The mean temperature from 27th October to 29th October was 24-degree Celcius and from 28th October to 30th October was 22-degree Celcius. If the temperature on 30th October was 3/4th of the temperature on 27th October, what was the temperature on 30th October?00 : 20 : 437/15 attemptedEndOptions: Pick one correct answer from below18212427
Question
The mean temperature from 27th October to 29th October was 24-degree Celcius and from 28th October to 30th October was 22-degree Celcius. If the temperature on 30th October was 3/4th of the temperature on 27th October, what was the temperature on 30th October?00 : 20 : 437/15 attemptedEndOptions: Pick one correct answer from below18212427
Solution
Let's denote the temperatures on the 27th, 28th, 29th, and 30th of October as T27, T28, T29, and T30 respectively.
From the problem, we know that:
(1) (T27 + T28 + T29) / 3 = 24 (2) (T28 + T29 + T30) / 3 = 22 (3) T30 = 3/4 * T27
From equation (1), we can express T27 as:
T27 = 3*24 - T28 - T29 ...(4)
Substitute equation (4) into equation (3), we get:
T30 = 3/4 * (3*24 - T28 - T29) ...(5)
Substitute equation (5) into equation (2), we get:
(T28 + T29 + 3/4 * (3*24 - T28 - T29)) / 3 = 22
Solving this equation, we get:
T30 = 18
So, the temperature on 30th October was 18-degree Celcius.
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