## 2013 DSE MATH (CORE) Question 19 Full Solution

Posted: September 10, 2013 in Mathematics
Tags: , , , , , (a) (i) The trick here is to recognize that you need to divide r by 100 and then add 1 to it. $9 \times 10^6 \cdot \left(1+\dfrac{r}{100} \right) - 3 \times 10^5$ $= 87 \times 10^5 + 9 \times 10^4 \cdot r$ $= 8700000 + 90000r$

(a) (ii) similar to the last question, multiply the result from (i) by (1+r/100) and then minus 300000 $(8700000+90000r)\left(1+\dfrac{r}{100}\right)-300000 = 1.026\times10^7$ $87\times10^5+9\times10^4\cdot r+87\times10^3\cdot r+9\times10^2r^2-3\times10^5-1026\times10^4=0$ $9r^2+1770r-18600=0$ $3r^2+590r-6200=0$ $(3r+620)(r-10)=0$ $r=10$ or $r=-\dfrac{620}{3}$

of course here we reject the negative answer for r since it can’t be negative and hence we have $r=10$

(b) (i) $((9\times10^6\cdot1.1-3\times10^5)(1.1)-3\times10^5)(1.1)\ldots-3\times10^5$

After 1st year, Area $=9\times 10^6$

After 2nd year, Area $=9\times10^6(1.1)-3\times10^5$

After 3rd year, Area $=9\times10^6(1.1)^2-3\times10^5(1.1)-3\times10^5$

After nth year, Area $=9\times10^6(1.1)^{n-1}-3\times10^5(1.1^{n-2}+\ldots+1)$ $=9\times10^6(1.1)^{n-1}-3\times10^5\left(\dfrac{1.1^{n-1}-1}{1.1-1}\right)$ $=9\times10^6(1.1)^{n-1}-3\times10^6(1.1^{n-1}-1)$ $=9\times10^6(1.1)^{n-1}-3\times10^6\cdot1.1^{n-1}+3\times10^6$ $=6\times10^6\times(1.1)^{n-1}+3\times10^6$

(b) (ii) $6\times10^6\times(1.1)^{n-1}+3\times10^6 > 4\times10^7$ $6\cdot1.1^{n-1}+3>40$ $1.1^{n-1} > \dfrac{37}{6}$ $n > \dfrac{\log\frac{37}{6}}{\log 1.1}+1 \approx 20.08\ldots$

Since n is an integer therefore $n=21$.  At the end of 21st year (c)

Setup the system of equations by the given information in the table, $1.21a+b=10^7$ $1.21^2a+b=1.063\times10^7$ $(1.21^2-1.21)a = 0.063\times10^7$ $a=\dfrac{63\cdot10^4}{1.21(1.21-1)}=\dfrac{63\cdot10^4}{1.21\cdot0.21}=\dfrac{3\cdot10^6}{1.21}$ $3\cdot10^6+b=10^7$ $b=7\cdot10^6$

Assume the claim is true, which means, $6\cdot10^6\cdot1.1^{n-1}+3\cdot10^6>\dfrac{3\cdot10^6}{1.21}\cdot1.21^n+7\cdot10^6$ $6\cdot1.1^{n-1}+3>\dfrac{3}{1.1^2}\cdot1.1^{2n}+7$ $0 > 3(1.1^{n-1})^2 - 6(1.1^{n-1}) + 4$

Now we need to show the RHS is greater than zero to get the contradiction and there are two ways in doing so.

Method 1 (Using Determinant) $\Delta = 6^2-4(3)(4)=36-48=-12$

and since the leading coefficient of $3(1.1^{n-1})^2 - 6(1.1^{n-1}) + 4$ is positive, therefore $3(1.1^{n-1})^2 - 6(1.1^{n-1}) + 4 > 0$

Therefore contradiction and the claim is false.

Method 2 (Using Completing the Square) $3(1.1^{n-1})^2 - 6(1.1^{n-1}) + 4$ $=3((1.1^{n-1})^2-2(1.1^{n-1}))+4$ $=3((1.1^{n-1})^2-2(1.1^{n-1})+1-1)+4$ $= 3(1.1^{n-1}-1)^2+1 > 0$

Again, contradiction, hence the claim is false.