Quadratic equation | ISI-B.stat | Objective Problem 240

Try this beautiful problem based on Quadratic equation, useful for ISI B.Stat Entrance.

Quadratic equation | ISI B.Stat Entrance | Problem 240

The equations \(x^2 + x + a = 0\) and \(x^2 + ax + 1 = 0\)

  • (a) cannot have a common real root for any value of a
  • (b) have a common real root for exactly one value of a
  • (c) have a common root for exactly two values of a
  • (d) have a common root for exactly three values of a.

Key Concepts

Algebra

Quadratic equation

Roots

Check the Answer

Answer: (b)

TOMATO, Problem 240

Challenges and Thrills in Pre College Mathematics

Try with Hints

Let the equations have a common root \(α\).Therefore \(α\) must satisfy two given equations.......

Therefore,

Now, \(α^2 + α + a = 0\)...................(1)
And, \(α^2 + aα + 1 = 0\).......................(2)

Can you find out the value of \(a\)?

Can you now finish the problem ..........

Therefore,

Using cross-multiplication betwwen (1) & (2) we will get.......

\(\frac{α^2}{(1 – a^2)} =\frac{ α}{(a – 1)} = \frac{1}{(a – 1)}\)
\(\Rightarrow {α}^2 = \frac{(1 – a^2)}{(a – 1) }=- (a + 1)\) & \(α=\frac{(a-1)}{(a-1)}=1\)

Now \({α}^2=(α)^2\)
\(\Rightarrow -(a+1)=1\)
\(\Rightarrow a = -2\)

Therefore (b) is the correct answer....

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Problem on Balls | ISI-B.stat | Objective Problem 128

Try this beautiful problem on balls based on Number theory, useful for ISI B.Stat Entrance.

Problem on Balls | ISI B.Stat Entrance | Problem 128

A bag contains colored balls of which at least 90% are red. Balls are drawn from the bag one by one and their color noted. It is found that 49 of the first 50 balls drawn are red. Thereafter 7 out of every 8 balls are red. The number of balls in the bag can not be

  • (a) \(170\)
  • (b) \(210\)
  • (c) \(250 \)
  • (d) \(194\)

Key Concepts

Number theory

Percentage

Inequility

Check the Answer

Answer: (b) \(210\)

TOMATO, Problem 128

Challenges and Thrills in Pre College Mathematics

Try with Hints

Let the number of balls in the bag is n.Let, m number of times 8 balls are drawn. Therefore, \(n = 50 + 8m\). Red balls =\( 49 + 7m\)

Can you now finish the problem ..........

Percentage of red balls = \(\frac{(49 + 7m)}{(50 + 8m)}≥ \frac{90}{100}\)

\(\Rightarrow \frac{(49 + 7m)}{(50 + 8m)}≥0.9\)

\(\Rightarrow 49 + 7m ≥ 45 + 7.2m\)

\(\Rightarrow 0.2m ≤ 4\)

\(\Rightarrow m ≤ 20\)

\(\Rightarrow n ≤ 50 + 8 \times 20 = 210\)

Therefore we can say that at most \(210\) balls can be drawn. So there must be at most \(210\) balls in the bag, and so there cannot be \(250\) balls in the bag (because \(250>210\)).

Therefore option (C) is the correct answer

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Math Olympiad

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