Question

the compound interest on a certain sum for 2 years at 6% per annum is ruppes. 90 find the sum???????​

Answer 1

Answer:

25000

Step-by-step explanation:

Solution

Verified by Toppr

Given:

Interest rate

=

6%

per annum

Time

=

2

years

Simple interest

(SI)=

PTR/100

where P is principle amount, T is time taken, R is rate per annum

Let sum is P

Now,

SI= (P× 2× 6)/100

⇒SI=

(12P)/100

⇒ SI= (3P)/25

1

To find the amount we have the formula,

Amount

(A)=

P(1+

(r/100))

n

wherePis present value,r

is rate of interest,nis time in years.

Also,

CI=

A−

P

Now substituting the values in above formula we get,

∴ CI= P(1+ 6/100)

2

− P

⇒ CI= P(1+ 3/50)

2

− P

⇒ CI= P(53/50)

2

− P

⇒CI=

P(2809)/(2500)–P

⇒CI=

309P/2500

——– equation 2

Now the difference is

⇒

90=

309P/2500–(3P)/25

⇒

90=

309P–(300P)/2500

(CI–SI)= 309P/2500–(3P)/25

⇒ 90= 9P/2500

⇒ P= 90× 2500/9

⇒ P= 10× 2500

⇒ P= 25000

∴ Sum

=

25000

Answer 2

• As per the data given in the question, we have to find the value of the expression.

          Given data:- the compound interest on a certain sum for $2$ years at $6\%$ per annum is ruppes $90.$

        To find:- Value of the expression.

         Solution:-

• Interest can be compounded on any given frequency schedule, from continuous to daily to annually.

          so that for calculating the sum

         let,

      $\text { Interest rate }=6 \% \text { per annum }\\\text { Time=2 years }\\\text { Simple interest }(\mathrm{SI})=\frac{PTR}{100}.$

Where P is the principal amount, T is the time taken, R is the rate per annum.

      Let sum is P

      $\begin{array}{l}\mathrm{SI}=(\mathrm{P} \times 2 \times 6) / 100 \\\Rightarrow \mathrm{SI}=(12 \mathrm{P}) / 100 \\\Rightarrow \mathrm{SI}=(3 \mathrm{P}) / 25 \longrightarrow_{} \text {equation } 1\end{array}$

    To find the amount we have the formula,

     $\text { Amount }(A)=P(1+(R / 100))^{n}$

   Where P is the present value, r is the rate of interest, n is time in years.

  Now substituting the values in the above formula we get

    $\begin{array}{l}\therefore C \mid=P(1+6 / 100)^{2}-P \\\Rightarrow C I=P(1+3 / 50)^{2}-P \\\Rightarrow C I=P(53 / 50)^{2}-P \\\Rightarrow C I=P(2809) /(2500)-P \\\Rightarrow C I=309 P / 2500\longrightarrow \text {equation } 2\end{array}$

         Now the difference is

     $\begin{array}{l}\Rightarrow 90=309 P / 2500-(3 P) / 25 \\\Rightarrow 90=309 P-(300 P) / 2500 \\\Rightarrow 90=9 P / 2500 \\\Rightarrow P=90 \times 2500 / 9 \\\Rightarrow P=10 \times 2500 \\\Rightarrow P=25000\end{array}$

      Hence we will get the value $p=2500.$