The amount of solute present in a saturated solution at a certain temperature is called ………………….. 

B) unsaturated solution

Correct option is A) diluted

B) concentrated

C) concentration

A) \(\cfrac{mass \,of \,solute}{mass \,of \,solution}\) × 100 

Correct option is C) 25%

Correct option is D) 20%

Correct option is (A) 8 Let the amount taken be Rs P. \(\therefore\) S.I. = \(\frac{16}{25}\times\) P = \(\frac{16P}{25}\) R = T \(\therefore\) S.I. = \(\frac{PRT}{100}\) \(\Rightarrow\) \(\frac{16P}{25}\) = \(\frac{P\times R\times R}{100}\)  \((\because\)  R = T) \(\Rightarrow\) \(R^2=\) \(\frac{16P}{25}\times\frac{100}P\) \(=16\times4=64=8^2\) \(\Rightarrow\) R = 8 \(\Rightarrow\) T = R = 8

Correct option is (C) ₹ 2166.66 Let the amount be P. R = \(12\frac{1}{2}\%=\frac{25}{2}\%\) T = 1 year Amount after 1 year \(=P(1+\frac R{100})^1\) \(\Rightarrow\) \(P(1+\frac R{100})\) = 2437.50   (Given) \(\Rightarrow\) \(P(1+\frac{25}{200})\) = 2437.50 \(\Rightarrow\) \(P(1+\frac18)\) = 2437.50 \(\Rightarrow\) \(\frac{9P}8\) = 2437.50 \(\Rightarrow\) P = 2437.50 \(\times\frac89=\frac{19500}9\) = 2166.66 \(\therefore\) Amount is Rs 2166.66