Question
Mohit is a teacher and he teach some students in two
(Morning & evening) slots. There are 5 boys and ‘x’ girls in morning slot, while there are ‘x’ boys and 8 girls in evening slot. If one student is chosen at random from any slot, then the probability of being girl is 5/9. Which of the following can be the value of ‘x’? I. 10 II. 4 III. 3 IV. 5 V. 6Solution
Probability for being girl from morning slot = (1/2) × (x/(x+5)) Probability for being girl from evening slot = (1/2) × (8/(x+8)) ATQ – 1/2 [x/(x+5)+(8/(x+8))]=5/9 10/9=(x(x+8)+8(x+5))/(x+8)(x+5)  10/9=(x2+8x+8x+40)/(x2+13x+40) 10x2+130x+400=9x2+144x+360 x2–14x+40=0 x2–10x–4x+40=0  x(x –10) –4(x –10)=0  x=10,4  So, both I and II follow
I. Â 3y2Â + 13y - 16 = 0
II. 3x2 – 13x + 14 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 31x² - 170x + 216 = 0
Equation 2: 22y² - 132y + ...
I. 20y² - 13y + 2 = 0
II. 6x² - 25x + 14 = 0
I. 2y2Â + 11y + 15 = 0
II. 3x2Â + 4x - 4= 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...
I. 5x² = 19x – 12
II. 5y² + 11y = 12
I. x2 – 13x + 40 = 0
II. 2y2 – 15y + 13 = 0Â
I. x² + 3x – 154 = 0
II. y² + 5y – 126 = 0
I. x² + 11x + 24 = 0
II. y² + 17y + 72 = 0
I. 4p² + 17p + 15 = 0
II. 3q² + 19q + 28 = 0
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