A table is a set of data arranged in rows and columns and is one of the most common ways of putting information across to people.
A Table
Example 1: Questions 1−4 are based on the table below, which is a record of the performance of a baseball team for the first seven weeks of the season.
Q.1.2: How many games did the team win during the first seven weeks?
(A) 2
(B) 21
(C) 29
(D) 50
(E) 58
Answer: C
Solution: To find the total number of games won, add the number of games won for all the weeks.
5 + 4 + 5 + 6 + 4 + 3 + 2 = 29
The correct answer is choice (C).
Q.1.2: What percent of the games did the team win?
(A) 21%
(B) 29%
(C) 42%
(D) 58%
(E) 72%
Answer: D
Solution:
The team won 29 out of 50 games.
29/50 = 58/100 = 58%
The correct answer is choice (D).
Q.1.3: Which week was the worst for the team?
(A) Second week
(B) Fourth week
(C) Fifth week
(D) Sixth week
(E) Seventh week
Answer: E
Solution: Compare the numbers in the wonlost columns.
The seventh week was the only week that the team lost more games than it won.
The correct answer is choice (E).
Q.1.4: Which week was the best week for the team?
(A) first week
(B) third week
(C) fourth week
(D) fifth week
(E) sixth week
Answer: B
Solution:
Compare the numbers in the win–loss columns. Since there is more than 1 week with more wins than losses, you need to do a quick calculation.
Method 1
Compare the values of the ratio wins/losses. The greatest value will be the best week.
5/3 ≈ 1.67
5/2 = 2.5
6/3 = 2
4/2 = 2
The value for the third week is 2.5.
The correct answer is choice (B).
Method 2
Find the percentage of games won using the number of wins and the number of games each week.
5/8 = 62.5%
5/7 ≈ 70%
6/9 ≈ 67%
4/6 ≈ 67%
The team won 70% of its games in the third week.
The correct answer is choice (B).
Example 2: Here is an example consisting of tabular data:
Q. 2.1: The category receiving the least percentage help from the centre (in the entire data) is:
(A) Category B in 1995
(B) Category C in 1996
(C) Category B in 1996
(D) Category D in 1995
Correct Answer is Option (B).
 In this type of question, it is better to examine the alternatives given rather than trying to find the least percentage from the table.
 Let us now calculate the required percentage of the given alternatives:
Q. 2.2: The difference between the average costs paid by the Centre during 1995 and 1996 is:
(A) Rs. 66 lakh
(B) Rs. 13.2 crore
(C) Rs. 132 lakh
(D) Rs. 13.2 lakh
Correct Answer is Option (C).
 Adding all the cost figures in the 1995 column, i.e.
18.4 + 2.6 + 13.0 + 6.6 + 55.0, you get 95.6. The average in 1995:
=95.6/Number of categories = 95.6/5
= Rs. 19.12 Similarly, the average in 1996:
=(17.4+1.6+10.0+10.6+62.6) / 5
= Rs. 20.44 The difference = Rs. (20.44−19.12)
= Rs. 1.32 cr
= Rs. 132 lakh
(Note how the answer needed conversion from crores to lakhs).
Q. 2.3: Monthly cost to the city receiving E category assistance in 1996 is most nearly:
(A) Rs. 1.8 crore less than that in 1995
(B) Rs. 2.1 crore more than that in 1995
(C) Rs. 2.1 crore less than that in 1995
(D) Rs. 1.8 crore more than that in 1995
Correct Answer is Option (B).
 Here, straight calculation is only needed. We need to look at the total assistance figures.
Q. 2.4: Assuming that 50% of the persons receiving category B help in 1995 were adults caring for minor children, but the city’s contribution towards maintaining these adults was 40% of the total contribution to B program in 1995, average amount paid by the city for each adult per year in 1995 is most nearly:
(A) Rs. 5900
(B) Rs. 6000
(C) Rs. 7500
(D) Rs. 3000.
Correct Answer is Option (B).
Q. 2.5: Monthly costs to the city of category D during 1995 and 1996 bear a ratio (most nearly):
(A) 2 : 3
(B) 5 : 3
(C) 3 : 2
(D) 3 : 5
Correct Answer is Option (D).
 Again, we can straightaway determine the answer through simple calculation.
 Since a ratio is required to be calculated, we can avoid the division by 12.
 Directly from the table we have, total assistance in 1995 and 1996 for Category D as 26.4 and 42.6.
 Hence the ratio is 26.4 : 42.6 = 3 : 5 nearly.
Example 3: Directions: Study the table carefully to answer the following questions.
Q. 3.1: How many marks did Himani get in all the subjects together?
A. 505
B. 496
C. 525
D. 601
E. None of these
Correct Answer is Option (C)
Solution: Required number = (58 ×150)/100 + (60 ×150)/100 + (64 ×150)/100 + (54 ×125)/100 + (70 ×75)/100 + (60 ×75)/100 + (72 ×75)/100 + (66 ×50)/100
= 87 + 90 + 96 + 67.5 + 52.5 + 45 + 54 + 33 = 525
Q. 3.2: What are the average marks obtained by all students together in Physics?
A. 75.33
B. 56.5
C. 64.25
D. 48.88
E. None of these
Correct Answer is Option (B)
Required average marks obtained = 452 × (75/100) × (1/6) = 56.5
Q. 3.3: How many students have scored the highest marks in more than one subject?
A. Three
B. Two
C. One
D. None
E. None of these
Correct Answer is Option (A)
Bakul, Chaitanya, and Gauri scored the highest marks in more than one subject.
Q. 3.4: Marks obtained by Ankita in Sanskrit is what percent of marks obtained by Gauri in the same subject? (rounded off to two digits after decimal)
A. 91.43
B. 94.29
C. 103.13
D. 109.38
E. None of these
Correct Answer is Option (D)
Required percentage = (70 × 100)/64 = 109.38
Q. 3.5: Who has scored the highest marks in all the subjects together?
A. Chaitanya
B. Himani
C. Deepali
D. Gauri
E. None of those.
Correct Answer is Option (E)
Total marks scored by Ankita = 90 + 96 + 100.5 + 73.75 + 52.5 + 48.75 + 51 + 35 = 547.5
Similarly
Total marks cored by Bakul=726.25
Total marks scored by Chaitanya = 647.25
Total marks scored by Deepali =554.75
Total marks scored by Gauri =614.5
Total marks scored by Himani = 525
Thus, highest marks are scored by Bakul.
Example 4: Directions: Answer the questions based on the information given below.
The given table shows the data regarding the number of air conditioners and air coolers sold by a shop in 5 different months and also percentage of those installed.
If total number of air conditioners installed in all the months together is 1540 units, then find out the number of air coolers installed in the month of June.
A – 120
B – 170
C – 140
D – 210
E – 150
Solution
In March, Number of air conditioners installed = 2x * 0.8 = 1.6x
Number of air coolers installed = (3x + 90) * 1/3 = x + 30
Similarly, we can find out the number of air conditioners and air coolers installed in the given 5 months.
Total number of air conditioners installed in all the months together
= (1.6x) + (2x + 400) + (x + 300) + 2x + (0.9x + 90)
= 7.5x + 790
According to the question,
7.5x + 790 = 1540
Or, 7.5x = 750
Or, x = 100
Number of air coolers installed in the month of June = 1.5 x 100 = 150
Hence, option E.
Example 5: Directions: Answer the questions based on the information given below.
The given table shows the data regarding the number of air conditioners and air coolers sold by a shop in 5 different months and also percentage of those installed
Find the difference between number of air conditioners installed in April and June together and the number of air coolers installed in March and May together.
A – Can’t be determined
B – 970
C 870
D – 210
E – 530
Solution
In March, Number of air conditioners installed = 2x * 0.8 = 1.6x
Number of air coolers installed = (3x + 90) * 1/3 = x + 30
Similarly, we can find out the number of air conditioners and air coolers installed in the given 5 months.
Number of air conditioners installed in April and June together = (2x + 400) + 2x = 4x + 400
Number of air coolers installed in March and May together = (x + 30) + (3x – 600) = 4x – 570
Required difference = (4x + 400) – (4x – 570) = 970
Hence, option B.
198 videos165 docs152 tests

1. What is the purpose of a table in data analysis? 
2. How can tables be helpful in solving problems related to data analysis? 
3. What are some common types of tables used in data analysis? 
4. How can one effectively interpret the information presented in a table chart? 
5. How can one improve their skills in solving table chart problems? 

Explore Courses for Mechanical Engineering exam
