If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:12:27

GMAT: Data sufficiency 3

Video transcript

we're on problem number 10 and it says in a certain class one student is to be selected at random to read what is the probability that a boy will read okay so I guess we can assume that everyone in the class can read because they don't really talk much about that and we can I think we can assume that everyone in the class is either a boy or girl and we don't have any third genders in this in this reality okay so what are the statements we essentially just want to know what what's probably we pick a boy in the class right whether or not they read is kind of so what's the probability that we pick a boy so to figure out whether we're going to pick a boy we have to know essentially what what percentage of the class are boys so what are the statements they give us statement number one is two thirds of the students in the class are boys well that's just what I asked for two thirds boys well that exactly tells me the probability that I pick a boy it's two out of three or sixty 6.66 you know repeating percentage so this alone is enough so what does the see if statement - is it all useful statement - ten in the stands ten of the students in the class are girls 10 girls well this especially by itself is a fairly useless statement because maybe they're they're ten girls maybe there are zero boys in which case this probability would be zero but maybe there are 10 girls and ten boys in which case this would be fifty percent so ten girls by itself really doesn't tell us much so in this case all we need a statement 1 and statement 2 alone is not sufficient so that is what I always forget these a statement one alone is sufficient with statement two is not sufficient that's a problem 11 let me switch colors to green problem 11 if 5x plus 3y is equal to 17 so 5x plus 3y is equal to 17 they want to know what X is equal to X is equal to what okay the first statement they give us is X is a positive integer X is a positive integer well that doesn't help us solve X I mean X could be any positive integer and then it gave us no constraints on Y so we really could pick X as any positive integer so that doesn't help us with that statement number two is y is equal to 4x well this is a little bit more interesting y is equal to 4x well here we could substitute this in for Y and then solve for X and remember these problems you really don't have to solve for X you just have to know that you could so I immediately just recognizing it I could say well this is all I need to do to solve for X and just to prove it to you let's do it although you shouldn't have to do this on the GMAT so 5x plus 3 times y well Y is equal to 4x is equal to 17 so you get 5x plus 12 X is equal to 17 or 17 X is equal to 17 X is equal to 1 so all you needed was statement 2 and statement 1 was fairly useless so that is B statement two alone is sufficient but statement 1 alone is not sufficient problem 12 problem 12 problem 12 does the prod so they're asking whether the product j/k m n equal 1 does that equal 1 so the first statement today give us let me scroll down a little bit the first statement they give us is J K over m n is equal to 1 let's see we could do a little algebraic manipulation multiply both sides by MN you get J K is equal to MN right I just multiplied both sides by MN but that really doesn't doesn't help me I mean that just tells me that this part is equal to this part you know but I doesn't really tell me that they when you multiply them all together you equal 1 let's see a statement to is it all helpful statement 2 tells us that J is equal to 1 over K and M is equal to 1 over n so let's see if we could do some interesting algebraic manipulation so let's do multiply both sides of this by K you get J K is equal to 1 and do the same thing here multiply both sides times n you get M n is equal to 1 well this is useful because if J times K is equal to 1 and M times n is equal to 1 so then J times K this is equal to 1 and M times n is equal to 1 so the whole products is going to be 1 times 1 well that's definitely going to be equal to 1 so using just the information statement 2 we were able to verify that J times K times n times n is equal to 1 so this is all we needed statement 1 was fairly useless so that is what that is B statement two alone is sufficient statement 1 alone is not sufficient do problem number 13 problem 13 a certain expressway has exits JK L and M in that order all right I already my guess is that it's going to be some type of a number line problem if I had to the intuition let's see the exits j k l and m l and m that's the highway in that order and then they say what is the road distance from exit k to exit l so they want to know this distance right here that's what we have to see if we have sufficient data to figure out so statement 1 they tell us the road distance from exit j 2l is 21 so we could just write that in so line segment 2j L is equal to 21 so that tells us that this distance right here is 21 well that by itself because we don't know where K is within that 21 miles so that alone doesn't help us let's see if statement 2 helps figure out things the road distance from K to M is 26 K to M is equal to 26 so this distance is equal to 26 this doesn't really help us at all because if you think about it imagine so the district a and M is fixed right but K could be here which would put em here right or and then of course that would change the KL distance or K could be really close to L which would put em here right if you viewed you know M is definitely 26 miles or whatever it is 26 ya kilometers further from K so we don't know where K is this this second constraint still doesn't tell us where K is relative to you know sit closer to J is it closer to L we don't know so both of them combined are useless so that is e problem number 14 unless I miss something problem number 14 is the integer K a prime number so K prime K Prime so the first statement they give us is that 2 K is equal to 6 well we could just do a little algebra here divide both sides by 2 and we get K is equal to 3 and 3 is prime so this is good that's all we needed we're done well let's see what 2 gives us maybe 2 also gives us that C 2 says that 1 1 is less than K which is less than 6 well this is kind of useless because this this is just telling us that K we know it's an integer because they say is the integer K prime number but K could be 2 in which case it wouldn't be well 2 is prime but K could be 4 in which case it's not prime or it can still be 3 so we don't know so this is kind of useless so all we need the statement 1 and statement 2 is not sufficient so that's a problem number 15 these switch colors 15 okay so they've they've drawn some some type of XY axes so that's the y axis let's the x-axis alrighty X Y and then they they have this they drew 1-1 and then they label the quadrants quadrant one 102 quadrant three we do it in Roman numerals and they say 104 and they say if a B does not equal zero so a times B does not equal zero which essentially tells us that neither a or B neither a nor B is zero okay a B does not equal zero in what quadrant of the coordinate system does does Point a belie so they want to know where a B is so quadrant of a B okay let's look at the data all right so we know a B can't well we know it's not along the axes because if one of these were 0 then we'd be along the axes either A or B or 0 let's see it says this is interesting it says B a is in is in Quadrant four B a so this is interesting so quadrant four quadrant four tells us what if something is in quadrant four let me do it in a different color this is AB is I think the first interesting problem we've encountered quadrant four tells us that X is greater than zero right because we're in the positive x-axis and tells us that Y is less than zero right because we're in the negative y-axis right we're here so if B a is in Quadrant four in this case B is playing the role of X right this is a point B a so that tells us that B has to be greater than zero B is greater than zero and this is telling us that a because it's playing the function of the y-coordinate here a has to be less than zero a has to be less than zero so if we talk about the point a comma B a comma B well now we're in a situation where we know that a is less than zero so we know that X is less than zero and we know that B is greater than 0 and now B is playing the y-coordinate so we know that Y is greater than zero so if X is less than zero where the plot negative x and y is greater than zero over here so a comma B is going to have to be there someplace so statement number one was all we needed to know that it's going to be in quadrant two so statement number one is good now let's take a look at statement number two well statement number two is kind of the same thing so we actually don't even have to work it out that statement two alone is also sufficient just eyeballing it but let me work it out but we can apply the exact state same logic statement two says a comma minus V in quadrant three well if we're in quadrant three that means that X is less than zero right and a plays the X role so this tells us that a is less than zero and quadrant three tells us that Y is also less than zero and the y coordinate here is minus B so it tells us that minus B is less than zero right this point is minus B so minus B has to be less than zero which tells us you can multiply both sides by negative 1 or divide and that tells us that B is greater than zero so you you get the same information from statement two as you get from statement 1 that a is less than 0 and B is greater than 0 and then you can use that to figure out where that a B has to be in quadrant two so each of them independently are sufficient to answer the questions so the answer is d see you in the next video