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And problem 124. According to a car dealer sales report, 1/3 of the cars sold during a certain period were sedans, and 1/5 of the other cars sold were station wagons. So 1/3 of the cars sold, so the total, were equal to sedans. And 1/5 of the other cars sold were station wagons, so 1/5 of the remainder of this. So that's 1/5 of the 2/3 cars that weren't sedans. So 1/5 of the 2/3 of the total were station wagons. Let's call it W for wagons. If N station wagons were sold during that period-- so W is equal to N-- how many sedans in terms of n were sold? So we know that 1/5 times 2/3 times the total is equal to N. Let's just substitute S for T. If we multiply both sides of this top equation times 3, we know that the total is equal to 3 times the number of sedans. So let's substitute that here for T. So you get 1/5 times 2/3 times 3S is equal to N. 3 cancels out. 3 divided by 3. So you have 2/5 S is equal to N, or-- let me just write it up here. So multiply both sides by 5/2. You have S is equal to 5/2 N. And that is choice D. Problem 125. If p/q is less than 1, and p and q are positive integers, which of the following must be greater than 1? So think about this. If p/q is less than 1, that tells us that p is less than q. You can multiply both sides of this times q, and you get that. p has to be less than q. And they're both positive integers. So there's nothing tricky here with negative numbers. So just the instincts, which of the following must be greater than 1? I don't know if this is going to be a choice, but if this is less than 1, if you invert it it's going to be greater than 1. So divide both sides of this by p. You'll get 1 is less than q/p. And that makes sense. If 1/2 is less than 1, then 2/1 is going to be greater than 1. If 2/3 is less than 1, then 3/2 is going to be greater than 1. So if p/q is less than 1, then q over p is going to be greater than 1. And we did it algebraically by just multiplying both sides by q and then multiplying both sides by p. That is a choice. E, q/p. That made it easy. I thought I might have to do more work if that didn't show up. So q/p is definitely greater than 1. Next question, 126. It would take 1 machine 4 hours to complete a large production order. So machine 1, it would take 4 hours. So you could say 1 order every 4 hours, or 1/4 order per hour. That's machine 1. And another machine, 3 hours to complete the same order. So machine 2 does 1 order in 3 hours, or 1/3 orders per hour. How many hours would it take both machines working simultaneously at their respective constant rates to complete the order? If they're working together, you're going to have one machine producing at 1/4 order per hour, plus another machine producing at 1/3 order per hour. So their combined rate is going to be what over 12? And then 1/4 is the same thing as 3/12, and 1/3 is the same thing as 4/12. So their combined rate is going to be 7/12 order per hour. And you want to know how long it's going to take them to make 1 order. So you can think of production is equal to rate times time. And time is equal to production divided by rate. Or if you don't want to think about them, just think if I can make 7/12 orders per hour, if I have to make 1 order, I'm just going to be 1 order divided by 7/12 order per hour. This might seem a little confusing to you, but if these were constant numbers, try to play with it. If I had to make 10 orders, and I can make 2 orders per hour, then I just divide 10 by 2, which would tell me 5 hours. But it's just a little more confusing when I have to make 1 order and my rate is 7/12 orders for hour. But hopefully if you substitute whole numbers in your head it makes little bit of sense. So this is the same thing as 12/7. And then of course the orders cancel out, and you end up with hours in the numerator. So 12/7 hours or 1 5/7 hours. What do they have? Yes, that's choice C. Question 127. To mail a package, the rate is x cents for the first pound, and y cents for each additional pound, where they tell us that x is greater than y. 2 packages weighing 3 pounds and 5 pounds respectively can be mailed separately or combined as one package, so 3 pounds and 5 pounds. That's one package. Which method is cheaper, and how much money is saved? Let's think about it. If you keep them separate, how much are you going to pay? For the 3 pound packet, you're going to pay x for the first pound. And then you're going to have 2 pounds left that you are going to pay y at. And then for this one, you are going to pay x for the first pound. You're going to have 4 pounds left, so plus 4y. So when you add them together, your total expense is going to be 2x plus 6y. That's if you mail them separately. Now if you combine them into one 8 pound package, what do you have? You have to pay x for the first pound, and then you have 7 pounds left that you're going to pay for y. So it's going to cost x plus 7y. So first of all, we know that this is going to be more. Why do we know this is more? Because essentially we're taking one of the y's-- and when you pay them separately for that one extra pound that here you're paying it as in x cents, instead of y cents that you're paying in this case. And x is more than y. And that will hold out when we actually subtract the 2 out. Actually, hopefully you understand the reasoning why this is going to be larger because x is greater than y. So let's figure it out. What's the difference? This is the higher number. And let's subtract out the lower number, minus x minus 7y. I just subtracted x plus 7y. And that you end up with x minus y. 6 minus 7y is minus 1y. So x minus y. And the fact that x is greater than y lets you know this is positive. So that also tells you that this is the more expensive of the alternatives. Separate is the more expensive. And the difference is x minus y. So they ask which method is cheaper? So the combined method is cheaper and you save x minus y cents. And that's choice A. Choice B, they say combined is cheaper with a savings of y minus x. But y minus x is going to be a negative number. So you can't say you saved a negative number. They tell us that x is greater than y, so you could cancel that out. And the other ones don't make any sense at all. Next problem, 128. If money is invested at r% interest compounded annually, the amount of the investment will double in approximately 70/r years. If Pat's parents invested $5,000 in a long-term bond that pays 8% interest compounded annually, what will be the approximate total amount of the investment 18 years later as Pat is ready for college? It doubles every 70/r years, where r is 8. They gave us that information, so let's try to use it. So how long does it take to double at an 8% rate? So it's 70 divided by 8 is equal to what? It's equal to 8.5? 8 goes into 70 eight times, 64. 6, 0. 8 goes into 60 seven times. 7 times 8 is 56. And you have a 40. 8.75 years. This isn't going to be exact, but you're going to double your money roughly every 8.75 years if using this formula they give you at an 8% interest rate. And they say what will be the total amount of the investment 18 years later? So 18 years, that's roughly two doublings. So you're going to start with $5,000, and after 8.75 years you're going to be at $10,000. You're going to double in 8.75 years. And then after another 8.75 years, you are going to be at $20,000. This isn't exactly 18 years. This is 17.5, but that's about as close as we're going to get. So in 17.5 years, you get to $20,000. They are asking 18 years, and this is all about approximation because they don't want you to do some hardcore exponent math here. And so the answer is A. And if you look at that 8.75, that's roughly you're doubling every 9 years. They ask us how long. So you could say after 9 years I'm at $10,000, after another 9 years I'm at $20,000. So after 18 years, I'll be at $20,000. And I'm all out of time. See you in the next video.