A number is to be selected at random from the set above. What is the

Solutions of the equation \((x-5)(x+10)(2x-5)=0\) are: \(x=5\), \(x=-10\) and \(x=2.5\).

There are total of 12 numbers in the set {-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10} and it contains 2 roots: -10 and 2,5, so the probablitiy that the number selected will be a solution of the given equation is \(P=\frac{2}{12}=\frac{1}{6}\).

Answer: B.
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it contains 2 roots: -10 and 2.5 ??

Roots = Solution to the equation

We know the only solutions are -10,2.5,5
And of these, two are present in the set at hand
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Aaaaaa OMG, I actually didn't notice that there isn't number 5(positive) in this bunch of numbers. I assumed it, and I got it wrong!

My advice to everyone: Don't assume anything!!!

Hi All,

We're asked for the probability of choosing a number (from a given set of 12 numbers) that will "fit" a given equation. That equation is set equal to 0, which makes solving the equation rather easy. Since we're dealing with the PRODUCT of three "terms", if ANY of those terms equals 0, then the product will equal 0.

IF... we select X = -10 from the set, then we have (-15)(0)(-25) = 0, so X = -10 IS a solution.
IF... we select X = 0 from the set, then we have (-5)(10)(-5) = 250, which is NOT a solution.

The given equation has 3 solutions: +5, -10 and +5/2. However, only two of those solutions appear in the set (-10 and 2.5). Thus, the probability of selecting a number that fits the equation is 2/12 = 1/6

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi pushpitkc

can you please explain what i did wrong

i did this \(x^2+10-5x-50(2x-5)\) now left with it

\(x^2+10x-5x-100x+250 = 0\)

\(x^2+10x-5x-100x+250 = 0\)

\(x^2 -95x+250 = 0\) now what

thank you

Hi dave13

You have 3 values of x possible and we need to check how many of these values are part
of the set. I don't seem to understand why are you trying to form a quadratic equation

Btw, as for your question, the highlighted portion is wrong

\((x-5)(x+10)(2x-5) = 0\) -> \((x^2 - 5x +10x - 50)(2x - 5) = 0\) -> \((x^2 + 5x - 50)(2x - 5) = 0\)
The entire equation will need to be multiplied and we will have a cubic equation

Hope this clears your confusion!

pushpitkc thanks can you please show how do we find 2 roots: -10 and 2,5, i am kind of confused what to do with highlighted part
\((x^2 + 5x - 50)(2x - 5) = 0\)

Count there are 12 items
x = 5 or x = -10 or x = 2.5
Two are in the set
B.

Hey dave13

If 2x - 5 = 0 -> 2x = 5 -> x = 5/2 = 2.5
If x - 5 = 0 -> x = 5
If x + 10 = 0 -> x = -10

Therefore, the roots of the quadratic equation are -10, 5, and 2.5

According to the given equation, x can equal the following:

x - 5 = 0

x = 5

or

x + 10 = 0

x = -10

or

2x - 5 = 0

2x = 5

x = 2.5

Since -10 and 2.5 are in the set, the probability of selecting one of those solutions is 2/12 = 1/6.

Answer: B
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Clearly,
There are two solutions = -10,2.5.
So P(X) = 2/12 = 1/6

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