Simple Math Question

[Maverick56]

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Can someone show me the quickest step to turn

(0.275/0.225) into (11/9)

 

[Shjanzey]

Can someone show me the quickest step to turn

(0.275/0.225) into (11/9)

multiply by 100/100

275/225 factor out 5 ==> 55/45 == 11/9

To do this quick you have to try to get a common factor that you can almost do in your head.

The best is if you are close to something divisible by 10s, 100s, or 1000s. The next best thing (for me anyways) are fives. I can still do those in my head.

So if this was something like .296/.225 I would have done the following

multiply by 100/100 == 296/225 < 300/225, factor out 5 == 60/45 == 12/9 == 4/3 . So you just found a quick way to know that your value is clost to 4/3 (A little bit less). Knowing that is good enough for most multiple choice.

 

[Postictal Raiden]

Can someone show me the quickest step to turn

(0.275/0.225) into (11/9)

multiply top and bottom by 4

 

[Shjanzey]

multiply top and bottom by 4

I don't see how this is better. The following are my impressions.

1) 4 wouldn't just pop into my head right off the bat. If you can auto-see that, you are pretty good, but I didn't see it...and the OP obviously didn't see it right off.
2) Doing the multiplication would require me to longhand both the top and bottom, which would end up taking more time. Even though it multiplies out nicely, I would still longhand it to avoid mental mistakes.
3) This leads me to conclude that we should propose a method that would allow the OP to quickly find any answer, if it isn't obvious to them from the start.

 

[gettheleadout]

Honestly I would just divide both top and bottom by 25. Its apparent right away that both are multiples of 25. For bottom, 225 is one multiple above 200. At four 25's per 100, we get 8 + 1 = 9. For top, either do 8 + 3 or 12 - 1 and get 11.

 

[circulus vitios]

I don't mean to hijack the thread, but what about 10^-3.7?

TBR says "10^-3.7 = (10^0.3)*10^-4"

What in the ****? What is the method behind that? It's nice how they don't even mention this technique but they thought it was appropriate to have a page-long table that painstakingly confirms log(ab)=log(a)+log(b) and log(a/b)=log(a)-log(b).

 

[milski]

I don't mean to hijack the thread, but what about 10^-3.7?

TBR says "10^-3.7 = (10^0.3)*10^-4"

What in the ****? What is the method behind that? It's nice how they don't even mention this technique but they thought it was appropriate to have a page-long table that painstakingly confirms log(ab)=log(a)+log(b) and log(a/b)=log(a)-log(b).

That's the same as 10^(a+b)=10^a*10^b. It follows directly from log(ab)=log(a)+log(b).

10^.3 is about the same as cube root of 10 which should be about 2.1 or something like that - 2^3 is 8, so you need a bit more than 2. 10^-4=0.0001. =>10^-3.7=2.1*0.0001 = 0.00021

 

[gettheleadout]

I don't mean to hijack the thread, but what about 10^-3.7?

TBR says "10^-3.7 = (10^0.3)*10^-4"

What in the ****? What is the method behind that? It's nice how they don't even mention this technique but they thought it was appropriate to have a page-long table that painstakingly confirms log(ab)=log(a)+log(b) and log(a/b)=log(a)-log(b).

That's the same as 10^(a+b)=10^a*10^b. It follows directly from log(ab)=log(a)+log(b).

10^.3 is about the same as cube root of 10 which should be about 2.1 or something like that - 2^3 is 8, so you need a bit more than 2. 10^-4=0.0001. =>10^-3.7=2.1*0.0001 = 0.00021

Wanna know a trick for this? If you're not familiar with what I call the "log trick of 3's" then read this first: http://forums.studentdoctor.net/showpost.php?p=6412828&postcount=3

So knowing that -log(3 x 10^-q) = ~ (q-1).5 such that -log(3 x 10^-5) = ~4.5
I messed around in Wolfram|Alpha one day and realized this trick works another way:

If you have an expression where 10^-q, then the trick of 3's holds, just in reverse.

10^-5 = 1 x 10^-5
10^-5.5 = ~3 x 10^-6
10^-6 = 1 x 10^-6
10^-6.5 = ~3 x 10^-7

and so on. So for a value like 10^-3.7, I would note that the decimal 7 is greater than 5, meaning the value of the mantissa is below 3, and the exponent is 10^-4. Conclusion is that 10^-3.7 = <3 x 10^-4

This should be accurate enough to identify the answer needed, if this is a terminal step in solving a problem.

 

[Postictal Raiden]

I don't mean to hijack the thread, but what about 10^-3.7?

TBR says "10^-3.7 = (10^0.3)*10^-4"

What in the ****? What is the method behind that? It's nice how they don't even mention this technique but they thought it was appropriate to have a page-long table that painstakingly confirms log(ab)=log(a)+log(b) and log(a/b)=log(a)-log(b).

the way I approach this is as the following:

10^-3.7 is between 10^-3 and 10^-4

so, the number has to be between 0.001 and 0.0001

Furthermore, -3.7 is closer to -4 as it is to -3, so my final number should be less than 0.0005.

Without using logarithmic or a calculator, I would conclude that 10^-3.7 is 0.0003.

Remember, MCAT is all about estimation, so you almost always treat 0.0003 and 0.00021 the same way.

 

[Postictal Raiden]

Wanna know a trick for this? If you're not familiar with what I call the "log trick of 3's" then read this first: http://forums.studentdoctor.net/showpost.php?p=6412828&postcount=3

So knowing that -log(3 x 10^-q) = ~ (q-1).5 such that -log(3 x 10^-5) = ~4.5
I messed around in Wolfram|Alpha one day and realized this trick works another way:

If you have an expression where 10^-q, then the trick of 3's holds, just in reverse.

10^-5 = 1 x 10^-5
10^-5.5 = ~3 x 10^-6
10^-6 = 1 x 10^-6
10^-6.5 = ~3 x 10^-7

and so on. So for a value like 10^-3.7, I would note that the decimal 7 is greater than 5, meaning the value of the mantissa is below 3, and the exponent is 10^-4. Conclusion is that 10^-3.7 = <3 x 10^-4

This should be accurate enough to identify the answer needed, if this is a terminal step in solving a problem.

This is a cool trick!