Compound interest
- Thread starter Saumyojit
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I would write:
[MATH]615=P(1+r)^2[/MATH]
[MATH]600=P(1+r)[/MATH]
Now you have two unknowns in two equations. I would use the second equation, which implies:
[MATH]1+r=\frac{600}{P}[/MATH]
And substitute into the first to get:
[MATH]615=P\left(\frac{600}{P}\right)^2=\frac{600^2}{P}[/MATH]
Hence:
[MATH]P=\frac{600^2}{615}=\frac{24000}{41}[/MATH]
[MATH]615=P(1+r)^2[/MATH]
[MATH]600=P(1+r)[/MATH]
Now you have two unknowns in two equations. I would use the second equation, which implies:
[MATH]1+r=\frac{600}{P}[/MATH]
And substitute into the first to get:
[MATH]615=P\left(\frac{600}{P}\right)^2=\frac{600^2}{P}[/MATH]
Hence:
[MATH]P=\frac{600^2}{615}=\frac{24000}{41}[/MATH]
P is 6000
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Okay, I misread the problem...the two number given are interest, not the total amount. The interest earned is:
[MATH]I=P\left(\left(1+\frac{r}{n}\right)^{nt}-1\right)[/MATH]
How many times per year is the interest compounded in the first scenario?
[MATH]I=P\left(\left(1+\frac{r}{n}\right)^{nt}-1\right)[/MATH]
How many times per year is the interest compounded in the first scenario?
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If we assume it is once per year, then we get the following system:
[MATH]P((1+r)^2-1)=615[/MATH]
[MATH]P(1+2r-1)=600[/MATH]
Or:
[MATH]Pr(2+r)=615[/MATH]
[MATH]2Pr=600\implies r=\frac{300}{P}[/MATH]
And so we find:
[MATH]300\left(2+\frac{300}{P}\right)=615[/MATH]
[MATH]2+\frac{300}{P}=\frac{41}{20}[/MATH]
[MATH]\frac{300}{P}=\frac{1}{20}[/MATH]
[MATH]\frac{P}{300}=20[/MATH]
[MATH]P=6000\quad\checkmark[/MATH]
[MATH]P((1+r)^2-1)=615[/MATH]
[MATH]P(1+2r-1)=600[/MATH]
Or:
[MATH]Pr(2+r)=615[/MATH]
[MATH]2Pr=600\implies r=\frac{300}{P}[/MATH]
And so we find:
[MATH]300\left(2+\frac{300}{P}\right)=615[/MATH]
[MATH]2+\frac{300}{P}=\frac{41}{20}[/MATH]
[MATH]\frac{300}{P}=\frac{1}{20}[/MATH]
[MATH]\frac{P}{300}=20[/MATH]
[MATH]P=6000\quad\checkmark[/MATH]
D
Deleted member 4993
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The compound interest on a sum of money for 2 years is rs 615
Let the principal = P and interest rate = r. Then:
615 = Ic= P(1+r)2 - P ........................................(1)
The simple interest on a sum of money for 2 years is rs 600
Let the principal = P and interest rate = r. Then:
600 = Is= 2 r * P ........................................(2)
2Pr = 600........................................................................(3)
from (1)
2Pr + Pr2 = 615 ................................................................(4)
from (3) & (4)
Pr2 = 15 ..................................................................(5)
using (3) and (5)
r = 1/20 .....................................................................................(6)
Now use (6) in (2) to calculate P
Check your answer using calculated values of P and r in equations (1) and (2)
r/100 u have not written. how did u know r is in decimal ?
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If we let:
[MATH]r=\frac{R}{100}\implies R=100r[/MATH]
where \(R\) is the interest rate given as a percentage, the outcome will be unchanged.
[MATH]A=P\left(1+\frac{R}{100n}\right)^{nt}=P\left(1+\frac{100r}{100n}\right)^{nt}=P\left(1+\frac{r}{n}\right)^{nt}[/MATH]
[MATH]r=\frac{R}{100}\implies R=100r[/MATH]
where \(R\) is the interest rate given as a percentage, the outcome will be unchanged.
[MATH]A=P\left(1+\frac{R}{100n}\right)^{nt}=P\left(1+\frac{100r}{100n}\right)^{nt}=P\left(1+\frac{r}{n}\right)^{nt}[/MATH]
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Okay, for some reason you wish to make the calculations more cumbersome, so you begin with:
[MATH]P\left(\left(1+\frac{R}{100}\right)^2-1\right)=615[/MATH]
[MATH]P\left(1+2\frac{R}{100}-1\right)=600[/MATH]
Now, the second equation implies:
[MATH]\frac{R}{100}=\frac{300}{P}[/MATH]
Putting this into the first equation, we obtain:
[MATH]P\left(\left(1+\frac{300}{P}\right)^2-1\right)=615[/MATH]
[MATH]P\left(\frac{600}{P}+\frac{300^2}{P^2}\right)=615[/MATH]
[MATH]600+\frac{300^2}{P}=615[/MATH]
[MATH]\frac{300^2}{P}=15[/MATH]
[MATH]P=\frac{300^2}{15}=\frac{15^2\cdot20^2}{15}=15\cdot400=6000[/MATH]
[MATH]P\left(\left(1+\frac{R}{100}\right)^2-1\right)=615[/MATH]
[MATH]P\left(1+2\frac{R}{100}-1\right)=600[/MATH]
Now, the second equation implies:
[MATH]\frac{R}{100}=\frac{300}{P}[/MATH]
Putting this into the first equation, we obtain:
[MATH]P\left(\left(1+\frac{300}{P}\right)^2-1\right)=615[/MATH]
[MATH]P\left(\frac{600}{P}+\frac{300^2}{P^2}\right)=615[/MATH]
[MATH]600+\frac{300^2}{P}=615[/MATH]
[MATH]\frac{300^2}{P}=15[/MATH]
[MATH]P=\frac{300^2}{15}=\frac{15^2\cdot20^2}{15}=15\cdot400=6000[/MATH]
SI for 2 years=600
2(p*r)/100=600
pr = 30000
R/100=300/P
IGNORE this STEP
pr = 30000 do it with this .
CI=A-p= {p(1+r/100)^2 }-p
615={p(1+(r^2)/10000+2r/100) -p
615={p+ p(r^2)/10000+2pr/100}-p
615=(pr^2)/(10^4) + pr/50
substituting pr=30000
615={(3*10^4)r } / (10^4 )+ {3* (10^4) }/ 50
LCM OF (10^4 , 50)= 10^4
615= {(3*(10^4) )r +600*(10^4) } / 10^4
6150000=30000 r +6000000
6150000 - 6000000 = 30000 r
r=150000 / 30000 =5
p =6000
2(p*r)/100=600
pr = 30000
R/100=300/P
IGNORE this STEP
pr = 30000 do it with this .
CI=A-p= {p(1+r/100)^2 }-p
615={p(1+(r^2)/10000+2r/100) -p
615={p+ p(r^2)/10000+2pr/100}-p
615=(pr^2)/(10^4) + pr/50
substituting pr=30000
615={(3*10^4)r } / (10^4 )+ {3* (10^4) }/ 50
LCM OF (10^4 , 50)= 10^4
615= {(3*(10^4) )r +600*(10^4) } / 10^4
6150000=30000 r +6000000
6150000 - 6000000 = 30000 r
r=150000 / 30000 =5
p =6000
in the question it is not written that both the rates and principle of si and ci are same but why we are taking same?
@
MarkFL
@Dr.Peterson @Jomo
in the question it is not written that both the rates and principle of si and ci are same but why we are taking same?
I saw in internet that some are doing like this :
The additional CI of Rs 15 is interest on Rs 300 interest received in first year. So rate of interest is 15/300 x 100 = 5%.
Now, interest of Rs 300 at the end of first year @ 5% pa means principal at the beginning of the year is :
P = 300/5 x 100 = Rs 6000
The additional CI of Rs 15 is interest on Rs 300 interest received in first year - please explain this line
MarkFL
@Dr.Peterson @Jomo
in the question it is not written that both the rates and principle of si and ci are same but why we are taking same?
I saw in internet that some are doing like this :
The additional CI of Rs 15 is interest on Rs 300 interest received in first year. So rate of interest is 15/300 x 100 = 5%.
Now, interest of Rs 300 at the end of first year @ 5% pa means principal at the beginning of the year is :
P = 300/5 x 100 = Rs 6000
The additional CI of Rs 15 is interest on Rs 300 interest received in first year - please explain this line
Dr.Peterson
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When a problem is not clear, and you can't ask the author what it means, you can generally assume that it can be solved when interpreted as intended. Since if the rates were not the same, there would not be enough information (and since they would have mentioned that they were different), you assume that they are the same.
This kind of thinking would eliminate many of your concerns. Word problems are not real life; we traditionally make the simplest possible assumptions.
As for your second question, do you understand that the difference between simple and compound interest is that in the latter, interest is added in, so that subsequent interest is paid on that interest?
But this method is a "trick" suitable for those who understand the topic well. It is equivalent to the algebraic method. If you are not ready to think beyond formula, don't worry about it. Focus on understanding the subject more deeply.
This kind of thinking would eliminate many of your concerns. Word problems are not real life; we traditionally make the simplest possible assumptions.
As for your second question, do you understand that the difference between simple and compound interest is that in the latter, interest is added in, so that subsequent interest is paid on that interest?
But this method is a "trick" suitable for those who understand the topic well. It is equivalent to the algebraic method. If you are not ready to think beyond formula, don't worry about it. Focus on understanding the subject more deeply.
@Dr.Peterson " subsequent interest is paid on that interest" yes i know it .BUt it is not individually interest but on the both interest + p that is amt .
315 is Ci on the previous year amt or that year principal. 15 is additonal on 300 + p not only 300 right?
315 is Ci on the previous year amt or that year principal. 15 is additonal on 300 + p not only 300 right?
Last edited:
Dr.Peterson
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Let's look at the answer, and see what is happening:
Simple interest for the first year is 5/100 * 6000 = 300. This is also the amount of compound interest, but this is added into the account, so the new principal is 6300.
Simple interest for the second year is 5/100*6000 = 300 again, since the principal has not changed. Compound interest for the second year is the same as simple interest on the increased principal, 6300: 5/100*6300 = 315.
Now where did the difference in the second year come from? We can write the new principal as 6000 + 300, and distribute the multiplication: 5/100*(6000 + 300) = 5/100*6000 + 5/100*300 = 300 + 15. So the difference in interest is exactly the interest on the interest (5% of the 300 that was earned the first year and added in to the principal.
Is this what you were saying? Or, if not, do you understand now?
The principal, as has been shown, is 6000, and the rate is 5% per annum. We are assuming, for lack of other information, that compound interest is compounded annually.
Simple interest for the first year is 5/100 * 6000 = 300. This is also the amount of compound interest, but this is added into the account, so the new principal is 6300.
Simple interest for the second year is 5/100*6000 = 300 again, since the principal has not changed. Compound interest for the second year is the same as simple interest on the increased principal, 6300: 5/100*6300 = 315.
Now where did the difference in the second year come from? We can write the new principal as 6000 + 300, and distribute the multiplication: 5/100*(6000 + 300) = 5/100*6000 + 5/100*300 = 300 + 15. So the difference in interest is exactly the interest on the interest (5% of the 300 that was earned the first year and added in to the principal.
Is this what you were saying? Or, if not, do you understand now?