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June 23rd, 2016, 11:19 PM  #1 
Banned Camp Joined: May 2016 From: earth Posts: 703 Thanks: 56  How to solve this temperature question?
Hello, In the summer of 2012, in New Delhi, the mean temperature of Monday to Wednesday was 41Â°C and of Tuesday to Thursday was 43Â°C. If the temperature on Thursday was 15% higher than that of Monday, then the temperature in Â°C on Thursday was (A) 40 (B) 43 (C) 46 (D) 49 
June 24th, 2016, 01:04 AM  #2 
Senior Member Joined: Apr 2014 From: UK Posts: 954 Thanks: 342 
let A, B C and D represent MonThurs [1] (A+B+C)/3 = 41 [2] (B+C+D)/3 = 43 [3] A*1.15 = D from [1] and [2] we can see that (DA)/3 = 2 rearranging we get D = 6 + A or A = D  6 substituting into [3] we get (D6)*1.15 = D 0.15D = 6.9 D = 46 
June 24th, 2016, 01:07 AM  #3 
Banned Camp Joined: May 2016 From: earth Posts: 703 Thanks: 56 
OK, That what i am doing i left thanks

October 11th, 2016, 12:23 AM  #4 
Newbie Joined: Jun 2016 From: india Posts: 24 Thanks: 4 
Temperature on, mon + tue + wed =41 *3 = 123 ....(i) tue + wed + thu = 43 *3 = 129....(ii) Subtract equation(i) from equation(ii), thu  mon = 6 ...(iii) It is given that, Temperature on thu = 15% more than on mon = 15/100 mon + mon = 115mon/100 put this value in equation (3), 115mon/100  mon = 6 mon = 40 Put this value in equation (iii) thu = 46 

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