Question

Jharana got Rs 1881 interest of certain sum for 2 years at 9% compounded yearly.Find the sum​

Answer 1

Answer:

Let the sum of money be

′

x

′

.

T=2 years

C.I=2700. , Rate=R

A=P(1+

100

R

)

2

⇒

(x+2700)=x(1+

100

R

)

2

⟶(I)

Now,

S.I=2500

⇒2500=

100

x×R×2

⇒

xR=125000

⇒

R=

x

125000

→(II)

Putting

′

x

′

in eq (I)

(

R

125000

+2700)=

R

125000

(1+

100

R

)

2

⇒(

R

125000+2700R

)=

R

125000

(1+

100

R

)

2

⇒(125000+2700R)=125000(1+

100

R

)

2

⇒1250+27R=1250(1+

100

R

)

2

⇒1250+27R=1250(1+2(0.0R)+(0.0R)

2

)

⇒1250+27R=1250+25R+0.125R

2

⇒0.125R

2

−2R=0

⇒R(0.125R−2)=0

⇒R=0 or 0.125R−2=0

But R can't be 0.

∴0.125R−2=0

⇒R=

0.125

2

=16

∴

R=16

%

Answer 2

Answer:

${1}^{st} \: year \: \\ \: \: \: \: \: \: \: \: \: \: principal = 1881 \\ rate \: \: of \: interest \: = \: 9\%p.a. \: \\ time \: taken \: = \: 1yr \\ \\ interest = \frac{1881 \times 9 \times 1}{100} \\ \: \: = 169.29 \\ final \: amount \: = \: 1881 + 169.29 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2050.29 \\ \\ {2}^{nd} year \: \\ principal \: = 2050.29 \\ rate \: of \: interest \: = 9\%p.a. \\ time \: taken \: = 1yr \\ interest \: = \frac{2050.29 \times 9 \times 1}{100}\\ = 184.5261 \\ final \: amount \: = 2050.29 + 184.5261 \\ = 2234.8161 \\ compound interest = 2234.8161 - 1881 = 353.8161 \\ \\ hope \: this \: helped \: uhh \: \\ \\ \\ \\ \\ thank \: uhh$