Was the average (arithmetic mean) of the temperature at noon in degree

the thinking you are applying may suit in either increasing or decreasing set but here when any of the variable can have any possible values .
a1+a2+a3+a4=116
a2+a3+a4+a5=30*4=120,only possible result is a5-a1=4
now see a1,a2,a3,a4,a5 can take any value,no boundry whatsoever.so average doesn't matter at all

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As for your question,

Yes you can have a case such as 26,30,30,30,30 but you can also have a case of 4,40,40,32,8 satisfying the 2 statements at the same time but in the 2nd case, the average is <28.

You can recognize this from the given statements combined together, to get e-a=4. So once you get this information, you can have a big range of values that may or may not have average >28.

Hence E is the correct answer.

Hope this helps.

Asked: Was the average (arithmetic mean) of the temperature at noon in degrees Celsius in town A during the first 5 days of last month at least 28 degrees Celsius?
\(x_1+ x_2+ x_3+x_4+ x_5 >=28*5 = 140\)

Let the temperatures at noon in town A during 5 days be \(x_1, x_2, x_3, x_4, x_5\)respectively
(1) The sum of the temperatures at noon in town A during the first 4 days was 116 degrees Celsius.
\(x_1+x_2+x_3+x_4=116\)
Since \(x_5\) is not known but should be >=24 to answer yes to \(x_1+ x_2+ x_3+x_4+ x_5 >=28*5 = 140\)
NOT SUFFICIENT

(2) The average of the temperatures at noon in town A during the second, third, fourth, and fifth days was 30 degrees Celsius.
\(x_2+x_3+x_4+x_5=30*4 =120\) but should be >=20 to answer yes to \(x_1+ x_2+ x_3+x_4+ x_5 >=28*5 = 140\)
Since \(x_1\) is not known
NOT SUFFICIENT

Combining (1) & (2) together
(1) The sum of the temperatures at noon in town A during the first 4 days was 116 degrees Celsius.
\(x_1+x_2+x_3+x_4=116\)
(2) The average of the temperatures at noon in town A during the second, third, fourth, and fifth days was 30 degrees Celsius.
\(x_2+x_3+x_4+x_5=120\)
(2) - (1) gives
\(x_5 - x_1 = 120 -116 = 4\)

\(x_1 & x_5\) can take multiple values
\(If\ x_5 = 25 \ and\ x_1 = 21\ then\ x_1+ x_2+ x_3+x_4+ x_5 >=140\)
But if
\(If\ x_5 = 23\ and\ x_1 = 19\ then\ x_1+ x_2+ x_3+x_4+ x_5 <140\)
NOT SUFFICIENT

IMO E

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