Avg of two terms & largest possible value

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The discussion revolves around finding the largest integer value of 'a' given that the average of 3a and 4b is less than 50, with the condition that 'a' is twice 'b'. The equation simplifies to 3a + 4b < 100. Participants suggest expressing everything in terms of 'b' since 'a' equals 2b, leading to the conclusion that 'a' must be less than 20. The largest integer value for 'a' is determined to be 19, assuming 'b' is also an integer. The conversation emphasizes the importance of correctly substituting values and maintaining clarity in variable definitions.
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Homework Statement


If the average (arithmetic mean) of 3a and 4b is less than 50, and a is twice b, what is the largest possible integer value of a?

Homework Equations


Avg of two variables is:
(a+b)/2

The Attempt at a Solution


(3a + 4b)/2 < 50
3a + 4b <100
Now try out values such a = 2b
I) a= 10 & b=20
3(10) + 4(20) <100
30 + 80 <100
110<100 false
2) a=9 & b = 18
3(9) + 4(18)
27 + 72
99
but this answer is not correct. Some body please guide me.
Zulfi.
 
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You seem to have flipped the meaning of a and b when you chose numbers.
 
zak100 said:

Homework Statement


If the average (arithmetic mean) of 3a and 4b is less than 50, and a is twice b, what is the largest possible integer value of a?

Homework Equations


Avg of two variables is:
(a+b)/2

The Attempt at a Solution


(3a + 4b)/2 < 50
3a + 4b <100
Now try out values such a = 2b
I) a= 10 & b=20
3(10) + 4(20) <100
30 + 80 <100
110<100 false
2) a=9 & b = 18
3(9) + 4(18)
27 + 72
99
but this answer is not correct. Some body please guide me.
Zulfi.

If you know that a = 2b, why do you not just express everything in terms of b alone?
 
zak100 said:
i.e a =19
Yes, unless b is also supposed to be an integer. I agree it does not say so.
 
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