Math, asked by suryasathwik11, 2 months ago

B.
On what date will * 1,950 lent on 5th Januar
2011 amount to 2,125-50 at 5 per cent
annum simple interest ?​

Answers

Answered by omsahu9422
0

Step-by-step explanation:

Date = 5th Jan, 2011

P = 1950

A = 2125.50

R = 5

I = A-p = 175.5

I=\frac{P T R}{100}I=

100

PTR

175.5=1950 \times T \times \frac{5}{100}175.5=1950×T×

100

5

T=\frac{351}{195}T=

195

351

T=\frac{351}{195} \times 365T=

195

351

×365

T=73 \times 9=657 DaysT=73×9=657Days

657 Days = 1 year 292

5th Jan, 2011 to 1 year = 5th jan, 2012

292- near to one year. Less than 73 days. Calculate from December.

Dec = 31, Nov = 30, Oct = 12 Days which means 19th Oct , 2012 we will get interest.

P_{1}=2000P

1

=2000

T_{1}=3\ yearsT

1

=3 years

R_{1}=rR

1

=r

I_{1}=\frac{P_{1} T_{1} R_{1}}{100}I

1

=

100

P

1

T

1

R

1

=\frac{(2000 \times 3 \times r)}{100}=

100

(2000×3×r)

P_{2}=2400P

2

=2400

T_{2}=3\ yearsT

2

=3 years

R_{2}=rR

2

=r

I_{2}=\frac{P_{2} T_{2} R_{2}}{100}I

2

=

100

P

2

T

2

R

2

=2400 \times 3 \times \frac{r}{100}=2400×3×

100

r

And as given I_{2}=I_{1}+60I

2

=I

1

+60

\frac{I_{1}}{I_{2}}=\frac{2000 \times 3 \times r}{2400 \times 3 \times r}

I

2

I

1

=

2400×3×r

2000×3×r

\frac{I_{1}}{I_{1}+60}=\frac{5}{6}

I

1

+60

I

1

=

6

5

6 I_{1}=5\left(I_{1}+60\right)6I

1

=5(I

1

+60)

I_{1}=300I

1

=300

Substitute I_{1}I

1

in the above

300=\frac{2000 \times 3 \times R}{100} \Rightarrow R=5300=

100

2000×3×R

⇒R=5

Rate percent is 5 per annum.

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