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# A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00

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A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00  [#permalink]

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21 Sep 2019, 05:00
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64% (02:36) correct 36% (02:44) wrong based on 138 sessions

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A certain manufacturer uses the function $$C(x) = 0.04x^2 – 8.5x + 25,000$$ to calculate the cost, in dollars, of producing x thousand units of its product. The table above gives values of this cost function for values of x between 0 and 50 in increments of 10. For which of the following intervals is the average rate of decrease in cost less than the average rate of decrease in cost for each of the other intervals?

A. From x = 0 to x = 10
B. From x = 10 to x = 20
C. From x = 20 to x = 30
D. From x = 30 to x = 40
E. From x = 40 to x = 50

PS18871.01

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2019-09-21_1558.png [ 8.05 KiB | Viewed 1487 times ]

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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00  [#permalink]

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21 Sep 2019, 05:51
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This question is an example of how gmat might test us on calculation intensive questions!
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A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00  [#permalink]

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Updated on: 17 Oct 2019, 07:29
gmatt1476 wrote:

A certain manufacturer uses the function $$C(x) = 0.04x^2 – 8.5x + 25,000$$ to calculate the cost, in dollars, of producing x thousand units of its product. The table above gives values of this cost function for values of x between 0 and 50 in increments of 10. For which of the following intervals is the average rate of decrease in cost less than the average rate of decrease in cost for each of the other intervals?

A. From x = 0 to x = 10
B. From x = 10 to x = 20
C. From x = 20 to x = 30
D. From x = 30 to x = 40
E. From x = 40 to x = 50

PS18871.01

Attachment:
2019-09-21_1558.png

Hello Bunuel chetan2u other experts,
What is the exact meaning of 'the average rate of decrease in cost'?
And the meaning of 'the average rate of decrease in cost for each of the other intervals'?

Originally posted by RK007 on 13 Oct 2019, 01:42.
Last edited by RK007 on 17 Oct 2019, 07:29, edited 1 time in total.
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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00  [#permalink]

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17 Oct 2019, 05:37

Can you help with the above question? Thank you
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Joined: 02 Aug 2009
Posts: 8281
A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00  [#permalink]

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17 Oct 2019, 09:25
3
gmatt1476 wrote:

A certain manufacturer uses the function $$C(x) = 0.04x^2 – 8.5x + 25,000$$ to calculate the cost, in dollars, of producing x thousand units of its product. The table above gives values of this cost function for values of x between 0 and 50 in increments of 10. For which of the following intervals is the average rate of decrease in cost less than the average rate of decrease in cost for each of the other intervals?

A. From x = 0 to x = 10
B. From x = 10 to x = 20
C. From x = 20 to x = 30
D. From x = 30 to x = 40
E. From x = 40 to x = 50

PS18871.01

Attachment:
2019-09-21_1558.png

Hi RK007,
The average rate of decrease in cost --This means decrease in cost per unit and as the number of units is same in each, that is 10,000 units, we have to divide the decrease in cost by 10,000.

Now, Since the change in cost will be $$\frac{previous - this}{10,000}$$ . We are looking for the least value of $$\frac{previous - this}{10,000}$$ and the denominator is the SAME so it will depend on numerator( previous - this)..
So basically the answer will be the minimum drop from the previous, so it will
0-10........81
10-20......73
20-30....65
30-40....57

Another way..
We are looking at a quadratic equation $$C(x)=0.04x^2-8.5x+25000$$, and the coefficient of x^2 is positive, 0.04, so it will give us a U parabola, which becomes less steeper as we move up..
Hence more the value of x, lesser will be the decrease. Here 40-50 is the max interval given, so the average rate of decrease in cost will be least here..

E

However, a bit disappointed with the wordings on an official question..
A certain manufacturer uses the function $$C(x) = 0.04x^2 – 8.5x + 25,000$$ to calculate the cost, in dollars, of producing x thousand units of its product.
This tells us that against 50, the cost of 50,000 units is given by 24,675 and 40,000 units is given by 24724...
But, ofcourse the question is meant to tell us that as the number of the items produced increases, the cost per 10,000 unit decreases and that is given by the table..
so what we have C(X) is the cost of 10,000 items when the production is increased to x thousand items..
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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00  [#permalink]

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17 Oct 2019, 20:32
gmatt1476 wrote:

A certain manufacturer uses the function $$C(x) = 0.04x^2 – 8.5x + 25,000$$ to calculate the cost, in dollars, of producing x thousand units of its product. The table above gives values of this cost function for values of x between 0 and 50 in increments of 10. For which of the following intervals is the average rate of decrease in cost less than the average rate of decrease in cost for each of the other intervals?

A. From x = 0 to x = 10
B. From x = 10 to x = 20
C. From x = 20 to x = 30
D. From x = 30 to x = 40
E. From x = 40 to x = 50

PS18871.01

Attachment:
2019-09-21_1558.png

Since each interval has the same length (10 units), the interval that has the lowest average rate of decrease in cost is the interval that has the lowest decrease. Therefore, we can just calculate the decrease in each interval.

A. From x = 0 to x = 10: the decrease is 25,000 - 24,919 = 81.

B. From x = 10 to x = 20: the decrease is 24,919 - 24,846 = 73.

C. From x = 20 to x = 30: the decrease is 24,846 - 24,781 = 65.

D. From x = 30 to x = 40: the decrease is 24,781 - 24,724 = 57.

E. From x = 40 to x = 50: the decrease is 24,724 - 24,675 = 49.

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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00  [#permalink]

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05 Nov 2019, 23:48
1
Let x be lower value of the interval and y be the higher (given diff of interval =10) then
Avg decease in cost= [C(y)-C(x)]/10
Solving the equation
= [0.04(y^2-x^2)-8.5(y-x)]/10 but y-x=10
so Avg= [0.04(y^2-x^2)-85]/10
For Avg will minimum if y^2-x^2 is maximum that will only happen for interval 40-50
Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00   [#permalink] 05 Nov 2019, 23:48
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