Logical Reasoning and Data Interpretation (LRDI) in CAT or any other non-CAT exam, is about analyzing and interpreting data and bringing it to a meaningful conclusion. The huge pile of data needs to be sanitized, categorized, and then simplified into some information that can help in decision making. Line Graphs are one of the important types in the Graphs and Tables category.

LRDI in CAT, XAT, and other exams have questions on Line Graphs every year. A line graph will be given, you will have to analyze and extract information from that and answer the questions that follow.

There are two types in which Line Graph questions are prepared.

**Simple Line Graph** – Questions in this type has a Line Graph that depicts the data present in the question, and you are supposed to analyze the graph

**Line Graph couple with other graphs** –In this type, Line graphs are usually mixed with histograms or dot chart

Question 1 – Based on the information answer the questions which follow. The data was collected for the industry to analyze the impact and importance of select parameters. The Figure represents the performance of the industry on select parameters which arc Fixed Capital. Materials, Value-added, and Number of Factories from the year 2008-09 to 2015-16. Total inputs – (Output — Value added). The table represents select performance indicators which are Output, Number of Workers, and Emoluments from the year 2008-09 to 2015-16. [IIFT-2018/Set C]

Year | Output in Rs (billion) | No. of workers | Emoluments in Rs (billion) |

2008-09 | 11442 | 239966 | 65 |

2009-10 | 12241 | 250009 | 81 |

2010-11 | 14993 | 289965 | 102 |

2011-12 | 18250 | 325000 | 135 |

2012-13 | 19249 | 330000 | 147 |

2013-14 | 21493 | 380000 | 177 |

2014-15 | 23251 | 369996 | 202 |

2015-16 | 25506 | 399988 | 245 |

Q.1) In which year fixed capital per factory is lowest?

Q.2) In which year Material as a proportion of ‘Total inputs’ is highest?

Q.3) For how many years annual percentage growth in fixed capital is greater than annual percentage growth in number of factories?

1) In which year fixed capital per factory is lowest?

2008 – 09: 5280/3800 = 1.38/ factory

2011-12: 6000/4175 = 1.43/ factory

2013-14: 8500/5000 = 1.7/ factory

2015-16: 10000/5000 = 2/factory

2) In which year Material as a proportion of ‘Total inputs’ is highest?

2008 – 09: 4500/(11442-4750) = 4500/6692 = 0.672

2009-10: 5200/7241 = 0.7181

2010-11: 6500/9493 = 0.68

2011-12: 8200/11750 = 0.69

3) For how many years annual percentage growth in fixed capital is greater than annual percentage growth in number of factories?

Year | Fixed Cap | No. of factors | ||

2008-09 | 5200 | 15.3 | 3800 | 7.30% |

2009-10 | 6000 | 3.3 | 4100 | 9.70% |

2010-11 | 6200 | 12.9 | 4500 | 6.60% |

2011-12 | 7000 | 0 | 4800 | 2.00% |

2012-13 | 7000 | 21.4 | 4900 | 2.00% |

2013-14 | 8500 | 5.85 | 5000 | -2% |

2014-15 | 9000 | – | 4900 | – |

2015-16 | 10000 | 11.11 | 5000 | 2% |

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Question 1 The birth and death rates of a particular city are given below. Study the chart carefully and answer the given questions:

**Calculate the average number of births in the city between 2001 to 2005.**

a. 4.4 million

b. 4 million

c. 2.8 million

d. 3.2 billion

From the graph, calculate the values of total births between 2001 to 2005. Then find the average of them.

So, average births = (3.5 + 4 + 5 + 3 + 6.5)/ 5 = 4.4.

Therefore, the average births between 2001 to 2005 is 4.4 million (since units are in millions).

Hence, option (c) is the answer.

**By what percent (approx.) did the number of births increased in 2005 from the previous year?**

a. 180

b.150

c.90

d.117

Solution:

Increase in deaths from 2004 to 2005 = 6.5 – 3 = 3.5 million

Percentage increase in deaths = (3.5 / 3) × 100 = 116.66 %.

The closest option is 117 %, hence the answer will be option (d).

**Calculate the difference between the number of deaths and the number of births over the years.**

a.5million

b.6million

c.1billion

d.3million

Solution:

Total deaths over the years = 27

Total births over the years = 22

The difference in deaths and births = 27 – 22 = 5.

Hence, the correct answer will be option (a) i.e. 5 million.

**In which year the percentage increase in birth rate over the previous year was the maximum?**

a.2001

b.2002

c.2003

d.2004

e. 2005

Solution:

Percentage increase or decrease for any given year = (difference of values from previous year/ value in that year) × 100.

For 2002, percentage increase = (4 – 3.5) × 100 = 50 %.

For 2003, percentage increase = 100 %.

For 2004, percentage increase = 200 % (decrease).

For 2005, percentage increase is = 350 %.

Therefore, percentage increase was highest in 2005 and so, option (e) is the correct answer.

Question 2 The following graph shows the quantity of milk and the food grains consumed by the male and female population, annually. Read the graph and answer the questions:

**i) Per capita production of milk was least in which year**

Solution The overall production is increasing steadily but the milk produced decreased a sudden decline in 1996. Hence, to calculate least capita production we should consided 1990 and 1996

In 1990, Per capita production = Total milk produced/ Total population = (5 milion gallons/ 69 million)

Thus, per capita production = 0.072 gallon/ person

In 1996, Per capita production = Total milk produced/ Total population = (6 milion gallons/ 82 million)

Thus, per capita production = 0.073 gallon/ person

Hence, the year with least year capita production is 1991.

Preparing any topic from the scratch requires patience, hard work, and above all commitment. Line Graphs are usually easy, to prepare for them follow the steps:

**Level – 1**

- Learn Speed calculation: For an effective and quick calculation, be thorough with tables till 20, memorize squares till 30, and cubes till 15. Extracting data from Line Graphs into a table consumes most of the time; speedy calculation tricks can help you analyze data in the table quickly
- Fraction – Percentage relation: Most of the questions require approximate calculations, knowing the percentage – fraction conversion will help solve the problem quickly
- Practice beginner level questions on Line Graphs

**Level – 2**

- Move on to more complex problems, attempt beginner and intermediate level mock on Line Graphs
- Solve previous year questions on Line Graphs and keep a watch on time
- Keep attempting mocks to check your performance
- Topic-wise mocks, provided by MBAP, can be utilized to enhance your performance

**Level – 3**

- For advanced level preparation, start practising questions from “How to prepare for Data Interpretation” by Arun Sharma
- Questions in the book are categorized into Level of Difficulty (LOD), based upon your preparation level, start attempting 3 or 4 questions daily

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