Question

MATHS MATHS MATHS


here is. a question for maths geniuses....

Answer 1

Hi frnds.........

principle=Rs50000

rate=8%

time=3years

for first year:

P=Rs50000

R=8%

T=1 year

SI=PxRxT/100

=50000x8x1/100

=Rs4000

Amount=SI+P

=Rs(4000+50000)

=Rs54000

for second year:

P=Rs54000

R=8%

T=1 year

SI=PxRxT/100

=54000x8x1/100

=Rs4360

Amount=SI+P

=Rs(4360+54000)

=Rs58360

for third year:

P=Rs58360

R=8%

T=1 year

SI=PxRxT/100

=58360x8x1/100

=Rs4268.8

Hope it helps ....... :D

Answer 2

Solution:

Given

Principal (p) = Rs50000

Rate of interest (r) = 8% p.a

Let the amount at the end of the second year = A

$\boxed{Amount = p\times(1+\frac{r}{100})^{n}}$

i)

Let the number of times interest paid in 2 years ( n)=2

A = 50000[1+8/100]²

= 50000 × (108/100)²

= 5×108×108

= Rs58320

Therefore,

The amount standing to her

credit at the end of the second

year (A) = Rs58320

ii ) Finding interest for third year:

P = Rs 58320,

R = 8%

T = 1 year

Simple Interest (I) = (PTR)/100

= ( 58320 × 1 × 8 )/100

= Rs 4665.60

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