Question

compound interest for a certain amount at 13%p.a for 2 years exceeds the interest on the same amount at 12%p.a for two years by Rs 225. The principal amount is?​

Answer 1

Answer:

• Therefore, 3200=P(1+
• 100
• 10
• )
• 2
• −P
• ⇒3200=P(1.1)
• 2
• −P
• ⇒1320=1.21P−P
• ⇒P=
• 0.21
• 3200
• ⇒P= Rs. 15238
• Simple interest =
• 100
• PRT
• ⇒12000=
• 100
• P×10×2
• ⇒P=
• 20
• 1000×100
• ⇒P= Rs. 60,000
• Sum of the principal =15238+60000= Rs. 75238

Answers is 75238

Answer 2

Given:

Compound interest for a certain amount at 13%p.a for 2 years exceeds the interest on the same amount at 12%p.a for two years by Rs 225.

Solution:

Know that,

$\[A = P{\left( {1 + \frac{r}{{100}}} \right)^n}\]$

Write the following equation using given information,

$\[ \Rightarrow P{\left( {1 + \frac{{13}}{{100}}} \right)^2} - P{\left( {1 + \frac{{12}}{{100}}} \right)^2} = 225\]$

$\[ \Rightarrow P\left\{ {{{\left( {\frac{{113}}{{100}}} \right)}^2} - {{\left( {\frac{{111}}{{100}}} \right)}^2}} \right\} = 225\]$

Apply, $\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]$

$\[ \Rightarrow P\left\{ {\left( {\frac{{113 + 112}}{{100}}} \right)\left( {\frac{{113 - 112}}{{100}}} \right)} \right\} = 225\]$

$\[\begin{array}{l} \Rightarrow P \times 225 \times 1 = 225 \times 100 \times 100\\ \Rightarrow P = 10,000\end{array}\]$

Hence the principal is Rs. 10000.