B

Math problem

6 Answers
I'm currently stuck on 2 math problems: 1. Suppose a population of 250 crickets doubles in size every 6 months. How many crickets will there eve after 2 years? a). 4000 crickets b). 6000 crickets c). 2000 crickets 2. Suppose an investment of $6600 doubles in value every 8 years. How much is the investment worth after 40 years? a). $211,200 b). $105,000 c). $66,000 d). $528,000 thanks!
F

Start with 250 crickets.
After 6 months you have 500
After 1 years (next 6 months), you have 1,000
After 1 years and 6 month (next 6 months), you have 2,000
After 2 years (next 6 months), you have 4,000
So the answer is a). 4000 crickets



Investment of $6600 doubles in value every 8 years
Amount after 8 years from date of investment is 2*$6600 =$13,200
Amount after 16 years from date of investment is 2*$13,200 =$26,400
Amount after 24 years from date of investment is 2*$26,400 =$52,800
Amount after 32 years from date of investment is 2*$52,800 =$105,600
Amount after 40 years from date of investment is 2*$105,600 =$211,200

SO the answer is $211,200

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J

The answer to first question is: a) 4000 crickets & second is: a) $211,200

Steps:
1. You need to find how many iterations of doubling are occurring. In first case it is 4 times. 2years i.e. 24 months/6months in second case 40years/8years=5
2. Now find factor which in case is 2 if tripling it will be 3.
3. You need to multiply base/initial number with iterations (calculated in step 1) power factor( calculated in step 2)
So for first problem answer is: 250 x 2^4 & for second problem it is: 6600 x 2^5

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C

1 Start with 250 crickets...after 6 months you have 500....another 6 months (a year), you have 1,000. Another 6 months (one and half years) you have 2,000 and after a full two years you have 4,000.
So the answer is a - 4,000

2 Every 8 years in terms of 40 years is five lots of doubling (40 divided by 8)
So $6600 becomes $13200 after 8 years
$26,400 after 16 years
$52,800 after 24 years
$105,600 after 32 years

...and $211,200 after 40 years - again the answer is a!

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S

Everytime something is doubled it means a factor 2. So at the first example you have to double 4 times wich means four factor 2. In math you can write it as 2^4 wich is equal to a factor 16 (2^4=2x2x2x2). so the answer is 16x250=4000
At the second example following the same process: You have to double 5 times, wich leads to a factor 2^5= 32 ( 2^5=2x2x2x2x2) so the answer is 32x6600=211200

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A

Original population is 250. Since it doubles every six months so rate of increase is 50% per half year, which means 100% = 100/100 = 1.00 per year. Time t = two years and number of conversion periods n = 2.
A = P(1 + r)^nt
A = 250(1 + 1.00)^(2*2)
A = 4000
So the population of crickets after two years is 4000

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J

Yes, I checked both answers and the guy is right.

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