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31 Oct 2014, 02:43
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If [x] denotes the greatest integer less than or equal to x for any number x, is [a]+[b]=1?
(1) ab=2.
(2) 0<a<b<2.
Source: GCT M28-56
(1) ab=2.
(2) 0<a<b<2.
Source: GCT M28-56
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31 Oct 2014, 03:45
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Nice question!
[a]+[b]=1 can happen if one value is negative and one positive OR if one value is 0 and the other is 1.
A - ab = 2
if both a and b are negative - then [a]+[b]=1 is not possible
If a and b are positive - can it result in one value being 0 (i.e. 0<a<1) and the other 0 (1<b<2)
This is impossible as if 1<b<2 then it has to be multiplied by a number larger than 1 to give 2.
But if a is larger than one then [a]+[b]=2.
So if ab=2 then it cannot happen that [a]+[b]=1 and A is SUFFICIENT
B - 0<a<b<2
If 0<a<1 and 1<b<2 then [a]+[b]=1 is true
if 0<a<1 and a<b<1 then [a]+[b]=1 is not true
So we cannot conclude and B is NOT SUFFICIENT
[a]+[b]=1 can happen if one value is negative and one positive OR if one value is 0 and the other is 1.
A - ab = 2
if both a and b are negative - then [a]+[b]=1 is not possible
If a and b are positive - can it result in one value being 0 (i.e. 0<a<1) and the other 0 (1<b<2)
This is impossible as if 1<b<2 then it has to be multiplied by a number larger than 1 to give 2.
But if a is larger than one then [a]+[b]=2.
So if ab=2 then it cannot happen that [a]+[b]=1 and A is SUFFICIENT
B - 0<a<b<2
If 0<a<1 and 1<b<2 then [a]+[b]=1 is true
if 0<a<1 and a<b<1 then [a]+[b]=1 is not true
So we cannot conclude and B is NOT SUFFICIENT
31 Oct 2014, 03:54
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If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?
Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...
(1) ab = 2. First of all this means that a and b are of the same sign.
If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.
If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.
We have that the answer to the question is NO. Sufficient.
(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.
Answer: A.
THIS QUESTION IS DISCUSSED HERE: new-set-number-properties-149775-60.html#p1205389
Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...
(1) ab = 2. First of all this means that a and b are of the same sign.
If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.
If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.
We have that the answer to the question is NO. Sufficient.
(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.
Answer: A.
THIS QUESTION IS DISCUSSED HERE: new-set-number-properties-149775-60.html#p1205389
