GMAT: Math 2

We're on problem 7. And they drew this graph, so I guess I'll draw it as well. That's the x-axis. That's the y-axis. Let's see, they draw 1, 2, and they draw 1, 2, 3. And then the graph goes through here and then curves up like that and actually goes through this point right there. It goes through y is equal to 3, at least the way they drew it. On the graph before, when x equals 1/2, y is equal to 2. So they're saying that this point right here is 1/2 comma 2. Fair enough. And when x is equal to 1, y is equal to 1. So they're saying that this point right here-- I'll do it in a different color, just to not make it too confusing-- that point right there is the point 1 comma 1. The graph is symmetrical with respect to the vertical line at x equals 2. So that means that it's a mirror image around that vertical line. According to the graph when x is equal to 3, y should be equal to? So when x is equal to 3, we are 1 away from the line x is equal to 2, or what the line is a mirror image of, right? So when we were 1 to the left of it, that's when x is equal to 1. And we learned y is equal to 1. So for 1 to the right-- and this is symmetric around this line-- then we're going to have the same height, right? So when we were 1 to the left, the height was 1. When we're 1 to the right, the height should also be 1. So this should be the point 3 comma 1. So the choice is E. All that other information was kind of a waste of time. Next problem. Problem 8. When 1/10% of 5,000-- I always like to get these into decimals. 1/10% that equals this divided by 100, so it's 1/1,000. So it's 0.001. So 0.001 times 5,000 is subtracted from 1/10 of 5,000. The difference is-- OK. So they're saying 1/10 of 5,000 minus 1/1,000 of 5,000. Right? That's 1/10 of a percent is 1,000. So you get 500 minus 5, right? That's 495, which is choice D. Problem 9. Which of the following is the value of the square root of the cube root. So let's see. How many 0's do they have? 1, 2, 3, 4. 1, 2, 3, 4, 6, 4. OK. So there's-- how many total numbers are behind the decimal point? 1, 2, 3, 4, 5, 6. So it's a cube root. So some number times itself 3 times is equal to 0.0064. And you know whenever you multiply decimals, you add up the number of spots behind the decimal point. If I have two numbers that have two numbers to the right of the decimal point and I multiply them, I end up with 4. So if you just wanted to eyeball this, you'd say, well, first of all, what number to the third power is equal to 64? Well 4 to the third power is equal to 64, right? It's equal to 16 times 4, which is equal to 64. So this should have a 4 some place in it. It should look like a 4. It's going to be a decimal number because obviously the number has gotten smaller when I took the exponent of this number. And the clue is how many places are there behind the decimal? 1, 2, 3, 4, 5, 6. And I'm multiplying whatever this number is times itself 3 times. So in order to get 6 digits behind the decimal point, that original number probably has two digits behind the decimal point. So my initial kind of estimation is going to have a 4 in it. And it's going to be 0.04. And I'm pretty sure I'm right here. Well, you could try it out for yourself, but 0.04 times 0.04. You say, oh, 4 times 4 is 16. But I'm going to have now 1, 2, 3, 4 spaces behind the decimal point. And then if you multiply that times 0.04, 4 times 16 is 64. And now I'm going to have 1, 2, 3, 4, 5, 6 spaces behind the decimal point. And I get 0.000064, which is exactly there. So the problem now boils down to square root of 0.04. And you use the exact same logic. That is equal to 0.2. And try to square 0.2, you'll get 0.04 right? 0.2 times 0.2. You say 2 times 2 is 4. And I have 1, 2 spaces behind the decimal point. 0.04, and that's D. Problem 10. Raffle tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random will have a number with a hundreds digit of 2? So we're essentially saying, out of this collection how many of them have a hundreds digit of 2. So essentially it's going to be anything from 200 to 299. Well that's essentially 100 digits, right? Those are 100 numbers that have a hundreds digit of 2, right? Think about it. 200 by itself would be 1. From 200 to 1 would be 2. So 200 to anything would be that anything plus 1. So there's 100 numbers that satisfy it. And then how many numbers are we picking from? Well, if it was from 100 to 350 it be 250 numbers. But since it's from 101 to 350, we're picking from a pool of 249 numbers. Right? Although let me think about that. Let me think about that. Right, we're picking from a pool of-- I just want to make sure. If it was 101 to 102, we would be picking from a pool of-- Right, we're picking from a pool of 249 numbers, so the odds of having a number with a 2 in the hundreds digit will be 100/249. So that is choice E. I'm just making sure that I haven't made a-- Actually no, I'm going to take that back. If I said from 300 to 350, that's 51 digits. If I say 101 to 350, that's not 249 digits. I made a mistake. Because we're including 101, right? So it would actually be 250 digits. Sorry, it's 250 numbers. We're picking from a bag of 250 numbers. Think about it. If it was from 101-- you have to make sure you get your boundaries right on these-- if we said from 101 to 105, how many are this? This would be 101, 102, 103, 104, and 105. That's 1, 2, 3, 4, 5. It's 5 digits. So it's 105 minus 101 plus 1. So similarly, the bag, the number of digits we can pick from, is going to be 350 minus 101 plus 1, which is equal to 250. So anyway, out of the bag of 250, 100 numbers satisfy it. So we're going to have 100 in 250 chance of picking the right number. And if we divide the top and the bottom by 50-- that's the greatest common divisor-- we get 2/5. And that is choice A. And I got a little cautious because I saw that they had all these choices that were really close to each other. And they all seemed to test, like choices D and E, they all seemed to test whether you got the boundaries right, whether you really counted the pool correctly. And whether you really counted the number of digits correctly. And I always have to go through the exercise of let me make sure whether or not I need to count 101, whether I can just subtract them, or whether it's the fractions plus 1. I always have to clarify that in my own brain. Maybe that was easier for you. Question 11. On Monday, a person mailed 8 packages weighing an average of 12 and 3/8 pounds. And on Tuesday, 4 packages. So this is 8. Weighed that much. And on Tuesday, 4 packages weighing an average of 15 and 1/4 pounds. What was the average weight in pounds of all packages the person mailed on both days? So essentially you have to take the sum of the weights of all of them and divide by the total, right? So what's the sum of the weights of all of them? So the sum of the weights of these 8 packages-- we know their average. So the sum of the weights is going to be 8 times 12 and 3/8-- that's the sum of all of these-- plus 4 times the weight of these. Plus 4 times 15 and 1/4. And we're going to divide that by the total number of packages. Divided by 12. So this is equal to, let's see. What's 8 times 12? 8 times 12 is 96. And what's 8 times 3/8? So it's 96 plus 3. You could view 12 and 3/8 as 12 plus 3/8. So I kind of just did a distributive property there. You could turn this into an improper fraction, if that's easier. And then plus-- what's 4 times 15? It's 60, right? And then 4 times 1/4 is 1. All of that over 12. And so we have 99 plus 61. All of that over 12. Which is equal to 160/12. Let's see. Does 12 go into 160? 12 times 12 is 144. Let me think about this. So let me just divide top and bottom by 4. So you get 40/3. And let's see. All of the answers they have it in terms of fractions. So 3 goes into 40 how many times? It goes into 13, right? 13 times 3 is 39. So it's 13 with remainder 1. So 13 and 1/3 times, so A is the answer. So the average of all of the packages on both days are 13 and 1/3 pounds. See you in the next video.