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# Tutor profile: Ashfak S.

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Ashfak S.
Developer in Goldman Sachs and a math tutor
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## Questions

### Subject:GMAT

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Question:

If two distinct integers a and b are picked from {1, 2, 3, 4....100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

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Ashfak S.

Any positive integer will have '1' and the number itself as factors. That makes it a minimum of 2 factors (except '1' which has only one factor). If the positive integer has only one more factor, then in addition to 1 and the number, the square root of the number should be the only other factor. There are two key points in the above finding. The number has to be a perfect square. And the only factor other than 1 and the number itself should be its square root. Therefore, if a positive integer has only 3 factors, then it should be a perfect square and it should be the square of a prime number. Let us look at an example. 4 has the following factors: 1, 2, and 4 (exactly 3 factors). It is the square of '2' which a prime number. Let us look at one more example. 25 has the following factors: 1, 5, and 25 (exactly 3 factors). It is the square of '5' which is a prime number. Squares of numbers that are not prime numbers will have more than 3 factors. For instance, 36 is a perfect square. But it has 9 factors. So, the first step to solving this question is to count the number of squares of prime numbers from 1 to 100 4, 9, 25, and 49 are the 4 numbers that have exactly 3 factors. The second step is to compute the number of ways of selecting two distinct integers from the set of first 100 positive integers. Two distinct integers can be selected from the set of 100 positive integers in 100C2 ways. This value is the denominator of the required probability. Now, we need to compute the numerator to arrive at the required probability. We have to compute the number of outcomes in which the product of two numbers will result in a number that has 3 factors Or, we have to compute the number of outcomes in which the product of the two numbers will be 4 or 9 or 25 or 49. The product of two numbers 'a' and 'b' will be 4 when one of the numbers is 1 and the other is 4. There is only one set that will result in this product. The same holds good for the other 3 numbers as well. Product of 'a' and 'b' will be 9 when one of the numbers is 1 and the other is 9 and so on. Therefore, there are 4 outcomes in which the product of the two numbers will result in a number that has exactly 3 factors. Hence, the required probability = favorable outcomes / total outcomes. i.e., 4 / 100C2 Or the required probability = 2 / 25*99

### Subject:Statistics

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Question:

Data Sufficiency : If m, s are the average and standard deviation of integers a, b, c, and d, is s > 0? Statement 1 : m > a Statement 2 : a + b + c + d = 0

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Ashfak S.

Statement 1 : If a = b = c = d, the average m will be the same as a. Since m > a, all the elements in the set cannot be the same, and therefore, s > 0. hence statement 1 is sufficient. Statement 2 : Approach: Look for a counter example Example: When a = b = c = d = 0, s = 0 Counter Example: When a = -4, b = 0, c = 0, and d = 4, s > 0 Statement 2 ALONE is NOT sufficient.

### Subject:Algebra

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Question:

the equation 24^2 + 25x -47 /ax-2 = -8x-3- 53/ax-2 is true for all values of x ≠ 2/a, where a is a constant. what is the value of a?

Inactive
Ashfak S.

There are two ways to solve this question. The faster way is to multiply each side of the given equation by ax−2 (so you can get rid of the fraction). When you multiply each side by ax−2, you should have: 24x2+25x−47=(−8x−3)(ax−2)−53 You should then multiply (−8x−3) and (ax−2) using FOIL. 24x2+25x−47=−8ax2−3ax+16x+6−53 Then, reduce on the right side of the equation 24x2+25x−47=−8ax2−3ax+16x−47 Since the coefficients of the x2-term have to be equal on both sides of the equation, −8a=24, or a=−3. The other option which is longer and more tedious is to attempt to plug in all of the answer choices for a and see which answer choice makes both sides of the equation equal. Again, this is the longer option

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