class 9 science physics chapter motion and force and laws of motion ka notes
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4
Answer:
Answer:-
Given:-
Principal (P) = Rs. 3,10,000
Rate of interest (R) = 10 %
Time period (T) = 3 years.
We know that;
\begin{gathered} \boxed{\sf \: Amount(A) = P \bigg(1 + \frac{R}{100} \bigg) ^{T}} \\ \end{gathered}
Amount(A)=P(1+
100
R
)
T
Hence,
\begin{gathered} \implies \sf \: A = 310000 \bigg( 1 + \frac{10}{100} \bigg) ^{3} \\ \\ \\ \implies \sf \: A = 310000 \bigg( \frac{100 + 10}{100} \bigg) ^{3} \\ \\ \\ \implies \sf \: A = 310000 \times \bigg( \frac{110}{100} \bigg) ^{3} \\ \\ \\ \implies \sf \: A = 310000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \\ \\ \\ \implies \boxed{\sf \: A = Rs. \: 4,12,610} \\ \end{gathered}
⟹A=310000(1+
100
10
)
3
⟹A=310000(
100
100+10
)
3
⟹A=310000×(
100
110
)
3
⟹A=310000×
10
11
×
10
11
×
10
11
⟹
A=Rs.4,12,610
∴ Neha receives Rs. 4,12,610 at the end of 3 years.
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