% out of initial calls) of C = (300z × 88/100) = 264z
Final received calls (% out of lead calls) of A = (320z × 90/100) = 288z
Final received calls (% out of lead calls) of B = (3zx × 80/100) = 3zx × (4/5)
Final received calls (% out of lead calls) of C = (264z × y/100)
Total number of initial calls in the company = 24000
⇒ (400z + 300z + 300z) = 24000
⇒ 1000z = 24000
⇒ z = (24000/1000)
⇒ z = 24
Total number of calls not received calls by C = (300h – 264z × y/100)
⇒ (300 × 24 – 264z × y/100)
⇒ (7200 – 264z × y/100)
According to the question
Total number of calls not received by company C is 4464 less than the total number of calls received by company A
⇒ (7200 – 264z × y/100) + 4464 = 288z
⇒ (7200 – 264 × 24 × y/100) + 4464 = 288 × 24
⇒ 7200 – 6336y/100 + 4464 = 6912
⇒ (720000 – 6336y + 446400)/100 = 6912
⇒ (720000 – 6336y + 446400) = 6912 × 100
⇒ (1166400 – 6336y) = 691200
⇒ (-6336y) = (691200 – 1166400)
⇒ (-6336y) = -475200
⇒ y = [-475200/(-6336)]
⇒ y = 75
∴ The value of y is 75
The table shows the daily income (in Rs.) of 50 persons.
Study the table and answer the question:
|
Income (Rs) |
No. of persons |
|
Less than 200 |
12 |
|
Less than 250 |
26 |
|
Less than 300 |
34 |
|
Less than 350 |
40 |
|
Less than 400 |
50 |
How many persons earn Rs. 200 or more but less than Rs. 300?
Answer (Detailed Solution Below)
Tabulation Question 7 Detailed Solution
Download Solution PDFCalculation:
Number less than 200 = 12
Number less than 250 = 26
Number less than between 250 and 200 = (26 – 12)
⇒ 14
Again,
Number less than 250 = 26
Number less than 300 = 34
Number less than between 300 and 250 = (34 – 26)
⇒ 8
Persons earn Rs. 200 or more but less than Rs. 300 = (14 + 8)
⇒ 22
∴ Required persons is 22
Study the given table and answer the question that follows.
The table shows the classification of 100 students based on the marks obtained by them in History and Geography in an examination.
|
Subject |
Marks out of 50 |
||||
|
40 and above |
30 and above |
20 and above |
10 and above |
0 and above |
|
|
History |
9 |
32 |
80 |
92 |
100 |
|
Geography |
4 |
21 |
66 |
81 |
100 |
|
Average (Aggregate) |
7 |
27 |
73 |
87 |
100 |
Based on the table, what is the number of students scoring less than 20% marks in aggregate?
Answer (Detailed Solution Below)
Tabulation Question 8 Detailed Solution
Download Solution PDFCalculation
We have 20% of 50 = 10
Therefore Required number:
Number of students scoring less than 10 marks in aggregate
= 100 - Number of students scoring 10 and above marks in aggregate
= 100 - 87
= 13.
The number of students scoring less than 20% marks in aggregate is 13.
The following table shows the number of different items in different shops and their respective selling prices per unit.
|
Shops |
Total No. of Items |
AC ∶ Cooler ∶ Fan |
Selling Price per unit |
||
|
Cooler |
AC |
Fan |
|||
|
A |
5000 |
4 ∶ 5 ∶ 1 |
8000 |
25000 |
8500 |
|
B |
1800 |
3 ∶ 2 ∶ 4 |
10000 |
20000 |
16000 |
|
C |
3400 |
6 ∶ 4 ∶ 7 |
6000 |
42000 |
15000 |
|
D |
3600 |
4 ∶ 2 ∶ 3 |
12000 |
32000 |
8000 |
|
E |
4000 |
5 ∶ 1 ∶ 4 |
8000 |
26500 |
12200 |
|
F |
1210 |
2 ∶ 4 ∶ 5 |
11000 |
28000 |
11100 |
Find the percentage of total revenue which comes from Cooler from shop E, considering all given items are being sold from shop E and from all the given shops only given three items are being sold. (Rounded off to three decimal places)
Answer (Detailed Solution Below)
Tabulation Question 9 Detailed Solution
Download Solution PDFCalculation:
Total cooler sold by shop E = 4000 × (1/10)
⇒ 400
Selling price of 400 coolers = 400 × 8000
⇒ 3200000
Total ACs sold by shop E = 4000 × (5/10)
⇒ 2000
Selling price of 2000 ACs = 2000 × 26500
⇒ 53000000
Total Fans sold by shop E = 4000 × (4/10)
⇒ 1600
Selling price of 1600 Fans = 1600 × 12200
⇒ 19520000
Now,
Required % = [3200000/(3200000 + 53000000 + 19520000)] × 100
⇒ [3200000/(75720000)] × 100
⇒ 4.226 ≈ 4.23%
∴ The required answer is 4.23%.
study the given table and answer the question that follows.
The table shows the classification of 100 students based on the marks obtained by them in Statistics and Mathematics in an examination out of 50.
| Subject | 40 and above | 30 and above | 20 and above | 10 and above | 0 and above |
| Mathematics | 8 | 33 | 90 | 92 | 100 |
| Statistics | 5 | 22 | 60 | 87 | 100 |
If at least 60% marks in Mathematics are required for pursuing higher studies in Mathematics, then how many students will be eligible to pursue higher studies in Mathematics?
Answer (Detailed Solution Below)
Tabulation Question 10 Detailed Solution
Download Solution PDFCalculation:
Total marks = 50
Eligible for higher studies in mathematics marks = 50 × 60% = 30
Total students who are eligible to pursue higher studies in mathematics = 33
∴ The correct answer is 33.
The percentage marks obtained by 6 students in different subjects are given below. The maximum marks for each subject have been indicated in the table.
|
Subject Student |
Physics | Mathematics | Hindi | Geography | English | History |
| Maximum marks → | 80 | 150 | 100 | 75 | 120 | 50 |
| P | 70 | 44 | 88 | 88 | 70 | 38 |
| Q | 90 | 40 | 54 | 92 | 65 | 40 |
| R | 85 | 32 | 70 | 64 | 55 | 30 |
| S | 75 | 70 | 58 | 80 | 60 | 35 |
| T | 65 | 60 | 45 | 88 | 50 | 42 |
| U | 60 | 50 | 60 | 72 | 25 | 48 |
What are the average marks obtained by all students in Geography?
Answer (Detailed Solution Below)
Tabulation Question 11 Detailed Solution
Download Solution PDFGiven:
The percentage marks obtained by 6 students in different subjects are given below. The maximum marks for each subject have been indicated in the table.
|
Subject Student |
Physics | Mathematics | Hindi | Geography | English | History |
| Maximum marks → | 80 | 150 | 100 | 75 | 120 | 50 |
| P | 70 | 44 | 88 | 88 | 70 | 38 |
| Q | 90 | 40 | 54 | 92 | 65 | 40 |
| R | 85 | 32 | 70 | 64 | 55 | 30 |
| S | 75 | 70 | 58 | 80 | 60 | 35 |
| T | 65 | 60 | 45 | 88 | 50 | 42 |
| U | 60 | 50 | 60 | 72 | 25 | 48 |
Concept used:
Average = \(\frac{Total\ marks}{Total \ students}\)
Calculation:
According to the question,
Marks of P in geography = 88% of 75 = 0.88 × 75 = 66 marks
Marks of Q in geography = 92% of 75 = 0.92 × 75 = 69 marks
Marks of R in geography = 64% of 75 = 0.64 × 75 = 48 marks
Marks of S in geography = 80% of 75 = 0.80 × 75 = 60 marks
Marks of T in geography = 88% of 75 = 0.88 × 75 = 66 marks
Marks of U in geography = 72% of 75 = 0.72 × 75 = 54 marks
Total marks obtained by students in Geography = 66 + 69 + 48 + 60 + 66 + 54 = 363
Average marks obtained by students in Geography = \(\frac{363}{6}\) = 60.5
∴ The average marks obtained by all students in Geography is 60.5.
Mistake Points
The maximum mark in Geography is 75. Please note that the marks of students are given in the percentage.
For example, P's marks in Geography = 88% of 75 = 66 marks
So the sum of marks of all students in geography = 363
Average = 363/6 = 60.5
The following table shows the amount of annual rainfall in inches in 8 different states of India.
|
|
T |
A |
M |
N |
K |
P |
D |
R |
|
2015 |
80 |
70 |
98 |
78 |
68 |
65 |
70 |
59 |
|
2016 |
85 |
70 |
95 |
77 |
69 |
60 |
71 |
59 |
|
2017 |
86 |
71 |
96 |
76 |
66 |
67 |
71 |
59 |
|
2018 |
84 |
70 |
96 |
75 |
67 |
66 |
69 |
61 |
|
2019 |
80 |
74 |
97 |
74 |
67 |
64 |
75 |
60 |
|
2020 |
81 |
75 |
98 |
75 |
68 |
65 |
74 |
65 |
|
2021 |
82 |
72 |
98 |
73 |
70 |
65 |
73 |
65 |
How many states have witnessed an increase in the amount of annual rainfall continuously for two times in immediate years?
Answer (Detailed Solution Below)
Tabulation Question 12 Detailed Solution
Download Solution PDFCalculation:
State T has witnessed two times increase in 2016 and 2017 or 2020 and 2021.
State A has witnessed two times increase in 2019 and 2020
State M has witnessed two times increase in 2019 and 2020
State K has witnessed two times increase in 2020 and 2021
So, total 4 states witnessed an increase in the amount of annual rainfall continuously for two times in immediate years
∴ The required answer is 4.
Study the given table and answer the question that follows.
The table shows the number of candidates who appeared (App), qualified (Qual) and selected (Sel) in a competitive examination from four states Delhi, Goa, Karnataka, and Maharashtra over the years 2012 to 2016.
|
Years |
Delhi |
Goa |
Karnataka |
Maharashtra |
||||||||
|
|
App |
Qual |
Sel |
App |
Qual |
Sel |
App |
Qual |
Sel |
App |
Qual |
Sel |
|
2012 |
8000 |
850 |
94 |
7800 |
810 |
82 |
7500 |
720 |
78 |
8200 |
680 |
85 |
|
2013 |
4800 |
500 |
48 |
7500 |
800 |
65 |
5600 |
620 |
85 |
6800 |
600 |
70 |
|
2014 |
9500 |
850 |
90 |
8800 |
920 |
86 |
7000 |
650 |
70 |
7800 |
720 |
84 |
|
2015 |
9000 |
800 |
70 |
7200 |
850 |
75 |
8500 |
950 |
80 |
5700 |
485 |
60 |
|
2016 |
7500 |
640 |
82 |
7400 |
560 |
70 |
4800 |
400 |
48 |
6500 |
525 |
65 |
The number of candidates selected from Maharashtra during the period under review is approximately what percentage of the number selected from Delhi during this period?
Answer (Detailed Solution Below)
Tabulation Question 13 Detailed Solution
Download Solution PDFCalculation:
Total selected from Delhi = 94 + 48 + 90 + 70 + 82 = 384
Total selected from Maharastra = 85 + 70 + 84 + 60 + 65 = 364
Required percentage = (364 × 100)/384 = 94.79%
∴ The correct answer is 94.79%.
The following table gives information of panchayat elections held in six villages P, Q, R, S, T and U.
| Village |
Total number of votes (in hundreds) |
Votes polled (in %) |
Valid votes (in %) |
| P | 50 | 80 | 80 |
| Q | 60 | 75 | 80 |
| R | 100 | 65 | 65 |
| S | 80 | 60 | 70 |
| T | 60 | 80 | 90 |
| U | 40 | 90 | 60 |
Hint:
1) percentage of votes polled = \(\rm {{Total \ votes \ polled} \over Total \ number \ of \ votes}\) × 100
2) percentage of valid votes = \(\rm {{Total \ valid \ votes} \over Total \ votes \ polled}\) × 100
Find the ratio of invalid votes of village R to that of village U?
Answer (Detailed Solution Below)
Tabulation Question 14 Detailed Solution
Download Solution PDFCalculation:
Ratio of invalid votes of village R to that of village U = 100 × 65% × 35% : 40 × 90% × 40%
⇒ 65 × 35 : 36 × 40
⇒ 65 × 7 : 36 × 8
⇒ 455 : 288
∴ The required answer is 455 : 288.
Study the following table and answer the question below.
| School name | Total number of students enrolled | Percentage of enrolled students who opted for Biology | Ratio of male to female students who opted for Biology |
| A | 900 | 30% | 7 ∶ 8 |
| B | 400 | 38% | 9 ∶ 10 |
| C | 1000 | 24% | 5 ∶ 19 |
| D | 800 | 18% | 5 ∶ 7 |
What is the ratio of the total number of male students to that of female students who opted for Biology in schools A and D together?
Answer (Detailed Solution Below)
Tabulation Question 15 Detailed Solution
Download Solution PDFCalculation:
Number of male students who opted biology in school A = (900 × 30% × 7)/15 = 126
Number of male students who opted biology in school D = (800 × 18% × 5)/12 = 60
Number of female students who opted biology in school A = (900 × 30% × 8)/15 = 144
Number of female students who opted biology in school D = (800 × 18% × 7)/12 = 84
Male students who opted for biology in schools A and D : Female students who opted for biology in schools A and D
⇒ (126 + 60) : (144 + 84)
⇒ 186 : 228 = 31 : 38
∴ The correct answer is 31 : 38.