Line Graph DI (Data Interpretation) Short-Cut Tricks & Tips :

Line Graph DI (Data Interpretation) Short-Cut Tricks & Tips Plays a vital role in Exam. In every exam you will get at least 5-10 questions from this topic. So candidates must focus on this topic and download this Line Graph DI (Data Interpretation) Short-Cut Tricks & Tips pdf to get important questions with best solution regarding Line Graph DI (Data Interpretation) Short-Cut Tricks & Tips. We have put all Previous Year Questions of Line Graph DI (Data Interpretation) Short-Cut Tricks & Tips that are Asked in various Govt & Private Exam.

Let us look at a shortcut method to solve a certain type of line graph problems using an example. If we attempt to solve it with complete complicated calculations, we will end up wasting a lot of precious time. Remember that the examiners don’t expect you to solve everything on paper. They want to see if you are capable of applying your logic to evaluate the graph, eliminate options, and arrive at the answer through estimation.

Once you master this technique, we will, in subsequent articles, deal with more complicated sets of line graphs.

Line Graphs

Line graph represents data in the form of straight lines that connect various data values. Both line graphs and bar graphs are used to convey same things and hence can be used inter-changeably. For example, a line graph can be generated by joining the tip of the bar graph.
Positives
1. Trends can be even better established in Line graphs than Bar graphs.   
2. Questions pertaining to percentage change and growth rates become easier to solve using line graphs.    

Negatives
1. It has a similar disadvantage as the Bar graph in terms of establishing the exact values
2. Line graphs can only indicate the value at the end of a certain period and not between any two values.

Shortcuts to crack DI sets containing Line Graphs

1. Use the advantage of looking at slope of the line in questions pertaining to growth rates.
The growth rate or the decline rate is calculated as the growth or the decline as a percentage of the initial value. For this rate to be highest, the growth or the decline has to be the highest on a very small initial value. While for this rate to be the lowest, the growth or the decline has to be the lowest on a very high initial value.

For the first question, we need to compare the values pertaining to the ‘0th month’ and the ‘2nd month’. Now looking at the slope (steepness) of the graph. For her the growth is the highest and that has happened on a very small initial value.

For the second question, we need to compare values pertaining to the ‘0th month’ and the ‘5th month’. . The growth is the least in his case and that too on a very high initial value.

2. Beware that in a Line graph you can only know value at the end of a certain time period and not the values in between two time periods.

Directions ( 1 – 5 ):Study the following Graph carefully and answer the questions given below :

Units of Raw Material Manufactured and Sold by a Company Over the Years ( Units in Crores )

Q-1) What is the average number of units sold over the years ?
a)440000000
b)4400000
c)440000
d)44000000
e)None of these

Q-2) What is the respective ratio of the difference between the number of units manufactured and sold in the year 2013 to the difference between the number of units manufactured and sold in the year 2014 ?
a)2 : 3
b)1 : 2
c)1 : 4
d)3 : 5
e)None of these

Q-3) What is the respective ratio of the number of units manufactured in the year 2011 to the number of units manufactured in the year 2015 ?
a)7 : 11
b)9 : 14
c)7 : 9
d)9 : 11
e)None of these

Q-4) What is the approximate percent increase in the number of units sold in the year 2015 from the previous year ?
a)190
b)70
c)60
d)95
e)117

Q-5) What is the difference between the number of units manufactured and the number of units sold over the years?
a)50000000
b)5000000
c)500000000
d)500000
e)None of these

SOLUTION:
Ans :(4)  44000000

QUESTION 2:EXPLANATION
STEP 1 : Number of units manufactured and sold in the year 2013
= ( 5.5 – 5 ) crore
=  0.5 crore
STEP 2 : Number of units manufactured and sold in the year 2014
= ( 5 – 3 ) crore
= 2 crore
Required ratio = 0.5 : 2

= 5:20

= 1:4

QUESTION 3 : EXPLANATION
STEP 1 :Required ratio = 4.5 : 7

= 45:70

= 9:14

Ans :(2) 9 : 14

QUESTION 4 : EXPLANATION
STEP 1 : Increase    =    6.5 – 3
=3.5 crore
STEP 2 : Increase % =  35 / 3 x 100 = 117 % 

QUESTION 5 : EXPLANATION
STEP 1 : Number of sold units = 22 crore
STEP 2 : Number of manufacturing units = ( 4.5 + 5 + 5.5 + 5 + 7 ) crore
= 27 crore

STEP 3 : Required Difference  = ( 27 – 22 ) crore

= 5 crore

= 50000000

Ans :(1) 50000000