The average of temperatures at noontime from Monday to Friday is 50;

Question

The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?

A. 20
B. 25
C. 40
D. 45
E. 75

abcd12345 · Accepted Answer

there are 5 days so the sum of temperature can be 50*5=250
lowest is 45. to find the maximum range we can say the temperature was the lowest for 4 of the 5 days
so 4*45=180. 
on the fifth day it is 250-180=70
range is therefore 70-45=25
answer B
If 70 was one of the answer choices I am afraid I would have overlooked the range and chosen it

pike · Answer

45 + a + b + c + d = 250

We want to maximise a single variable, which will maximise the range.

Assume, a = b = c = 45
250 - 4*45 = 70

Range = 70 - 45 = 25

zuberahmed · Answer

Monday : 45
Tuesday : 45
Wednesday : 45
Thursday : 45
Friday : 70

The avg : 50 
The range : 25

B is the answer.

stonecold · Answer

Nice Question.
Here is what i did in this one =>

Let the values in increasing order be -->
w1
w2
w3
w4
w5

Using -->

Mean =\frac{Sum}{#}

Sum(5)=50*5=250
Now w1=45

To maximise the range we must maximise w5.
To maximize w5,we must minimise all others.
Since no median constraint is provided ->Set will look like -> w1,w1,w1,w1,w5 => 4w1+w5=250
w5=70
Hence Range = 70-45=25

Hence B

saurabhsavant · Answer

Hi, Nic explanation, i have just one question though, w5 can be maximised by assuming any hypothertical value...say 1000000 deg....other variables shall get automatically adjusted Is this question complete? Thanx Anxiously awaiting reply

stonecold · Answer

Absolutely not.
You cannot assume W5 to be 1000000000

Look at the minimum value => 45
The other values must be greater than or equal to 45 since 45 is the minimum.

SO basically there are two constraints =>
One => Sum = 250
Two => Least temperature is 45.

I hope that helps

Regards
Stone Cold

ScottTargetTestPrep · Answer

Since the average of 5 daily temperatures is 50, the sum is 250. To maximize the range, we want one value to be as large as possible. This is done by minimizing the other 4 values. We can minimize the temperatures for 4 of the days, so the sum of the temperatures for those 4 days is 45 x 4 = 180. Thus, the final, maximum temperature, is 250 - 180 = 70 and the maximum range is 70 - 45 = 25.
 
Answer: B

Bunuel · Answer

Given: T(min) = 45 and average = 50 --> sum of the temperatures = 50*5 = 250.

We want to maximize the range --> in order to maximize the range we need to maximize the highest temperature --> in order to maximize the highest temperature we should make all other temperatures as low as possible, so equal to 45 (lower limit) --> T(max) = 250 - 4*45 = 70 --> Range = T(max) - T(min) = 70 - 45 = 25.

Answer: B.

Krabhay · Answer

Maximum possible temp would occur when except for the warmest day when the temp is the highest all other four days record the lowest temp i.e. 45 degrees.
So, maximum possible temperature = (50 x 5) – (45 x 4) = 70 degrees.
So, maximum possible range = maximum – minimum = 70 – 45= 25 degrees.

MHIKER · Answer

Don't look for the maximum possible temperature. The question asked the difference between the maximum possible temperature and the lowest temperature.

Total temperature = 50*5=250
The least possible temperature for 4 days =45*4=180
The maximum possible temperature = 250-180=70

The range = 70-45=25

The answer is B.

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The average of temperatures at noontime from Monday to Friday is 50;

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