Rs. (y-1500) was invested in scheme J on (R-1)% per annum on compound interest. Rs. ‘y’ was invested in scheme K on (R+1)% per annum on simple interest. After four years the interest obtained from scheme K is Rs. 2308 more than the interest obtained from scheme J after three years. If the value of ‘R’ is the product of two prime numbers and each of those prime numbers is a single digit number and the difference between each of these prime numbers is four, then find out the value of ‘y’.

Solution

If the value of ‘R’ is the product of two prime numbers and each of those prime numbers is a single digit number and the difference between each of these prime numbers is four. We know that single digit prime numbers are 2, 3, 5 and 7. Now the difference between each of these prime numbers is four. So these prime numbers will be 7 and 3. So the value of ‘R’ = 7x3 = 21 Rs. (y-1500) was invested in scheme J on (R-1)% per annum on compound interest. Rs. ‘y’ was invested in scheme K on (R+1)% per annum on simple interest. After four years the interest obtained from scheme K is Rs. 2308 more than the interest obtained from scheme J after three years. y x (R+1)% x 4 = [(y-1500) x (100+R-1)% x (100+R-1)% x (100+R-1)% - (y-1500)] + 2308 Put the value of ‘R’ in the above equation. y x (21+1)% x 4 = [(y-1500) x (100+21-1)% x (100+21-1)% x (100+21-1)% - (y-1500)] + 2308 y x 22% x 4 = [(y-1500) x 120% x 120% x 120% - (y-1500)] + 2308 0.88y = [(y-1500) x 1.2 x 1.2 x 1.2 - (y-1500)] + 2308 0.88y = [1.72800(y-1500) - (y-1500)] + 2308 0.88y = (y-1500)x[1.728 - 1] + 2308 0.88y = (y-1500)x0.728 + 2308 0.88y = 0.728y - 1092 + 2308 0.88y-0.728y = -1092+2308 0.152y = 1216 Value of ‘y’ = 8000