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According to the formula F = 9/5*C + 32, if the temperature in degrees

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usre123
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According to the formula F = 9/5*C + 32, if the temperature in degrees [#permalink] New post  Updated on: 04 Feb 2019, 04:49
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According to the formula \(F=\frac{9}{5}*C +32\), 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) \(48 \frac{3}{5}\)

(E) 59
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Originally posted by usre123 on 27 Sep 2013, 23:25.
Last edited by Bunuel on 04 Feb 2019, 04:49, edited 2 times in total.
Renamed the topic and edited the question.
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KarishmaB
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Re: According to the formula F = 9/5*C + 32, if the temperature in degrees [#permalink] New post  03 Jan 2017, 04:49
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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 in degrees [#permalink] New post  28 Sep 2013, 04:09
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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.

Answer: B.
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Re: According to the formula F = 9/5*C + 32, if the temperature in degrees [#permalink] New post  27 Sep 2013, 23:52
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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


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

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

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, if the temperature in degrees [#permalink] New post  28 Sep 2013, 04:11
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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.

Answer: B.


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a-train-traveling-at-a-constant-speed-down-a-straight-track-87124.html
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Re: According to the formula F = 9/5*C + 32, if the temperature in degrees [#permalink] New post  01 Oct 2013, 10:54
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.

Answer: B.


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, if the temperature in degrees [#permalink] New post  02 Oct 2013, 03:35
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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.

Answer: B.


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


If F=59, then \(C = \frac{5}{9}*(F-32) = \frac{5}{9}*(59-32) = \frac{5}{9}*27 = 15\).

Does this make sense?
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Re: According to the formula F = 9/5*C + 32, if the temperature in degrees [#permalink] New post  02 Jan 2017, 14:37
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 in degrees [#permalink] New post  19 Aug 2017, 13:26
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VeritasPrepKarishma wrote:
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


hi mam

since the change is defined by the relation between F and C only, we can say directly that the "change in F" is equal to 9/5 times the "change in C". That's it...

thanks mam .. :thumbup:
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Re: According to the formula F = 9/5*C + 32, if the temperature in degrees [#permalink] New post  24 Dec 2020, 00:17
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Given expression is in terms of F but we need the value of C, hence we will express the expression in terms of C.

F = \(\frac{9}{5}\) * C + 32

At C = 0, F = 32

Increased value: F = 32 + 27 = 59

=> C = \(\frac{5}{9}\) * ( F - 32)

=> C = \(\frac{5}{9}\) * ( 59 - 32)

=> C = \(\frac{5}{9}\) * 27

=> C = 15

Answer B
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