Answer:
Rs. 7000
Step-by-step explanation:
n=3 years,r=10% and C.I.=Rs.2317
∴2317=P[(1+10010)3−1]
∵C.I.=P[(1+100r)n−1]
⇒2317=P(1.331−1)
⇒P=0.3312317=7000
∴ Required sum =Rs.7000
If you understand so comment me .
SUM = RS 2003
GIVEN :
RATE = 10 %
TIME = 3 YEARS
COMPOUND INTEREST = RS 663
TO PROVE :
FIND THE SUM
PROOF :
A =
[tex]p {(1 + R / 100 )}^{t} [/tex]
CI = A - P
CI =
[tex]p {( 1 + R / 100 )}^{t} - p[/tex]
663 = P
[tex]ci = p( {( 1 + 10/ 100 )}^{3} - 1) \\ \\ 663 = p( {(1 + 0.1)}^{3} - 1) \\ \\ p = \frac{663}{ {1.1}^{3} - 1} = \frac{66 3}{1.331 - 1} = \frac{663}{0.331} = 2003.[/tex]
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Answers & Comments
Answer:
Rs. 7000
Step-by-step explanation:
n=3 years,r=10% and C.I.=Rs.2317
∴2317=P[(1+10010)3−1]
∵C.I.=P[(1+100r)n−1]
⇒2317=P(1.331−1)
⇒P=0.3312317=7000
∴ Required sum =Rs.7000
If you understand so comment me .
Answer:
SUM = RS 2003
Step-by-step explanation:
GIVEN :
RATE = 10 %
TIME = 3 YEARS
COMPOUND INTEREST = RS 663
TO PROVE :
FIND THE SUM
PROOF :
A =
[tex]p {(1 + R / 100 )}^{t} [/tex]
CI = A - P
CI =
[tex]p {( 1 + R / 100 )}^{t} - p[/tex]
663 = P
[tex]ci = p( {( 1 + 10/ 100 )}^{3} - 1) \\ \\ 663 = p( {(1 + 0.1)}^{3} - 1) \\ \\ p = \frac{663}{ {1.1}^{3} - 1} = \frac{66 3}{1.331 - 1} = \frac{663}{0.331} = 2003.[/tex]