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# ps-corn

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VP
Joined: 30 Sep 2004
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ps-corn [#permalink]

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30 May 2005, 14:43
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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 5*x cents per day while the price of wheat is decreasing at a constant rate of squareroot2*(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
_________________

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Kudos [?]: 411 [0], given: 0

Director
Joined: 18 Apr 2005
Posts: 543

Kudos [?]: 37 [0], given: 0

Location: Canuckland
Re: ps-corn [#permalink]

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30 May 2005, 15:29
christoph wrote:
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 5*x cents per day while the price of wheat is decreasing at a constant rate of squareroot2*(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

320 + 5x = 580 - x*2^(1/2) + x

x = 260/(4+2^(1/2))

plugging this back into equiation

my guess is E (I don't have a calculator)

Kudos [?]: 37 [0], given: 0

Senior Manager
Joined: 21 Mar 2004
Posts: 445

Kudos [?]: 76 [0], given: 0

Location: Cary,NC
Re: ps-corn [#permalink]

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30 May 2005, 16:56
sparky wrote:
christoph wrote:
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 5*x cents per day while the price of wheat is decreasing at a constant rate of squareroot2*(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

320 + 5x = 580 - x*2^(1/2) + x

x = 260/(4+2^(1/2))

plugging this back into equiation

my guess is E (I don't have a calculator)

why did you assume that only 1 day is involved ...i.e. prices decrease & increase for 1 day only ?

This is a nasty question or something is missing here

Assume prices become same after N days

320 + N(5x) = 580 - N(xsqrt2 - x )

N = 260 / x(4+sqrt2)

There are 2 variables and only 1 equation.

Something seems to be wrong or missing.
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ash
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Director
Joined: 18 Apr 2005
Posts: 543

Kudos [?]: 37 [0], given: 0

Location: Canuckland
Re: ps-corn [#permalink]

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30 May 2005, 19:34
ashkg wrote:
sparky wrote:
christoph wrote:
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 5*x cents per day while the price of wheat is decreasing at a constant rate of squareroot2*(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

320 + 5x = 580 - x*2^(1/2) + x

x = 260/(4+2^(1/2))

plugging this back into equiation

my guess is E (I don't have a calculator)

why did you assume that only 1 day is involved ...i.e. prices decrease & increase for 1 day only ?

This is a nasty question or something is missing here

Assume prices become same after N days

320 + N(5x) = 580 - N(xsqrt2 - x )

N = 260 / x(4+sqrt2)

There are 2 variables and only 1 equation.

Something seems to be wrong or missing.

I think you can only solve this problem if you assume that x represents the number of days (that's what the problem probably implies)

In terms of 1 day assumption, I didn't assume that, I was simply looking for x - the number of days that need to pass so the prices equialize.

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Director
Joined: 01 Feb 2003
Posts: 842

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Location: Hyderabad

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30 May 2005, 21:56
It is same as time-distance problem - relative distance between prices = 260 cents. and constant rates of both are given

=> price will be equal after (260)/x(4+sqrt(2)) = 260(2.6)/x(14) days

=> 320 + (26^2)*5/14 = 26*13*5/700 + 3.2 = 3.2 + .65*3.6 =
3.2 + 3.6(1-0.25-0.1) = 3.2 + 3.6 - 0.9 - .36 = 6.8 - 1.26 = 5.6 (approx)

Ans: E

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Senior Manager
Joined: 17 Apr 2005
Posts: 372

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Location: India

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31 May 2005, 11:31
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Vithal wrote:
It is same as time-distance problem - relative distance between prices = 260 cents. and constant rates of both are given

=> price will be equal after (260)/x(4+sqrt(2)) = 260(2.6)/x(14) days

=> 320 + (26^2)*5/14 = 26*13*5/700 + 3.2 = 3.2 + .65*3.6 =
3.2 + 3.6(1-0.25-0.1) = 3.2 + 3.6 - 0.9 - .36 = 6.8 - 1.26 = 5.6 (approx)

Ans: E

I too vote for E.

Vithal , classic analogy. Bushed@5x , Wheat@ .14x
So when the prices meet the Bushel would have increased by (5/5.14)*260 ~= 2.47

So the meeting price 3.20 + 2.47 = 5.67

HMTG.

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31 May 2005, 11:31
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# ps-corn

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