Learn for the GMAT Math 36 Video

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Learn for the GMAT Math 36

Before moving on, I want to address from 180 again because we got it right to the last video but then as soon and I finished the video I realized there was a much, much simpler way of doing that. This in the nationwide poll, N people were interviewed. ¼ 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 how many people, what fraction of the entire population said yes to both? It would be 1/3 of ¼ which is equal to 1/3 of ¼ just equal to 1/12. 1/12 of the population said yes to both or you can say 1/12 of N said 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 just subtract that from one. So 1-1/12 = 12/12 -1/2 = 11/12 so the population did not say yes to both. And so at 11/12 and then the population is N, so it's 11N/12 and that’s how we got choice E. Problem 181: The ratio of two quantities is 3 to 4. So let's say quantity, one is X one is Y and equals 3 to 4. If each of the quantities is increased by 5, what is the ratio of these new quantities, x + 5 to y + 5. Well it really depends on what multiple these are of 3 and 4. I mean I could and write; well I already have a sense that’s its choice E. It cannot be determined from the information given and let's prove it. I’m going to show you two different axis on Y’s and when you add 5 to both, your going to get complete different answers. So you cannot determine from the information until they snuck in a data’s efficiency question until the problem solving. So one, x could be 3. If x=3 and y=4, that definitely satisfies this condition or we could have x=6 and y=8, right? That also satisfies the 3 to 4 ration. But what happens when we add 5 to both of this? X+5 would be equal to 8 and y+5 would be equal to 9. So the new ratio, it becomes 8:9. But what about in this case? You would have x+5 is equals to 6+5 which is 11 and then y+5 is equal to 8+5 which is 13. And 8, 9 is a very different factor than 11, 13, it's not like you can reduce one into the other. So I can find an x and y that 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’s, sneaky data sufficiency problem. Problem 182: If the average arithmetic means of x and y is 60 and the average— So essentially, you’re saying x+y/2 is equal to 60, thus the average of them is 60 and the average of y and z is 80. So y+z/2 is equal to 80, what s the value of z-x. So let's see if we can do that. So rewrite this, let's solve for x in terms of y. So we get—under a different color. We get x+y=120 just multiplied both sides by 2, you get x=120-y. And I’ll solve for z in terms of y here. Multiply both sides by 2 you get y+z=160, z=160-y. So what’s z-x? So z is 160-y, 160–y-(120-y) so that is equal to let's see 160-y–120+y and its good that y’s are canceling out. Whenever they give you something like this, you can almost—it's usually safe guess that if you just go forth with the algebra that nice things will happen like this. So this y cancel—I have left with 160–120 which is 40 and that is choice B. Problem 183: If ½ of the air on a tank is remove with heat stroke of a vacuum pump. So we have ½ tanks per stroke. I’m guessing this going to be way different. What fraction of the original amount of air has been removed after 4 strokes? Oh know this is interesting, okay. So after one stroke, so what fraction of the original amount of air has been removed? So it says here how much has been left, right? So left, we could do both actually, left and then removed. So after one stroke, ½ of the air of the tank is removed with the heat stroke, right. So after one stroke, you have ½ left and you had ½ removed right? Then after two strokes, it takes that half of the air. So it takes that half—this half, so if you take at a half of this half, you have a 4 left over and you’ve also removed another 4th, because this add up to what was left before. After 3 strokes, you take out half of this so you have an 8th left over and you took out another 8th and then after 4 strokes, you have a 16th of the air left over. And actually you don’t have to do this column. That’s a 4 stroke of a 16th of the original air, right. 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 one minus this. So 1–1/16, well that’s 16/16 -1/16 = 15/16, choice A, I didn’t have to do worrying about that column. Problem 184: If the two 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? Alright, the same digits for reverse order. So M could be AB and N could be BA and so which cannot be the sums. So let's just—I’m just experimenting. Let's see what if you add AB and BA, (AB+BA). So if B+C, if B+A, I guess there’s two assumptions. If B+A<10, let's do this one. B+A<10 then you would have—this digit would be B+A and then this digit over here would also be B+A, right? So definitely, choices D and E work. So we’re left with choices A, B, and C because none of this fit this paradigm, where we had the same—twice. So all of those assumed that the B+A>10. So if B+A>10 then what happens? A+B, B+A—so essentially, let's think about it a little bit. If B+A you’ll have the ones digit from B+A and then you’ll have one up here and you had 1+A+B so you’d have B+A+1 right. And so this will be a two digit number and if you think about it—so if you do the three digits of the number this is going to be the ones digit from B+A, right? This is going to be the ones digit from B+A+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 satisfied that right? 165, 6 is 1 more than 5 and it should work out. We should be able to figure out B in an A to satisfy that and then C also satisfies that. 121, two is 1 more than one. Now I want to make sure you understand because this actually pretty interesting but the doctor of reasoning A is the answer. But if you think about it, we’re assuming B+A > 0. So here you’d have the one digit of B+A and you’d carry a 1. Now you add A+B and that 1—the ones digit is going to be one greater. This is going to be the ones digit of B+A+1 and that’s why I’m saying that the 10 digits should has to be one greater than the ones. That was an interesting problem. Anyway, see in the next video.

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Before moving on, I want to address from 180 again because we got it right to the last video but then as soon and I finished the video I realized there was a much, much simpler way of doing that. This in the nationwide poll, N people were interviewed. ¼ 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 how many people, what fraction of the entire population said yes to both? It would be 1/3 of ¼ which is equal to 1/3 of ¼ just equal to 1/12.

1/12 of the population said yes to both or you can say 1/12 of N said 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 just subtract that from one. So 1-1/12 = 12/12 -1/2 = 11/12 so the population did not say yes to both. And so at 11/12 and then the population is N, so it's 11N/12 and that’s how we got choice E.

Problem 181: The ratio of two quantities is 3 to 4.

So let's say quantity, one is X one is Y and equals 3 to 4. If each of the quantities is increased by 5, what is the ratio of these new quantities, x + 5 to y + 5. Well it really depends on what multiple these are of 3 and 4. I mean I could and write; well I already have a sense that’s its choice E. It cannot be determined from the information given and let's prove it. I’m going to show you two different axis on Y’s and when you add 5 to both, your going to get complete different answers. So you cannot determine from the information until they snuck in a data’s efficiency question until the problem solving.

So one, x could be 3. If x=3 and y=4, that definitely satisfies this condition or we could have x=6 and y=8, right? That also satisfies the 3 to 4 ration. But what happens when we add 5 to both of this? X+5 would be equal to 8 and y+5 would be equal to 9. So the new ratio, it becomes 8:9. But what about in this case? You would have x+5 is equals to 6+5 which is 11 and then y+5 is equal to 8+5 which is 13. And 8, 9 is a very different factor than 11, 13, it's not like you can reduce one into the other.

So I can find an x and y that 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’s, sneaky data sufficiency problem.

Problem 182: If the average arithmetic means of x and y is 60 and the average—

So essentially, you’re saying x+y/2 is equal to 60, thus the average of them is 60 and the average of y and z is 80. So y+z/2 is equal to 80, what s the value of z-x. So let's see if we can do that. So rewrite this, let's solve for x in terms of y. So we get—under a different color. We get x+y=120 just multiplied both sides by 2, you get x=120-y.

And I’ll solve for z in terms of y here. Multiply both sides by 2 you get y+z=160, z=160-y. So what’s z-x? So z is 160-y, 160–y-(120-y) so that is equal to let's see 160-y–120+y and its good that y’s are canceling out. Whenever they give you something like this, you can almost—it's usually safe guess that if you just go forth with the algebra that nice things will happen like this. So this y cancel—I have left with 160–120 which is 40 and that is choice B.

Problem 183: If ½ of the air on a tank is remove with heat stroke of a vacuum pump.

So we have ½ tanks per stroke. I’m guessing this going to be way different. What fraction of the original amount of air has been removed after 4 strokes? Oh know this is interesting, okay. So after one stroke, so what fraction of the original amount of air has been removed? So it says here how much has been left, right? So left, we could do both actually, left and then removed.

So after one stroke, ½ of the air of the tank is removed with the heat stroke, right. So after one stroke, you have ½ left and you had ½ removed right? Then after two strokes, it takes that half of the air. So it takes that half—this half, so if you take at a half of this half, you have a 4 left over and you’ve also removed another 4th, because this add up to what was left before. After 3 strokes, you take out half of this so you have an 8th left over and you took out another 8th and then after 4 strokes, you have a 16th of the air left over. And actually you don’t have to do this column. That’s a 4 stroke of a 16th of the original air, right. 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 one minus this. So 1–1/16, well that’s 16/16 -1/16 = 15/16, choice A, I didn’t have to do worrying about that column.

Problem 184: If the two 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?

Alright, the same digits for reverse order. So M could be AB and N could be BA and so which cannot be the sums. So let's just—I’m just experimenting. Let's see what if you add AB and BA, (AB+BA). So if B+C, if B+A, I guess there’s two assumptions. If B+A<10, let's do this one. B+A<10 then you would have—this digit would be B+A and then this digit over here would also be B+A, right?

So definitely, choices D and E work. So we’re left with choices A, B, and C because none of this fit this paradigm, where we had the same—twice. So all of those assumed that the B+A>10. So if B+A>10 then what happens? A+B, B+A—so essentially, let's think about it a little bit. If B+A you’ll have the ones digit from B+A and then you’ll have one up here and you had 1+A+B so you’d have B+A+1 right.

And so this will be a two digit number and if you think about it—so if you do the three digits of the number this is going to be the ones digit from B+A, right? This is going to be the ones digit from B+A+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 satisfied that right? 165, 6 is 1 more than 5 and it should work out. We should be able to figure out B in an A to satisfy that and then C also satisfies that. 121, two is 1 more than one. Now I want to make sure you understand because this actually pretty interesting but the doctor of reasoning A is the answer. But if you think about it, we’re assuming B+A > 0. So here you’d have the one digit of B+A and you’d carry a 1.

Now you add A+B and that 1—the ones digit is going to be one greater. This is going to be the ones digit of B+A+1 and that’s why I’m saying that the 10 digits should has to be one greater than the ones. That was an interesting problem.

Anyway, see in the next video.

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