Question
Find the rate of simple interest at which Rs. 10000
should be invested for 2 years so that the interest earned will be same as the interest received when the same amount is invested at 20% at compound interest compounded annually for same time period.Solution
Compound interest earned in 2 years = 10000 × [(1.2)^2 – 1] = 0.44 × 10000 = Rs. 4400 Let the rate of simple interest is R% According to question, (10000 × R × 2)/100 = 4400 200 × R = 200 R = 22%
I. 2y2 – 19y + 35 = 0
II. 4x2 – 16x + 15 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 42x + 392 = 0
Equation 2: y² - 46y + 480 = 0
Equation 1: x² - 120x + 3500 = 0
Equation 2: y² - 110y + 3025 = 0
Find the coefficient of x³ in (2x − 3)⁶.
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 22x + 120 = 0
Equation 2: y² - 25y + 144 = 0
Find the value of 'x' and 'y' in the following equation:
7x - 2y = 46
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 ...
l. 3x2 + 17x + 24 = 0
II. 2y2 + 15y + 27 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y.
I. x
I. 22x² - 97x + 105 = 0
II. 35y² - 61y + 24 = 0
