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Answer:

P = 5,500, r=2 1/3 , I = 1,155,n =?

[tex] \sf \: I \: = \frac{PRN}{100} \\ \sf 1155= \frac{5500 \times ( \frac{7}{2}) \times n }{100} \\ \sf \: n = \frac{1155 \times 100}{2750 \times 7} \\ \sf \: n = 6[/tex]

•°• The time period is 6 yrs.

Answer:

Given :

• Principal is Rs 5500, rate of interest is 2 1/3% and simple interest is Rs 1155.

To Find :-

• What is the time period (n).

Formula Used :-

[tex]\clubsuit[/tex] Time Period (n) Formula :

[tex]\bigstar \: \: \sf\boxed{\bold{\pink{n =\: \dfrac{S.I \times 100}{P \times r}}}}\: \: \: \bigstar\\[/tex]

where,

• n = Time Period
• S.I = Simple Interest
• P = Principal
• r = Rate of Interest

Solution :-

Given :

• Principal = Rs 5500
• Rate of Interest = 2 1/3%
• Simple Interest = Rs 1155

According to the question by using the formula we get,

[tex]\implies \bf n =\: \dfrac{S.I \times 100}{P \times r}[/tex]

[tex]\implies \sf n =\: \dfrac{1155 \times 100}{5500 \times 2\dfrac{1}{3}}[/tex]

[tex]\implies \sf n =\: \dfrac{115500}{5500 \times \bigg(2 + \dfrac{1}{3}\bigg)}[/tex]

[tex]\implies \sf n =\: \dfrac{115500}{5500 \times \bigg(\dfrac{2 \times 3 + 1}{3}\bigg)}[/tex]

[tex]\implies \sf n =\: \dfrac{115500}{5500 \times \bigg(\dfrac{6 + 1}{3}\bigg)}[/tex]

[tex]\implies \sf n =\: \dfrac{115500}{5500 \times \dfrac{7}{3}}[/tex]

[tex]\implies \sf n =\: \dfrac{115500}{\dfrac{5500 \times 7}{3}}[/tex]

[tex]\implies \sf n =\: \dfrac{115500}{\dfrac{38500}{3}}[/tex]

[tex]\implies \sf n =\: \dfrac{115500}{1} \times \dfrac{3}{38500}[/tex]

[tex]\implies \sf n =\: \dfrac{3465\cancel{00}}{385\cancel{00}}[/tex]

[tex]\implies \sf n =\: \dfrac{\cancel{3465}}{\cancel{385}}[/tex]

[tex]\implies \sf\bold{\red{n =\: 9\: years}}[/tex]

[tex]\therefore[/tex] The time period (n) for which the money is borrowed is 9 years .