Question

In how many years will the interest on RS.2500 will be RS.1350 at 9 p.c.p.a

Answer 1

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathsf{ \: \: Given \: : \mapsto }} \\ \\ </p><p></p><p> \bold{Principal \: \: amount, P = Rs. 2500}</p><p> \\ \\ \bold{Interest, S.I. = Rs. 1350} \\ \\ \bold{</p><p>Rate \: \: of \: \: interest, r = 9\%}</p><p> \\ \\ \\ \underline{\mathsf{ \: \: To \: \: find \: : \mapsto }} \\ \\ \bold{ \: \: \: No. \: \: of \: \: \: years = \: ?}$

$\bold{Let,} \\ \: \: \: \: \: \: \: \: \bold{</p><p>No. \: \: \: of \: \: \: years \: = \: n }</p><p></p><p> \\ \\ \underline{ \bold{We \: \: \: know \: \: that,}} \\ \\ \: \: \: \: \: \mathcal{</p><p>S.I. = \frac{ Prn}{1 00} } \\ \\ \mathcal{ \Rightarrow \: \: 1350 = \frac{2 \: 5 \: \cancel{0} \cancel{0} \times 9 \times n}{1 \: \cancel{0} \cancel{0}} } \\ \\ \mathcal{ \Rightarrow \: \:1350 = 25 \times 9 \times n } \\ \\ \mathcal{ \Rightarrow \: \: 1350 = 225 \times n \: } \\ \\ \mathcal{ \Rightarrow \: \:n = \frac{1350}{225} } \\ \\ \: \bold{ \therefore \: \: \: n \: = \: 6}$

$\mathsf{The \: \: required \: \: number \: \: of \: \: years \: = \: 6 \: \: years.}$

Answer 2

Answer:

Principal amount =2500

Intrest=1350

Rate of intrest=9 p.c.p.a

N=?

Intrest=P×N×R/100

1350=2500×N×9/100

1350=225

N=1350÷225

N=6 years

6 years will the intrest on RS. 2500 Will be Rs. 1350 at 9 p.c.p.a