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Current time:0:00Total duration:10:28

GMAT: Data sufficiency 27

Video transcript

alright we're on problem 113 113 on page 287 so they've drawn this thing let me see if I could draw it so we have a right triangle in the center so one side that's the other side and then the base that's right triangle and then they drew some squares on every side of that triangle let me see if I can oh well I can do that one square well I don't know I'm gonna draw the other square cuz I can I draw it an angled one I'll just draw the lines the other square and then the last one I'll just draw it some lines like something like this comes off like that like that and like that so that's what they have and they call this a this B C D and they tell us that this is a right angle they say in the figure above if the area of triangular region D is 4 so the area of triangular region D is equal to 4 what is the length of a side of square region a so we need to know the length of the side of square region a so all of these are let's see all of these are well they don't tell us that this and this our square region just yet but they tell us this is a square region all right so the length of this side if we know the length of this side and that side then we can figure out the length of the hypotenuse of this right triangle so that's all we need and let's see if we call let's call this length right here let's call this little B little B and let's call this length little C so we already know that little B if we know the area of this triangle is 4 we know that little B times a little C so height times base times 1/2 is equal to 4 or that also tells us that it'll be times it'll see is equal to 8 I don't know what that does for us let's read the statements statement number 1 the area of square region B is 9 so B is equal to 9 well that's useful right if the area of this region and it's a square is 9 what's each of the sides going to be the side is going to be the square root of that right little B is going to be equal to 3 all right if little B is equal to 3 then the area of this is 3 times 3 we can do that because it's a square which would be 9 so if little B is 3 then what is C well then we get that C is equal to was that 8 over 3 using this information and we got that remember just based on the fact that the area of this triangle right here is 4 right it's a little B times the little C times 1/2 would be equal to 4 so if we know what little B is and we know what little C is then we can solve for let's call that little a and little a is what what the whole point was about what is the length of a side of the square region a so we could just solve for you know B squared plus C squared is equal to a squared right this squared plus this squared is equal to the hypotenuse squared and we just solve for a and we'd be done so statement one alone is sufficient and this is really messy especially taking an 8/3 squared and all that so I'm not going to do it and it would be a waste of time on the GMAT anyway let's look at statement number two statement to the area of square region C is 64 over 9 so C is equal to 64 over 9 so this is kind of very similar to the information that gave us the statement 1 if the area of this is 64 over 9 and this is a square region then the length of each each of its sides or in this case little C the length of Little Caesar is going to be the square root of that so little C is going to be 8 over 3 which is exactly what we figured out here and then we could use this information that we got from the problem statement that B times C is going to be equal to 8 and we got that from the fact that the area of this triangle is 4 so we use that and we could figure out that B so B times 8 thirds is equal to 8 and then you get B is equal to 3 and then you apply this again so each and to solve for a so each of the statements 1 & 2 are each individually sufficient to answer this question next problem 114 if X is to be selected at random from set T what is the probability that so they want to say what is the probability at 1/4 X minus 5 is less than or equal to 0 well simplify this real fast that means that what's the probability that 1/4 X is less than 5 and that means what's the probability let's multiply both sides by 4 I just added 5 both sides here and let's multiply both sides by 4 and we don't have to change the inequality since it's 4 is a positive number so it's the same thing as what's the probability that X is less than 20 and I like this a lot more because it's just simple and it lets me my brain grasp for the problem better and what are the statements they give us statement 1 T is a set of 8 integers T has 8 integers as has 8 integers that by itself doesn't help me because I don't know what those eight integers look like I mean maybe mate all eight could be larger than 20 in which case it would be impossible to find one that's less than or equal to 20 or maybe all eight are less than 20 in which case you would have a hundred percent chance so statement 1 by itself is fairly useless statement 2 T is inclusive sorry T is contained in the set of integers from 1 to 25 T is contained in the set of integers from 1 to 25 so that also doesn't help me so T within 1 to 25 so they're essentially telling us that every number and T is between this range but if we just take statement 2 alone they're not telling us first of all how many numbers are in T statement 1 tells us that but statement two tells us and then even if we knew how many numbers were in T we don't know where they're distributed maybe all of the numbers in T are less than 20 in which case there's a 100% chance that you would select one that's less than 20 but maybe maybe four of them are greater than 20 right maybe we have 21 22 23 24 and 4 of them are less than 20 or maybe something in between so given that you still don't know where T is distributed relative to 20 so I say that we still don't have enough information even with both of these statements to solve this problem next question 115 it takes the most confidence to choose that e to say that one cannot solve this problem 115 well I also I'm always second-guessing myself I'm like well I'm gonna make sure I haven't missed some piece of information no but I'm pretty confident here all right 115 if Sarah's age is exactly twice Bill's age so Sarah is age is twice Bill's age what is Sarah's age they want to know Sarah's age s for Sarah's age let me scroll down a little bit problem statement number one says four years ago Sarah's age was exactly three times bills age so Sarah's age four years ago so Sarah minus 4s is Sarah's age today so Sarah minus four is exactly three times bills was exactly three times bills age so was three times bills age then so B minus 4 right Sarah's age four years ago with three times bills age four years ago fair enough so just what this and this can I solve for s well sure I have two linear equations and two unknowns so I can solve for either variable and to prove it to you I'll do it all of this once again is a waste of time from the point of view of the GMAT S is equal to 3b minus 8 and then B from there we get B is equal to s over 2 so s is equal to three all right I just divide 2 into both sides so 3 s over 2 minus 8 what can I do here so I have 3 one and a half s I could subtract it from both sides so I have minus I'm doing it fast just 2 minus 1/2 s is equal to minus 8 and then multiply both sides by minus 2 and you get Sarah is 16 years old and you're done but you didn't have to do any of this that was all a waste of time you just have to see that you had two linear equations and two unknowns which is enough to solve for either of the variables statement 2 tells us 8 years from now Sarah's age will be exactly 1.5 times bills age so the same thing Sarah's age 8 years from now it's going to be equal to one and a half times bills age 8 years from now same thing we have two linear equations one that they gave in the problem statement and one they gave in statement two and two unknowns no reason why you can't solve for both variables and you might want to do it for yourself just as an exercise but you do it the same way you did it right here with I always find substitution to be the easiest especially when one of the equations is really easy to manipulate next problem so each statement individually is sufficient to solve this problem next problem 116 oh I just realized I'm out of time I'll do 116 in the next video see you soon