H2 Maths 2012 A Level Paper 2 Q4 Solution; H2 Maths Tuition

(i)

1 Jan 2001 –> $100

1 Feb 2001 —> $110

1 Mar 2001 –> $120

Notice that this is an AP with $a=100$ ; $d=10$

$\displaystyle\begin{array}{rcl}S_n&=&\frac{n}{2}(2a+(n-1)d)\\ &=&\frac{n}{2}(200+10(n-1))>5000 \end{array}$

$\frac{n}{2}(200+10(n-1))-5000>0$

From GC, $n>23.5$

$n=24$ (months)

This is inclusive of 1 Jan 2001!!!

Thus, 1 Jan 2001 + 23 months —> 1 Dec 2002

(ii)

1 Jan 2001 –> 100

end of Jan 2001 –> 1.005(100)

1 Feb 2001 –> 1.005(100)+100

end of Feb 2001 –> 1.005[1.005(100)+100]= $1.005^2 (100)+1.005(100)$

From the pattern, we can see that

$\displaystyle\begin{array}{rcl}S_n&=&1.005^n(100)+1.005^{n-1}(100)+\cdots+1.005(100)\\ &=&\frac{a(r^n-1)}{r-1}\\ &=&\frac{1.005(100)[1.005^n-1]}{1.005-1}\\ &=&\frac{100.5(1.005^n-1)}{0.005}\\ &=&20100(1.005^n-1) \end{array}$

$5000-$100=$4900

$20100(1.005^n-1)>4900$

$20100(1.005^n-1)-4900>0$

From GC, $n>43.7$

So $n=44$ months (inclusive of Jan 2001 !!!)

1 Jan 2001+36 months —> 1 Jan 2004

1 Jan 2004+7 months —> 1 Aug 2004

Then on 1 Sep 2004, Mr B will deposit another $100, making the amount greater than $5000.

Hence, answer is 1 Sep 2004.

(iii)

Let the interest rate be x %.

Note that from Jan 2001 to Nov 2003 is 35 months. (Jan 2001 to Dec 2001 is 12 months, Jan 2002 to Dec 2002 is 12 months, Jan 2003 to Nov 2003 is 11 months :))

$5000-$100=$4900

Modifying our formula in part ii, we get

$\displaystyle S_n=\frac{(1+x/100)(100)[(1+x/100)^n-1]}{(1+x/100)-1}=4900$