Question no 12 . The difference between compound interest and simple interest at 5 percent per annum in 2 years ₹30 find the sum
Answers
Answer:
\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}
∣SOLUTION∣
\begin{gathered} \mathfrak{Suppose,} \\ \: \: \: \: \mathtt{ \: \: \: \: \: \: The \: \: sum \: \: is = P} \\ \\ \underline{ \mathfrak{ \: Given, \: }} \\ \\ \: \: \: \: \mathtt{Rate \: \: of \: \: interest, r = 5 \% \: p.a.} \\ \\ \: \: \: \: \mathtt{No. \: \: of \: \: years, n = 2 \: \: years.} < /p > < p > \end{gathered}
Suppose,
Thesumis=P
Given,
Rateofinterest,r=5%p.a.
No.ofyears,n=2years.</p><p>
\begin{gathered} \underline{ \mathfrak{ \: \: We \: \: know \: \: that,}} \\ \\ \mathtt{S.I. = \frac{P \times r \times n}{100} } \\ \\ \mathtt{ = \frac{P \times 5 \times 2}{100} } \\ \\ \mathtt{ = \frac{P \times 10}{100} } \\ \\ \mathtt{ = \frac{P}{10} }\end{gathered}
Weknowthat,
S.I.=
100
P×r×n
=
100
P×5×2
=
100
P×10
=
10
P
\begin{gathered} \underline{ \mathfrak{ \: \: Again, \: }} \\ \\ \mathtt{C.I. = A - P } \\ \\ \mathtt{ = P(1 + \frac{r}{100}) {}^{n} - P } \\ \\ \mathtt{ = P \:[ (1 + \frac{r}{100}) {}^{n} - 1 ] } \\ \\ \mathtt{ = P \: [(1 + \frac{5}{100} ) {}^{2} - 1] } \\ \\ \mathtt{ = P \times [( \frac{105}{100} ) {}^{2} - 1 ] } \\ \\ \mathtt{ = P \times [( \frac{11025}{10000}) - 1 ]} \\ \\ \mathtt{ = P \times ( \frac{11025 - 10000}{10000} )} \\ \\ \mathtt{ = P \times \frac{1025}{10000} }\end{gathered}
Again,
C.I.=A−P
=P(1+
100
r
)
n
−P
=P[(1+
100
r
)
n
−1]
=P[(1+
100
5
)
2
−1]
=P×[(
100
105
)
2
−1]
=P×[(
10000
11025
)−1]
=P×(
10000
11025−10000
)
=P×
10000
1025
\begin{gathered}\mathfrak{Now, \: } \\ \: \: \: \: \: \: \: \: \underline{ \mathfrak{We \: \: have \: \: given,}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{Difference \: \: between \: \: Compound \: \: } \\ \bold{Interest \: \: and \: \: Simple \: \: Interest \: \: is \: \: Rs. 30.}\end{gathered}
Now,
Wehavegiven,
DifferencebetweenCompound
InterestandSimpleInterestisRs.30.
\begin{gathered} \underline{ \bold{ \: \: A.T.Q., \: \: }} \\ \\ \: \: \: \: \: \mathtt{C.I. - S.I. = 30 } \\ \\ \mathtt{\Rightarrow P \times \frac{1025}{10000} - \frac{P}{10} = 30}\: \\ \\ \mathtt{ \Rightarrow \frac{1025 \times \: P -1000 \times \: P }{10000} = 30 } \\ \\ \mathtt{\Rightarrow \frac{25 \times \: P }{10000} = 30} \\ \\ \mathtt{\Rightarrow 25 \times \: P = 300000} \\ \\ \mathtt{\Rightarrow P = \frac{300000}{25} } \\ \\ \mathtt{ \therefore \: \: P = 12000}\end{gathered}
A.T.Q.,
C.I.−S.I.=30
⇒P×
10000
1025
−
10
P
=30
⇒
10000
1025×P−1000×P
=30
⇒
10000
25×P
=30
⇒25×P=300000
⇒P=
25
300000
∴P=12000
\begin{gathered} \bold{Hence, } \\ \bold{ < /p > < p > \: \: \: \: \: \: \: The \: \: sum \: \: of \: \: money = \underline {\red{Rs. \: 12000}}} < /p > < p > \end{gathered}
Hence,
</p><p>Thesumofmoney=
Rs.12000
</p><p>
\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: VERIFICATION \: \: } \mid}}}}}
∣VERIFICATION∣
\begin{gathered} \underline{ \mathfrak{ \: \: We \: \: have, \: \: }} \\ \\ \mathtt{P = Rs. 12000} \\ \\ \mathtt{S.I. = \frac{ P}{10} } \\ \\ \mathtt{ = \frac{12000}{10} } \\ \\ \mathtt{ = 1200}\end{gathered}
Wehave,
P=Rs.12000
S.I.=
10
P
=
10
12000
=1200
\begin{gathered} \underline{ \mathfrak{ \: \: < /p > < p > Again, \: }} \\ \\ \mathtt{ < /p > < p > < /p > < p > C.I. = P \times \frac{1025}{10000} } \\ \\ \mathtt{ = 12000 \times \frac{1025}{10000} } \\ \\ \mathtt{ = 12 \times \frac{1025}{10} } \\ \\ \mathtt{ = \frac{12300}{10} } \\ \\ \mathtt{ = 1230}\end{gathered}
</p><p>Again,
</p><p></p><p>C.I.=P×
10000
1025
=12000×
10000
1025
=12×
10
1025
=
10
12300
=1230
\begin{gathered} \mathfrak{Now,} \\ \\ \mathtt{C.I. - S.I. =1230 - 1200 = \underline{30} }\end{gathered}
Now,
C.I.−S.I.=1230−1200=
30