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Sajjad1994
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During the time periods in which City Y’s average daily temperature 30 Jan 2020, 11:03
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C
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95% (hard)

Question Stats:

42% (02:12) correct 58% (02:26) wrong based on 156 sessions

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During the time periods in which City Y’s average daily temperature was increasing while City X’s was decreasing, the average daily temperature in City Y exceeded that in City X by approximately

(A) 0°
(B) 4°
(C) 10°
(D) 15°
(E) 19°

Source: Master GMAT
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C
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rajatchopra1994
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During the time periods in which City Y’s average daily temperature 30 Jan 2020, 21:15
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Explanation:

For City Y temperature was increasing while city X was decreasing were sept-oct and Nov- dec.
Sept-Oct.: City X average = 50, City Y = 46
Nov-Dec.: City X average = 36, City Y = 60.

City X average: (50+36)/2= 43
City Y average: (36+60)/2= 53

Average increase of 10 between City Y and City X.

IMO-C
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During the time periods in which City Y’s average daily temperature 10 May 2024, 07:58
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Hello Experts,

Can someone please give a solution to this.
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Lord_Biplab
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During the time periods in which City Y’s average daily temperature 30 Jun 2024, 08:13
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­During the months of \(September\) through \(January\), the average temperature of \(X\) is reducing while that of \(Y\) is increasing.

Average daily temperature of \(X\) during this period (Sept.-Jan.) = \(\frac{(60+30)}{2}\) = 45 \(F\).
Average daily temperature of \(Y\) during this period (Sept.-Jan.) = \(\frac{(40+65)}{2}\) = 52.5 \(F\).

Difference of the averages = 7.5 \(F\).
Closest option is 10 \(F\).

So, I chose 10 \(F\).­
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During the time periods in which City Y’s average daily temperature 01 Jul 2024, 05:48
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Lord_Biplab
­During the months of \(September\) through \(January\), the average temperature of \(X\) is reducing while that of \(Y\) is increasing.

Average daily temperature of \(X\) during this period (Sept.-Jan.) = \(\frac{(60+30)}{2}\) = 45 \(F\).
Average daily temperature of \(Y\) during this period (Sept.-Jan.) = \(\frac{(40+65)}{2}\) = 52.5 \(F\).

Difference of the averages = 7.5 \(F\).
Closest option is 10 \(F\).

So, I chose 10 \(F\).­
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Why did you take the first and last temperatures and calculate average ? There was another temperature ' in the middle

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During the time periods in which City Y’s average daily temperature 01 Jul 2024, 06:04
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kingbucky

Lord_Biplab
­During the months of \(September\) through \(January\), the average temperature of \(X\) is reducing while that of \(Y\) is increasing.

Average daily temperature of \(X\) during this period (Sept.-Jan.) = \(\frac{(60+30)}{2}\) = 45 \(F\).
Average daily temperature of \(Y\) during this period (Sept.-Jan.) = \(\frac{(40+65)}{2}\) = 52.5 \(F\).

Difference of the averages = 7.5 \(F\).
Closest option is 10 \(F\).

So, I chose 10 \(F\).­
Why did you take the first and last temperatures and calculate average ? There was another temperature ' in the middle

Posted from my mobile device
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­I know, but as the plots are pretty much straight I took the short cut.
If you take the average of the differences of the temperatures of \(X\) and \(Y\) for these four months it turns out to be approx. 10.5 \(F\).

Difference for September to November  = 50 - 46 = 4 \(F\).
Difference for November to January = 58 - 35 = 17 \(F\).

Average difference = \(\frac{(4+17)}{2}\) = 10.5 \(F\).





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During the time periods in which City Y’s average daily temperature 01 Jul 2024, 19:54
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Sajjad1994
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1.jpg

During the time periods in which City Y’s average daily temperature was increasing while City X’s was decreasing, the average daily temperature in City Y exceeded that in City X by approximately

(A) 0°
(B) 4°
(C) 10°
(D) 15°
(E) 19°

Source: Master GMAT
Show more
­simply calculate average for Sep to Dec of Y i.e. (40+ 55+ 65)/3 = 53, and same for X for Sept to Dec, i.e. 60+ 40+30)/3=43
Difference is 10
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