\(\frac{a+b+c+d+e+f}{6
}\)

\(f= a+b+c+d+e\)

substitute

\(\frac{a+b+c+d+e+a+b+c+d+e}{6}\)

Sove and you have \(\frac{f}{3}\) which in terms of fractcio is always GREATER than \(\frac{f}{5}\)

Of course f is positive otherwise the average would not make sense

Ok thanks. However, I don't think that the average is automatically positive. There are cases where a negative average makes sense.

For instance, a, b, c, d, e could represent temperatures in C. The average temperature in C could very well be negative, 0, or positive. Therefore, the answer is D as we need more information on the nature of the variables a, b, c, d, e.

@gpowderdog is absolutely right.

f can be negative (or zero for that matter), which means the correct answer is D

a + b + c + d + e = f
avg = f /5

thus A: f + f/ 2 = f

hence f> f/5

Given, f=a+b+c+d+e

The average of a, b, c, d, e, and f= (a+b+c+d+e+f)/6 =(f+f)/6=2f/6=f/3

if f is positive, f/3>f/5
if f is negative, f/3<f/5

I would go for D.

Imagine,
First Situation: a=0, b=0, c=0, d=0, e=0. So, f=0. then, Quantity A= (0+0+0+0+0+0)/6=0 and Quantity B= 0/5 (So, Quantity A=Quantity B)
Second Situation: a=1, b=1, c=1, d=1, e=1. So, f=5. then, Quantity A= (1+1+1+1+1+5)/6=10/6 and Quantity B= 5/5=1 (So, Quantity A>Quantity B)

So, I would go for D

I also think answer would go D if we do not have conditions that a,b,c,d,e are all positive numbers.

Case 1:
simply put 0 to all numbers.
both (A) and (B) will be 0.

Case2:
put 1 to all numbers
(A) f = 5
f+a+b+c+d+e = 5 + 5 = 10
10/6 = 5/3

(B) 1

So (A) > (B)

Given above 2 cases, I think the answer should be D)

A = 2F/6 = F/3

B= F/5

If F is positive A is greater but F could be negative so answer is D.

Hello guys,

Can we conclude that whenever given comparison question without a restriction for the integer values, The relationship can never always be determined since integers can be positive and negative.

Posted from my mobile device

No

always we have to be careful what the stem tells you about the question. It is not enough to say that the solution is D based on your assertion

regards

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