H2 Maths 2008 P1 Q10
(i) A student saves $10 on 1 January 2009. On the first day of each subsequent month she saves $3 more than in the previous month, so that she saves $13 on 1 February 2009, $16 on 1 March 2009, and so on. On what date will she first have saved over $2000 in total? [5]
(ii) A second student puts $10 on 1 January 2009 into a bank account which pays compound interest at a rate of 2% per month on the last day of each month. She puts a further $10 into the account on the first day of each subsequent month.
(a) How much compound interest has her original $10 earned at the end of 2 years? [2]
(b) How much in total is in the account at the end of 2 years? [3]
(c) After how many complete months will the total in the account first exceed $2000? [4]
Answer:
(i)
first term = 10
2nd term = 13
3rd term = 16
so this is an AP, first term 10, common difference 3
When sum of AP ≥ 2000
n/2 (2*10 + (n-1)*3) = 2000
n (20 + 3n - 3) = 4000
3n² + 17n - 4000 = 0
Use your graphic calculator: n=33.79 or -39.46 (reject)
So on the 34th month, she will save over $2000
Date is 1st Oct 2011
(ii)
(a) end of 2 years = 24 months
Amount of compound interest = 10(1.02)24 - 10 = $6.08
(b) end of 2nd month = 10(1.02)2 + 10(1.02)
end of 3rd month = 10(1.02)3 + 10(1.02)2 + 10(1.02)
end of 4th month = 10(1.02)4 + 10(1.02)3 + 10(1.02)2 + 10(1.02)
So end of 24th month = 10(1.02)24 + 10(1.02)23 + ... + 10(1.02)2 + 10(1.02)
= sum of GP with first term 10(1.02) and common ratio 1.02 for 24 terms
= $10(1.02) (1.0224 - 1) / (1.02 - 1)
= $310.30
(c) Supposed at the end of n months, the account will first exceed $2000
sum of GP with first term 10(1.02) and common ratio 1.02 for n terms ≥ 2000
10(1.02) (1.02n - 1) / (1.02 - 1) ≥ 2000
(1.02n - 1) ≥ 3.92157
1.02n ≥ 4.92157
n ≥ lg (4.92157) / lg (1.02)
n ≥ 80.475
End of 81st month, amount = 10(1.02) (1.0281 - 1) / (1.02 - 1) = $2026.20
Start of 81st month = $2026.2 / 1.02 = $1986.47 (not needed, show for fun only)
The account first exceed $2000 after 81 complete months.
