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The principal at the end of the first year is $p$.

 $\frac{p\times 10 \times 1}{100} = 132$ $\Rightarrow p = 1320$

Let the actual principal $P$

so, after $1$ year amount will be,  $P +  \frac{P\times 10 \times 1}{100}$  $\Rightarrow$ $\frac{11P}{100} = 1320$

$P = 1200 /-$

 

Correct Ans: Option B

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