Question

a sum of money amounts to Rs. 17680 when invested for 3years at a certain rate of interest per annum and amounts to Rs.17600 when invested for 2years. at the rate pf 5% per annum simple interest. find the rate of interest in the first investment and the sum invested in both the cases.​

Answer 1

Answer:

Step-by-step explanation:

di avm aj malaya kunywa mterest MINOUSTA TE HALL, 1930 in 2 genea mid it».3.250 in 3 year

Anal Taterest 450-150es

atat e Seppay 2011-12-20

Rate of Simple Faterest-1150 2500100-0precof

Answer 2

Given :-

• A sum of money amounts to Rs. 17680 when invested for 3years at a certain rate of interest per annum and amounts to Rs.17600 when invested for 2years. at the rate pf 5% per annum simple interest

To Find :-

• Rate of interest in the first investment and the sum invested in both cases.

Solution :-

We know that

A = P(1 + R/100)ⁿ

But here, we may write it as

A = P(1 + TR/100)

Now

①

[tex]\sf A=P\bigg\{1+\dfrac{TR}{100}\bigg\}A=P{1+100TR}[tex]

\sf 17680=P\bigg\{1+\dfrac{3\times R}{100}\bigg\}17680=P{1+1003×R}

\sf 17680=P\bigg\{1+\dfrac{3R}{100}\bigg\}17680=P{1+1003R}

\sf 17680=P\bigg\{\dfrac{100+3R}{100}\bigg\}17680=P{100100+3R}[tex]

②

\sf A=P\bigg\{1+\dfrac{TR}{100}\bigg\}A=P{1+100TR}

\sf 17600=P\bigg\{1+\dfrac{2\times5}{100}\bigg\}17600=P{1+1002×5}

\sf 17600=P\bigg\{1+\dfrac{10}{100}\bigg\}17600=P{1+10010}

\sf 17600=P+\bigg\{1+\dfrac{1}{10}\bigg\}17600=P+{1+101}

\sf 17600=P\bigg\{\dfrac{10+1}{10}\bigg\}17600=P{1010+1}

\sf 17600=P\bigg\{\dfrac{11}{10}\bigg\}17600=P{1011}

\sf 17600\times10=P\times 1117600×10=P×11

\sf \dfrac{17600\times10}{11}=P1117600×10=P

\sf 1600\times 10=P1600×10=P

\sf 16000=P16000=P

Here

P(Case ①) = P(Case ②)

16000 = 16000

R a t E :-

\sf 17680=P\bigg\{\dfrac{100+3R}{100}\bigg\} \bigg\lgroup From \; 1\bigg\rgroup17680=P{100100+3R}⎩⎪⎪⎪⎧From1⎭⎪⎪⎪⎫

\sf 17680=16000\bigg\{\dfrac{100+3R}{100}\bigg\}17680=16000{100100+3R}

\sf \dfrac{17680}{16000}=\dfrac{100+3R}{100}1600017680=100100+3R

\sf 1.105=\dfrac{100+3R}{100}1.105=100100+3R

\sf 1.105\times100=100+3R1.105×100=100+3R

\sf 110.5=100+3R110.5=100+3R

\sf 110.5-100=3R110.5−100=3R

\sf 10.5=3R10.5=3R

\sf\dfrac{10.5}{3}=R310.5=R

\sf\dfrac{105}{30}=R30105=R

\sf\dfrac{7}{2}=R27=R

\sf 3.5=R3.5=R