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If w and c are integers is w > 0 ? (1) w + c > 50 (2)

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If w and c are integers is w > 0 ? (1) w + c > 50 (2)  [#permalink]

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New post 16 Dec 2010, 05:28
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If w and c are integers is w > 0 ?

(1) w + c > 50
(2) c > 48

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Re: If w and c are integers is w > 0 ? (1) w + c > 50 (2)  [#permalink]

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New post 16 Dec 2010, 05:53
ajit257 wrote:
If w and c are integers is w > 0 ?

(1) w + c > 50
(2) c > 48

I have a doubt the ans.



Statement (1) is not SUFFICIENT

w + C > 50
Pick the numbers w=3 & C= 50 where W>0
Now if i pick W=-3 & C=60 , the statement still holds good but in this case W<0
Hence we cannot determine if W<0 or if W>0

Statement (2) is not SUFFICIENT
as it does not reflect anything about the value of W.

Even if we take these two statements together, we cannot determine if W>0 or not.
(i)Pick number since C>48 therefore pick C=49
which mean W+49>50 therefore W has to be greater than or equal to 2

(ii) If C = 52 then W could be W=0, W>0 or W=-1 hence insufficient.

As per me the answer should be E. Please do let me know your thoughts.

Cheers!!
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Re: If w and c are integers is w > 0 ? (1) w + c > 50 (2)  [#permalink]

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New post 16 Dec 2010, 06:12
2
ajit257 wrote:
If w and c are integers is w > 0 ?

(1) w + c > 50
(2) c > 48

I have a doubt the ans.


Obviously each statement alone is not sufficient.

When taken together we can still have both Yes and No answers: if c=100 and w=10 then the answer is Yes but if c=100 and w=-10 then the answer is No.

Answer: E.

I feel that it should be w+c<50 for (1). Then as \(c\) is an integer then the least value of it according to (2) will be 49, so max value of \(w\) in order \(c+w\) to be less than 50 is 0 (as \(w\) is also an integer), which means that the answer to the question is \(w>0\) is No. So, if it were \(w + c < 50\) the answer would be C and also the question would be much more interesting.

Hope it's clear.
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Re: If w and c are integers is w > 0 ? (1) w + c > 50 (2)  [#permalink]

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New post 17 Dec 2010, 05:32
Bunuel wrote:
ajit257 wrote:
If w and c are integers is w > 0 ?

(1) w + c > 50
(2) c > 48

I have a doubt the ans.


Obviously each statement alone is not sufficient.

When taken together we can still have both Yes and No answers: if c=100 and w=10 then the answer is Yes but if c=100 and w=-10 then the answer is No.

Answer: E.

I feel that it should be w+c<50 for (1). Then as \(c\) is an integer then the least value of it according to (2) will be 49, so max value of \(w\) in order \(c+w\) to be less than 50 is 0 (as \(w\) is also an integer), which means that the answer to the question is \(w>0\) is No. So if it were \(w + c < 50\) the answer would be C and also the question would be much more interesting.

Hope it's clear.


Bunuel, taking the qtn as you specified i.e. w+c<50 Please clarify why am i getting w < 2
1) \(w+c<50\)
2) \(c > 48\)

now 2) can be written as \(-c < -48\)
as both the inequalities have the same sign, 1) + 2)
==> \(w < 50-48\)
==> \(w< 2\)
why am i not getting \(w <= 0\) for the answer to be C

Regards,
Murali.
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Re: If w and c are integers is w > 0 ? (1) w + c > 50 (2)  [#permalink]

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New post 17 Dec 2010, 06:04
1
muralimba wrote:
Bunuel wrote:
ajit257 wrote:
If w and c are integers is w > 0 ?

(1) w + c > 50
(2) c > 48

I have a doubt the ans.


Obviously each statement alone is not sufficient.

When taken together we can still have both Yes and No answers: if c=100 and w=10 then the answer is Yes but if c=100 and w=-10 then the answer is No.

Answer: E.

I feel that it should be w+c<50 for (1). Then as \(c\) is an integer then the least value of it according to (2) will be 49, so max value of \(w\) in order \(c+w\) to be less than 50 is 0 (as \(w\) is also an integer), which means that the answer to the question is \(w>0\) is No. So if it were \(w + c < 50\) the answer would be C and also the question would be much more interesting.

Hope it's clear.


Bunuel, taking the qtn as you specified i.e. w+c<50 Please clarify why am i getting w < 2
1) \(w+c<50\)
2) \(c > 48\)

now 2) can be written as \(-c < -48\)
as both the inequalities have the same sign, 1) + 2)
==> \(w < 50-48\)
==> \(w< 2\)
why am i not getting \(w <= 0\) for the answer to be C

Regards,
Murali.


First of all:
You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So, you can directly subtract \(c > 48\) from \(w+c<50\): \(w+c-c<50-48\) --> \(w<2\) (no need to rewrite and then add). But this approach won't work for this question: \(w<2\) would be correct if we were not told that both \(w\) and \(c\) are integers then for example c=48.5>48 and w=1>0 would be valid solutions but since both must be integers then \(c\) can be 49, 50, 51 ... and \(w\) can be 0, -1, -2, ... (so c=integer>48 and w=integer<1)

Hope it's clear.
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Collection of Questions:
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 w and c are integers is w > 0 ? (1) w + c > 50 (2)  [#permalink]

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New post 17 Dec 2010, 06:38
oops. Those are integers. Yes this will not apply in case of integers.
I missed it out.

Thanks Bunuel. Kudos for u...

regards,
Murali.
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Re: If w and c are integers is w > 0 ? (1) w + c > 50 (2)  [#permalink]

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