Tabulation MCQ Quiz - Objective Question with Answer for Tabulation - Download Free PDF
Last updated on Jun 13, 2025
Latest Tabulation MCQ Objective Questions
Tabulation Question 1:
Comprehension:
Directions: Study the table given below and answer the questions carefully.
The table below shows the percentage of appeared and qualified candidates in SBI PO from different states.
|
Appeared Candidates = 24000 |
Qualified Candidates = 4000 |
States |
Percentage of appeared candidates |
Percentage of qualified candidates |
Andhra Pradesh |
25% |
18% |
Gujarat |
10% |
12% |
Delhi |
15% |
18% |
Madhya Pradesh |
12% |
16% |
Maharashtra |
18% |
20% |
Rajasthan |
20% |
16% |
Find the ratio of qualified candidates from Madhya Pradesh, Maharashtra, and Rajasthan together to the appeared candidates from Andhra Pradesh, Gujarat, and Delhi?
Answer (Detailed Solution Below)
Tabulation Question 1 Detailed Solution
Calculation:
Qualified candidates from Madhya Pradesh, Maharashtra, and Rajasthan = (16 + 20 + 16)% of 4000
⇒ 52 / 100 × 4000
Appeared candidates from Andhra Pradesh, Gujarat, and Delhi = (25 + 10 + 15)% of 24000
⇒ 50 / 100 × 24000
∴ Required Ratio = (52 × 4000) : (50 × 24000)
⇒ 52 × 4 : 50 × 24
⇒ 13 : 75
The ratio of qualified candidates from Madhya Pradesh, Maharashtra, and Rajasthan together to the appeared candidates from Andhra Pradesh, Gujarat, and Delhi is 13 : 75.Tabulation Question 2:
Comprehension:
Study the table and answer the following questions:
Number of products sold by 5 companies during 5 months
Month |
Company |
||||
P |
Q |
R |
S |
T |
|
January |
130 |
80 |
155 |
85 |
75 |
February |
155 |
110 |
88 |
110 |
90 |
March |
180 |
130 |
115 |
100 |
120 |
April |
160 |
145 |
240 |
140 |
165 |
May |
140 |
120 |
130 |
100 |
165 |
Out of the total number of products sold by company 'T' in March, April and May together, 32% were made of mild steel. What was the total number of mild steel products sold by company 'T' in March, April and May together ?
Answer (Detailed Solution Below)
Tabulation Question 2 Detailed Solution
Calculation:
Total products sold by T in March, April & May = 120 + 165 + 165 = 450
Mild steel products = 0.32 × 450 = 144
∴ The total number of mild steel products sold by company 'T' in March, April and May together are 144.
Tabulation Question 3:
Comprehension:
Study the table and answer the following questions:
Number of products sold by 5 companies during 5 months
Month |
Company |
||||
P |
Q |
R |
S |
T |
|
January |
130 |
80 |
155 |
85 |
75 |
February |
155 |
110 |
88 |
110 |
90 |
March |
180 |
130 |
115 |
100 |
120 |
April |
160 |
145 |
240 |
140 |
165 |
May |
140 |
120 |
130 |
100 |
165 |
What is the average number of products sold by company 'P', 'S' and 'T' in February ? (Approximate)
Answer (Detailed Solution Below)
Tabulation Question 3 Detailed Solution
Given:
Products sold in February by
P = 155
S = 110
T = 90
Concept used:
Average = Sum of observation / Number of observations
Calculation:
Total = 155 + 110 + 90 = 355
Average = 355/3 = 118.3 = 118
∴ The average number of products sold by company 'P', 'S' and 'T' in February is 118.
Tabulation Question 4:
Comprehension:
Study the table and answer the following questions:
Number of products sold by 5 companies during 5 months
Month |
Company |
||||
P |
Q |
R |
S |
T |
|
January |
130 |
80 |
155 |
85 |
75 |
February |
155 |
110 |
88 |
110 |
90 |
March |
180 |
130 |
115 |
100 |
120 |
April |
160 |
145 |
240 |
140 |
165 |
May |
140 |
120 |
130 |
100 |
165 |
The number of products sold by company 'S' approximately increased by what percent from January to April?
Answer (Detailed Solution Below)
Tabulation Question 4 Detailed Solution
Given:
Products sold by S in January = 85
Products sold by S in April = 140
Calculation:
Difference = 140 - 85 = 55
Required % = \(\dfrac{55}{85}\) × 100 = 64.71%
∴ The number of products sold by company 'S' approximately increased by 64.71 percent from January to April
Tabulation Question 5:
Comprehension:
Study the table and answer the following questions:
Number of products sold by 5 companies during 5 months
Month |
Company |
||||
P |
Q |
R |
S |
T |
|
January |
130 |
80 |
155 |
85 |
75 |
February |
155 |
110 |
88 |
110 |
90 |
March |
180 |
130 |
115 |
100 |
120 |
April |
160 |
145 |
240 |
140 |
165 |
May |
140 |
120 |
130 |
100 |
165 |
What is the ratio of the total number of products sold by company 'P' and 'R' in March to the total number of products sold by the same companies in April?
Answer (Detailed Solution Below)
Tabulation Question 5 Detailed Solution
Calculation:
Total number of products sold by company 'P' and 'R' in March = 180 + 115 = 295
Total number of products sold by the same companies in April = 160 + 240 = 400
Required Ratio = 295 : 400 = 59 : 80
∴ The ratio of the total number of products sold by company 'P' and 'R' in March to the total number of products sold by the same companies in April is 59 : 80.
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 given table carefully and answer the following question.
The table shows the percentage of students of four departments-Mechanical, Civil, Computer Science and Applied - with each student being in only one department. The table also shows the number of students of these four departments in five different colleges, with the total number of students being 2080.
College |
Number of Students |
Mechanical |
Civil |
Computer Science |
Applied |
IIT Delhi |
430 |
- |
20% |
- |
10% |
IIT Kanpur |
350 |
20% |
- |
25% |
- |
IIT Bombay |
- |
20% |
18% |
- |
32% |
IIT Madras |
- |
- |
25% |
18% |
35% |
IIT Guwahati |
400 |
20% |
22% |
- |
20% |
If the number of students in IIT Bombay is 20% less than the number of students in IIT Madras, then the number of students in IIT Bombay is:
Answer (Detailed Solution Below)
Tabulation Question 15 Detailed Solution
Download Solution PDFCalculation
The number of students at IIT Bombay and in IIT Madras = 2080 - (430 + 350 + 400)
⇒ 2080 - 1180
⇒ 900
Let the number of students at IIT Madras be 5a.
The number of students at IIT Bombay = 5a - 5a × 20%
⇒ 4a
⇒ 5a + 4a = 900
⇒ 9a = 900
⇒ a = 100
4a → 400
The number of students at IIT Bombay is 400.