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# According to the formula F=9/5 (C) +32, if the temperature

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According to the formula F=9/5 (C) +32, if the temperature [#permalink]

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27 Sep 2013, 22:25
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According to the formula F=9/5 (C) +32, if the temperature in degrees Farenheit (F) increases by 27, by how much does the temperature in degrees Celsius (C) increase?

A) 9
B) 15
C) 47
D) 48 3/5
E) 59
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Re: According to the formula F=9/5 (C) +32, [#permalink]

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27 Sep 2013, 22:52
usre123 wrote:
According to the formula F=9/5 (C) +32, if the temperature in degrees Farenheit (F) increases by 27, by how much does the temperature in degrees Celsius (C) increase?

A)9
B)15
C)47
D) 48 3/5
E) 59

$$F_i = \frac{9}{5}* (C_i) +32$$

$$F_i+27= \frac{9}{5}* (C_f) +32$$

On Subtracting, we get$$27 = \frac{9}{5}* (C_f-C_i) \to (C_f-C_i) = 3*5 = 15$$

B.
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Re: According to the formula F=9/5 (C) +32, [#permalink]

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28 Sep 2013, 03:09
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usre123 wrote:
According to the formula F=9/5 (C) +32, if the temperature in degrees Farenheit (F) increases by 27, by how much does the temperature in degrees Celsius (C) increase?

A)9
B)15
C)47
D) 48 3/5
E) 59

You can plug in values.

C = 5/9*(F-32)

F=32 --> C=0;
F=32+27=59 --> C=5/9*27=15.

Increase = 15 degrees.

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Re: According to the formula F=9/5 (C) +32, [#permalink]

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28 Sep 2013, 03:11
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Re: According to the formula F=9/5 (C) +32, [#permalink]

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01 Oct 2013, 09:54
Bunuel wrote:
usre123 wrote:
According to the formula F=9/5 (C) +32, if the temperature in degrees Farenheit (F) increases by 27, by how much does the temperature in degrees Celsius (C) increase?

A)9
B)15
C)47
D) 48 3/5
E) 59

You can plug in values.

C = 5/9*(F-32)

F=32 --> C=0;
F=32+27=59 --> C=5/9*27=15.

Increase = 15 degrees.

I dont understand why you only multiplied 5/9 by 27. what about F-32?
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Re: According to the formula F=9/5 (C) +32, [#permalink]

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02 Oct 2013, 02:35
usre123 wrote:
Bunuel wrote:
usre123 wrote:
According to the formula F=9/5 (C) +32, if the temperature in degrees Farenheit (F) increases by 27, by how much does the temperature in degrees Celsius (C) increase?

A)9
B)15
C)47
D) 48 3/5
E) 59

You can plug in values.

C = 5/9*(F-32)

F=32 --> C=0;
F=32+27=59 --> C=5/9*27=15.

Increase = 15 degrees.

I dont understand why you only multiplied 5/9 by 27. what about F-32?

If F=59 --> C = 5/9*(F-32) = 5/9*(59-32) = 5/9*27 = 15.

Does this make sense?
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Re: According to the formula F=9/5 (C) +32, if the temperature [#permalink]

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25 Dec 2014, 05:24
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Re: According to the formula F=9/5 (C) +32, if the temperature [#permalink]

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21 Jan 2016, 12:42
Hello from the GMAT Club BumpBot!

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Re: According to the formula F=9/5 (C) +32, if the temperature [#permalink]

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02 Jan 2017, 13:37
Could someone please explain what happens to the F=9/5 (C) +32? In all of the provided solutions, no one does anything with this, but I'm confused as to why that is.
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Re: According to the formula F=9/5 (C) +32, if the temperature [#permalink]

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03 Jan 2017, 03:49
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skimmingit wrote:
Could someone please explain what happens to the F=9/5 (C) +32? In all of the provided solutions, no one does anything with this, but I'm confused as to why that is.

32 is constant. It doesn't bring about the change. The change is defined by the relation between F and C only. 32 just gets cancelled out.

F1 = (9/5)*C1 + 32 ......(I)
F2 = (9/5)*C2 + 32 ......(II)

(II) - (I)

F2 - F1 = (9/5)*(C2 - C1) + 32 - 32

F2 - F1 = (9/5)*(C2 - C1)

27 = (9/5)*(C2 - C1)

(C2 - C1) = 15
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Re: According to the formula F=9/5 (C) +32, if the temperature   [#permalink] 03 Jan 2017, 03:49
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# According to the formula F=9/5 (C) +32, if the temperature

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