simple and compound interest question

[gre_girl]

hey friends

1. The difference between the second and third year's interest on a certain sum at 8% compound interest is $8.64 , find the sum.

 

2. A sum of money placed at compound interest doubles itself in 4 year.In how many years will it amount to 8 times itself?

some one try for this

regards

[rocky80]

1. 100

 

2. 12 yrs

[mishuk]

1. 1250

[hotsummertime]

Is the ans to first one 92.6?

 

How do you get the 2nd?

[rocky80]

Is the ans to first one 92.6?

 

How do you get the 2nd?

 

yes it is

i misread the question. :doh:

[proteansubterfuge]

1) X*1.08*1.08 - x*1.08 = 8.64

x=100

the interest for the second year is:

108 -100 = 8

the interest for the third year is:

116.64 -100 = 16.64

therefore the sum of the interests are 8+16.64 = 24.64

is it correct?

Am i reading the question correctly?

hotsummertime how did you get 92.6?

 

thanks,

Leo

[proteansubterfuge]

question 2:

x = interest rate

y = amount of money

x^4*y = 2y (eliminate y)

x^4 = 2

x^4*x^4 =x^8 (2*2=4 (they asked for 8 times the money))

x^8*x^4=x^12 (4*2=8 (now we have 8 times the money))

 

as Rocky says 12 years

[joy_Montreal]

i think, for the 1st one, mishuk did well. the correct answer should be 1250. HOW?

2nd yrs int=xr(1+r)

3rd yrs int=xr(1+r)^2

so the differrence=xr(1+r){1+r-1}=xr^2(1+r)=x*(.08^2)*(1.08)=8.64

hence, x should be = 1250.

[gre_girl]

ya guys

you are great friends.

first ans. 1250 and second ans.12 yrs

someone plz explain que.one

hey rocky and mishuk how can you get the ans. for both question?

plz explain other method

regards

[hotsummertime]

Hello joy,

 

Why do you have the equations

 

2nd yrs int=xr(1+r)

3rd yrs int=xr(1+r)^2

 

I think the second year amount should be x(1+r/100)^2 and third year x(1+r/100)^3.

 

Please clarify. Thanks.

 

With my x(1+r/100)^3 - x(1+r/100)^2 = 8.64 I get 92.6 which apparently is not the right answer.

[gre_girl]

hey joy

plz can u explain both example.i cant understand.

regards

[joy_Montreal]

let x be the amount, and r be the rate of interest.

so,

end of year|| running amount || interest

1st || x+rx=x(1+r) || rx

2nd || x(1+r)+rx(1+r)=x(1+r)^2 ||rx(1+r)

3rd ||x(1+r)^2+rx(1+r)^2 || rx(1+r)^2

 

so, now taking the interest difference, as explained in the previous reply, i think you can get to the 1st answer.

 

ragards

[KBTA]

For Q1. we should be careful about word "between" and "sum".

Here, sum is investment.

For the word between we are counting 2nd yr interest after the first yr and 3rd yr interest after the 2nd yr.

Please check and inform about my in case of Q1.

 

Q2.

2p=p(1+i/100)^4

=> i=100*[2^(1/4)-1]

8p=p[1+i/100]^n

=> 8=[1+2^(1/4)-1]^n=2^(.25n)

=>2^3=2^(0.25n)

=> n=3/0.25=12

n=12

[rajneelam]

actually interest difference between 2nd year and 3rd year is due to applicable interest on the interest of 2nd year. Let's calculate interest of 2nd year = x(say).

I = PTR/100 => 8.64=x * 1 * 8/100 => x = 864/8 = 108RS

if 108 was interest in 2nd year. which is

p*r*(1+r) = 108

p=108/r*(1+r) = 108/0.08(1.08) =100/0.08 = 10000/8 = 1250RS

Above trick is useful in solving many problems......

[rajneelam]

Lets's give a look to the 2nd problem it is saying money is getting doubled in each of two year

Year AMOUNT

Initially x

4 2x

4+4 2*2x=4x (Money is getting doubled in every four year)

4+4+4 2*4x = 8x (Again money is getting doubled in next 4 year)

 

=> 12years

[Prathyu557]

Any one help me with these problems

1.Compound interest earned on a sum for second and third years are Rs.1440 and Rs.1656 respectively . Find the rate of interest?

2.A borrowed a sum of money from B at SI at a rate of 10%p.a for the first 2years,12%p.a for the next 3years and 15%p.a for beyond 5years.If A paid Rs.5332 as interest after 7years then find the sum

[armarm]

1)if sum=s,interest rate=r, then you have the following system of equations

s*( (1+r)^3-(1+r)^2 )=1656

s*( (1+r)^2-(1+r) )=1440

replacing 1440 instead of s*( (1+r)^2-(1+r) ) in the first equation we get 1+r=1656/1440 then r=15%

 

2) s+5332=s*1.10^2*1.12^3*1.15^2

s+5335=2.248s

s=4272