If C is the temperature in degrees Celsius and F is the temperature in

9C = 5(F – 32)
--> C = 5F/9 - 32/9

Let \(F_1\) & \(F_2\) be the recorded temperatures in Fahrenheit and
\(C_1\) & \(C_2\) be the recorded temperatures in Celsius

--> \(C_1\) = 5\(F_1\)/9 - 32/9 ....... (1)
&
\(C_2\) = 5\(F_2\)/9 - 32/9 ....... (2)

Given, \(F_2\) - \(F_1\) = 45

(2) - (1) --> \(C_2\) - \(C_1\) = 5\(F_2\)/9 - 5\(F_1\)/9 = 5/9*(\(F_2\) - \(F_1\)) = 5/9*45 = 25

IMO Option C

Pls Hit kudos if you like the solution

given
C=5/9F-32/9
F1-F2=45
C1=5/9F1-32/9
C2=5/9F1-32/9
subtract
c1-c2= 5/9*45 ; 25
IMO C

Hi All,

We're told that C is the temperature in degrees Celsius and F is the temperature in degrees Fahrenheit, and the relationship between temperatures on the two scales is expressed by the equation 9C = 5(F - 32). On a day when the temperature extremes recorded at a certain weather station differed by 45 degrees on the Fahrenheit scale, we're asked by how many degrees did the temperature extremes differ on the Celsius scale. This question can be solved rather easily by TESTing VALUES (as Brent has shown in his post). In addition, the answer choices are sufficiently 'spread out' that we can use a little logic and a little math to answer this question without too much work.

The equation 9C = 5(F - 32) can be 'rewritten' as...

C = 5(F - 32)/9

5/9 is a little more than 1/2.... so we can 'rewrite' the equation again as....

C = (a little more than 1/2)(F - 32)

Thus, for every degree that F 'changes', the value of C will change by a little more than 1/2 of a degree.

In this question, F changes by 45 degrees, so the value of C will change by (a little more than 1/2)(45) degrees. Half of 45 is 22.5, so we're looking for an answer that's a little more than 22.5 degrees. There's only one answer that matches...

Final Answer:

GMAT assassins aren't born, they're made,
Rich

I just plugged in numbers.
It was 50 -Farenheite

That became 5 Farenheit.

Found out celcius for each and than simply found how much was the difference.

Posted from my mobile device

I think this is a simple problem:

9C1= 5F1-150

9C2= 5F2-150

9 (C1-C2)= 5 (F1-F2)- 150+150

9 (C1-C2)= 5 (45) { SINCE F1-F2= 45 as per the question)

C1-C2= 25

Hence the answer is C.

Please correct if my working is wrong.

Plug 45 and 0 to get 25.

\(9C = 5(F – 32)\)

\(9C = 5(32 – 32)\)
\(0C = 32F\)

\(9C = 5(77 – 32)\)
\(9C = 225\)
\(C = 25\)

Answer is C.

when comparing 2 values of C and F, you can see the relation that the constant just adds to both recordings, and hence its effect gets nullified.

Thus the only thing of relevance is 9C = 5F
9*change in C = 5*change in F

9Change in C = (5*45)/9

thus Change in C = 25 (C)

Hi,

Plugging in F=45 in the formula would have been the easiest resort which results in option 65/9. THIS is a big TRAP.

the question clearly mentions, the difference or the change is 45F and hence I have assumed values.
I took day 1, F=100 hence after applying the formula C becomes 340/9= 37 (just taking the integer values)
Day 2 F=145 and C becomes 565/9= 62
the difference then becomes 62-37=25 degrees Celsius for a change in 45 degrees F.

option C

Hope this is useful:)
Please hit Kudos, if you like it.

Given: If C is the temperature in degrees Celsius and F is the temperature in degrees Fahrenheit, then the relationship between temperatures on the two scales is expressed by the equation 9C = 5(F – 32).

Asked: On a day when the temperature extremes recorded at a certain weather station differed by 45 degrees on the Fahrenheit scale, by how many degrees did the temperature extremes differ on the Celsius scale?

9C1 = 5(F1-32)
9C2 = 5(F2-32)

9(C1-C2) = 5(F1-F2) = 5*45
C1-C2 = 5*45/9 = 25

IMO C

chetan2u, Why can we use the difference between other units and not the difference between Fs?
For eg, the difference between 4 meters and 10 meters is 6 meters. We don't need to convert 4 meters and 10 meters separately, we can just convert 6 meters to cms.
Similarly, we don't need to calculate the kilogram values separately or any other unit's value separately and we can simply convert the difference to the other unit. Why doesn't it work here?
I don't understand. If you could please help me understand this, would be of great help.
Thanks a lot

Hi 0Lucky0,

You CAN approach this question in the way that you're asking about, but you have to do a little work first (to manipulate the given formula so that you better understand the relationship between 1 degree of F and 1 degree of C). As an aside, this question can be solved rather easily by TESTing VALUES (as Brent has shown in his post). In addition, the answer choices are sufficiently 'spread out' that we can use a little logic and a little math to answer this question without too much work.

The equation 9C = 5(F - 32) can be 'rewritten' as...

C = 5(F - 32)/9
C = (5/9)(F - 32)

5/9 is a little more than 1/2.... so we can 'rewrite' the equation again as....

C = (a little more than 1/2)(F - 32)

Thus, for every degree that F 'changes', the value of C will change by a little more than 1/2 of a degree.

In this question, F changes by 45 degrees, so the value of C will change by (a little more than 1/2)(45) degrees. Half of 45 is 22.5, so we're looking for an answer that's a little more than 22.5 degrees. There's only one answer that matches...

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

Hi Rich,
Thanks for your reply. Really appreciate it. However, I did see yours and Brent's explanations above and my doubt remained so the reason why I asked chetan2u. As far as I can tell, It looks like a copy/paste from above, so unfortunately my doubt still remains.

Hi

The reason is the formula involved.
For m and cm, 1m is 100cm, so increase of 45 m will give 4500 as the answer.

But the relation in C and F is not straightforward. But we can still use it here.
9C = 5(F – 32) Or \(C=\frac{5}{9}(F-32)\)
Now, we are increasing F by 45, so whatever value F-32 was has also increased by 45.
So, \(C=\frac{5}{9}(45)=5*5=25\)

I hope this addresses the query