What is the difference between the compound interest and simple interest for 2 years on an amount of Rs 15000 at the rate of 12 1 2 per annum?

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RD Sharma Solutions Class 8 Mathematics Solutions for Compound Interest Exercise 14.2 in Chapter 14 - Compound Interest

Question 36 Compound Interest Exercise 14.2

The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs. 360. Find the sum.

Answer:

Given,

Time = 2 years

Rate = 7.5 % per annum

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 360

C.I – S.I = Rs 360

By using the formula,

P [(1 + R/100)^n - 1] – (PTR)/100 = 360

P [(1 + 7.5/100)^2 - 1] – (P(2)(7.5))/100 = 360

P[249/1600] – (3P)/20 = 360

249/1600P – 3/20P = 360

(249P-240P)/1600 = 360

9P = 360 × 1600

P = 576000/9

= 64000

∴ The sum is Rs 64000

Video transcript

hello everybody welcome to leader learning my name is rajna chaudhary and we have to write this statement in the equation form it is written that write equation for the statements for these statements so statement is one fourth of a number x minus g minus four gives four so one fourth of a number x would be one fourth of x that mean the value of this part is 1 by 4 of x then we have to minus 4 from it so let's minus 4 from it so minus 4 and gives gives means is equal to 4 so this is the equation for the statement we can write it like that at the place of off we can write multiply then minus 4 is equal to 4. we can also write it like x upon 4 minus 4 is equal to 4. so this is the form of equation for the statement i hope you understand the method see you in my next video don't forget to like comment and subscribe leader learning channel thank you for watching

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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

● Compound Interest

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Compound Interest with Growing Principal

Compound Interest with Periodic Deductions

Compound Interest by Using Formula

Compound Interest when Interest is Compounded Yearly

Compound Interest when Interest is Compounded Half-Yearly

Compound Interest when Interest is Compounded Quarterly

Problems on Compound Interest

Variable Rate of Compound Interest

Practice Test on Compound Interest

● Compound Interest - Worksheet

Worksheet on Compound Interest

Worksheet on Compound Interest with Growing Principal

Worksheet on Compound Interest with Periodic Deductions

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