GMAT: Math 36

Before moving on, I want to address problem 180 again. Because we got it right in the last video, but then as soon as I finished the video, I realized there was a much, much simpler way of doing that. They said in a nationwide poll, n people were interviewed. 1/4 of them answered yes to question one. And of those, 1/3 answered yes to question two. So then you have another 1/3. So what fraction of the entire population said yes to both? Well, it would be 1/4 of-- or 1/3 of 1/4 which is equal to 1/12. 1/12 of the population said yes to both. Or you could say 1/12 of n said yes to both. And they ask us which of the following expressions represents the number of people interviewed who did not answer yes to both? So it's everybody else. These are the people who said yes to both. So essentially you just subtract that from 1. So 1 minus 1/12 is equal to 12/12 minus 1/12, which is equal to 11/12 of the population did not say yes to both. And so 11/12 and then the population is n, so it's 11n over 12. And that's how we got choice E. Problem 181. The ratio of two quantities is 3:4. So, let's say one is x, one is y, and it equals 3:4. If each of the quantities is increased by 5, what is the ratio of these new quantities? x plus 5 to y plus 5. Well, it really depends on what multiple these are of 3 and 4. Well, I already have a sense that it's choice E. It cannot be determined from the information given. And let's prove it. I'm going to show you two different x's and y's, and when you add 5 to both you're going to get a completely different answer. So you cannot determine it from the information. So they snuck in a data sufficiency question into the problem solving. So, x could be 3. If x is equal to 3 and y is equal to 4, that definitely satisfies this condition. Or we could have x is equal to 6 and y is equal to 8. Right? That also satisfies the 3 to 4 ratio. What happens when we add 5 to both of these? x plus 5 would be equal to 8. And y plus 5 would be equal to 9. So, the new ratio becomes 8:9. What about this case? You would have x plus 5 is equal to 6 plus 5, which is 11. And then y plus 5 is equal to 8 plus 5, which is 13. And 8/9 is a very different fraction than 11/13. It's not like you can reduce one into the other. So I can find an x and a y to satisfy this, but when I add 5 to the top and bottom I get two different answers. So you cannot determine this with the information given. So the choice is E. Sneaky data sufficiency problem. 182. If the average arithmetic mean of x and y is 60-- so essentially you're saying x plus y over 2 is equal to 60. The average of them is 60. And the average of y and z is 80. So, y plus z over 2 is equal to 80. What is the value of z minus x? So let's see if we can do that. So let's rewrite this. Let's solve for x in terms of y. I'll do it in a different color. We get x plus y is equal to 120. Just multiplied both sides by 2. You get x is equal to 120 minus y. Now let's solve for z in terms of y here. Multiply both sides by 2, you get y plus z is equal to 160. z is equal to 160 minus y. So what's z minus x? So z is 160 minus y, minus x, 120 minus y. So that is equal to 160 minus y, minus 120 plus y. And it's good that the y's are canceling out. Whenever they give you something like this, it's usually a safe guess that if you just go forth with the algebra that nice things will happen like this. So these y's cancel out and we're left with 160 minus 120 which is 40. And that is choice B. Problem 183. If 1/2 of the air in a tank is removed with each stroke of a vacuum pump-- so we have 1/2 tank per stroke. I'm guessing this is going to be a rate of change. What fraction of the original amount of air has been removed after four strokes? Oh no. This is interesting. OK. So, after one stroke, what fraction of the original amount of air has been removed? Let's just figure out how much has been left. We could do both actually. Left and then removed. So, after one stroke-- 1/2 of the air in the tank is removed with each stroke. So after one stroke, you have 1/2 left and you have 1/2 removed. But, after two strokes, it takes out half of the air. Right. So it takes out 1/2 of this half. So if you take out 1/2 of this half you have 1/4 left over. And you've also removed another 1/4. Right? Because these add up to what was left before. After three strokes, you take out 1/2 of this. So you have 1/8 left over. And you took out another 1/8. And then after four strokes, you have 1/16 of the air left over. Actually I didn't even have to do this column. After four strokes you have 1/16 of the original air. But they want to know what fraction of the original amount of air has been removed. So this is what you have left. So what's been removed is 1 minus this. So, 1 minus 1/16. Well, that's 16 over 16 minus 1 over 16. Which is equal to 15 over 16. Choice A. I didn't even have to worry about that column. Problem 184. If the 2 digit integers M and N are positive and have the same digits, but in reverse order, which of the following cannot be the sum of M and N? All right, same digits in reverse order. So M could be AB. And N could be BA. So which cannot be the sum? I'm just experimenting. Let's see you if you add AB and BA. So, if B plus A-- I guess there's two assumptions. If B plus A is less 10, let's do this one. B plus A is less than 10. Then you would have-- this digit would be B plus A. And then this digit over here would also be B plus A. So, definitely choices D and E work. So we're left with choices A, B, and C. Because none of these fit this paradigm where we have the same digit twice. So, all of those assume that B plus A is greater than 10. So, if B plus A is greater than 10, then what happens? A plus B. B plus A. So let's think about it a little bit. If B plus A, you'll have the ones digit from B plus A. And then you'll have a 1 up here. You add 1 plus A plus B. So, you'd have B plus A plus 1. And so this will be a two-digit number. And if you think about it-- if we do the three digits of the number, this is going to be the ones digit from B plus A right? This is going to be the ones digit from B plus A plus 1. So this number is going to be one more than this number. That's an interesting problem. This one has to be one more than that one. And so both B and C satisfy that, right. 165. 6 is one more than 5. And it should work out. We should be able to figure out a B and an A that satisfy that. And then C also satisfies that. 121. 2 is one more than 1. I want to make sure you understand that because it's actually pretty interesting. Deductive reasoning, A is the answer. But if you think about it, we're assuming B plus A is greater than 0. So, here, you'd have the ones digit of B plus A and you'd carry a 1. Now when you add A plus B and that 1, the ones digit is going to be 1 greater. This is going to be the ones digit of B plus A plus 1. And that's why I'm saying that the tens digit has to be 1 greater than the ones. That was an interesting problem. Anyway, see you in the next video.