Question

I will surely mark you as brainliest
Please solve on notebook plz guys please i am really in a difficult situation.

➡The difference between the compound interest and simple interest on a sum of 15000 for 2 years is 96 what is the rate of interest
|The answer given is 8%|

Answer 1

Method - 1:

Given Principal = 15000, Time n = 2 years.

(i)

We know that SI = PTR/100

                           = (15000 * 2 * r)/100

                           = 30000r/100

                          = 300r.

(ii)

We know that CI = [P(1 + r/100)^n - P]

⇒ [15000(1 + r/100)^2 - 15000]

⇒ 15000 + 300r + (3r^2/2) - 15000

⇒ 300r + (3r^2/2).

Now,

Given that the difference between CI and SI = 96.

⇒ 300r + (3r^2/2) - 300r = 96

⇒ 3r^2/2 = 96

⇒ 3r^2 = 192

⇒ r^2 = 64

⇒ r = 8.

Therefore, rate of interest = 8%.

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Method - 2:

We know that Difference between SI and CI for 2 years:

D = pr^2/100^2

⇒ 96 = (15000 * r^2)/10000

⇒ 10000 * 96 = 15000r^2

⇒ 64 = r^2

⇒ r = 8.

Therefore, rate of interest is 8%.

Note: I don't have pen and paper with me at present.

Hope it helps!

Answer 2

Solution : -

$\bold{ Given, }$

Principal ( P )= ₹ 15000  

Difference between CI and SI = ₹ 96

Time ( T ) = 2 years  

Rate ( R ) = ?  

Let the rate be x \% ,

Difference between CI and SI = ₹ 96  

C.I. – S.I. = 96

$= > P \large{ \: [( \: } \small{1 + \frac{r}{100} }\large{)}^{n} - 1 \large{ ]} - \small{\frac{ P \times R \times T}{ 100 } } = 96$

substituting the values from the question,

$= > 15000 \large{ [(} \small{1+ \frac{r}{100} } \large{)} ^{2} - \small{1}] - \small{\frac{15000 \times r \times 2}{100} } = 96$

$= > \small{ 15000}\large{[ } \small{\: \: ( \cancel{1} + \frac{r}{100} - \cancel{1})(1 + \frac{r}{100} + 1)} \large{]} - \small{ \frac{15000 \times 2r}{100} } = 96$

$= > \small{ 15000} \large{[(} \small{\frac{r}{100}} \large{)( } \small{\frac{r}{100} + 2} \large{)]} - \small{\frac{15000 \times 2 \times r}{100} } = 96$

$= > \small{ 15000} \large{[( }\small{\frac{r}{100} } \large{) } ^{2} \small{ + \frac{2r}{100} } \large{] } - [\small{ 15000 \times \frac{ 2r}{ 100}} ] = 96$

$= > \small{15000} \large{[ ( }\small{\frac{r}{100} + \cancel{ \frac{2r}{100}} - \cancel{ \frac{2r}{100} }} \large{)]} ^{2} = \small{96}$

$= > \small{15000} \large{(} \small{ { \frac{r}{100}} \large{ )}}^{2} = 96 \\ \\ \\ \\ = > 15000 \times \frac{r^{2} }{100 \times 100}= 96$  

$=>\dfrac{3}{2} \times {r}^{2} = 96 \\ \\ \\ \\ = > {r}^{2} = \frac{ \cancel{96} \: \: ^{32} \times 2}{ \cancel{3}}$

$=> {n}^{2} = 32 \times 2 \\ \\ \\ \\ = > {r}^{2} = 64 \\ \\ \\ \\ = > r = \pm \: \sqrt{64} \\ \\ \\ \\ = > r = \pm \: 8$

Rate cann't be negative, therefore r or rate % = 8 %