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What is the average (arithmetic mean) of 8a and 4b? (1) a + b = 5 (2

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What is the average (arithmetic mean) of 8a and 4b? (1) a + b = 5 (2  [#permalink]

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New post 02 May 2019, 03:32
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Question Stats:

78% (00:46) correct 22% (01:23) wrong based on 60 sessions

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Re: What is the average (arithmetic mean) of 8a and 4b? (1) a + b = 5 (2  [#permalink]

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New post 02 May 2019, 03:37
Bunuel wrote:
What is the average (arithmetic mean) of 8a and 4b?

(1) a + b = 5

(2) 2a + b = 11


average (arithmetic mean) of 8a and 4b
re written as
4a+2b ; 2*(2a+b)

#1
a+b=5
a & b not known insufficeint
#2
2a+b=11
so 2*11; 22
IMO B ; sufficient
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Re: What is the average (arithmetic mean) of 8a and 4b? (1) a + b = 5 (2  [#permalink]

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New post 02 May 2019, 04:06
Archit3110 wrote:
Bunuel wrote:
What is the average (arithmetic mean) of 8a and 4b?

(1) a + b = 5

(2) 2a + b = 11


average (arithmetic mean) of 8a and 4b
re written as
4a+2b ; 2*(2a+b)

#1
a+b=5
a & b not known insufficeint
#2
2a+b=11
so 2*11; 22
IMO B ; sufficient

Can't we put the value as "A= 5-B" and solve the equation for the first option?

Posted from my mobile device
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Re: What is the average (arithmetic mean) of 8a and 4b? (1) a + b = 5 (2  [#permalink]

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New post 02 May 2019, 04:26
cantaffordname wrote:
Archit3110 wrote:
Bunuel wrote:
What is the average (arithmetic mean) of 8a and 4b?

(1) a + b = 5

(2) 2a + b = 11


average (arithmetic mean) of 8a and 4b
re written as
4a+2b ; 2*(2a+b)

#1
a+b=5
a & b not known insufficeint
#2
2a+b=11
so 2*11; 22
IMO B ; sufficient

Can't we put the value as "A= 5-B" and solve the equation for the first option?

Posted from my mobile device


You do not have value of B to calculate the final result.
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Re: What is the average (arithmetic mean) of 8a and 4b? (1) a + b = 5 (2  [#permalink]

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New post 02 May 2019, 05:00
cantaffordname wrote:
Archit3110 wrote:
Bunuel wrote:
What is the average (arithmetic mean) of 8a and 4b?

(1) a + b = 5

(2) 2a + b = 11


average (arithmetic mean) of 8a and 4b
re written as
4a+2b ; 2*(2a+b)

#1
a+b=5
a & b not known insufficeint
#2
2a+b=11
so 2*11; 22
IMO B ; sufficient

Can't we put the value as "A= 5-B" and solve the equation for the first option?

Posted from my mobile device



we have 2 variables in total. if we just substitute one for the other still we are left with a variable. Thus st1 is not sufficient.

best things u can do is to to substitute nd try to solve. then u see why substitution is not working here.
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Re: What is the average (arithmetic mean) of 8a and 4b? (1) a + b = 5 (2  [#permalink]

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New post 22 Oct 2019, 10:27
Rewrite 8a + 4b = ? as 2a + b

1) a + b = 5, 2V1E
2) 2a + b = 11, can multiply the entire equation by 4 = 8a + 4b = 44, sufficient.
Answer is B
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Re: What is the average (arithmetic mean) of 8a and 4b? (1) a + b = 5 (2  [#permalink]

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New post 23 Oct 2019, 14:05
cantaffordname wrote:
Archit3110 wrote:
Bunuel wrote:
What is the average (arithmetic mean) of 8a and 4b?

(1) a + b = 5

(2) 2a + b = 11


average (arithmetic mean) of 8a and 4b
re written as
4a+2b ; 2*(2a+b)

#1
a+b=5
a & b not known insufficeint
#2
2a+b=11
so 2*11; 22
IMO B ; sufficient

Can't we put the value as "A= 5-B" and solve the equation for the first option?

Posted from my mobile device


I went down the same rabbit hole. Someone please correct me if I am wrong!

We cannot do this, because we have 1 EQUATION and 1 EXPRESSION. We do not know what the expression in the prompt is equal to. Thus, statement 1 leaves us with 1 equation and 2 variables (and 1 expression)
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Re: What is the average (arithmetic mean) of 8a and 4b? (1) a + b = 5 (2   [#permalink] 23 Oct 2019, 14:05
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