Question

Ramesh borrowed from suresh certain sum for 2 years at simple interest. He lent this sum to dinesh at the same rate for 2 years compound interest. At the end of 2 years Ramesh received rupees 110 as compound interest from Dinesh but paid rupees 100 as simple interest to Suresh. Find the sum and the rate of interest.

Answer 1

Sum = Rs 250  & Rate of interest = 20%

Step-by-step explanation:

Let say sum = P

Rate of interest = R

Time = 2 Years

Simple interest = P * R  * 2/100   = 100

=> PR = 5000

=> P = 5000/R

Compound interest = P * (1 + R/100)²  - P  = 110

=> P (  (1 + R/100)² - 1) = 110

=> (5000/R)  (  1 + R²/10000 + 2R/100 - 1) = 110

=> 500 ( R²/10000 + 2R/100) = 11 R

=>  R²/20  + 10R = 11R

=>  R²/20 = R

=> R = 20

P = 5000/R

=> P = 5000/20 = 250

Sum = Rs 250

Rate of interest = 20%

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Answer 2

Answer:

Sum = ₹250

Rate of interest = 20%pa

Step By Step Explaination:

S.I. = ₹100

Time given = 2yrs

therefore, $\frac{PRT}{100}$ = 100

=> PR x $\frac{2}{100\\}$ = 100

=> PR = $\frac{100 x 100}{2}$    =>100x50    =>5000

∴ P = $\frac{5000}{R\\}$

and, C.I. = ₹110

Time given = 2yrs

∴C.I. = P [(1 + $\frac{R}{100} ^{2}$) - 1 ]

Taking LCM

=> 110 =  $\frac{5000}{R}$ [ $(\frac{100 + R}{100} )^{2}$ - 1]

Using formula for-

$(a+b)^{2}$ = $a^{2}$ + 2ab + $b^{2}$

=> 110 = $\frac{5000}{R}$ [($\frac{100^{2} +2 * R * 100 + R^{2} }{100^{2} }$) -1]

=> 110 = $\frac{5000}{R}$ [($\frac{10000 + 200R + R^{2} }{10000}$)-1]

Taking LCM for 1

=> 110 = $\frac{5000}{R}$  [$\frac{10000 + 200R + R^{2} - 10000 }{10000}$]

10000 -10000 = 0

=> 110 = $\frac{5000}{R}$ X $\frac{200R + R^{2} }{10000}$

Taking R common

=> 110 = $\frac{5000}{R}$ X R x $\frac{200 +R}{10000}$

=> 110 = $\frac{200 + R }{2}$

Taking 2 to the other side-

=> 110 x 2 = 200 + R

=> 220 = 200 + R

Taking 200 to the other side

=> 220 - 200 = R

= 20 = R

∴ R = 20%

Now,

Principal  = $\frac{5000}{R}$

=> $\frac{5000}{20}$

=> 250

therefore, P = ₹250

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thank you :)