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 Post subject: GMAT FunctionsPosted: Fri Jun 17, 2011 4:27 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424
Suppose the number of fans at a baseball game is given by the function b(x) = 150x – 6 and the number of fans at football game is given by f(b) = 2b + 12. What is the value of x when there are 1200 fans at the football game?
A. 2
B. 3
C. 4
D. 5
E. 7

(C) The first thing to do is to substitute f = 1200 into the second equation to solve for b:
b = (f – 12)/2 = (1200 – 12)/2 = 1188/2 = 594. Now, we substitute b = 594 into the first equation to solve for x:
x = (b + 6)/150 = (594 + 6)/150 = 4. The correct answer is C.
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I could not understand the second step of this question, when we have to put the value of b into the second equation. Why do we put the value of b in the place of b(x)?

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 Post subject: Re: math (t.2, qt. 18): functionsPosted: Fri Jun 17, 2011 4:58 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
We need to find the value of x, which is a variable of the function b(x).

What is the function b(x)? The function sets up a relation between b and x, in other words any value of x defines the corresponding value of b.
b(1) = 150 × 1 – 6 = 144
b(4) = 150 × 4 – 6 = 594

But what if we choose the value of a function first, i.e. set the value of b? Will it define the corresponding value of x? In our case it will.
If b = 144, then 144 = 150x – 6, so x = 1.
If b = 594, then 594 = 150x – 6, so x = 4.

So, basically, the second step is the same as the first one. In the first step the specific value of f, 1200, defined the corresponding specific value of b, 594, which in its turn defined the corresponding specific value of x, 4 (in the second step).

Note, that in any function its variable (argument of a function) defines one value of that function, but NOT any value of a function defines ONE SPECIFIC value of its variable (argument of a function).
For example, b(x) = x². If b = 4, then x can be -2 or +2. Both these values define b = 4.
b = -4 does NOT yield any value of the variable x at all, because any square is a non-negative value.

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 Post subject: Re: math (t.2, qt. 18): functionsPosted: Fri Aug 12, 2011 9:01 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424
The first thing to do is to substitute f = 1200 into the second equation to solve for b:

b = (f – 12)/2 = (1200 – 12)/2 = 1188/2 = 594.

Can you explain how you got from the given two equations to the above equation? I don't understand how f = 1200 in the second equation came about, nor do i understand how that yielded the above equation.

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 Post subject: Re: math (t.2, qt. 18): functionsPosted: Fri Aug 12, 2011 9:17 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
Can you explain how you got from the given two equations to the above equation?
First of all, note, that the original expressions are not equations at first, but formulas of the functions.

How did we get the formula "b = (f – 12)/2"?
1. We took the original formula, f = 2b + 12.
2. Subtract 12 from the both sides, i.e. "move" it from right side to the left one: f – 12 = 2b
3. Divide by 2: (f – 12)/2 = b

Then we just plugged in the value of f, 1200.
b = (1200 – 12)/2 = 1200/2 – 12/2 = 600 – 6 = 594.

Quote:
I don't understand how f = 1200 in the second equation came about
The specific value of f, 1200, was given in the original question statement.

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 Post subject: Re: math (t.2, qt. 18): functionsPosted: Sat Sep 03, 2011 4:16 pm

Joined: Sun May 30, 2010 3:15 am
Posts: 424
where is the formula for b coming from?
b = (f – 12)/2?
also for the second formula?
could you walk me in more detail through the algebraic manipulation?

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 Post subject: Re: math (t.2, qt. 18): functionsPosted: Sat Sep 03, 2011 4:28 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
where is the formula for b coming from?
b = (f – 12)/2?
1. We took the original formula, f = 2b + 12.
2. Subtract 12 from the both sides, i.e. "move" it from right side to the left one: f – 12 = 2b
3. Divide by 2: (f – 12)/2 = b

questioner wrote:
also for the second formula?
1. We took the original formula, b = 150x – 6.
2. Add 6 to the both sides, b + 6 = 150x.
3. Divide by 150, (b + 6)/150 = x.

NOTE, that in each case we can plug the value first and then calculate the desired variable:
The first formula, f = 2b + 12, transforms into 1200 = 2b + 12.
The second formula, b = 150x – 6, transforms into 594 = 150x – 6.

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