Question
The difference between compound interest and simple
interest on a sum for 2 years at 10% per annum, when the interest is compounded annually is Rs. 25. If the interest were compounded half yearly, the difference in two interests would be:Solution
S.I – C.I = (P r²)/((100)²) 25 = (P (10)²)/((100)²) P = (25 ×100 ×100)/(10 × 10 ) P = Rs 2,500 S.I = (2500×10 ×2)/(100 ) = Rs 500 C.I = 2,500(1+ 5/100)^4- 2,500 C.I = 2,500 ×105/100×105/100×105/100×105/100 – 2,500 = Rs 538.76 ∴ Required difference = 538.76 – 500 =Rs 38.76 Alternate method: S.I – C.I (for annually basis in %) = r^2/100 % = 10^2/100 % = 1% which is given= RS. 25 But when it is half yearly, r will be 5% ad it will be applicable for 4 years. For 1 year , CI = 5 + 5 + (5×5)/100=10.25% For 2 years , CI = 10.25+10.25 +(10.25×10.25)/100 = 20.5+(41/4×41/4)/100 = (20.5+1681/1600)% & SI for 2 years on half yearly = 5+5+5+5 = 20% So CI – SI = 20.5 + 1681/1600 - 20 = 1681/1600+0.5=2481/1600% So Now 1% = 25 So 2481/1600% = 25×2481/1600% = 38.76 Rs.
I. x2 – 39x + 360 = 0
II. y2 – 36y + 315 = 0
I. 40 x² - 93 x + 54 = 0
II. 30 y² - 61 y + 30 = 0
What will be the product of smaller roots of both equations.
I. 3x2 - 14x + 15 = 0
II. 15y2 - 34 y + 15 = 0
I. x2 – 13x + 36 = 0
II. 3y2 – 29y + 18 = 0
Equation 1: 2x2 - 21x + 54 = 0
Equation 2: 4y2 - 23y + 15 = 0
Difference between the roots of equation 1 is approx...
I. 4x² - 15x + 9 = 0
II. 20y² - 23y + 6 = 0
- For what value of a does the quadratic equation x² + ax + 81 = 0 have real and identical roots?
In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer
I. x² - 8x + 15 = 0 ...
I. 3x2 + 3x - 60 = 0
II. 2y2 - 7y + 5 = 0