Question
A borrows Rs 7000 from B at 10% p.a compound interest
compounded annually. At the end of every year he pays Rs 2200 and at the end of 3rd years he pays all the amount, how much amount does he pay at the end of the 3rd year?Solution
At the end of the 1st year: Amount = 7000 × 110% = Rs. 7700 Rest amount = 7700 – 2200 = Rs. 5500 At the end of 2nd year: Amount = 5500 × 110% = Rs. 6050 Rest amount = 6050 – 2200 = Rs. 3850 At the end of 3rd year: Amount = 3850 × 110% = Rs. 4235
If x² + 2x + 9 = (x – 2) (x – 3), then the resultant equation is:
What will be the product of smaller roots of both equations.Â
I. 40 x² - 93 x + 54 = 0
II. 30 y² - 61 y + 30 = 0
I. 27x6 - 152x3 + 125 = 0
II. 216y6 - 91y3 + 8 = 0
I. x2 – 9x + 18 = 0
II. y2 – 5y + 6 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 3x² + 6x - 9 = 0
Equation 2: 2y² - 16y + 32 = 0
I. 117x² + 250x + 117 = 0
II. 54y² -123y + 65 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 19x² - 92x + 73 = 0
Equation 2: 17y² - 76y + 59 = 0
I. x2 – 13x + 40 = 0
II. 2y2 – 15y + 13 = 0Â
In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer Â
I. x2 – ...
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