Question
If '38k721' is divisible by 3, then what is the highest
possible value of 'k'?Solution
ATQ,
A number is divisible by 3 when the sum of its digits is divisible by 3.
Sum of digits of '38k721' = 3 + 8 + k + 7 + 2 + 1 = (k + 21)
(k + 21) is divisible by 3 when k = 0, 3, 6 or 9.
Therefore, maximum value of 'k' = 9
I. x2 - 11x + 24 = 0
II. y² - 5y + 6 = 0
I. 25p + 2(2p2 – 1) = 8(p + 5)
II. 8q2 + 35q – 78 = 0
I. 3q² -29q +18 = 0
II. 9p² - 4 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 41x + 400 = 0
Equation 2: y² - 41y + 420 = 0
I. 20x² - 93x + 108 = 0
II.72y² - 47y - 144 = 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...
If the roots of the quadratic equation 5x² + 4x + 6 = 0 are α and β, then what is the value of [(1/α) + (1/β)]?
...I. 15b2 + 26b + 8 = 0
II. 20a2 + 7a - 6 = 0
I. x2 – 10x + 21 = 0
II. y2 + 11y + 28 = 0
I). p2 + 22p + 72 = 0,
II). q2 - 24q + 128 = 0
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