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    Question

    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’.
    A 7500 Correct Answer Incorrect Answer
    B 9000 Correct Answer Incorrect Answer
    C 6000 Correct Answer Incorrect Answer
    D 5500 Correct Answer Incorrect Answer
    E None of the above Correct Answer Incorrect Answer

    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

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