Is x:2 a positive number? (1) x – y is a positive number. (2) x + y is a positive number.
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I never came across such equation anywhere else except Learnhub tests. Please explain. Thanks.
IMO ans is C. Reasoning of same is given below.
1. Statment 1. Result of X – Y depends on values of both the nos. Not only when x >0, and x > y then result will be postive, but also when X<0 and X>Y then also result will be +ive. Thus we can not say wheather X is +/—ive, thus insufficient.
2. Statment 2
Result of X + Y depends on values of both the nos. Not only when Y >0, and x > y then result will be postive, but also when X<0 and lXl < lYl then also result will be + ive. Thus we can not say wheather X is +/—ive, thus insufficient.
3. If we add both the statements we get 2x= +ive intiger. If we divide a postive intiger by by 4(2×2, so as to get X/2) will also be +ive, as +ive/ +ive is a +ive intiger.
Thus correct ans is C
Let me know wheather I am correct alongwith flaw in my logic / reasoning.
I agree regarding the first question. It will be corrected on the test shortly, thanks for pointing this out.
Regarding the second question- statement 1 says that q > 2 and s > 4. Therefore those 2 sections are clearly positive, and we’re left with (p – 1)(r – 3) ^ 3. But we don’t know what p and r are, so they can fall in any range- for example, if p = 0 but r = 5, the product is negative, but if p = 1 the product is 0. Therefore the first statement is insufficient.
Statement 2 is sufficient, since we know that x ^ 2 or x ^ 4 are always > = 0 regardless of x. Thus the answer should be B. Let me know if there’s any error here.
This Q is on Kaplan CAT no.1 on CD provided in the book. oLahav your explanation is correct except that the fact that p q r s are non zero no.s. So expln for statement 1 should hv no conditions based on p=0. But otherwise it seems correct. One tip for all GMAT takers, KAPLAN CD has tough CATs (4no.s). Try n get these. I’m normally gr8 at Quant but here I didn’t do well as it was tough, good learning though. Quant Qs are pretty tough than given in any of the books e.g. KAPLAN, OG, PRINCETON, BARRONS. Get CD for Kaplan. I am giving my exam next week. Any more last min tips ?
I got one more interesting: If on coordinate plane, pt. A has coordinates (-3,4). How far is A from E? 1. E on Y axis, 4 units away from origin. 2. If A was twice as far from E, it wud be same distance from E as Cis at coordinates (0,-2)
Ans. is C. Pls explain.
For the pqrs question- you’re right, I forgot that p cannot be 0. My explanation still stands though, if you use 1/2 for p the product would still be negative.
For the coordinate question- statement 1 is not sufficient, since it gives 2 points for E (0,4) which is 3 units away, and (0, – 4) which is more than 3 units away. B isn’t sufficient either (pretty clear since we don’t know E, and if you use distance formula you’ll get more than 1 answer). Together though, we can just plug in the two points from A into the condition in B- you get that (0, 4) is 3 units away from A and 6 from C, so it fits. (0, – 4) doesn’t fit, so the answer must be (0,4) and therefore you need both statements.
This one’s pretty easy, as long as you know that if 2 ^ x = 2 ^ y, x=y (can be proved using log rules). So we got 2m – n = 1 straight away. Then since 256 = 2 ^ 8, statement 1 gives us that n = 3 so we get just plug in and solve for m, so it’s clearly sufficient. Statement 2 says m + 2n = 8, and then you can use the 2 equations to solve for m and n, so it’s sufficient too. Therefore answer is D.
FIRST STATEMENT SAYS , E on Y axiz , 4 units away from origin.. But in which direction ? It could ( 0,+4 ) or ( 0,-4) .. so A is alone not sufficient.
THE SECOND STATEMENT is also not sufficient , since we dont know E.But as OLahah explain , if we use both the statement , we can cross check E’s distance form C with two points i.e. (0,+4) and (0,-4). Out of these , only ( 0,+4 ) sutisfy the second condition.
So answer is C
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