Series and Sequences Q's (1 Viewer)

[-pari-]

these are the kinda questions that make me certain i'll flop the hsc. pft.


1) a fish farmer began business on 1 January 1998 with a stock of 100 000 fish. he had a contract to supply 15400 fish at a price of $10 per fish to supply 15400 fish at a price of $10 per fish to a retailer in December each year. in the period between January and the harvest in December each year the number of fish increases by 10%.

(i) find the number of fish just after the second harvest in december 1999.

answer: 88660

BUT that is after subtracting another 15400 assuming that straight after harvest he supplied to his contractor. but i didn't subtract 15400....how are you meant to know which way they meant?!

(ii) (i've tried this one a billion times) show that Fn, the number of fish just after the nth harvest is given by:
Fn = 154 000 - 54000 (1.1)^n

(iii) when will the farmer have sold all his fish, and what will be his total income? (got the first bit: approx. 11yrs - couldn't do the income bit)

(iv) each december the offers to buy the farmer's business by paying $15 per fish for his entire stock. when should the farmer sell to maximise his total income?

(i'm guessin this would involve geometrical app. of calc...?) answer: approx 7.245 years

 

[Forbidden.]

I have seen exactly the same questions (Q10 of a past HSC Maths paper I bet), I have the SuccessOne book at home.

 

[watatank]

1) a fish farmer began business on 1 January 1998 with a stock of 100 000 fish. he had a contract to supply 15400 fish at a price of $10 per fish to supply 15400 fish at a price of $10 per fish to a retailer in December each year. in the period between January and the harvest in December each year the number of fish increases by 10%.

(i) find the number of fish just after the second harvest in december 1999.

answer: 88660

1st year.
starts off with 100 000 fish. during the year it increases by 10% (100 000 * 1.1) = 110000 fish. then the harvest. at the end of the first year there is (110000 - 15400) = 94600 fish

2nd year.
starts off with 94600 fish. during the year it increases by 10% (94600 * 1.1) = 104060 fish. then the harvest. at the end of the second year there is (104060 - 15400) = 88660 fish

 

[watatank]

[/I](ii) (i've tried this one a billion times) show that Fn, the number of fish just after the nth harvest is given by:

Fn = 154 000 - 54000 (1.1)^n

first year. (100000*1.1) - 15400

second year.
= [ (100000*1.1) - 15400 ]*1.1 - 15400
= 100000*1.1² - 15400(1+1.1)

you should see a pattern of some sort?

third year = 100000*1.1³ - 15400(1+1.1+1.1²)

nth year = 100000*1.1ⁿ - 15400(1+1.1+1.1²+....+1.1^n-1)

= 100000*1.1ⁿ - 15400 [ (1.1ⁿ - 1) / (1.1-1) ]

sorry that's as far as i got

 

[-pari-]

bump! [thanks for tryin watatank]

1st year.
starts off with 100 000 fish. during the year it increases by 10% (100 000 * 1.1) = 110000 fish. then the harvest. at the end of the first year there is (110000 - 15400) = 94600 fish

2nd year.
starts off with 94600 fish. during the year it increases by 10% (94600 * 1.1) = 104060 fish. then the harvest. at the end of the second year there is (104060 - 15400) = 88660 fish

mhmm...but how did you know to subtract that last 15400? i just left it at 104060....
__________________

 

[watatank]

bump! [thanks for tryin watatank]


mhmm...but how did you know to subtract that last 15400? i just left it at 104060....
__________________

uhm, there's a harvest every december...after the number of fish increase by 10% every year....they ask for the total just after the second harvest, so that means you have to subtract the number of fish harvested... ie 15400?

 

[watatank]

first year. (100000*1.1) - 15400

second year.
= [ (100000*1.1) - 15400 ]*1.1 - 15400
= 100000*1.1² - 15400(1+1.1)

you should see a pattern of some sort?

third year = 100000*1.1³ - 15400(1+1.1+1.1²)

nth year = 100000*1.1ⁿ - 15400(1+1.1+1.1²+....+1.1^n-1)

= 100000*1.1ⁿ - 15400 [ (1.1ⁿ - 1) / (1.1-1) ]

sorry that's as far as i got

sorry i lied. i can continue. i was on the right track, only just realise i could keep going

= 100000*1.1ⁿ - 15400 [ (1.1ⁿ - 1) / (0.1) ]

dividing by 0.1 is like multiplying by 10 right?

= 100000*1.1ⁿ - 154000 [ (1.1ⁿ - 1)]

expand the bracket

= 100000*1.1ⁿ - 154000*(1.1ⁿ) +154000

1.1ⁿ is a common factor so

= 154000 + 1.1ⁿ(100 000 - 154000)

= 154000 - 54000(1.1ⁿ)

 

[-pari-]

aahhhh....

genius.

though it gets me thinking...you can do that a year after your hsc, i still can't do that less than a month before my trials.

great.

thanks, though..