Question
A sum of 74000 is divided into two parts such that the
simple interest on the first part for 3 years at 15% per annum is equal to the simple interest on the second part for 5.5 years at 12% per annum, the greater part is?Solution
Let the sum invested for 3 years (first part) be P.
As per given condition:
P * 3 * 15/100 = (74000 - P) * 5.5 * 12/100
=> 15P = (74000 - P) * 22
=> 37P = 74000 * 22
=> P = 44000
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
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