Verified by Ajay Kumar
Find the rate of interest if ₹1,000 amounts to ₹1,210 in 2 years, compounded annually.
7%
15%
13%
10%
Answer: D
Solution: Using \(A=P(1+\frac{r}{100})^n\), we get \(1210=1000(1+\frac{r}{100})^2\). Thus, \((1+\frac{r}{100})^2=1.21\Rightarrow1+\frac{r}{100}=1.1\Rightarrow r=10\%\). Trick: The compound growth factor over two years is 1.21, whose square root is 1.1.
यदि ₹1,000 की राशि वार्षिक चक्रवृद्धि ब्याज पर 2 वर्षों में ₹1,210 हो जाती है, तो ब्याज की दर ज्ञात कीजिए।
7%
15%
13%
10%
उत्तर: D
हल: \(A=P(1+\frac{r}{100})^n\) का प्रयोग करने पर \(1210=1000(1+\frac{r}{100})^2\)। इसलिए \((1+\frac{r}{100})^2=1.21\Rightarrow1+\frac{r}{100}=1.1\Rightarrow r=10\%\)। ट्रिक: दो वर्षों का चक्रवृद्धि वृद्धि गुणक 1.21 है, जिसका वर्गमूल 1.1 है।
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