CAT DI LR section has become increasingly tough beginning from 2015. However, Understanding the basics of Bar graphs, Pie Charts, Multiple graphs, Line Graphs etc forms an integral part of solving tougher CAT level DI LR questions for the CAT Exam. This question is from CAT 2017 Slot 2 CAT DILR.

There were seven elective courses – E1 to E7 - running in a specific term in a college. Each of the 300 students enrolled had chosen just one elective from among these seven. However, before the start of the term, E7 was withdrawn as the instructor concerned had left the college. The students who had opted for E7 were allowed to join any of the remaining electives. Also, the students who had chosen other electives were given one chance to change their choice. The table below captures the movement of the students from one elective to another during this process. Movement from one elective to the same elective simply means no movement. Some numbers in the table got accidentally erased; however, it is known that these were either 0 or 1.

Further, the following are known:

1. Before the change process there were 6 more students in E1 than in E4, but after the reshuffle, the number of students in E4 was 3 more than that in E1.

2. The number of students in E2 increased by 30 after the change process.

3. Before the change process, E4 had 2 more students than E6, while E2 had 10 more students than E3.

The table is given below -

Question 3: After the change process, which course among E1 to E6 had the largest change in its enrolment as a percentage of its original enrolment?

- E1
- E2
- E3
- E6

Starts Sat, April 27th, 2019

**General Solution**

This question is not as difficult as it appears at first sight. Let us give the sum of all the cells in each row and column in the final row/column.

Now, what does the last column represent? What does the overall total 292 mean? If we crack this, we are through.

The last column gives the total number of students who took each elective. Wait, is that the case? Now, the sum given right now does not include the elements that have gotten erased. That is still a missing piece of the jigsaw. We need to worry about the empty cells also. But even without worrying about these, what can we infer? What can we say about 31, 101 and 292 in the last column?

What can we say about 31, 101 and 292 in the last column?

For E1, we can say that there were 31 students who left E1 (including the 9 who stayed in E1). Since all the cells in the first row are filled. This is straight-forward. Similarly, the last row tells us that there were 101 who had originally taken E7. There are totally 300 students. So, the 292 total tells us that 8 students are yet to be accounted for. Or, of the empty cells, 8 have to be 1 and the rest have to be 0. There are totally 12 empty cells. So, 8 have to be 1’s and 4 zeroes.

Now, let us move on to other constraints. There are 1 or 2 very juicy ones here.

Before the change process there were 6 more students in E1 than in E4 – What does this tell?

There are 31 in elective E1. So, the total number of students originally in E4 should have been 31 - 6 = 25. This is brilliant. This gives us two cells straightaway.

Now, let us move on to other constraints. There are 1 or 2 very juicy ones here.

Before the change process there were 6 more students in E1 than in E4 – What does this tell?

There are 31 in elective E1. So, the total number of students originally in E4 should have been 31 - 6 = 25. This is brilliant. This gives us two cells straightaway.

Before the change process, E4 had 2 more students than E6

Or, two more cells get filled. The row with E6 should have empty cells now filled with 1.

Before the change process, E4 had 2 more students than E6

Now, let us move to the other constraints. We have filled 6 cells. We need to fill 6 more. Of the remaining 6, four have to be 1’s and 2 have to be zeroes. Only then will our total be 300.

But after the reshuffle, the number of students in E4 was 3 more than that in E1.

Currently, E4 is at 21. The maximum E1 can get to is 18 ( it is currently at 17 with one cell remaining to be filled). Or, E1 should go to 18 and E4 should remain at 21. Fabulous.

There are three cells remaining. All three should be 1 for the total to reach 300.

We did not even need to use the last constraint.

E1 changes from 31 to 18 . A percentage change of {\\frac{31-18}{31}\\)} * 100 % = {\\frac{13}{31}\\)}*100 % = 41.93 %

E2 changes from 46 to 76. A percentage change of {\\frac{76-46}{46}\\)} * 100 % = {\\frac{30}{46}\\)}*100 % = 65.21 %

E3 changes from 36 to 78. A percentage change of {\\frac{78-36}{36}\\)} * 100 % = {\\frac{42}{36}\\)}*100 % = 116.67%

E6 changes from 23 to 61. A percentage change of {\\frac{61-23}{23}\\)} * 100 % = {\\frac{28}{23}\\)}*100 % = 165.21%

Therefore, E6 had the largest change in its enrolment as a percentage of its orginal enrolment.

The question is **"After the change process, which course among E1 to E6 had the largest change in its enrolment as a percentage of its original enrolment?"**

Choice D is the correct answer.

Learn from 4 time CAT 100 percentiler

Register in 2 easy steps and start learning in 30 seconds!

Batches filling fast

Personalised attention, Small batch sizes.

Copyrights © 2019 All Rights Reserved by 2IIM.com - A Fermat Education Initiative.

Privacy Policy | Terms & Conditions

CAT^{®} (Common Admission Test) is a registered trademark of the Indian Institutes of Management. This website is not endorsed or approved by IIMs.

2IIM Online CAT Coaching

A Fermat Education Initiative,

10-C, Kalinga Colony, Bobbili Raja Salai

K.K.Nagar, Chennai. India. Pin - 600 078

**Phone:** (91) 44 4505 8484

**Mobile:** (91) 99626 48484

**WhatsApp:** WhatsApp Now

**Email: **prep@2iim.com