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secondary 4 | A Maths
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Good evening!!!
I need some help to find the value of “k”.. can someone kindly help me please? :D
Thank you for your time and effort!:)
Date Posted: 4 years ago
Views: 517
This question is is similar to those involving compound interest.
Increase by k% per year means we can write in the form A(1 + k/100)^x , just like our compound interest formula P(1 + r/100)ⁿ
So A(1.11)^x = 20000
A(1 + 0.11)^x = 20000
A(1 + 11/100)^x = 20000(1 + k/100)^0
(We take end of 2017 to be the start so number of years here = 0. And anything to them power of 0 equals 1)
So comparing the two,
A = 20000, k = 11
Increase by k% per year means we can write in the form A(1 + k/100)^x , just like our compound interest formula P(1 + r/100)ⁿ
So A(1.11)^x = 20000
A(1 + 0.11)^x = 20000
A(1 + 11/100)^x = 20000(1 + k/100)^0
(We take end of 2017 to be the start so number of years here = 0. And anything to them power of 0 equals 1)
So comparing the two,
A = 20000, k = 11
Ahh.. i see.. thank you so much!! :-) This is a bit similar to emaths now.. haha
Yes, very similar
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Good evening Sonia! Here are my workings for this question.
Date Posted:
4 years ago
Hi Sonia! We take the initial time of data measurement to start from end 2017 so that x = 0 at the end of 2017.
A growth rate of k% can be thought of as an “interest” rate of k%. This tags on to the “1” in the formula
Pfinal = P (1 + k%)^n.
A growth rate of k% can be thought of as an “interest” rate of k%. This tags on to the “1” in the formula
Pfinal = P (1 + k%)^n.
Thank you so much! i understand it now :-)
I doubt addition of numbers to percentage is allowed in O levels. Might be marked down. Better to rewrite as k/100 first, then conclude k% is 11% and therefore k is 11
Sonia, for an alternative approach to this question you can refer to my comment in the main section
Sonia, for an alternative approach to this question you can refer to my comment in the main section
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