Question

*In how many years the amount for Rs. 10000 becomes Rs. 12100 at 10% compound interest?*

1️⃣ 1
2️⃣ 2
3️⃣ 3
4️⃣ 4

Answer 1

2 years is the answer Thankyou

Answer 2

$\Large {\underline { \sf \orange{Explication \: of \: Steps :}}}$

• Principal = Rs. 10000
• Amount = Rs. 12100
• Rate = 10 %
• Time (n years) = ?

We know that,

$\bigstar \: \boxed{\sf {A = { P \Bigg \lgroup 1 + \dfrac{R}{100} \Bigg \rgroup}^{n} } }$

$\longrightarrow \sf {12100 = {10000 \Bigg \lgroup 1 + \dfrac{10}{100} \Bigg \rgroup }^{n} }$

$\longrightarrow \sf {12100 = /{10000 \Bigg \lgroup \dfrac{100 + 10}{100} \Bigg \rgroup}^{n} }$

$\longrightarrow \sf {12100 = {10000 \Bigg \lgroup \dfrac{110}{100} \Bigg \rgroup}^{n} }$

$\longrightarrow \sf {\cancel{\dfrac{12100}{10000}} = {\Bigg \lgroup \dfrac{11}{10} \Bigg \rgroup}^{n} }$

$\longrightarrow \sf {\dfrac{121}{100} = {\Bigg \lgroup \dfrac{11}{10} \Bigg \rgroup}^{n} }$

We can write,

• 121 = 11 × 11 ⇒ 11²
• 100 = 10 × 10 ⇒ 10²

$\longrightarrow \sf { {\Bigg \lgroup \dfrac{11}{10} \Bigg \rgroup}^{2} = {\Bigg \lgroup \dfrac{11}{10} \Bigg \rgroup}^{n} }$

As the bases are equal, another powers must be equal.

We get that,

$\longrightarrow \boxed { \large \sf {2 = n } } \: \: \orange { \bigstar}$

❝ Therefore, in 2 years the amount for Rs. 10000 becomes Rs. 12100 at 10% compound interest.❞