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




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


(b) (i)


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)





(b) (ii)

6\times10^6\times(1.1)^{n-1}+3\times10^6 > 4\times10^7


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



Setup the system of equations by the given information in the table,



(1.21^2-1.21)a = 0.063\times10^7




Assume the claim is true, which means,



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}-1)^2+1 > 0

Again, contradiction, hence the claim is false.


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