Question:
A sum of ₹12,000 is borrowed at 10% compound interest per annum. If it is to be repaid in two equal annual installments, what is the value of each installment?
₹12,000 की राशि 10% वार्षिक चक्रवृद्धि ब्याज पर उधार ली जाती है। यदि इसे दो समान वार्षिक किस्तों में चुकाया जाना है, तो प्रत्येक किस्त का मान क्या है?
Options:
- ₹6,914.28 (approx ₹6,914)
- ₹6,600
- ₹7,200
- ₹6,450
- ₹6,914.28 (लगभग ₹6,914)
- ₹6,600
- ₹7,200
- ₹6,450
Correct Answer: a)
Explanation:
Formula for equal annual installment \(x\):
\(P = \frac{x}{1 + r/100} + \frac{x}{(1 + r/100)^2}\).
Here \(r = 10\% \implies 1 + r/100 = \frac{11}{10}\).
\(12000 = \frac{10x}{11} + \frac{100x}{121} = x \left(\frac{110 + 100}{121}\right) = \frac{210x}{121}\).
\(x = \frac{12000 \times 121}{210} = \frac{400 \times 121}{7} = \frac{48400}{7} \approx ₹6,914.28\).
\(P = \frac{x}{1 + r/100} + \frac{x}{(1 + r/100)^2}\).
Here \(r = 10\% \implies 1 + r/100 = \frac{11}{10}\).
\(12000 = \frac{10x}{11} + \frac{100x}{121} = x \left(\frac{110 + 100}{121}\right) = \frac{210x}{121}\).
\(x = \frac{12000 \times 121}{210} = \frac{400 \times 121}{7} = \frac{48400}{7} \approx ₹6,914.28\).
समान वार्षिक किस्त का सूत्र:
\(P = \frac{x}{1.1} + \frac{x}{(1.1)^2} = x \left(\frac{10}{11} + \frac{100}{121}\right) = \frac{210x}{121}\)
\(x = \frac{12000 \times 121}{210} = \frac{48400}{7} \approx ₹6,914.28\)।
\(P = \frac{x}{1.1} + \frac{x}{(1.1)^2} = x \left(\frac{10}{11} + \frac{100}{121}\right) = \frac{210x}{121}\)
\(x = \frac{12000 \times 121}{210} = \frac{48400}{7} \approx ₹6,914.28\)।