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What is the absolute difference between the cubes of two [#permalink]
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15 Mar 2012, 23:23
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What is the absolute difference between the cubes of two different nonnegative integers? (1) One of the integers is 2 greater than the other integer. (2) The square of the sum of the integers is 49 greater than the product of the integers.
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Re: DS What is the absolute difference between the cubes of two [#permalink]
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15 Mar 2012, 23:31
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Hi, I got the answer as A. Let the 2 nos be x & y We need to find x^3  y^3 Statement 1 : x = 2+y substitute in quest stem => (2+y)^3  y^3 from the above you can find y & hence x Statement 2 : (x+y)^2  xy > 49 In statement 2 we cant eliminate x or y hence Statement 2 is insufficient Thus answer is A Hope this helps
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Re: DS What is the absolute difference between the cubes of two [#permalink]
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16 Mar 2012, 04:15
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boomtangboy wrote: Hi,
I got the answer as A.
Let the 2 nos be x & y
We need to find x^3  y^3
Statement 1 : x = 2+y
substitute in quest stem => (2+y)^3  y^3 from the above you can find y & hence x
Statement 2 : (x+y)^2  xy > 49 In statement 2 we cant eliminate x or y hence Statement 2 is insufficient
Thus answer is A
Hope this helps Statement 2 : (x+y)^2  xy= 49 on simplification we get \(x^2+y^2+xy=49\). Not sufficient. But when we combine 1&2 \(x^3  y^3=(xy)(x^2+y^2+xy)\) =(2)(49) so we can answer by 1&2 together. So the answer should be C. 



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What is the absolute difference between the cubes of two different nonnegative integers?Given: \(x\) and \(y\) are different nonnegative integers. Question: \(x^3y^3=?\) (1) One of the integers is 2 greater than the other integer > \(xy=2\). There exist infinitely many such \(x\) and \(y\), thus we'll have infinitely many different values of \(x^3y^3\). For example consider \(x=2\), \(y=0\) and \(x=3\), \(y=1\). Not sufficient. (2) The square of the sum of the integers is 49 greater than the product of the integers > \((x+y)^2=xy+49\) > \(x^2+xy+y^2=49\). Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are nonnegative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient. Notice that if we were told that \(x\) and \(y\) are positive integers (instead of nonnegative integers) then this statement would be sufficient, since only one pair would remain: (5,3).(1)+(2) Since from (1) \(xy=2\) then from (2) only one pair is valid: (5,3). Sufficient. Answer: C. Hope it's clear.
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Re: What is the absolute difference between the cubes of two [#permalink]
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16 Mar 2012, 05:04
IMO the correct answer is C first the requirement is to find the value of x^3  y^3 or (xy)( x^2 + y^2 + xy) 1) x = y + 2 or x y = 2, Not sufficient 2) (x+y)^2 = 49 + xy, as we simplify we get x^2 + y^2 + xy = 49 insufficient because we don't know the value of x  y from this statement combine 1 and 2, we get the value of (xy) and ( x^2 + y^2 + xy) hence sufficient.
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Re: What is the absolute difference between the cubes of two [#permalink]
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16 Mar 2012, 05:08
Hey Bunnel, thanx for a nice explanation . I have learned something new today with ur explanation. Every time i read ur posts, it further reimposes my faith that ur a true legend. Thanx a ton.
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Re: What is the absolute difference between the cubes of two [#permalink]
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16 Mar 2012, 18:30
Thanks a ton Bunuel. Your explanations are really helpful. Thanks a lot again!



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Re: What is the absolute difference between the cubes of two [#permalink]
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16 Mar 2012, 20:56
Ya thanks Bunuel.... I forgot the expansion of (a^3  b^3)
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cubes of x and y [#permalink]
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04 Apr 2012, 12:05
Could you please help me to understand where is the flaw in my reasoning: I chose the answer D for this problem, since the first statement gives us the relationship between two variables : if the second item is two more than the first one, doesn't it mean x=x+2, where x+2 =y
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Re: cubes of x and y [#permalink]
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Re: What is the absolute difference between the cubes of two [#permalink]
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04 Apr 2012, 12:21
But, Bunuel, why can't i set x=x+2, where x+2=y? Since the first statement tells us "one of the integers is 2 greater than the other". The only thing confused me  "ONE of the integers", what means i don't know the exact order May be my question is stupid one but i just need to grasp it fully



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Re: What is the absolute difference between the cubes of two [#permalink]
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04 Apr 2012, 12:35
Galiya wrote: But, Bunuel, why can't i set x=x+2, where x+2=y? Since the first statement tells us "one of the integers is 2 greater than the other". The only thing confused me  "ONE of the integers", what means i don't know the exact order May be my question is stupid one but i just need to grasp it fully First of all x=x+2 doesn't make any sense, because x's cancel out and we have that 0=2, which is not true. Next, "one of the integers is 2 greater than the other integer" CAN be expressed as x+2=y, (assuming that y is larger), but even in this case we cannot get the single numerical value of y^3x^3, for example consider: y=2, x=0 and y=3, x=1. Hope it's clear.
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Re: What is the absolute difference between the cubes of two [#permalink]
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05 Apr 2012, 04:09
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IMO C..
nice work bunnel.....
But are we expected to know 5,3 is the pair which solves given condition if integers are only positive?



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Re: What is the absolute difference between the cubes of two [#permalink]
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09 Jul 2014, 02:57
Bunuel wrote: vdbhamare wrote: IMO C..
nice work bunnel.....
But are we expected to know 5,3 is the pair which solves given condition if integers are only positive? You can get that the statements taken together are sufficient even if you don't know that the only pair possible is (5, 3): \(x^3y^3=(xy) (x^2+x y+y^2)\) > (1) says that \(xy=2\) and (2) says that \(x^2+x y+y^2=49\) > \(x^3y^3=(xy) (x^2+x y+y^2)=2*49=98\). Hope it's clear. It was difficult to see that \(x^3y^3\) can we written as \((xy) (x^2+x y+y^2)\) is there an easier way to understand this or must it be learnt by heart.



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Re: What is the absolute difference between the cubes of two [#permalink]
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09 Jul 2014, 03:17
qlx wrote: Bunuel wrote: vdbhamare wrote: IMO C..
nice work bunnel.....
But are we expected to know 5,3 is the pair which solves given condition if integers are only positive? You can get that the statements taken together are sufficient even if you don't know that the only pair possible is (5, 3): \(x^3y^3=(xy) (x^2+x y+y^2)\) > (1) says that \(xy=2\) and (2) says that \(x^2+x y+y^2=49\) > \(x^3y^3=(xy) (x^2+x y+y^2)=2*49=98\). Hope it's clear. It was difficult to see that \(x^3y^3\) can we written as \((xy) (x^2+x y+y^2)\) is there an easier way to understand this or must it be learnt by heart. It's good to memorize that formula. But even if you don't know it, you can still the correct answer as shown here: whatistheabsolutedifferencebetweenthecubesoftwo129157.html#p1059501
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Re: What is the absolute difference between the cubes of two [#permalink]
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11 Jul 2014, 06:05
Bunuel wrote: What is the absolute difference between the cubes of two different nonnegative integers?
Given: \(x\) and \(y\) are different nonnegative integers. Question: \(x^3y^3=?\)
(1) One of the integers is 2 greater than the other integer > \(xy=2\). There exist infinitely many such \(x\) and \(y\), thus we'll have infinitely many different values of \(x^3y^3\). For example consider \(x=2\), \(y=0\) and \(x=3\), \(y=1\). Not sufficient.
(2) The square of the sum of the integers is 49 greater than the product of the integers > \((x+y)^2=xy+49\) > \(x^2+xy+y^2=49\). Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are nonnegative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.
Notice that if we were told that \(x\) and \(y\) are positive integers (instead of nonnegative integers) then this statement would be sufficient, since only one pair would remain: (5,3).
(1)+(2) Since from (1) \(xy=2\) then from (2) only one pair is valid: (5,3). Sufficient.
Answer: C.
Hope it's clear. Hello Bunuel . After solving stem 2 and coming to the point of \(x^2+xy+y^2=49\) , I deduced that since it is an equation in 2 variables it will not be solvable. I substituted from 1 ( y=x+2) and then realized that it transforms to a simple quadratic and then with the mentioned restriction , came to C. Would this be a right approach ?



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Re: What is the absolute difference between the cubes of two [#permalink]
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11 Jul 2014, 11:38
himanshujovi wrote: Bunuel wrote: What is the absolute difference between the cubes of two different nonnegative integers?
Given: \(x\) and \(y\) are different nonnegative integers. Question: \(x^3y^3=?\)
(1) One of the integers is 2 greater than the other integer > \(xy=2\). There exist infinitely many such \(x\) and \(y\), thus we'll have infinitely many different values of \(x^3y^3\). For example consider \(x=2\), \(y=0\) and \(x=3\), \(y=1\). Not sufficient.
(2) The square of the sum of the integers is 49 greater than the product of the integers > \((x+y)^2=xy+49\) > \(x^2+xy+y^2=49\). Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are nonnegative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.
Notice that if we were told that \(x\) and \(y\) are positive integers (instead of nonnegative integers) then this statement would be sufficient, since only one pair would remain: (5,3).
(1)+(2) Since from (1) \(xy=2\) then from (2) only one pair is valid: (5,3). Sufficient.
Answer: C.
Hope it's clear. Hello Bunuel . After solving stem 2 and coming to the point of \(x^2+xy+y^2=49\) , I deduced that since it is an equation in 2 variables it will not be solvable. I substituted from 1 ( y=x+2) and then realized that it transforms to a simple quadratic and then with the mentioned restriction , came to C. Would this be a right approach ? Have you read this: Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are nonnegative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient. Notice that if we were told that \(x\) and \(y\) are positive integers (instead of nonnegative integers) then this statement would be sufficient, since only one pair would remain: (5,3).
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