sud21 wrote:At the end of n years, an investment p and its interest totally is s, where s=p*(1.05)^n. If p is $2000, is n greater than 4?
1). S > 2200
2). S < 3000
Be ready for rough estimation of few positive integer powers over the number 1.05 or 21/20 here. Remember, when a positive integer power more than 1 is put over a number which is already more than 1, the number's value increases, and if 0 < n (the power) < 1, then the number's value decreases. Those who are good at number properties won't take much time in estimating that (1.05) ^4 is approximately around 1.2.
Given is the rephrased 'Amount' formula when interest is compounded annually, it reads
Amount (s) = Investment (p) × [1 + (Rate/100)] ^n.
In here, we have s = $2000 × [1 + (5/100)] ^n; and the question, is n > 4?
(1) If s > 2200, then 2000 × [1 + (5/100)] ^n > 2200, or (21/20) ^n > 22/20, it only proves that n > 1. This statement is hence, not sufficient.
(2) If s < 3000, then 2000 × [1 + (5/100)] ^n < 3000, or (21/20) ^n < 30/20, it only proves that n could be any number less than a number which is fairly more than 4. Hence, n could be less than, equal to, or greater than 4. This statement is hence, not sufficient.
When the two statements are taken together, we have 22/20 < (21/20) ^n < 30/20, and still, n could be less than, equal to, or greater than 4. [spoiler]Not Sufficient.
E[/spoiler]
