The difference between the compound interest and the simple interest f

1.	The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is 120. Find the sum
Answer» GIVEN :- CI - SI = RS. 120. Rate ( R ) = 8 %. Time ( n ) = 2 years. TO FIND :- The Sum. SOLUTION :- LET the Principal be "x". $\\ : \implies \displaystyle \sf \: SI = \dfrac{P \times R \times T}{1 00 } \\ \\ \\$ $: \implies \displaystyle \sf \: SI = \frac{x \times 8 \times 2}{100} \\ \\ \\$ $: \implies \underline{\boxed{\displaystyle \sf SI = \frac{16x}{100} }} \\ \\$ Now, $\\ \\ : \implies \displaystyle \sf \: CI = P \Bigg[ \bigg(1 + \dfrac{R}{100}\bigg)^{n} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg(1 + \dfrac{8}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{100 + 8}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{108}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{108 \times 108}{100 \times 100} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664}{10000} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664 - 10000}{10000}\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{1664 }{10000}\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \times \frac{1664}{10000} \\ \\ \\$ $: \implies \underline{ \boxed{ \displaystyle \sf \: CI = \frac{1664x}{10000} }} \\ \\$ Now According to the question, $\\ \\ : \implies \displaystyle \sf \: CI - SI = 120 \\ \\ \\$ $: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{16x}{100} = 120 \\ \\ \\$ $: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{1600x}{10000} = 120 \\ \\ \\$ $: \implies \displaystyle \sf \: \frac{1664x - 1600x}{10000} = 120 \\ \\ \\$ $: \implies \displaystyle \sf \: \frac{64x}{10000} = 120 \\ \\ \\$ $: \implies \displaystyle \sf \:64x = 120 \times 10000 \\ \\ \\$ $: \implies \displaystyle \sf \:64x = 1200000 \\ \\ \\$ $: \implies \displaystyle \sf \:x = \frac{1200000}{64} \\ \\ \\$ $: \implies \underline{ \boxed{ \displaystyle \sf \:x = 18750}} \\ \\$ Hence the REQUIRED sum is Rs. 18750.

The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is 120. Find the sum

Answer»

GIVEN :-

CI - SI = RS. 120.
Rate ( R ) = 8 %.
Time ( n ) = 2 years.

TO FIND :-

The Sum.

SOLUTION :-

LET the Principal be "x".

$\\ : \implies \displaystyle \sf \: SI = \dfrac{P \times R \times T}{1 00 } \\ \\ \\$

$: \implies \displaystyle \sf \: SI = \frac{x \times 8 \times 2}{100} \\ \\ \\$

$: \implies \underline{\boxed{\displaystyle \sf SI = \frac{16x}{100} }} \\ \\$

Now,

$\\ \\ : \implies \displaystyle \sf \: CI = P \Bigg[ \bigg(1 + \dfrac{R}{100}\bigg)^{n} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg(1 + \dfrac{8}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{100 + 8}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{108}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{108 \times 108}{100 \times 100} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664}{10000} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664 - 10000}{10000}\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{1664 }{10000}\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \times \frac{1664}{10000} \\ \\ \\$

$: \implies \underline{ \boxed{ \displaystyle \sf \: CI = \frac{1664x}{10000} }} \\ \\$

Now According to the question,

$\\ \\ : \implies \displaystyle \sf \: CI - SI = 120 \\ \\ \\$

$: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{16x}{100} = 120 \\ \\ \\$

$: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{1600x}{10000} = 120 \\ \\ \\$

$: \implies \displaystyle \sf \: \frac{1664x - 1600x}{10000} = 120 \\ \\ \\$

$: \implies \displaystyle \sf \: \frac{64x}{10000} = 120 \\ \\ \\$

$: \implies \displaystyle \sf \:64x = 120 \times 10000 \\ \\ \\$

$: \implies \displaystyle \sf \:64x = 1200000 \\ \\ \\$

$: \implies \displaystyle \sf \:x = \frac{1200000}{64} \\ \\ \\$

$: \implies \underline{ \boxed{ \displaystyle \sf \:x = 18750}} \\ \\$

Hence the REQUIRED sum is Rs. 18750.

The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is 120. Find the sum

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The difference between the compound interest and the simple interest for 2 years at 8% per annum on a﻿ certain sum of money is 120. Find the sum​

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The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is 120. Find the sum