Question
An amount becomes six times of itself in 50 years at
simple interest. Find the value of (2R + 30).Solution
Let the amount invested be Rs. 'p'.
Interest earned = 6p − p = Rs. 5p
Simple interest = Principal × Rate × Time ÷ 100
So, 5p = p × R × 50 ÷ 100
Or, R = (5 × 100) ÷ 50 = 10
Required value = 2 × 10 + 30 = 20 + 30 = 50
I. 2p² - 11p + 12 = 0
II. 2q² - 17q + 36 = 0
- Determine the remainder when equation 4p³- 5p² + 2p + 1 is divided by (4p - 3).
I. 99x² + 161 x + 26 = 0
II. 26 y² + 161 y + 99 = 0
I. x2 - 4x – 21 = 0
II. y2 + 12y + 20 = 0
I. 3q² -29q +18 = 0
II. 9p² - 4 = 0
I. 8x2 - 2x – 15 = 0
II. 12y2 - 17y – 40 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between 'p' and 'q' and choose...
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 29x + 210 = 0
Equation 2: y² - 27y + 182 = 0
I. 15y2 + 4y – 4 = 0
II. 15x2 + x – 6 = 0
I. p2 +7p + 10 = 0 II. q2 - q – 6 = 0
