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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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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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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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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, [#permalink]

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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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if-the-speed-of-x-meters-per-second-is-equivalent-to-the-127005.html

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

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New post 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.

Answer: B.


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

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

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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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 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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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]

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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   [#permalink] 19 Aug 2017, 12:26
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