One of the hardest GMAT questions

This is from a practice GMAT test. #math #maths #mathematics #shorts
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[@gf0878]

Your questions are getting easier and easier

[@ashusharma5572]

I guess its because its a shorts video?

[@luckyblo82]

@@ashusharma5572 nah some questions used to be real difficult

[@mateothode5603]

You* are getting better and better

[@pixelapse9613]

It's just like nothing, isn't

[@ceebad8985]

@@mateothode5603 Nah this is like elementary level maths let’s be real

[@psibarpsi]

One of those confidence-builders, huh?

[@asaeltongri7090]

Yeah that’s what I was thinking 😭

[@evann5900]

all his shorts are like this lmao

[@ELdASenSei]

Lmao this is so true

[@ch.s2510]

As someone who got 675(96th) on GMAT focus edition and 90(100th) on quantitative reasoning This question is very easy to answer. Just simply replace those algebra with -1, 1, 2

[@dradoode]

Nice flex 🫡

[@jamshiddilmurodov4925]

Wow, that's really cool. Any advice you can give those about to take the test? Would really appreciate it

[@whistlewithishaan]

Thats correct, ii used the same ideology

[@ItzMatthewPlayz]

thats a great line of thought.

[@coralreef2329]

I thought I was a genius for getting this... then I realized it was a CONFIDENCE BUILDER

[@protondium_8927]

Commenting before seeing the end: The answer must be E because 1. A negative number (

[@vinayakmarar4891]

What if y and z are negative numbers and x is positive then xyz>0

[@protondium_8927]

@@vinayakmarar4891 True, I should have noticed that the direction that positive numbers went in wasn't specified. I assumed that we were following the standard number line conventions that I have been taught but this ambiguity would change the answer.

[@animezoneamv9116]

@@vinayakmarar4891 always from viewer's point of view(And by CONVENTION), right side is positive, centre is centre and left side is always negative.

[@matthieupaty5032]

Yeah but III would be 2(1-1) which is just 0 and there is no option for only one and 2

[@fuerdichundmich]

@@matthieupaty5032Therefore no answer is correct.

[@Thirdbase9]

Assuming that the number line reads left (negative) to right (positive).

[@maxhagenauer24]

I simply just said the space between them all were 1 and so I replaced x, y, and z with -1, 1, and 2 like an actual number line. Then just put those in for the variables to see if they work.

[@ib5299]

Me too

[@Phantom_Deluxe7]

Same

[@animezoneamv9116]

Hmm.. I did like the video. But everything is same. Just playing with + and -

[@lordbeany1]

It does not state which direction would make a value greater or less than zero. One could easily argue x>0 making the first statement false since (+)*(-)*(-) is greater than 0.

[@estebanpelaezmunoz9655]

It's a number line. All number lines are arranged in away that everything to the left of 0 is less than 0, negative, and everything to the right of 0 is more than 0, positive, unless a specific number line states otherwise

[@mistercandle1389]

I agree, it's not stipulated that the numbers grow in one direction or another !

[@dirk.no-whisky.4u]

On the GMAT, it will be your standard left to right number line

[@jimray4210]

Too many (standard?) assumptions are necessary to solve this problem. If one does not assume the distances between the marks are equal, items II & III are necessarily true. As several individuals have noted item I is true only if values to the right of zero are positive. I believe this is a great problem for generating discussions on how to approach problem solving. (Imagine if the line represents a logarithmic plotting of values.) Thank you for presenting this problem for evaluation.

[@jimray4210]

*not necessarily true ... I regret any confusion I may have caused (Jim Ray)

[@TyDreacon]

If we dump the distance and negativity assumptions, isn't III still necessarily true? if y and z are positive, x is negative, therefore z(y-(-x)) = z(y+x) > 0-as explained in the video. But if y and z are negative, x is positive, therefore -z(-y-x) = -z(-1)(y+x) = z(y+x) > 0. Unless I missed something. EDIT: The problem explicitly states that the tick marks are evenly spaced-it's not an assumption in this one. So this means II ought to be true. z = 2y and x = -y. Thus x + z= -y + 2y = y. I is relying on left-to-right number line convention, however. Which I find kind of funny because, as far as my tired brain can tell, the other two maintain their answers if we abandon PEMDAS-another convention. Those statements seem awfully resilient!

[@supernovactc3283]

You do not have to assume equal distances, the question clearly states that 0:02 Also in a number line, negative is towards the left-hand side and positive towards the right-hand side.

[@jimray4210]

Boy talk about reading the whole problem: I glossed over the comment on evenly spaced ticks. I apologize for wasting electron and your time. Thank you for the clarifications and tutorial on reading.

[@nicholas31415]

So I get that standard number line notation always has negative numbers to the left of the zero, but this problem does not say to limit the problem to real numbers only. If this is a line of imaginary numbers then the answer would be "B. II only" so that should be the correct answer to this problem. Am I missing something?

[@guack5983]

That’s what I got at as well, except I didn’t consider imaginary numbers. All I thought about was that the number line could be positive to the left and negative to the right.

[@nicholas31415]

@@guack5983 As far as I know, in mathematics we can assume number lines use the Cartesian coordinates like we assume we should use PEMDAS, so on a horizontal line the left side of zero should always be the negative side. Though I do find it very interesting that flipping the direction of positive produces the same answer as using imaginary numbers.

[@RGP_Maths]

If positive is on the left, then III is also true, because z is negative and y-x is also negative. Basically III says 4y^2>0, so if it's a real number line, this is true.

[@kevinsly5134]

Nothing states the distance from x to 0 = y. It might be nitpicking but if it's not labeled, I see no reason you can assume x and y have the same absolute value.

[@VandarAlden]

​@Ovo Maltine what if it's a non linear scale? It says they are evenly spaced on number line not in values

[@tamimplayz]

I and III are definitely correct but confused about II. Alright gonna watch the video now

[@ayushkumarjha9921]

2nd is also true

[@pierrelorent3865]

I could agree with your solution if the axis drawing was not ambiguous. The two arrows either indicates that both directions are positive numbers or that the direction is unknown. Which changes the correct answer...

[@mirkhare1306]

I was thinking to myself earlier. "If only my math teacher explained this thoroughly rather than expecting me to just get it."

[@username-ur6dq]

1 isn't necessarily true since the direction of positivity wasn't stated

[@SpectreLgs]

Bro- it's always positive to the right, (Just an example) a>b>c is Incorrect c

[@brockobama257]

@@SpectreLgs you’re wrong, consider 3 dimensions and don’t label the axis. You have to define which side is positive.

[@SpectreLgs]

@@brockobama257 we are looking at a 2D plane and grid here, not 3D stuff, that's a whole different story Also from my perspective, going forward is positive for me, thus, going backwards is negative, upwards is also positive, and downwards is negative, left is negative and right is positive due to the upwards and downwards Also, thanks for making me think a little :)

[@gregoryanto3673]

Agree.

[@Paal2005]

@@SpectreLgs This is a question about mathematical logic. Nowhere in the the axioms or any math theorems does it say anything about positive being to the right. That's an assumption you make based upon standard practice, but all our mathermatics would function completely similar if we had negative go right and positive go left. Based only upon the information given and the qualitative information of the graphics, number one is incorrect.

[@DiarmuidHenry]

Can we necessarily assume that y,z>0? Aren't you making an assumption as to which direction is positive?

[@reddragon647]

Based on what you say,the only difference would be the first statement because it would be positive,but usually,it starts from left to right,but it's a good way of thinking that it could be a trick question

[@WirableCrown1]

FFS I had a panic attack when MATHs showed up in my shorts feed lol

[@worthysandwich5415]

option E is ok, but what if we invert the number line and e.g. x = 1, y = -1 and z = -2 then only statements 2 and 3 are correct

[@Bookphotocopier1288]

bro were u taught that number line to the right is positive and left is negative or what???

[@supernovactc3283]

In a number line, negative is towards the left-hand side and positive towards the right-hand side

[@Paal2005]

@@supernovactc3283 Not actually defined in mathematics, it's only a standard we use. All mathematical rules still function perfectly well with negative going right and positive going left. In a mathematical question meant to test rigorous logic, assumptions is the ultimate error, and you can't guarantee y being positive, only that it's of the opposite sign of x, and same sign as z. The more interesting thing is that you can still show II and III holds even without assuming the sign of the directions.

[@supernovactc3283]

@@Paal2005 In this problem, you are asked to subtract two numbers with opposite signs, so of course you need to know which one is positive and which is negative (positive - negative = positive) & (negative - positive = negative) Even though in mathematics we can take any direction to be positive and solve accordingly(mostly in geometry and complex numbers), we cannot do that here due to the reason stated above Here it says it is a number line which is equivalent to the x axis of a cartesian coordinate system. By convention (and not assumption) we have to take the rightward direction to be positive. This also works in polar coordinate system and not just cartesian.Say that equal distance between two successive points is 'a'.Since it represents distance, 'a' is a positive real number. y = a cos(0) = a z = 2a cos(0) = 2a x = a cos(pi) = -a So (x,y,z) = (-a,a,2a) You can put a to be any positive real number and therefore you cannot put 0. Hope this helps.

[@Paal2005]

@@supernovactc3283 You don't need to know the sign for the subtraction statement. No matter the sign, you add the value a (y), subtract -a, then multiply with a again. That gives 2a^2, which is positive independently of the value (and sign) of a.

[@ferociousfeind8538]

you can't know that the number line has positive numbers to the right- that's convention, but that's not guaranteed.

[@kulvirsingh5501]

I thought y-x would = 0 as y is the distance from 0 to Y and the distance from 0 to x so they should cancel right?Could you please explain why this is not the case

[@naopoint4096]

Ok lets say this. If Y was 1, then x would be -1. 1-(-1) would be the same thing as 1+1. You would be correct if it said X+Y though

[@Sir_Isaac_Newton_]

x = -y Not x = y basically

[@enteryournamehere6093]

Just because their distance from 0 is equal doesn't mean they're the same number. y - x is not the difference between the distances, but the difference between the numbers.

[@asparkdeity8717]

The average is given by the sum of the two endpoints which is halved, not their subtraction (regardless of sign)

[@asparkdeity8717]

Also y - x = 0 iff y = x , which is clearly not the case here

[@anurag123bagga]

I would agree with the entire solution except for when y=0 and the equidistant values is 0. then the answer would be B (II only)

[@elidoz7449]

that's why B was my answer

[@jasimmathsandphysics]

More problems like this please

[@exodus9669]

I dont think we can just assume that the numbers from left to right are negative to positive. x could equal 1 and y = -1 and z = -2. If this is the case statement 1 is false in this case (since it is not stated which side is positive and negative, there will always be a probability that statement 1 is false so we cant say it is true for definite). However, statements 2 and 3 remain true but no answers are available for this.

[@Lucas-ub7sl]

how do i mind my decisions

[@mirkhare1306]

You're awesome 👍

[@RichardRentrop]

Nowhere does it say that y>0. They can evenly be spaced with distance 0. If you allow this only II is true. The question is not clear

[@emmanueldestin2353]

Can you get equal increments of distance 0?

[@Kyanzes]

Finally, an easy question!

[@petzkuh]

The setup for this feels like a trick question. You can pretty easily work out that II and III must always be true, but I depends on x < 0 (assuming we're working with reals). Then, we can see from the available options that E is correct, but i feel like the problem statement should include that x < 0.

[@erikkonstas]

It does tho...?

[@williamsong7919]

This only holds if the distance between the marks are greater than 0. 1 and 3 doesn't hold in this case

[@rafaeldearaujo12]

If I choose the negatives to be the right side and the positive the left, the first statement isn’t true anymore…

[@supernovactc3283]

That is conventionally wrong

[@quigzinator]

Does evenly spaced doesnt mean the number line is linear? Like if the number line was exponential like a slide rule, these assumptions are incorrect right?

[@supernovactc3283]

Evenly spaced definitely means linear

[@quigzinator]

@@supernovactc3283 arent the marks on a slide rule evenly spaced?

[@supernovactc3283]

@@quigzinator The problem clearly states "evenly spaced number line" and not a slide rule

[@SamuelFreitas-vivant]

I couldn't believe that i got the answer in my head without paper.

[@SeaTurtle1122]

This assumes that y>0, which isn’t a given. The only one that’s necessarily correct if y can equal 0 is B.

[@thatGuySpectre]

There's a lot of unsound assumptions in there. For example, nowhere is it stated that the numbers are not zero. If x=y=z=0, they are evenly spaced, but only II is true. And the assumption that numbers on the right are positive is also not a given, you could have negative numbers on the right, and only x be positive. That would render I to be false, while keeping II and III true. The only one we know for sure to be correct is II

[@interestedbystander196]

You don't even have to worry about the third equation. Once you establish that I and II are correct, there is only one answer (E) that satisfies this option.

[@therealquinn28]

yes, but only if you say that the right side of 0 is positive and the left side is negative. But must that be the case ? Because it does not say it anywhere, so if you don't know which side is which you can only conclude that II is right

[@eatehbaby]

Actually the correct answer is b since you are not given the positive side of that line. As such both I and III could be false.

[@VandarAlden]

This is assuming the number line goes positive to the right...probably safe to assume but... Although that difference only changes the first one

[@tylerfowler4862]

Only thing I disagree with is that there's no restrictions on x,y and z not being equal to zero, in which case, 1 and 3 are false (since the inequalities less than/greater than, not less/greater than or equal to)

[@EdgedPixie]

First isn't necessary true: you've incorrectly assumed Y is positive just because it's to the right of zero, where no such context is given. If y0, which proves I is not true for all values of y.

[@koranoptreyawan9876]

acutually, you can assume x=-1 y=1 z=2

[@stkhan1945]

..distance is a modulus fn..hence always +ve

[@obvioustruth]

x is not negative! There is an arrow towards left, so numbers are positive in both directions from 0.

[@supernovactc3283]

In a number line, negative is towards the left-hand side and positive towards the right-hand side

[@bolivianem]

II only work if the number line has a linear scale if logarithmic scale then it won't

[@parthisMC]

E. Easy Just assign values to each. x = -1 0 = 0 y = 1 z = 2

[@fosskytheanswerer]

It should be a bit more complicated than that... How do we know which side of the line is positive/negative? Or maybe they are both positive/negative...

[@trombonetimmy1064]

Not sure what you mean, we know where zero is, so anything less is negative and anything more is positive

[@supernovactc3283]

In the cartesian sign convention, negative is towards the left-hand side and positive towards the right-hand side

[@fosskytheanswerer]

@@trombonetimmy1064 Is there anything that says where is negative or positive side, or whether it has two positive or negative sides? The fact that left side -> negative & right side -> positive is the most commonly used doesn't mean there is no other posibility

[@fosskytheanswerer]

@@supernovactc3283 Is there anything that says this is a case only for cartestian convention?

[@supernovactc3283]

Yes, a number line is the x axis of a cartesian coordinate system. You cannot proceed in any other way. Even in polar coordinates this holds true. Say the equal distance is a. So y = a cos 0 = a z = 2a cos 0 = 2a x = a cos(pi) = -a So y = a, z = 2a and x = -a

[@nutankumari3174]

Hey you are wrong, if y = y and as per instructions, x = - y then for the III statement it will result in 0 = 0 and not 0 > 0 .

[@supernovactc3283]

the distance is same as y but the direction is opposite, so x = -y

[@taylorrana-ruby391]

Are y and z necessarily positive? What if X is 1, y is -1 and z is -2? Nothing on the number line indicates which direction is positive…

[@supernovactc3283]

It is a convention that in a number line right-hand side is taken positive and the left-hand side is taken negative

[@grogmarsh-76]

You made an assumption that numbers on the right are positive which is not necessarily true. If it is possible for the right of 0 to be negative or positive then only 2 is true. By convention numbers left of 0 are negative.

[@SeegalMasterPlayz]

I used different way of assuming numbers and got it correct

[@ShivaniSingh-ph6sf]

This is a really helpful video! Which resoources did you use for prep?

[@Lillith.]

I doesn't have to be true. If left is positive and right is negative, I is false.

[@derradfahrer5029]

Where does it say, that negative is to the left and positive is to the right? It could also be the other way around so that y=-x and z=-2x Or even that y=x and z=2x

[@supernovactc3283]

In a number line, negative is towards the left-hand side and positive towards the right-hand side

[@derradfahrer5029]

@@supernovactc3283 Aha, so it's per definiton. Thanks.

[@supernovactc3283]

@@derradfahrer5029 Welcome 🙏

[@gideon_kang1]

you can just substitute x for -1, y for 1, and z for 2 because the ratio x : y : z will always equal -1 : 1 : 2

[@HarshalMore]

i did exactly the same thing..but the answer comes out to be different

[@anjalidwivedi2057]

E option

[@ignasicamps214]

That is unless y = 0

[@supernovactc3283]

Scale should never be 0 in a number line

[@MrJronson]

(1) isn't necessarily true because it depends on whether the number line is oriented with positive on the left or on the right. This is not stated in the problem. It would not be impossible for the positives to be on the left and negatives on the right, making (1) not always true

[@AlchemistOfNirnroot]

People trying to overcomplicate a very simple question. It's literally a number line in the question i.e a 1-D real line.

[@mayureshwarsontakke6187]

Why is this assumed that the points are equidistant? We can't assume it unless and until explicitly mentioned, right?

[@ralphy1054]

Technically statement 1 is false. There is nothing that implies which direction is positive and which is negative. If the right side is negative, xyz > 0 Statements 2 and 3 are still both true for the same reasons. Statement 3 is always true because (y-x) will always be the same sign as z, meaning z(y-x) is always positive

[@supernovactc3283]

In the Cartesian sign convention, negative is towards the left-hand side and positive towards the right-hand side

[@martynas1819]

There is no evidence to suggest that the distance from y to z is equal to the distance from 0 to y and that the distance from x to 0 is equal to the distance from 0 to y so the actual answer is I and III

[@adamfreed2291]

If y is allowed to be negative, then the answer is b.

[@DoxxTheMathGeek]

D. Oh! I thought the numbers were evenly distributed, so you wouldn't know which ones are what. Now i would say E.

[@TheKingBeyondEverything]

Am I delusional or did I actually managed to solve a Mind Your Decision Problem?

[@ruaridhusher4373]

Not if all the points are equal, which I think makes the most sense since this question is framed around graphs and it doesn't state otherwise. In that case, I think it's just 2, right?

[@bdfortes]

I thought it could be a trick question. If you changed the convention about which direction is positive, only number 2 would still be true.

[@cguy96]

I read it as there is a problem with this number line, not this is a problem about number lines. I kept waiting for the shoe to drop how the answer wasn’t E.

[@lilyblue3205]

Short answer. If they're equal, call all 1. X=-1. Y=1. Z=2. Have fun

[@exobytemonolith5339]

I think this is wrong actually. Only II and III are definitively correct. There is an assumption here that the left hand side is the negative side and the right hand is the positive. But, there is no information that confirms that, only our biases. We might be looking at it the wrong way. Meaning, if x is +ve and y and z are -ve then it could be true that xyz>0.

[@brianfox340]

These were all very obvious to me. So I stared at it for 5 minutes trying to figure out which one was a trick. None of them were tricks.

[@Mnaughten601]

since we have evenly spaced lines they all could be consider a unit. Thus we have -1(unit),0(unit),1(unit),2(unit), or -1,0,1,2. Now do math.

[@coolraviraj24]

where the question said that all the variables distance from 0 is same ?? as we are considering it same...so this question is coming in my mind.

[@Scrogan]

Ah but what if the number line goes from right to left? Nothing other than arbitrary convention suggests that y is positive and x is negative.

[@nicholas31415]

Is it arbitrary if it's the convention of mathematics and we are doing a math problem? If it was a science problem, I would agree, but as far as I've been able to find, mathematics always uses Cartesian coordinates.

[@darth21HFR]

@@nicholas31415 If the line is vertical, where do you put the negative numbers, up or down ? And what if it's neither horizontal nor vertical, how do you know ? The direction has to be specified and it's not the case in this problem. So instead of assuming, you should consider both cases, and some answers are not correct.

[@supernovactc3283]

@@darth21HFR This is a number line : a representation of all numbers, so it is meaningless to ask if the line is crooked or some such. What is significant, though, is the position of 0 and the position of positive and negative numbers relative to it. Conventionally in a number line, negative is towards the left-hand side and positive towards the right-hand side

[@nicholas31415]

@@darth21HFR That is a good question, but again, in mathematics we use Cartesian coordinates as a base for all number lines, so if the line is vertical then positive is up. There are as many different coordinate systems as you care to invent for your purpose in the same way that you can choose to ignore PEMDAS rules if you wish (I've seen that!) but this is a standard math problem. Personally, I think they should have made it more clear that they only want you to use real numbers and positive is to the right, but they didn't, so you and I would have both gotten this question wrong. If x = -(i) then the correct answer is B, but they assumed we wouldn't think of that either.

[@lenovalenova9940]

What if all the numbers are 0? Then answer is only 2

[@Studio_salesmen]

If y is equal to i than only 2 is correct, it never states that the numbers are on the real line

[@Dom-Nom-Nom]

You are making the not-necessarily-correct assumption that going to the right is positive. y could be a negative number which invalidates some of the properties

[@supernovactc3283]

In a number line, negative is towards the left-hand side and positive towards the right-hand side

[@beaclaster]

how is z(y-y)>0 when n•0=0

[@simonwitt2994]

What if the space is 0?

[@Inklow]

Instead of saying the the distance equals Y I said that it equals 0.5, makes it easier when counting

[@BetterSelfForever]

This is incorrect because y - x equals zero, and zero times 2y would still be zero, making E an incorrect choice

[@supernovactc3283]

x is negative and y is positive (y - x) is positive

[@Lodrik_Bardric]

Why are you taking for granted y is a positive number? The numbers could be arranged in descending order. Also you are assuming that the scale is linear while it could perfectly be logarithmic. So if the numbers are arranged in descending order statement I would be false because multiplying a positive and two negative Numbers would be a positive number. In case of a logarithmic scale, statement II would be false due to the fact that the difference between the numbers would not be the same. So actually, with the given information the only statement that’s always true would be III.

[@supernovactc3283]

Conventionally in a number line, negative is towards the left-hand side and positive towards the right-hand side

[@Lodrik_Bardric]

@@supernovactc3283 It doesn´t matter what the normal way to put numbers on a line is, but only the available information. There is no gerneric rule which forbids to put numbers in descending order nor a non linear scale. So if the intention was to have those restrictions to be taken into account, than you would need to state them as part of the querstion.

[@zacharyschwab400]

But you can’t assume that the number line is increasing from left to right? x could be positive, which would make 1 incorrect.

[@zacharyschwab400]

y-x, would be zero, which would make 3 incorrect as well, as anything times 0 is zero and not more than zero. The answer would be B

[@supernovactc3283]

In a cartesian sign convention, numbers increase left to right. Numbers to the right of 0 are positive and numbers to the left of 0 are negative x is negative and y is positive (y-x) is therefore positive

[@zacharyschwab400]

@@supernovactc3283 aight thanks I wasn’t sure the right to left negative to positive was something that was defined.

[@neilmorrone691]

In your Explanation of Statement 1 is INCORRECT. Since the Condition is the Product of xyz!!! You State (--) + ( +) < 0; It should be (--) × (+) < 0. YES, both Results are True, but your explanation misstates the Operation using + , NOT x.

[@RGP_Maths]

Nonsense. The parentheses convey implied multiplication, so he is simply saying negative x positive x positive < 0.

[@neilmorrone691]

@@RGP_Maths Negative NONSENSE! The Statement left to right is: ( -- ) + ( + ) < 0........ THERE IS NO IMPLIED MULTIPLICATION, THERE IS A " + " BETWEEN THE SETS OF PARENTHESES. AMEN.

[@animezoneamv9116]

@@neilmorrone691 Bruh are you a immature kid? He clearly by his voice said (+) multiplied by (-) by (+). What will you say for (1)2(3)? Same way he meant. It's not a mistake. You didn't understand him properly.

[@neilmorrone691]

@@animezoneamv9116 Nonsense, from a nonsensical sophistry!

[@Calmerism]

I'd argue that the first is actually not true since you do not know in which direction of the axis the negative part is. y and z could be negative and x positive, making the first statement false und thus none of the answers are correct.

[@supernovactc3283]

In the cartesian sign convention, negative is towards the left-hand side and positive towards the right-hand side

[@SnijtraM]

"y and z are a positive number". Says *nowhere* . They might as well be negative, no + sign at any arrow. It's II and III only, and that answer is not listed.

[@m.guypirate6900]

How do we know y is positive? That isnt given.

[@zhafrankhairy7028]

Finally i got it right

[@Saviour694]

I just put -1, 1, and 2 as a very gud move yoo

[@basil-vander-elst]

But what if y and z are negative and x is positive? Because it isn't marked what goes to positive infinity and what to negative infinity

[@aididdat1749]

If y is negative, it won't work

[@supernovactc3283]

In the cartesian sign convention, negative is towards the left-hand side and positive towards the right-hand side.

[@donutdude4174]

Is this what math was like before chatgpt? 😅

[@aliciahunt6004]

Can someone help explain why I is true? I thought -y + y + 2y would make xyz GREATER than 0?

[@axoltl1]

No, how is that possible? x (or -y if you want to give it that value) is

[@mushfiquealam6895]

Bro .. no way this is one of the hardest. Prolly one of the easiest

[@mohammadsaifaldeenabu-shaa5432]

1 & 2 & 3 are true .

[@loosens9999]

Stop making me feel smart

[@saraswatikatke4011]

Answer is E

[@Its-Thupten]

I am a class 10 student and I was only able to get the first one and other just confused me 😅

[@animezoneamv9116]

Thought you would use geometry to find this cuz this is an easy question.

[@KB-ln4vc]

i solved it 😢 iam so happy

[@zahirgizzi7009]

Couldn't it be that they all are 0? Then only II would be always true but I and III would only be true for nonzero x,y and z