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Bunuel
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Nice Question
Here is my solution to this one =>


\(Mean = \frac{Sum}{#}\)
Hence a+b/2<40
So a+b<80


We need to check if the sum of a and b is less than 80 or not

Statement 1-->
3a+3b=2*117
Hence a+b=2*39 => 78

Sufficient => YES

Statement 2-->
5a=b
1,5=> YES
10000,50000=> NO


Hence Not sufficient

Hence A
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i don't understand why ST 2 is insufficient. the way i see it:


(2) b=5a. --> \(\frac{6a}{2}\)<40 --> 6a <80 --> a~13. OK, plug in now: b=5(13); b=65.
- now, what is a+b? ~13 +65 = ~78. which is clearly under 80.

where am i going wrong here?
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i don't understand why ST 2 is insufficient. the way i see it:


(2) b=5a. --> \(\frac{6a}{2}\)<40 --> 6a <80 --> a~13. OK, plug in now: b=5(13); b=65.
- now, what is a+b? ~13 +65 = ~78. which is clearly under 80.

where am i going wrong here?


i had the exact same doubt
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3a+3b = 117*2
a+b/2 = 117/3

A Sufficient
From B we cannot conclude the average So answer should be A
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Bunuel
Is the average (arithmetic mean) of a and b less than 40?

(1) The average (arithmetic mean) of 3a and 3b is 117.
(2) b = 5a


We need to determine whether (a + b)/2 < 40, i.e., whether a + b < 80.

Statement One Alone:

The average (arithmetic mean) of 3a and 3b is 117.

Thus:

(3a + 3b)/2 = 117

3a + 3b = 234

a + b = 78

Statement one alone is sufficient to answer the question.

Statement Two Alone:

b = 5a

Since we do not know the values of a or b, we cannot determine whether a + b is less than 80.

Answer: A
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Bunuel
Is the average (arithmetic mean) of a and b less than 40?

(1) The average (arithmetic mean) of 3a and 3b is 117.
(2) b = 5a

Kudos for a correct solution.


frm stmn1 : we can find (a+b)= 78 and avg would be 39 which is <40 sufficient
and stmn2 ; not sufficient


IMO A
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