Let m be an integer such that 1 ≤ m ≤ 1000. The probability of selecting at random an integer m such that the quadratic equation 6 x2-5 m x+m2=0 has at least one integral solutions, is

Question

Let m be an integer such that 1 ≤ m ≤ 1000. The probability of selecting at random an integer m such that the quadratic equation 6 x2-5 m x+m2=0 has at least one integral solutions, is (A) (333/1000) (B) (500/1000) (C) (833/1000) (D) (667/1000)

Tardigrade Admin · Accepted Answer

The quadratic equation has two solutions

x = \frac{m}{2}; x = \frac{m}{3} (using x = \frac{- b \pm b ^{2} - 4 a c}{2 a})

There are 500 multiples of 2, 333 multiples of 3 , and 166 multiples of 6 between 1 and 1000 . Therefore, the probability is

p = \frac{500 + 333 - 166}{1000} = \frac{667}{1000}

Q. Let $m$ be an integer such that $1 \leq m \leq 1000$ . The probability of selecting at random an integer $m$ such that the quadratic equation $6 x^{2} - 5 m x + m^{2} = 0$ has at least one integral solutions, is

$\frac{333}{1000}$

$\frac{500}{1000}$

$\frac{833}{1000}$

$\frac{667}{1000}$

Q. Let m be an integer such that 1≤m≤1000. The probability of selecting at random an integer m such that the quadratic equation 6x2−5mx+m2=0 has at least one integral solutions, is

1000333​

1000500​

1000833​

1000667​

The quadratic equation has two solutions x=2m​;x=3m​ (using x=2a−b±b2−4ac​​ ) There are 500 multiples of 2, 333 multiples of 3 , and 166 multiples of 6 between 1 and 1000 . Therefore, the probability is p=1000500+333−166​=1000667​

Q. Let $m$ be an integer such that $1 \leq m \leq 1000$ . The probability of selecting at random an integer $m$ such that the quadratic equation $6 x^{2} - 5 m x + m^{2} = 0$ has at least one integral solutions, is

$\frac{333}{1000}$

$\frac{500}{1000}$

$\frac{833}{1000}$

$\frac{667}{1000}$