A number is to be selected at random....

the given polynomial would yield 0 for x = -5,-10,2.5
so required prob = 3/12 = 1/4

IMO C

total elements in the set = 12
total solution of the given equation (x+5)(x+10)(2x-5) = 0
-5,-10,2.5

all these exist in the set

probability = relevant cases/total cases = 3/12 = 1/4 (option 3)

Guys, this is wrong. The solutions are: +5, -10 and +2.5. However, from the list we only see 2 numbers: -10 and +2.5, therefore the probability is 2/12 which is equal to 1/6 (B)

indiantiger wrote:total elements in the set = 12
total solution of the given equation (x+5)(x+10)(2x-5) = 0
-5,-10,2.5

all these exist in the set

probability = relevant cases/total cases = 3/12 = 1/4 (option 3)

either one should be set equal to 0
x=-5, x=-10, x=2.5
three possibilities out of twelve => 1/4
c

this q. lacks symmetry in set, +1 is missing

The Jock wrote:{-10,-6,-5,-4,-2.5,-1,0,2.5,4,6,7,10}
A number is to be selected at the random from the set above.what is the probability that the number selected will be a solution of the equation (x+5)(x+10)(2x-5) = 0?
1. 1/12
2. 1/6
3. 1/4
4. 1/3
5. 1/2

Solutions to the equation are -5,-10,5/2

We have these solutions in the set, which has 12 elements.

Probability of picking any of the 3 solutions out of 12 is 3/12 = 1/4

He copied the equation wrong from the sample test. It should say


{-10,-6,-5,-4,-2.5,-1,0,2.5,4,6,7,10}
A number is to be selected at the random from the set above.what is the probability that the number selected will be a solution of the equation (x-5)(x+10)(2x-5) = 0?
1. 1/12
2. 1/6
3. 1/4
4. 1/3
5. 1/2



That is why the answer is B 1/6 and not C 1/4.

1/12+1/12+1/12=1/4

Answer Choice is (C)