DAT Destroyer Math #66

[FROGGBUSTER]

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A population of mice doubles every 10 days. If the original population was m mice, which is an equation for the number of mice after d days?

a. 2md/10
b. 20d/m
c. m*[2^(d/10)]
d. 2^(md/10)
e. 10^(2d/m)

Answer is C. Can someone break this one down for me? I don't understand the solution to this in the back of the book.

 

[mh0000]

c. m*[2^(d/10)]

(d/10) tells you how many doubling "periods" the system has gone through. If 20 days have elapsed, then you have gone through 2 doubling periods (20/10). So the generic form will be d/10. We use 2 as the base because we are doubling every period. Think about it, if 20 days have elapsed, we went through 2 periods of doubling, or in other words multiplied by a factor of 4 (2^2) because we doubled the original amount, then doubled this amount again. So [2^(d/10)] tells us by what factor the population has increased after a time of d. So after 20 days, we would expect the original population to have quadrupled, and we can just multiply the original population size (m) by this factor to get the total population size after 20 days (or whatever time period we're dealing with). This gives us a general formula:

m*[2^(d/10)]

Hope this makes sense, I probably didn't explain it that great

 

[FROGGBUSTER]

c. m*[2^(d/10)]

(d/10) tells you how many doubling "periods" the system has gone through. If 20 days have elapsed, then you have gone through 2 doubling periods (20/10). So the generic form will be d/10. We use 2 as the base because we are doubling every period. Think about it, if 20 days have elapsed, we went through 2 periods of doubling, or in other words multiplied by a factor of 4 (2^2) because we doubled the original amount, then doubled this amount again. So [2^(d/10)] tells us by what factor the population has increased after a time of d. So after 20 days, we would expect the original population to have quadrupled, and we can just multiply the original population size (m) by this factor to get the total population size after 20 days (or whatever time period we're dealing with). This gives us a general formula:

m*[2^(d/10)]

Hope this makes sense, I probably didn't explain it that great

Thanks man, that's a good explanation.

 

[toothhornet88]

Thanks man, that's a good explanation.

For questions where I have NO IDEA where to even start i just test couple of numbers and see which one it fits best...usually takes me 30 seconds and I get them correct!

 

[jwnichols21]

For questions where I have NO IDEA where to even start i just test couple of numbers and see which one it fits best...usually takes me 30 seconds and I get them correct!

I used the strategy you did, but I got answer choice A. to work too? I plugged in random values, so:

m=10 (starting amount of mice)
d=10 days
and then the end amount of mice would be 20 since initial amount gets doubled after 10 days.

So I plugged in 10 and 10 for m & d into the solutions and it works out for A. and C. What am I doing wrong here?

 

[FROGGBUSTER]

I used the strategy you did, but I got answer choice A. to work too? I plugged in random values, so:

m=10 (starting amount of mice)
d=10 days
and then the end amount of mice would be 20 since initial amount gets doubled after 10 days.

So I plugged in 10 and 10 for m & d into the solutions and it works out for A. and C. What am I doing wrong here?

Using this strategy could help save time during the test but when you're practicing it's best to just learn the concepts. Also, plugging in #s is risky because sometimes the #s work for more than 1 answer choice, like you're seeing now.

mh000 explained it really well above, you should read over that.

 

[jwnichols21]

Using this strategy could help save time during the test but when you're practicing it's best to just learn the concepts. Also, plugging in #s is risky because sometimes the #s work for more than 1 answer choice, like you're seeing now.

mh000 explained it really well above, you should read over that.

Is this a general formula, or one you had to just come up with based on the question? If so, how???

 

[FROGGBUSTER]

Is this a general formula, or one you had to just come up with based on the question? If so, how???

It's a general formula. It's the same idea as the compound interest formula as well. I think if you read mh000's post carefully a couple of times, you should understand it completely and have no problem with any of these types of problems.

 

[Streetwolf]

I used the strategy you did, but I got answer choice A. to work too? I plugged in random values, so:

m=10 (starting amount of mice)
d=10 days
and then the end amount of mice would be 20 since initial amount gets doubled after 10 days.

So I plugged in 10 and 10 for m & d into the solutions and it works out for A. and C. What am I doing wrong here?

If an answer choice said just "m + d" and you plugged your numbers into that, you'd get 20 but obviously it's not the right answer.

You didn't do anything wrong. Your choice in numbers just coincidentally works with the equation in A. At this point you need to eliminate B, D, and E, choose two new values for m and d, and plug them into A and C. Hopefully you'll be able to eliminate A. If not, try two new values until you're left with just C.

Same for any other problem with this approach. Eliminate all answers that don't work. Plug in new numbers to the remaining answers. Repeat until one answer is left.