Tabulation MCQ Quiz - Objective Question with Answer for Tabulation - Download Free PDF

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

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.

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.

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

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.

Calculation: 

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

Calculation:

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

Calculation

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.

Calculation:

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%.

Calculation:

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.

Given:

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.

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

Calculation:

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.

Calculation:

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%.

Calculation:

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.

Calculation

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.