Discussion

Explanation:

A = 1007, 
t = 5 years, 
r = 3%

∴ A = $x\left(1+\frac{R\times (T-1)}{100}\right)$ + $x\left(1+\frac{R\times (T-2)}{100}\right)$ + ... + x

⇒ 1007 = $x\left(1+\frac{3\times 4}{100}\right)$ + $x\left(1+\frac{3\times 3}{100}\right)$ + $x\left(1+\frac{3\times 2}{100}\right)$ + $x\left(1+\frac{3\times 1}{100}\right)$ + x

⇒ 1007= $x\left(5+\frac{3\times 10}{100}\right)$

⇒ x = $\frac{10070}{53}$ = 190

Hence, option (b).

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