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09 Mar 2016, 12:34
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Set A contains distinct integers: A = {2, 4, 6, -8, x, y}. When two numbers from this set are picked and multiplied, what is the probability that the product is less than zero?
(1) x*y is not equal to zero.
(2) |x| = |y|
(1) x*y is not equal to zero.
(2) |x| = |y|
10 Mar 2016, 06:50
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Bunuel
Hi,
INFO from Q
1) Most imp info that we should write down is they are DISTINCT integers..
2) the integers are 2, 4, 6, -8, x, y
3) 3 positive, 1 negative and 2 unknowns
Inference:-
the prob of product would depend on the signs of x and y and not so much on its NUMERIC value..
lets see the statements:-
(1) x*y is not equal to zero.
we just know none of x and y are 0, but does not help us in any way..
if only one is -ive, number of picking two integers such that their product is <0 will be 4 positive integers* 2 -iv e int=8 ways..
if both ar epositive, it will be 5*1=5..
so different answers
Insuff
(2) |x| = |y|
here we know x and y are having same numeric values..
But DISTINCT comes into play here..
x and y are different number but same numeric value, so one of them is +ive and other -ive..
so out of 6 integers 4 are +ive and 2 are -ive..
Suff
B
12 Mar 2016, 06:54
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Set A contains distinct integers: A = {2, 4, 6, -8, x, y}. When two numbers from this set are picked and multiplied, what is the probability that the product is less than zero?
(1) x*y is not equal to zero.
(2) |x| = |y|
In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), since x and y are distinct and one of them from x and y is negative, the number of positive numbers are determined by the total multiplication. It is unique and sufficient. However, this is an integer and probability question, which is one of the key questions. Apply the mistake type 4(A).
For 1), x>0, y>0 or x>0, y<0, which is not unique and not sufficient.
For 2), x=y or x=-y, which is distinct. So, only x=-y is possible and it determines the number of negative numbers, which is unique and sufficient.
Therefore, the answer is B not C.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
Set A contains distinct integers: A = {2, 4, 6, -8, x, y}. When two numbers from this set are picked and multiplied, what is the probability that the product is less than zero?
(1) x*y is not equal to zero.
(2) |x| = |y|
In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), since x and y are distinct and one of them from x and y is negative, the number of positive numbers are determined by the total multiplication. It is unique and sufficient. However, this is an integer and probability question, which is one of the key questions. Apply the mistake type 4(A).
For 1), x>0, y>0 or x>0, y<0, which is not unique and not sufficient.
For 2), x=y or x=-y, which is distinct. So, only x=-y is possible and it determines the number of negative numbers, which is unique and sufficient.
Therefore, the answer is B not C.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
