Question

Raghu a farmer borrowed a total of Rs.50000 from two of his friends, Sham and Ram.
He agreed to pay Sham an interest of 28%.p.a and to Ram an interest of 24%.p.a. both
at S.I. Raghu paid a total interest of Rs.12560 at the end of the year. How much did he
borrow from Ram?
(1) Rs. 28000 (2) Rs. 36000 (3) Rs.24000
Rs.30000

Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• Raghu borrowed 50000 Rs from Ram and Sham. Therefore principal is 50000 Rs.
• Raghu agreed to:

          ▣ Pay sham 28% rate of interest.                                                                                                                                                                                                        

          ▣ Pay ram 24% rate of interest.

• Raghu paid S.I. 12560 Rs. at the end of the year. Therefore, time taken was 1 year.

SUMMARY:

• Principle (P) = 50000 Rs.

• Rate (R):

          ▣ Pay sham 28% rate of interest.                                                                                                                                                                                                        

          ▣ Pay ram 24% rate of interest.

• Time (t) = 1 year.
• S.I. = 12560 rs.

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• How much did raghu borrow from ram?

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

Let raghu borrow x Rs. from Sham. Then, he will borrow (50000 - x) Rs. from ram.

Now, we can deduce that:

$\sf Total \ amount=Amount \ of \ ram+ Amount \ of \ sham.$

SI OF RAM:

• We have assumed that P = x Rs.
• We are given that rate is 28%
• We know that time is 1 year.

$\mapsto \sf S.I.=\dfrac{PRT}{100}$

$\sf \# Substitute.$

$\mapsto \sf S.I.=\dfrac{x \times 28 \times 1}{100}$

$\leadsto \sf S.I.=\dfrac{7x}{25}$

SI OF SHAM:

• We have assumed that P = 50000 - x Rs.
• We are given that rate is 24%
• We know that time is 1 year.

$\mapsto \sf S.I.=\dfrac{PRT}{100}$

$\sf \# Substitute.$

$\mapsto \sf S.I.=\dfrac{(50000-x) \times 24 \times 1}{100}$

$\leadsto \sf S.I.=\dfrac{6(50000-x)}{25}$

Then,

$\sf Total \ S.I=SI \ of \ ram+ SI \ of \ sham.$

$\to \sf 12560=\dfrac{7x}{25} +\dfrac{6(50000-x)}{25}$

$\to \sf 12560=\dfrac{7x+6(50000-x)}{25}$

$\to \sf 12560 \times 25=7x+6(50000-x)$

$\to \sf 314000=300000+x$

$\to \sf{\red{x=14000 \ Rs.}}$

Ram's amount is (50000-x)

⇒ 50000-14000

⇒ 36000

OPTION (2) IS CORRECT.

Regards,

Sujal Sirimilla

Ex-brainly star.