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June 23rd, 2016, 11:19 PM   #1
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How to solve this temperature question?

Hello,

In the summer of 2012, in New Delhi, the mean temperature of Monday to Wednesday was 41°C
and of Tuesday to Thursday was 43°C. If the temperature on Thursday was 15% higher than that of
Monday, then the temperature in °C on Thursday was
(A) 40 (B) 43 (C) 46 (D) 49
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June 24th, 2016, 01:04 AM   #2
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let A, B C and D represent Mon-Thurs

[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 (D-A)/3 = 2
rearranging we get D = 6 + A or A = D - 6
substituting into [3] we get (D-6)*1.15 = D
0.15D = 6.9
D = 46
Thanks from Country Boy and MMath
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June 24th, 2016, 01:07 AM   #3
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OK, That what i am doing i left thanks
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October 11th, 2016, 12:23 AM   #4
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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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