Question

a certain amount of money amounts to rupees 5500 in 2 years and 2 Rupees 5750 in 3 years find the principal and rate of interest​

Answer 1

Answer:

let, the principal is rs. p

and rate of profit is r

for the 1st case,

amount (a)= 5500

time (n) = 2 yrs

we know,

$p = \frac{a}{1 + nr} = \frac{5500}{1 +2r }$

for the 2nd case,

amount (a)= 5750

time (n)= 3 yrs

we know,

$p = \frac{a}{1 + nr} = \frac{5750}{1 + 3r}$

so,

$\frac{5500}{1 + 2r} = \frac{5750}{1 + 3r} \\ > 5500(1 + 3r) = 5750(1 + 2r) \\ > 22(1 + 3r) = 23(1 + 2r) \\ > 22 + 66r = 23 + 46r \\ > 20r = 1 \\ > r = \frac{1}{20} = \frac{1}{20} \times 100\% = 5\%$

putting r=1/20 in the 1st equation,

$p = \frac{5750}{1 + 3 \times \frac{1}{20} } = \frac{5750}{1 + 0.15} = \frac{5750}{1.15} = 5000$

so,

r=5%

p=5000

Answer 2

Step-by-step explanation:

Answer:

let, the principal is rs. p

and rate of profit is r

for the 1st case,

amount (a)= 5500

time (n) = 2 yrs

we know,

p = \frac{a}{1 + nr} = \frac{5500}{1 +2r }p=

1+nr

a

=

1+2r

5500

for the 2nd case,

amount (a)= 5750

time (n)= 3 yrs

we know,

p = \frac{a}{1 + nr} = \frac{5750}{1 + 3r}p=

1+nr

a

=

1+3r

5750

so,

\begin{gathered}\frac{5500}{1 + 2r} = \frac{5750}{1 + 3r} \\ > 5500(1 + 3r) = 5750(1 + 2r) \\ > 22(1 + 3r) = 23(1 + 2r) \\ > 22 + 66r = 23 + 46r \\ > 20r = 1 \\ > r = \frac{1}{20} = \frac{1}{20} \times 100\% = 5\%\end{gathered}

1+2r

5500

=

1+3r

5750

>5500(1+3r)=5750(1+2r)

>22(1+3r)=23(1+2r)

>22+66r=23+46r

>20r=1

>r=

20

1

=

20

1

×100%=5%

putting r=1/20 in the 1st equation,

p = \frac{5750}{1 + 3 \times \frac{1}{20} } = \frac{5750}{1 + 0.15} = \frac{5750}{1.15} = 5000p=

1+3×

20

1

5750

=

1+0.15

5750

=

1.15

5750

=5000

so,

r=5%

p=5000