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If the set S consists of all non-negative integers less than 7, what

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If the set S consists of all non-negative integers less than 7, what  [#permalink]

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If the set S consists of all non-negative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality \(3 > \frac{x^2}{3}\)?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 1

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If the set S consists of all non-negative integers less than 7, what  [#permalink]

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New post Updated on: 23 Oct 2018, 06:15
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Bunuel wrote:
If the set S consists of all non-negative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 1

Kudos for a correct solution.


Set, S = {0, 1, 2, 3, 4, 5, 6} [All Non-Negative Integers]

\(3 > \frac{x^2}{3}\)
i.e. \(x^2 < 3*3\)
i.e. \(x^2 < 9\)

Which is true for the values of x = 0, 1 and 2 out of 7 values available in Set S

i.e. Probability = Favourable Outcome / Total Outcome = 3/7

Answer: Option C
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Originally posted by GMATinsight on 08 Jul 2015, 03:08.
Last edited by GMATinsight on 23 Oct 2018, 06:15, edited 1 time in total.
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Re: If the set S consists of all non-negative integers less than 7, what  [#permalink]

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New post 08 Jul 2015, 03:44
1
Bunuel wrote:
If the set S consists of all non-negative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 1

Kudos for a correct solution.


Non negative numbers less than 7 will be = {0,1,2,3,4,5,6}

Given inequality : x^2<9 --> (x+3)(x-3)<0 ---> -3<x<3. Thus x can assume 0,1,2 as the values.

Thus the probability = 3/7, C is the correct answer.
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Re: If the set S consists of all non-negative integers less than 7, what  [#permalink]

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New post 08 Jul 2015, 05:56
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Numbers satisfying the criteria are 0, 1, and 2, so 3/7. (C)
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Re: If the set S consists of all non-negative integers less than 7, what  [#permalink]

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New post 08 Jul 2015, 21:01
1
Bunuel wrote:
If the set S consists of all non-negative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 1

Kudos for a correct solution.


Set S is set of NON-NEGATIVE INTEGER less than 7 , S = {0,1,2,3,4,5,6,7}
Total Number of element in set S = 7

We need X such that 3 > x^2/3
or 9 > x^2

Only three number from set S satisfy this condition , those are 0,1,2

Hence Required Probability = Number of Element satisfying condition / Total number of element = 3/7
Hence Option C should be correct.
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Re: If the set S consists of all non-negative integers less than 7, what  [#permalink]

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New post 13 Jul 2015, 01:27
Bunuel wrote:
If the set S consists of all non-negative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 1

Kudos for a correct solution.


800score Official Solution:

We are told that S = {0, 1, 2, 3, 4, 5, 6}, that is, all integers from 0 to 6. There are 7 integers in the set. Remember that zero is an integer that is neither negative nor positive.

Now, the expression can be simplified by multiplying both sides by 3. We can do this since the 3 in the denominator is a positive number.

Remember: When you multiply or divide both sides of an inequality by a negative number, you must change the direction of the inequality.

The expression is simplified to 9 > x^2.

The only numbers in S that can be squared and satisfy this condition are 0, 1, and 2. Therefore, 3 of the 7 integers in the set will satisfy the condition, and the desired probability is 3/7.

The correct answer is choice (C).
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If the set S consists of all non-negative integers less than 7, what  [#permalink]

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New post 13 Oct 2017, 08:28
Bunuel wrote:
If the set S consists of all non-negative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 1


We see that set S = {0, 1, 2, 3, 4, 5, 6}.

Let’s test each number:

3 > 0^2/3 = 0? yes

3 > 1^2/3 = 1/3? yes

3 > 2^2/3 = 4/3? yes

3 > 3^2/3 = 9/3 = 3? no

Since 3 of 7 numbers (namely, 0, 1, and 2) satisfy the inequality, the probability of getting these 3 numbers is 3/7.

Answer: C
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Re: If the set S consists of all non-negative integers less than 7, what  [#permalink]

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