Question
A and B together have total of Rs.6000 out of which they
donated 20% to the orphanage school. The remaining amount is to be then redistributed between them in such a manner that A gets 50% more amount than B. If the amount received by A is Rs.2X, then find the value of [(X/12) + 10].Solution
Let the amount received by B be Rs.P. Therefore, amount received by A = 1.50 x P = Rs.1.5P According to the question, => (P + 1.5P) = 6000 x 0.80 => 2.5P = 4800 => P = 1920 Share of A = 4800 – 1920 = Rs.2880 So, 2X = 2880 => X = 1440 Required value = [(X/12) + 10] = 120 + 10 = 130
In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer
I. 7x² - 19x + 10 = 0...
I. p² - 10p +21 = 0
II. q² + q -12 = 0
I. y/16 = 4/y
II. x3 = (2 ÷ 50) × (2500 ÷ 50) × 42 × (192 ÷ 12)
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. x= √(20+ √(20+ √(20+ √(20…………….∞)) ) )
II. y= √(5√(5√(5√(5……….∞)) ) )
...I. 7p + 8q = 80
II. 9p – 5q = 57
I. 2y2 – 19y + 35 = 0
II. 4x2 – 16x + 15 = 0
If x + 1/x = 4, find x⁵ + 1/x⁵.
I. 27(p + 2) = 2p(24 – p)
II. 2q2 – 25q + 78 = 0
I. x² + 11x + 24 = 0
II. y² + 17y + 72 = 0
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