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

Question

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?

(1) The sum of the temperatures at noon in town A during the first 4 days was 116 degrees Celsius.

(2) The average of the temperatures at noon in town A during the second, third, fourth, and fifth days was 30 degrees Celsius.

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Matthyrou · Answer

hi, 
looking for avg on five first day 
1) during the first 4 day, avg=29, insufficent bc if the fifth is 0°c, the avg will be below 28
2) same reasoning, insufficent 
both together, avg should be btw 29 and 30, sufficent,
ans c

Tammy01 · Answer

Q: 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?

(1) The sum of the temperatures at noon in town A during the first 4 days was 116 degrees Celsius.

(2) The average of the temperatures at noon in town A during the second, third, fourth, and fifth days was 30 degrees Celsius.

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Ans: E. Both are insufficient.

Explanation1: Both the statements doesn't mention whether the days are for last month.

Explanation2: Consider the statements to be giving values for first 5 days for last month.
Still, since the stem asks whether the avg of first 5 days is AT LEAST 28 degree C, we need to find the minimum possible value for sum of all 5 days.

Statement 1: a + b + c + d = 116. Consider a to be lowest i.e, a = 0; b + c + d = 116. (No value of e- Insufficient)
Statement 2: b + c + d + e = 120. Substitute b + c + d from Statement1 => 116 + e = 120 => e = 4. (No value of a- Insufficient)

Now consider both together,
a + (b + c + d) + e = 0 + (116) + 4 = 120. Avg = 120/5 = 24 degree C.

Thus, avg cant be at least 28 degree C.

bhanu29 · Answer

let a,b,c,d,e be temperature on 5 days
To find if avg >= 28, i.e (a+b+c+d+e)/5 >= 28
or
a+b+c+d+e >= 140 ?

Statement 1:
 The sum of the temperatures at noon in town A during the first 4 days was 116 degrees Celsius.
a+b+c+d = 116

a+b+c+d+e >= 140 ?
116+e >= 140?
e >= 24 ?

e can be any number (including negative numbers)
So we cannot definitely say e>=24
So  insufficient 
 AD /BCE

Statement 2:
 The average of the temperatures at noon in town A during the second, third, fourth, and fifth days was 30 degrees Celsius.

(b+c+d+e)/4 = 30
b+c+d+e = 120

a+b+c+d+e >= 140 ?
a+120 >= 140?
a>=20?
a can be any number (including negative numbers)
So we cannot definitely say a>=20
So  insufficient 
 AD / B CE

Combining Statement 1 and 2:
 a+b+c+d = 116
b+c+d+e = 120

Add both equations

a+ 2(b+c+d) + e = 236
b+c+d = (236 -e -a)/2

Substituting b +c + d
a+ (236 -e -a)/2 + e >= 140 ?
a/2 + e/2 + 118 >= 140 ?

a/2 + e/2 >= 22 ?
a + e >= 44 ?

a and e can be any numbers.
So we cannot definitely say a + e >= 44

So  insufficient 
 AD / B  C  E

Correct Answer E.

whatsinthename1 · Answer

1- Generally the range is indicative of the spread but it only focuses on extremes whereas the standard deviation encompasses everything. If the nature of members of the set is not consistent within the extremes and the middle ones we can’t know for sure if  SDA  is greater than  SDB . -  NS 
 
  2- Even if they are evenly spaced out the SD is determined by the common difference D. D would tell us how spread out the numbers are from the mean. Here we do not know if  DA  is larger than  DB . -  NS 
 
 1+2 - If the  rangeA  is higher than  rangeB , they have same number of members and have evenly spaced out numbers we can say that  DA  is larger than  DB  which gives us an yes/no for this.  Hence sufficient.
  
Ans C

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

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

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