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math help

at the beginning of the year 2000,a company bought a new machine for 15000 pound,each year the value of the machine decreases by 20 percent of its value at the start of the year.
when the value of the machine falls below 500 pound, the company will replace it. find the year in which the machine will be replaced.
my answer is in the year 2016,and this is what I wrote:
15000×0.8the power of n-1<500,n>16.24,n=17, so it is the year 2016.but the mark scheme is 2015,,,
Reply 1
Avatar for hoosie
hoosie
8
You are on the right track but made an error with the index in your working.
Try index n instead of n-1.
15000(0.8)⁰ = value Jan 2000
15000(0.8)¹ = value Jan 2001
Solve: 15000(0.8)ⁿ < 500
Original post by hoosie
You are on the right track but made an error with the index in your working.
Try index n instead of n-1.
15000(0.8)⁰ = value Jan 2000
15000(0.8)¹ = value Jan 2001
Solve: 15000(0.8)ⁿ < 500

but the formula is Un=a times r the power of n-1,right?where a is the first term,r is the common ratio,I got n is 17, and then U17 is 2000+17-1=2016.
(edited 4 years ago)
Reply 3
Avatar for hoosie
hoosie
8
Original post by Wuxinyu
but the formula is Un=a times r the power of n-1,right?where a is the first term,r is the common ratio,I got n is 17, and then U17 is 2000+17-1=2016.

The n I have used represents power not term number.
When n = 0 term number = 1 15000(0.8)⁰ = value Jan 2000
15000(0.8)² = value Jan 2002
when n(power) = 2 and term number (your n) = 3
15000(0.8)¹⁵= value Jan 2015
when n(power) = 15 and term number (your n) = 16
Original post by vbzl
Year 2016 is the correct answer.

so is the mark scheme wrong ?I do not think I have made any mistakes.:redface:
Reply 5
Avatar for Matureb
Matureb
19
It is 2015. At the start of 2016, the value is £422. So it dipped below 500 during 2015.
I do not think so,it asks which year to replace,so as at the start of 2016,the value is 422 and below 500,so it will be replaced in 2016,the value is above 500 at 2015,(U16=15000 × 0.8the power of 15=528,which is above500)
Reply 7
Avatar for hoosie
hoosie
8
Original post by Wuxinyu
I do not think so,it asks which year to replace,so as at the start of 2016,the value is 422 and below 500,so it will be replaced in 2016,the value is above 500 at 2015,(U16=15000 × 0.8the power of 15=528,which is above500)

Assuming the machine has to be in continuous use the company will replace the machine as soon as its value drops below 500 which is around the end of March 2015. Replacing it in Jan 2016 would the mean the machine would go unused for 9 months.
Original post by hoosie
Assuming the machine has to be in continuous use the company will replace the machine as soon as its value drops below 500 which is around the end of March 2015. Replacing it in Jan 2016 would the mean the machine would go unused for 9 months.

how did you get it is march 2015?
Reply 9
Avatar for hoosie
hoosie
8
Solving 15000(0.8)ⁿ < 500 gives
n > 15.2422 years

Check:
15000(0.8)¹⁵ = £527.77 Jan 2015
15000(0.8)¹⁵·²⁴²²= £499.9994 March 29 2015
15000(0.8)¹⁶ = £422.21 Jan 2016

0.2422 × 365 = 88 days into year 2015 (March 29)

Machine falls below £500 after 15 years and 88 days of use.
ahh,I got it! it seems that when we encountered sth like n>15.24, I just round it up to 16, which is not precise...so does our testbooks! thanks:smile:

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