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

Question

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

KarishmaB · Accepted Answer

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

Bunuel · Answer

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.

mau5 · Answer

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) 	o (C_f-C_i) = 3*5 = 15

B.

Bunuel · Answer

Other conversion problems to practice: if-an-object-travels-100-feet-in-2-seconds-what-is-the-103364.html the-rear-wheels-of-a-car-crossed-a-certain-line-0-5-second-135753.html a-train-traveling-at-a-constant-speed-down-a-straight-track-87124.html a-welder-received-an-order-to-make-a-1-million-liter-126154.html a-car-traveling-at-a-certain-constant-speed-takes-2-seconds-126658.html if-an-automobile-averaged-22-5-miles-per-gallon-of-gasoline-109083.html if-1-kilometer-is-approximately-0-6-mile-which-of-the-140731.html measuring-standards-conversion-106516.html if-10-millimeters-equal-1-centimeter-how-many-square-48538.html the-speed-of-light-is-approximately-149492.html the-rear-wheels-of-a-car-crossed-a-certain-line-0-5-second-135753.html if-the-speed-of-x-meters-per-second-is-equivalent-to-the-127005.html Hope this helps.

usre123 · Answer

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

Bunuel · Answer

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

Does this make sense?

skimmingit · Answer

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.

testcracker · Answer

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:

MathRevolution · Answer

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

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

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