On the number line, there are 4 points w,x,y and z, in an ascending

Question

On the number line, there are 4 points w,x,y and z, in an ascending order. If wxyz<0, which of the following must be positive?

1) wx
2) xy
3) yz

A) 1 only
B) 2 only
C) 3 only
D) 1 and 3 only
E) 2 and 3 only

Leo8 · Accepted Answer

FullSizeRender (7).jpg +B

rohit8865 · Answer

for wxyz<0 two cases on number line is possible case 1 :- w<0 & z>y>x>0 case 2:- w0 A) wx wx results >0 with case 2 & wx<0 with case 1.........Discard B)xy both cases we have xy>0.........Keep it C) yz yz results >0 with case 1 & yz<0 with case 2.........Discard Only B satiesfies Ans B

generis · Answer

For the product of two or more numbers to be negative, there must be an odd number of negative factors. Here there are four numbers. Either only one is negative, or three are negative. I assigned values. Ascending order, smallest to greatest, is w -1 < 2 < 3 < 4 Here, w is the one negative factor, and x, y, and z are positive Case B : If there are three negative factors, we have w -4 < -3 < -2 < 1 Here, w,x, and y are negative, and z is positive. Assess options for "must be positive." I. wx - In Case A, wx is negative. Reject. II. xy - In both Case A and B, xy is positive because x and y have the same sign. Keep. III. yz - In Case B, yz is negative. Reject. Only II is true. Answer B.

tapabrata · Answer

w,x,y and z are on the number line in ascending order. Since wxyz is <0, none of w,x,y,z can be 0. 0 can take 5 different places but we need to find cases when wxyz <0 Case 1 w wxyz>0 => discard case Case 2 w wxyz<0 => wx is +ve, xy is +ve yz is -ve Case 3 w wxyz>0 => discard case Case 4: w<0 wxyz<0 => wx is -ve, xy is +ve, yz is +ve Case 5 0 wxyz>0 => Discard Only for cases 2 and 4, wxyz<0. Now we will check the signs of wx, xy and yz wx -ve for case 4 => Not always +ve => 1) must not be true xy +ve for both cases => xy is always +ve => 2) must be true yz -ve for case 2 => not always +ve => 3) must not be true. Only 2) must be true Answer B

adkikani · Answer

chetan2u   pikolo2510   VeritasPrepKarishma   Bunuel   pushpitkc   niks18

Any other option than testing cases?

niks18 · Answer

hi adkikani You will have to work through case based scenarios as wxyz<0 so z>0 and a situation where x<0 and y>0 is not possible. because in that case wxyz>0 So Now only two cases remain Case 1: w,x,y<0 & z>0 This case will give Option 1 & Option 2 as positive solutions. Case 2: w<0 & x,y,z>0 This case will give Option 2 as the only Solution. As this is a Must be true question, so only Option 2 holds in all the scenarios.

pikolo2510 · Answer

I did it in this way

Only two possible cases satisfy the above condition

w x y z
1) - - - +
2) - + + +

In the both the above cases, only xy is positive. Hence answer is B

CEdward · Answer

The key words here are 'ascending order'.

Case 1: 
w = -ve, x = -ve, y = -ve, z = +ve

Case 2: 
w = -ve, x = +ve, y = -+e, z = +ve

We can see that only xy must always be positive (negative x negative = positive OR positive x positive = positive)

Answer is B.

Kinshook · Answer

Given: On the number line, there are 4 points w,x,y and z, in an ascending order.

Asked: If wxyz<0, which of the following must be positive?

wxyz < 0, when
Case1 : w<0 & x,y,z>0 or
Case 2: w,x,y<0 & z>0

1) wx: wx<0 in case 1 but wx>0 in case 2
2) xy: xy>0 in both cases: MUST BE POSITIVE
3) yz: yz>0 in case 1 but yz<0 in case 2

IMO B

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On the number line, there are 4 points w,x,y and z, in an ascending

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