Mock test Q.16

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How to solve q.16. I feel that there isn't sufficient info given for this.

Please clarify

[Malabika Guha Mustafi]

From 10a.m to 1p.m pollution increase (80-40) in 3 hours .

so slope is 40/3

so, at 8 a.m the pollution will be 40/3 and at 4 p. to 8 p.m(constant) it will be 120.

From 8 p.m to 12 p.m it comes down to 40/3.

Change or slope = (120-40/3 )/4

= 80/3

so. at 10 p.m, the pollution level = 120-160/3=200/3

On Wednesday, 11 November 2020 at 22:35:24 UTC+5:30 Malabika Guha Mustafi wrote:

rom 10a.m to 1p.m pollution increase (80-40) in 3 hours .

so slope is 40/3

so, at 8 a.m the pollution will be 40/3 and at 4 p. to 8 p.m(constant) it will be 120.

From 8 p.m to 12 p.m it comes down to 40/3.

Change or slope = (120-40/3 )/4

= 80/3

so. at 10 p.m, the pollution level = 120-160/3=200/3

[CG]

This is how I did it. 

Converted time to 24 hour format

Step 1 : Find the equation of the line passing through (10,40) and (13,80)

Step 2 : Get the y value for x = 8 and x = 16

Step 3 : Find the equation of the line passing through (20,120) and (24, 40/3)

Step 5 : Get the y value for x = 22 

[Malabika Guha Mustafi]

@ CG, very methodical representation..really like it.

[subhajit POD]

Hi all, 

This is Subhajit from the course support team.

I really like the solution done by Malabika. It is really precise and to the point. 

CG the explanation you provided is awesome. 

Thanks for all your efforts. 

Best

Subhajit

[Malabika Guha Mustafi]

Thank  you.

[Dhiman Sengupta]

Hi Malabika,

The solving technique you had given was spot on..

Though I did it a little different way but your process was proper way to solve this problem.

I found 8 PM it was 120 and 12 mid night it was 13.3., so diff is 106.7. So each hour change is 26.675..

Now to reach 13.3 after 4 hours , the possible  value after 2 hours could be, 

2*26.675 + 13.3 = 66.6 i.e. 200/3.

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[Malabika Guha Mustafi]

@ Sengupt

 same , just you changed to decimal. slope or change per hour = 80/3= 26.675.

13.3=40/3

Actually all the points on the same line[ pollution level at 8p.m=120,  '10 p.m and 12 p.m]. All solutions are proper according to their approach.

[rakesh ky]

good explanation by CG (SINCE NAME NOT MENTIONED BUT LIKE TO KNOW) ..THANKS FOR YOUR KIND HELP. I UNDERSTOOD THE SOLUTION PRACTICALLY AND EASILY..

[Cherian George]

Actually the method that Malabika explained takes lesser time to find the answers (~5minutes less) without finding the equation of the line. I'll explain in detail below:

Step 1 : Slope from (10,40) and (13,80) = (80-40)/3 = 40/3 (This means that between 0800hrs and 1600hrs the pollution increases by 40/3 every hour). We need to find the pollution value at 0500hrs to 0800hrs, 1600hrs to 2000hrs and 2400hrs onwards (same as 0500to0800)

Step 2 : At 0800hrs the pollution is 40 - 2*40/3 = (120-80)/3 = 40/3 [40/3 is multiplied by 2 since 1000hrs-0800hrs=2]

Step 3 : At 1600hrs the pollution is 80 + 3*40/3 = (240+120)/3 = 360/3 = 120 [40/3 is multiplied by 3 since 1600hrs-1300hrs=3]

Step 4 : At 2400hrs the pollution is the same as what we found in Step 2 i.e. 40/3

Step 5 : Find the slope from (20,120) to (24,40/3). = (40/3-120)/4 = (40-360)/12 = -320/12 = -80/3

Step 6 : Find the pollution at 10 PM or 2200hrs  : 120 - 2*80/3 = (360-160)/3 = 200/3

  [-80/3 is multiplied by 2 since 2200hrs-2000hrs=2]

This method is faster. No need to find the equation of the lines. Just understand slope as rise/run (Same way speed is distance/time)

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[Akshay Malik]

Approached the question the way @Malabika has done but did a silly calculation mistake and got stuck.