Question
A principal of Rs. 7,500 earns Rs. 1,125 as simple
interest in 3 years. What will be the simple interest on Rs. 9,000 for 5 years at the same rate?Solution
Let the rate of interest be 'R%' p.a. ATQ, 7500 × (R/100) × 3 = 1125 ⇒ R = 1125 ÷ 225 = 5 Now, required interest = 9000 × (5/100) × 5 = 9000 × 0.25 = Rs. 2,250
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
Relevant for Exams:
