math problem.... :(

[fashafosho]

SO I feel incredibly dumb after having successfully struggled with this....for a little more than 20 minutes. Needless to say, I will not be able to be this wasteful with time on the mcat...so can someone help me think of a faster way of doing problems involving ******ed fractions?

here's the problem:

1/1.9 - 1/2 = 1/x; x= 38 (it's an optics problem)

Thank you!

---

Members don't see this ad.

 

[Doctor Bagel]

SO I feel incredibly dumb after having successfully struggled with this....for a little more than 20 minutes. Needless to say, I will not be able to be this wasteful with time on the mcat...so can someone help me think of a faster way of doing problems involving ******ed fractions?

here's the problem:

1/1.9 - 1/2 = 1/x; x= 38 (it's an optics problem)

Thank you!

You should not have to do any arithmetic as detailed as this on the MCAT. Now if you did encounter a problem like this, you would just round 1.9 to 2 and wind up with 0 as the first part of the equation. The solution for x that would be the closest would be whatever value is greatest.

In general, round, round, round -- it's all about gross calculations instead of details.

 

[Blade28]

Agreed. This kind of stuff won't be on the MCAT.

But:

1/1.9 x (10/10) = 10/19
10/19 x (2/2) = 20/38

1/2 x (19/19) = 19/38

So 20/38 - 19/38 = 1/38 = 1/x

x = 38

 

[fashafosho]

Wow! thank you so much!!!
I didn't think of converting it to whole numbers....

 

[Blade28]

Oh I always do this for any "decimal fractions."

Eg. List a fraction x that satisfies:

1/8 < x < 1/9

So I would just do:

x = 1/8.5, which equals 2/17. Easy!

 

[hypnix]

I got a dumb math problem. I think...

I was solving a Stress problem and we're given that the cross section area is 2cm^2. According to the solution, they have the Area (A) as

A= 2cm^2 = 2*10^-4m^2. I'm not sure why it's to the negative 4th power. I did it to the negative 2nd power and so my final answer was off by 2 powers.

It's a lame question seeing how I had to realize that the answer is found by using the area. But yes, I'm not sure why it's not to the negative 2nd power.

 

[BloodySurgeon]

2 cm^-2 ( 1 m / 10^2 cm)^2 = 2 x 10^-4 m^2

 

[hypnix]

2 cm^-2 ( 1 m / 10^2 cm)^2 = 2 x 10^-4 m^2

Not sure why you need to square it again. The area is already given in its proper units. I guess if you think about it, the ^-4 makes more sense. I'm just confused cause I don't think I've ever had to do this on any problems dealing with the Area.

 

[tncekm]

You do have to square it again.

10cm&#178; is not equal to 10cm&#178; x 1m&#178;/100cm&#178;. That's essentially saying 10cm x 10cm = 1m&#178;. And that's not true because a square meter (1m x 1m) is 100cm x 100cm, or 10,000cm&#178;.

Hope that helps.

 

[BloodySurgeon]

( 1 m / 10^2 cm)^2 = (1 m^2 / 10^4 cm^2)

this allows for the conversion of cm^2 to m^2

hope this help.. best of luck

 

[Vihsadas]

Remember that the units are cm square. In order to find the proper conversion factor for square cm to square m, you have to square the entire conversion factor for cm to m, including the units!

So since 1cm = .01m
Then, for square cm we have to square everything:

--> (1cm)^2 = (.01m)^2
--> 1cm^2 = .0001m^2

That's the conversion factor you have to use.