Learn for the GMAT: Data Sufficiency 17

Khan Academy Presents: GMAT Prep: 77-79, pg. 284

Problem 77, 77 if Ms. Smith’s income was 20% more for her 1991 that it was for 1990 how much was their income in 1991? So 20% more for 1990 so income, let's call it income 1991 is 20% more than 1990 so that means 1.2 times income in 1990. I hope you get that right. If something is 20% more than something else is going to be 1.2 times that, 12 is 20% more than 10, 12 is 1.2 times 10 all right or you could view it as income plus 20% of income which is 1.2 times income either way, so let's see what they tell us. Now what do we have to figure out? What were trying to figure out this, income in 1991? Statement number one Ms. Smith’s income for the first six months of 1990 was $17,500.00 and the income for the last six months of 1990 was $20,000.00. what it seems. So as this seems, they're telling that she made $17,500.00 in the first six months of 1990 and her income for the last six months of 1990 and her income for the last six moths of 1990 were $20,000.00. Well they're essentially telling us the total income for 1990, the first six months and the last six months. There are 12 months in a year. So her total income for 1990 was $37,500.00 that equals income for 1990 so clearly if we know this is this, we just multiply that times 1.2 and we get the income for 1991, so this statement alone is sufficient. Let's see what they give us for statement number two. Ms. Smith’s income for 1991 was 7500 greater than for 1990 so they say income of 1991 was 7500 so it equals income for 1990 plus 7500. Well this alone does help us because what they already given us this so we have two linear equations, all right. This is one linear equation and two unknowns. This is another linear equation and two unknowns. So we have two linear equations and two unknowns. We can solve this probably the easiest way is just to substitute depending on what you want to solve for but we've done that multiple times. You can substitute 1.2 times 1990 here and then it's all for it or you could do the other way, you could do divided by 1.2 here and then substitute it there. But either way this is trivial algebra hopefully by this point to solve but this and this is definitely enough information to solve the problem. So two equations with two unknowns and you can do that in your spare time if you don’t believe me. So both statements alone are sufficient for this one, 78 in the figure above so I think I have to draw, let me see if I can it's the y axis and that’s the x axis and then they have a line and let's see what I can do. The line looks something like that and then they tell us it says, well it’s y. This is x and this is p. This right here is q and then draw this they call this point right here r. They drop that like that and this q is at point cd and p is at point ab and then they say the figure above segments PR and QR. So PR and QR so let me draw that out there. That’s PR and QR are each parallel to one of the rectangular coordinate axis, okay fair enough. This is parallel to the y axis. PR is parallel to the x axis fair enough. Is the ratio of the length of QR to PR equal to one? So they want to know QR over PR, is that equal to one? And immediately this should trigger something from algebra one. They're asking essentially is the slope of this line equal to one? Is the ratio of QR to PR, so rise over run is the slope of this line equal to one, so let's see what we can do? And slope is just you know you change in y over change in x and well you know this is change in, so what's this point first of all? You actually, you don’t have to know anything about slope. I don’t want to make you feel like you have memorized some formulas. What's this point going to be? So it's going to be—actually let's do it even better. What is the length of QR going to be? I haven’t looked at any of the data points right now. What is the length of QR? It was going to be this height, so it's this y which is d minus this y. This y is going to be b because all of this is y is equal to b right here. So QR is going to be equal to d minus b and PR is the length on the x axis. It's going to be this x, right. What is this z? Well this x is right here, C. X is equal to x. It's going to be this x minus this x. Well here x is equal to a and so the ratio is equal to d minus b over c minus a which is that if you remember the formula for the slope of the line. You just take the y1 minus y2 over x1 minus x2 but we didn’t have to memorize that. It is intuition because at this point right here R is the point to the x coordinate is c, the y coordinate is b. So hopefully that gives you an intuition. Now let's look at the statements. You wouldn’t have to do that on the real g method. It will be a waste of time. Statement number one tells us c is equal to three and d is equal to four. So that by it self that just gives us the first part of this that doesn’t help us figure out this entire ratio. So this by itself isn’t that useful maybe in conjunction of what else they give us. Statement two, a is equal to minus two and b is equal to minus one. Well if you use both of these statements together and then we have everything here. We have d, we have b, we have c and we have a so we can solve it. So if both statements together are sufficient for solving, for knowing whether the ratio of QR to PR is one or essentially it's the slope of this line equal to one. Next problem 79, 79 while on a straight road car x and car y traveling at different constant rates. If car x is now one mile ahead of car y, how many minutes from now will car x beat two miles ahead of car y? So right now, let me see, let see if I can do that. So x is here. Y is here and they’re going at constant rates. One mile so they’ve been traveling for some amount of time and x is one mile ahead and there's saying how long it is going to be before x is, how many minutes before x is two miles ahead. And they're constant rates, so if they start it off let's just think about it. They started off at the same point and it took 10 minutes for x to get one mile ahead. It would take another 10 minutes for it to get two miles ahead. Well that’s stop thinking about it, let's see what they give us for the statements. Statement number one, car x is traveling at 50 miles per hour and car y is traveling at 40 miles per hour. That seems to be pretty good information. Car x at 50 miles per hour, car y is for miles per hour. So essentially car y is moving away from car x at what? It's moving away at 50 minus 40 miles per hour. So if from car at y point of view car x is always pulling away at 10 miles per hour. All right does that make sense? If car x was going at 40 miles per hour they wouldn’t pulling away at all if it was going at 41 miles per hour it would be pulling away an incremental one mile per hour and you know as long as were not approaching the speed of light we can assume Newtonian classical physics and we could just take the difference between the two. So how long does it take for it to pull away another mile? Well how many minutes if you're going 10 miles per hour relative to something else, how many minutes does it take to go a mile? Well one you know you can figure that out but let me figure that out for you. So you know distance is equal to rate times time, so if your distance you want to know is one mile and you're rate is equal to 10 miles per hour times time, what's the time going to be equal? Time is going to be equal to one tenth of an hour or six minutes. So that’s the answer number one, one alone is sufficient. Well the answer number 79, one mile is sufficient. Let’s see what they give us for number two. Statement two, three minutes ago car x was one half mile ahead of car y, okay. So three minutes ago the state of affairs was this. Y was here. X was here and it was a half mile difference. So what does that tell us? That’s actually pretty good information too. Three minutes ago car x was half mile ahead of car y. Now car X is one mile ahead, so in three minutes x pulls away by half a mile and they're going at constant velocities, so the relative velocities between the two don’t change. So if it takes three minutes for extra pull away by half a mile it would take six minutes for extra pull away by a mile, right? You just multiply them by two they're all going at the same constant velocities. So at six minutes x pulls away by one mile and that’s actually what they're asking because they say, how many more minutes does it take x to pull away by another mile? They’ve probably been traveling for six minutes already and then another six minutes actually pull away by another mile. So two alone is also sufficient, so each of them independently is good enough to answer this question. See you in the next video.