bushel of corn

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bushel of corn

by vittalgmat » Mon Jan 19, 2009 9:30 am
Hi,
I searched BTG for solution to this, but could not find.
So I am posting it.

The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents per day, while the price of wheat is decreasing at a constant rate of sqrt2*x- x cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?


(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60
Source: — Problem Solving |

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Re: bushel of corn

by logitech » Mon Jan 19, 2009 9:41 am
320+5x=580-(sqrt2*x- x)

5x+ sqrt2*x- x = 260

4x+1.4X=260

5.4x=260

5x is around 240 cents , or 2 dollars 40 cents

Remember we are looking for a 5x increase over 320

320+240=560
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by vittalgmat » Mon Jan 19, 2009 9:50 am
Hi Logitech,
I read the problem differently. here is what I thought: Pls correct me.

corn is increasing at the rate of 5x cents/day while wheat is decreasing at the rate of sqrt(2)x -x cents/day. So in Y days the prices become equal.
So find y and then find x.
And this is where I got lost..

Thanks

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by logitech » Mon Jan 19, 2009 10:04 am
vittalgmat wrote:Hi Logitech,
I read the problem differently. here is what I thought: Pls correct me.

corn is increasing at the rate of 5x cents/day while wheat is decreasing at the rate of sqrt(2)x -x cents/day. So in Y days the prices become equal.
So find y and then find x.
And this is where I got lost..

Thanks
You are right and at the of the Yth day their prices will be equal so you can solve the both equations for the same DAY. Which means do not get lost with Y!

320+5XY = 580-Y(BLABLA)

Y(BLABLA)+5XY=260

Y(BLABLA+5X)=260

Y= 260/(BLABLA+5X)

and you get lost :)
LGTCH
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