Question

A person borrowed Rs2500 from two money lenders.For one loan he paid 8% per annum simple interest and for the other he paid 6% per annum simple interest.If he paid Rs180 at total interest for 1 year.How much did he borrow at 8% ?<br />1000<br />1200<br />1300<br />1500​

Answer 1

AnswEr :

⋆ First Part :

• Principal = Rs. n
• Rate = 8% p.a.
• Time = 1 Years

⋆ Second Part :

• Principal = Rs. (2500 - n)
• Rate = 6% p.a.
• Time = 1 Years

• According to the Question Now :

⇒ SI₁ + SI₂ = 180

• we know the formula for SI

⇒ prt₁ /100 + prt₂ /100 = 180

• plugging the values

⇒ (n × 8 × 1) + ((2500 - n) × 6 × 1) /100 = 180

⇒ 8n + 15000 - 6n = 180 × 100

⇒ 2n + 15000 = 18000

⇒ 2n = 18000 - 15000

⇒ 2n = 3000

⇒ n = 3000 /2

⇒ n = Rs. 1500

∴ He Borrowed Rs. 1500 at 8% p.a. SI

Answer 2

Answer:

Step-by-step explanation:

Given :-

Money borrowed by person = Rs 2500

Simple interest = 8 % and 6 %

Total interest paid in 1 year = Rs. 180

Solution :-

Let the money lent at 8% be Rs.x.

According to the question,

$\sf\implies \dfrac{x\times8\times1}{100}+\dfrac{(2500-x)\times6\times1}{100}=180$

$\sf\implies 8x + 15000 - 6x = 18000$

$\sf\implies 2x = 3000$

$\sf\implies x=\dfrac{3000}{2}$

$\bf\implies x=1500$

Hence, He borrow Rs. 1500 at 8%