Question

A sum of money lent at CI for 2 yr at 20% pa would fetch Rs. 964 more, if the interest was payable half yearly than if it was payable annually. What is the sum?

Note- Only quality and explained Answer needed :)​

Answer 1

Answer:

Step-by-step explanation:

Your Question :-

A sum of money lent at CI for 2 yr at 20% pa would fetch Rs. 964 more, if the interest was payable half yearly than if it was payable annually. What is the sum?

Given :-

• A sum of money lent at CI for 2 yr at 20% pa would fetch Rs.

• If the interest was payable half yearly than if it was payable annually.

To find out :-

• The sum.

Solution :-

Case I:- When CI compounded half-yearly

Rate of interest = 10%

Frequency of interest occuring in 2 yrs = 4

Applying the net% effect formula of 1st yr, (1st two frequencies)

$\rm Net\% \: effect = 10 + 10 + \dfrac{10 \times 10}{100} \\ \\= 10 + 10 + \dfrac{ \cancel{10 \times 10}}{ \cancel{100}}^{ \bf \: 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = 10 + 10 + 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \bf{21\% }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

For next half years a years,

$= 21 + 10 + \dfrac{21 \times 10}{100} \\ \\ = 21 + 10 + \frac{21 \times \cancel{10}}{10 \cancel{0}} \\ \\ = 21 + 10 + \frac{21}{10} \: \: \: \: \: \: \: \: \: \: \\ \\ = 31 + \frac{21}{10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \frac{31}{1} + \frac{21}{10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \frac{310 + 21}{10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \frac{331}{10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \boxed{ \bf33.1\%} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

For next half a years,

$= 33.1 + 10 + \dfrac{33.1 \times 10}{100} \\ \\ = 33.1 + 10 + \frac{33.1 \times \cancel{10}}{10 \cancel{0}} \\ \\ = 43.1 + \frac{33.1}{10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \frac{431}{10} + \frac{331}{10 \times 10} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \frac{431}{10} + \frac{331}{100} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \frac{4310 + 331}{100} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \frac{4641}{100} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \boxed{ \bf46.41\%} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Case II:-When CI compounded yearly

Rate of interest = 20%

Frequency of interest occuring in 2 yrs = 2

Applying the net% effect formula of 2 yrs,

We get,

$= 20 + 20 + \dfrac{20 \times 20}{100} \\ \\ = 20 + 20 + \frac{400}{100} \: \: \: \: \: \: \: \\ \\ = 20 + 20 + \frac{ \cancel{400}}{ \cancel{100}} ^{ \bf \: 4 } \: \: \: \\ \\ = 20 + 20 + 4 \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \boxed{ \bf44\%} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now, difference in % rate

= 46.41% – 44% =2.41%

According to question,

➡️ 2.41% ≡ 964

➡️ 100% ≡ x

$\rm\therefore x = \dfrac{964 \times 100}{2.41} \\ \\ \implies \rm x = \frac{964100}{2.41} \: \: \: \: \: \\ \\ \implies \rm x = \boxed{ \bf40000} \: \: \: \:$

Hence, The Sum is 40,000.

Answer 2

Answer:

The sum of money is Rs. 40000.

Step-by-step-explanation:

We have given that, for compound interest,

• Rate of interest ( R ) = 20 %
• Time period ( N ) = 2 years

We have to find the sum of money.

Let the sum be Rs. x.

Now, for interest compounded half yearly,

• Rate of interest ( R ) = 10 %
• Time period ( N ) = 4 years

Now, we know that,

$\displaystyle{\pink{\sf\:Compound\:interest\:=\:Amount\:-\:Principle}\sf\:\quad\:-\:-\:[\:Formula\:]}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:A\:-\:P}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:P\:\left(\:1\:+\:\dfrac{R}{100}\:\right)^N\:-\:P\:\quad\:-\:-\:-\:[\:Formula\:]}$

$\displaystyle{\implies\sf\:C\::\:I.\:=\:x\:\left(\:1\:+\:\dfrac{1\cancel{0}}{10\cancel{0}}\:\right)^4\:-\:x}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:1\:+\:\dfrac{1}{10}\:\right)^4\:-\:x}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:\dfrac{10\:+\:1}{10}\:\right)^4\:-\:x}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:\dfrac{11}{10}\:\right)^4\:-\:x}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:\dfrac{(\:11\:)^4}{(\:10\:)^4}\:-\:1\:\right)\:\quad\:-\:-\:-\:\left[\:\because\:\left(\:\dfrac{a}{b}\:\right)^m\:=\:\dfrac{a^m}{b^m}\:\right]}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:\dfrac{11\:\times\:11\:\times\:11\:\times\:11}{10\:\times\:10\:\times\:10\:\times\:10}\:-\:1\:\right)}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:\dfrac{14641}{10000}\:-\:1\:\right)}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:\dfrac{14641\:-\:10000}{10000}\:\right)}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\times\:\dfrac{4641}{10000}}$

$\displaystyle{\implies\boxed{\red{\sf\:C\:.\:I.\:=\:\dfrac{4641\:x}{10000}}}\sf\:\quad\:-\:-\:-\:(\:1\:)}$

Now, for interest compounded annually,

• Rate of interest ( R ) = 20 %
• Time period ( N ) = 2 years

$\displaystyle{\pink{\sf\:Compound\:interest\:=\:Amount\:-\:Principle}\sf\:\quad\:-\:-\:-\: [\:Formula\:]}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:A\:-\:P}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:P\:\left(\:1\:+\:\dfrac{R}{100}\:\right)^N\:-\:P}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:1\:+\:\cancel{\dfrac{20}{100}}\:\right)^2\:-\:x}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:1\:+\:\dfrac{1}{5}\:\right)^2\:-\:x}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:\dfrac{5\:+\:1}{5}\:\right)^2\:-\:x}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:\dfrac{6\:\times\:6}{5\:\times\:5}\:-\:1\:\right)}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:\dfrac{36}{25}\:-\:1\:\right)}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\left(\:\dfrac{36\:-\:25}{25}\:\right)}$

$\displaystyle{\implies\sf\:C\:.\:I.\:=\:x\:\times\:\dfrac{11}{25}}$

$\displaystyle{\implies\boxed{\red{\sf\:C\:.\:I.\:=\:\dfrac{11\:x}{25}}}\sf\:\quad\:-\:-\:-\:(\:2\:)}$

Now, from the given condition,

Interest compounded half yearly - Interest compounded annually = Rs. 964

$\displaystyle{\therefore\sf\:\dfrac{4641\:x}{10000}\:-\:\dfrac{11\:x}{25}\:=\:964\:\quad\:-\:-\:-\:[\:From\:(\:1\:)\:\&\:(\:2\:)\:]}$

$\displaystyle{\implies\sf\:\dfrac{4641\:x\:}{10000}\:-\:\dfrac{11\:x\:\times\:400}{25\:\times\:400}\:=\:964}$

$\displaystyle{\implies\sf\:\dfrac{4641\:x}{10000}\:-\:\dfrac{4400\:x}{10000}\:=\:964}$

$\displaystyle{\implies\sf\:\dfrac{4641\:x\:-\:4400\:x}{10000}\:=\:964}$

$\displaystyle{\implies\sf\:\dfrac{241\:x}{10000}\:=\:964}$

$\displaystyle{\implies\sf\:241\:x\:=\:964\:\times\:10000}$

$\displaystyle{\implies\sf\:241\:x\:=\:9640000}$

$\displaystyle{\implies\sf\:x\:=\:\cancel{\dfrac{9640000}{241}}}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:x\:=\:Rs.\:40000}}}}$

∴ The sum of amount is Rs. 40000.