What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?

We will discuss here how to find the difference of compound interest and simple interest.

If the rate of interest per annum is the same under both simple interest and compound interest  then for 2 years, compound interest (CI) - simple interest (SI) = Simple interest for 1 year on “Simple interest for one year”.

Compound interest for 2 years – simple interest for two years

= P{(1 + \(\frac{r}{100}\))\(^{2}\) - 1} - \(\frac{P × r × 2}{100}\)

= P × \(\frac{r}{100}\) × \(\frac{r}{100}\)

= \(\frac{(P × \frac{r}{100}) × r × 1}{100}\)

= Simple interest for 1 year on “Simple interest for 1 year”.

Solve examples on difference of compound interest and simple interest:

1. Find the difference of the compound interest and simple interest on $ 15,000 at the same interest rate of 12\(\frac{1}{2}\) % per annum for 2 years.

Solution:

In case of Simple Interest:

Here,

P = principal amount (the initial amount) = $ 15,000

Rate of interest (r) = 12\(\frac{1}{2}\) % per annum = \(\frac{25}{2}\) % per annum = 12.5 % per annum

Number of years the amount is deposited or borrowed for (t) = 2 year

Using the simple interest formula, we have that

Interest = \(\frac{P × r × 2}{100}\)

           = $ \(\frac{15,000 × 12.5  × 2}{100}\)

           = $ 3,750

Therefore, the simple interest for 2 years = $ 3,750

In case of Compound Interest:

Here,

P = principal amount (the initial amount) = $ 15,000

Rate of interest (r) = 12\(\frac{1}{2}\) % per annum = \(\frac{25}{2}\) % per annum = 12.5 % per annum

Number of years the amount is deposited or borrowed for (n) = 2 year

Using the compound interest when interest is compounded annually formula, we have that

A = P(1 + \(\frac{r}{100}\))\(^{n}\)

A = $ 15,000 (1 + \(\frac{12.5}{100}\))\(^{2}\)

   = $ 15,000 (1 + 0.125)\(^{2}\)

   = $ 15,000 (1.125)\(^{2}\)

   = $ 15,000 × 1.265625

   = $ 18984.375

Therefore, the compound interest for 2 years = $ (18984.375 - 15,000)

                                                             = $ 3,984.375

Thus, the required difference of the compound interest and simple interest = $ 3,984.375 - $ 3,750 = $ 234.375.

2. What is the sum of money on which the difference between simple and compound interest in 2 years is $ 80 at the interest rate of 4% per annum?

Solution:

In case of Simple Interest:

Here,

Let P = principal amount (the initial amount) = $ z

Rate of interest (r) = 4 % per annum

Number of years the amount is deposited or borrowed for (t) = 2 year

Using the simple interest formula, we have that

Interest = \(\frac{P × r × 2}{100}\)

           = $ \(\frac{z × 4  × 2}{100}\)

           = $ \(\frac{8z}{100}\)

           = $ \(\frac{2z}{25}\)

Therefore, the simple interest for 2 years = $ \(\frac{2z}{25}\)

In case of Compound Interest:

Here,

P = principal amount (the initial amount) = $ x

Rate of interest (r) = 4 % per annum

Number of years the amount is deposited or borrowed for (n) = 2 year

Using the compound interest when interest is compounded annually formula, we have that

A = P(1 + \(\frac{r}{100}\))\(^{n}\)

A = $ z (1 + \(\frac{4}{100}\))\(^{2}\)

   = $ z (1 + \(\frac{1}{25}\))\(^{2}\)

   = $ z (\(\frac{26}{25}\))\(^{2}\)

   = $ z × (\(\frac{26}{25}\)) × (\(\frac{26}{25}\))

   = $ (\(\frac{676z}{625}\))

So, the compound interest for 2 years = Amount – Principal

                                                    = $ (\(\frac{676z}{625}\)) - $ z

                                                    = $ (\(\frac{51z}{625}\))

Now, according to the problem, the difference between simple and compound interest in 2 years is $ 80

Therefore,

    (\(\frac{51z}{625}\)) - $ \(\frac{2z}{25}\) = 80

⟹ z(\(\frac{51}{625}\) - \(\frac{2}{25}\)) = 80

⟹ \(\frac{z}{625}\) = 80

⟹ z = 80 × 625

⟹ z = 50000

Therefore, the required sum of money is $ 50000

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Worksheet on Compound Interest with Periodic Deductions

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Discussion :: Compound Interest - General Questions (Q.No.14)

14. 

The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?

[A]. 8
[B]. 10
[C]. 12
[D]. Cannot be determined
[E]. None of these

Answer: Option A

Explanation:

What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
15000 x
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
1 + R
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
2 - 15000
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
-
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
15000 x R x 2
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
= 96
100 100

What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
15000
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
1 + R
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
2 - 1 - 2R
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
= 96
100 100

What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
15000
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
(100 + R)2 - 10000 - (200 x R)
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
= 96
10000

What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
R2 =
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
96 x 2
What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
= 64
3

What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
R = 8.

What will be the difference between simple interest and compound interest on a sum of 15000 for 2 years at the same rate of interest of 12.5 per annum?
Rate = 8%.

Anonymous said: (Nov 7, 2010)  
In that why -15000 is done in compound interest. As per formula its not there na?

Priyanka P. said: (Dec 2, 2010)  
@Anonymous for C.I. formula is=> p(1-r/100)^n-p without subtracting, which amount we get is "Principle amount+Interest"

so to get C.I. we have to subtract it..

Swap said: (Jan 18, 2011)  
Please tell something more about compound interest.

Ravin said: (Feb 6, 2011)  
Kindly explain in 3rd step howcum -1 would be showing...is this problem solve throuGh the LCM?

Narendra said: (Aug 6, 2011)  
SIMPLE FORMULA FOR TWO YEARS C.I.-S.I.=P(R/100)*(R/100) 96=15000(R/100)(R/100) R*R=960/15=64

R=8

Prawin said: (Jun 6, 2012)  
shortcut: p = d (100/r)^2 15000 = 96*100*100/r^2 r^2 = 96*100*100/15000

r = 8.

Vimal said: (Aug 20, 2012)  
Thanks prawin.

Mudasir said: (Sep 29, 2012)  
It's simply p(r/100)^2 p=1500,r=x,t=2yrs Thus 15000(x/100)^2 it finally comes out like x^2=64 x^2=8^2 x=8

Therefore rate=8%

Lucky said: (Jul 13, 2013)  
In 4th step how they done R^2=64 ?

If we solve it there should be R^2-200R=64 and this is a quadratic equation. Can anyone explain me how to solve this equation ?

Muthu said: (Nov 10, 2013)  
C.P - S.I [p(1-r/100)^n-p ]-[(P*R*N)/100] = 96. Where, p=1500, n=2, R=?, C.P and S.I difference is 96. [15000(1-r/100)^2-15000 ]-[(1500*R*2)/100] = 96 ...(1). Now, 15000 common in above so take it out, 15000[1(1-r/100)^2 - 1 ]-[(1*R*2)/100] = 96 ...(2). Take L.C.M. we get, 15000[(100 + R)^2 - 10000 - (200 x R)/10000] = 96 ...(3). Where (100 + R)^2 is like (a + b)^2 = (a^2 + b^2 + 2*a*b). (100^2 + R^2 + 2*100*R) = (10000 + R^2 + 200R). Sub (10000 + R^2 + 200R) in eqn(3) we get, 15000[ (10000 + R^2 + 200R) - 10000 - (200 x R)/10000] = 96. 15000[ (10000 + R^2 + 200R - 10000 - 200R) /10000] = 96. Now 10000 and 200R is cancelled, remaining. 15000[ R^2 /10000] = 96. [15000R^2/10000] = 96. [15R^2/10] = 96. divide by 5 we get, [3R^2/2] = 96. Now keep the R^2 in left hand side and move 3/2 to right side R^2 = 96(2/3) // 96/3 = 32. R^2 = 32*2. R^2 = 64 // square root of 64 is 8.

R = 8%.

Manish said: (Aug 25, 2014)  
Apply below simple formula and get the answer easily.

Difference = P(R)^2/(100)^2.

Pratima said: (Oct 31, 2014)  
Can any one provide shortcut for difference (CI-SI) questions for more than two years?

Muniyasoda said: (Mar 30, 2015)  
Explain more than two years?

Arshdeep said: (Jul 1, 2015)  
Simply use formula difference=p*r*r/100*100.

A Velavan said: (Jan 17, 2016)  
Difference between SI and CI when 2 years = p(r/100)^2. 96 = 15000(r/100)^2. R*r = 32*2. R^2 = 64.

R = 8.

Anshul said: (May 27, 2016)  
Thanks, @A Velavan.

Bhaumik said: (Aug 31, 2016)  
@Muthu.

You explained beautifully, Thanks.

Ravi said: (Oct 2, 2016)  
For the difference between CI and SI for 2 years formula is p = d * (100/r)^2.

Shine said: (Mar 23, 2017)  
C.I - S.I (FOR 2 YEARS)= P (R/100)^2. HENCE, 96 = 15000(R/100)^2.

Solve for R.

Ratan said: (Jul 11, 2017)  
Hi All , There is a shortcut method for this - C.I. - S.I. = P*R^2/(100)^2. so if we will put the values here it will be like this; 96 = 15000*R*R/10000, 64 = R*R,

So R =8.

Somnath said: (Jul 14, 2017)  
@Prawin and @Velavan.

Thank you very much for giving the simple trick for the solution.

Rahul said: (Sep 3, 2017)  
What will be solution, if time is 3years?

Anisa said: (Aug 31, 2018)  
Here 100^3*(CI-SI)/R^2(R+300).

Upasana said: (Nov 11, 2018)  
Can the difference be added to the principal to get the amount for compound interest formula?

And then by using CI formula rate can be calculated.

Wrick said: (Nov 12, 2018)  
Simply, the solution is; 15000 * [{1+(r/100)}^2-1]-(15000 * 2 * r)/100 = 96, => 15000[{1+(r/100)}^2-1-(2 * r)/100]= 96, => 15000[1+(r^2)/10000-1] = 96 where [(a+b)^2-2ab = a^2 + b^2], => 15000[r^2/10000] = 96, => r^2 = 64,

=> r = 8.

Himani said: (Aug 18, 2019)  
Simply: sum = difference * (100/R)^2 for 2 year.

sum = {difference * (100^3)}/{R^2*(300+R)} for 3 year.

Tamalika Roy said: (Aug 5, 2020)  
@All. The solution is; Difference = sum(R/100)^n => formula. Here;

96 = 15000(r/100)^2 = 8.

Siddu said: (Aug 11, 2020)  
P = si-ci/(r/100) 2 use this formula to get the answer.

Khyati Mehta said: (Mar 25, 2021)  
We can directly do this by using the formula. Difference between SI and CI for 2 years = principle *(Rate of Interest)^2 = P(R%)^2. So here the difference is 96, principle=15000. 96 = 15000*(R/100)^2.

r = 8%.

Naveen said: (Jul 15, 2021)  
CI-SI = P(R/100)^2. CI-SI = 96. P = 15000. 96 = 15000(R/100)^2, 96/15000 = R^2/100^2, 0.0064 = R^2/10000, 64 = R^2,

R = 8.

Ash said: (Jul 19, 2021)  
Hi guys, I will tell you the simplest method for the difference between Si and Ci sums. Use the formula:

Difference = P*(R^2)/100^2.