Learn for the GMAT: Math 15 Video

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Learn for the GMAT: Math 15

Were on problem 81 and let's see they gave this equation, y is equal to 248 minus 398 x, so which of the following values of x give the greatest value of y in the equation above. Which of the—okay, so were going to get as large of possible y. You see y is 248 minus this thing, minus this thing all right. So we want to do is since were subtracting this from 248 we want to minimize this right, as small as this can be and even if we can make, if we can make what were subtracting negative that would be awesome because then when you subtract the negative and that will get us above 248. And actually if we look at all of the choices they have one negative number there. You might say, “Oh, we put a zero and then y is going to be 248” but what if we actually put a negative number in 4x and that’s, what they have this choice E negative one. What happens when put negative one there and then y is equal to 248 minus negative one times 398, so it's minus, minus negative 398 which is the same thing as plus 398, so that’s actually the maximum point. Because were subtracting a negative right that’s even better than subtracting zero and it's a way better than subtracting a positive as were trying to maximize that, so the choice is E. 82, machine a produces bolts at the uniform rate of 120 volts every 40 seconds, every 40 seconds. So that’s the same as we you can see 120 volts every 40 seconds is the same thing as three volts per second right? And yeah, that’s right. And machine b okay that’s a, that’s a, the machine b produces volts of the uniform rate of so this b, of 100 volts every 20 seconds, 20 seconds. Well that’s equal to a five volt per second, if the two machines run simultaneously, how many seconds will it take for them to produce a total of 200 volts. So if you're running simultaneously what's your speed per second this was going to do three per second this was going to do five per second. So if you—if they run simultaneously you're going to be producing eight per second, three plus five. So eight volts per second is the combined rate and we want to know how many seconds will it take to do 200 volts. So we just take 200 volts divide it by eight seconds. All right, another way you know with the distance, your distance is equal to rate times time I guess you could view, this as you know production is equal to a rate times time, right. Normally rate is how fast you’re moving per second. Now it's how much you’re producing per second. And so your total production is going to be rate, your rate times time and then your time is going to be the production divided by the rate but I think sometimes we think too much in terms of form and you loose the intuition but I just wanted to show you that that there is a formula for it if you need one but the easiest way to think of it well I have to produce 200 volts I produce, if said, I have produce a 100 every seconds it's like it take me two seconds so if I wrap the volts eight every second I take 200 divided by eight. So 200 divided by eight and I can even do the unit’s. 200 volts divided by eight volt per second that’s equal to 200 over eight times volts times if this isn’t the denominator you just inverse it and you got, seconds per volt, the volt came sliding. So you know you got the unit’s right. So the sound degree that should give you conviction that you got the formula right and 200 divided eight is 25, so it is 25 seconds it choice B. Problem 83, what is the decimal equivalent of one fifth to the fifth power. One fifth to the fifth power, so let's think about this, let's try it one fifth as a decimal that’s .2, so it's .2 to the fifth power. So let's just I don’t know, let's just multiply it out, .2 times .2 is equal to what? Two times two and we have two digit’s behind the decimals, so it's .04 and then you multiply that times .2 times .2, two times four is eight and now we have three numbers behind the decimal point that’s going to be equal to .08, sorry, .008 right, one, two, three, three dot number is going to the decimal. So this is two to the third and then .008 times .2 to the right, this is to the third power that’s equal to two times eight is 16 and then how many numbers do we have behind the decimal point? One, two, three, four so were going to have .1234 so that’s to the fourth power, we have one left so .0016 times .2 is equal to, two times 16 is 32 and now we have one, two, three, four, five numbers behind the decimal point. So one, two, three, four, five that’s the answer and that is choice A. Another way actually the easiest way to think about it is without having even doing this, you can set it all, what's two to the fifth power. Two to the fifth power is 32, right, and had you that alone because there’s only one choice it even has a 32 behind the decimal point and you could said, “Oh!” And every time to I take to power then adding one digit behind the decimal point it's going to be 32 which going to have five number behind the decimal point, so .12345 I did it the small way. Next question 84, 90 minus eight times 20, all right the way they did, 20 divided by four all of that over one half this is just a fast simplification problem that becomes 90 minus, what 20 divided four, 20 divided four is five, five times eight is 40 all that over one half that becomes 50 over one half which is the same thing as 50 times two over one, which is equal to 100, that’s choice C. That’s just one you just have to do quickly and not make a careless mistake. I hope I did it quickly. I don’t know if I did a careless mistake and let's see 85 a dealer originally brought a 100 identically batteries at a total cost of q dollars. So q is equal to 100 batteries if each battery was sold at 50% above the original cost, so let b is the cost to each of the battery were sold at, so b is going to be 50% above the original cost. So it's going to be 1.5 times the original cost, right, this is 50% above the original cost for battery. And then in terms of q for who many dollars was each battery sold? Oh, they just drew us you know, let's see a dealer, a dealer originally brought a 100 identical batteries at a total cost of q dollars, if each battery was sold—oh, I'm sorry, so this is the cost right? So q is equal to 100 times what the dealer paid for it. Right, a dealer originally bought a 100 identical at a total cost of q dollars. Right, if each battery was sold at 50% above the original cost for back battery, so the selling price, this is the original cost, it's equal to 1.5 times the cost. Times the cost, in terms of q for how many dollars was each battery sold. So let's write q in terms of the cost, so cost is equal to q divided by a 100, it's divided both sides of that equation a hundred. And then we could just substitute C or the cost here, for here. The selling price is a 150% of this. So selling price is equal to 1.5 times c which q over a hundred. And let's see how they did, so selling price is equal to 1.5 q over a hundred I don’t expect to see that as a choice, but if we multiply, if we multiply the top and the bottom by two that equals three q over 200, three q over 200 that’s choice a. Then make sure with your variables right now. Problem 86, in increasing sequence, in an increasing sequence of 10 consecutive integers the sum of the first five integers is 560 what is the sum of the last five integers in the sequence. Okay, 10 consecutive integers, so x plus one plus x plus, we will right this way x plus one plus x plus two. Well actually the first one we must call it x, x plus three x plus four all right, this are the first—this right here is going to be the first five consecutive integers if we started x right. When you take the sum of those that well the sum of those is one two three four five x, five x let see one plus two is three, three plus three is six, six plus four is 10. 5x plus 10 is going to be equal to, they tell us it's equal to 560, fair enough or we could say 5x is equal to 550 I think we can use that information. Now what is the next five integers going to be, it's going to be x plus five x plus six x plus seven x plus eight and then x plus nine, right this are five. And what are those at up to be, that’s five x plus five plus six is 11 plus seven is 18 plus eight is 26 plus nine, 26 plus nine is 35, so it's 5x plus 35 and they want to know, they want to know what this is equal too right, that’s the next five. We know what 5x is it's equal to 550 from this setup. So it's going to be equal to 550 plus 35 which is too, 35 plus 550 is 585, 585 that’s choice A. And the trick to be able to do it fast is just to recognize that you didn’t have to solve all the way for x that you could reuse this 5x when you took the sum of the next five. And I'm all out of time see you in the next video.

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Were on problem 81 and let's see they gave this equation, y is equal to 248 minus 398 x, so which of the following values of x give the greatest value of y in the equation above. Which of the—okay, so were going to get as large of possible y. You see y is 248 minus this thing, minus this thing all right. So we want to do is since were subtracting this from 248 we want to minimize this right, as small as this can be and even if we can make, if we can make what were subtracting negative that would be awesome because then when you subtract the negative and that will get us above 248.

And actually if we look at all of the choices they have one negative number there. You might say, “Oh, we put a zero and then y is going to be 248” but what if we actually put a negative number in 4x and that’s, what they have this choice E negative one. What happens when put negative one there and then y is equal to 248 minus negative one times 398, so it's minus, minus negative 398 which is the same thing as plus 398, so that’s actually the maximum point. Because were subtracting a negative right that’s even better than subtracting zero and it's a way better than subtracting a positive as were trying to maximize that, so the choice is E.

82, machine a produces bolts at the uniform rate of 120 volts every 40 seconds, every 40 seconds. So that’s the same as we you can see 120 volts every 40 seconds is the same thing as three volts per second right? And yeah, that’s right. And machine b okay that’s a, that’s a, the machine b produces volts of the uniform rate of so this b, of 100 volts every 20 seconds, 20 seconds. Well that’s equal to a five volt per second, if the two machines run simultaneously, how many seconds will it take for them to produce a total of 200 volts.

So if you're running simultaneously what's your speed per second this was going to do three per second this was going to do five per second. So if you—if they run simultaneously you're going to be producing eight per second, three plus five. So eight volts per second is the combined rate and we want to know how many seconds will it take to do 200 volts. So we just take 200 volts divide it by eight seconds. All right, another way you know with the distance, your distance is equal to rate times time I guess you could view, this as you know production is equal to a rate times time, right.

Normally rate is how fast you’re moving per second. Now it's how much you’re producing per second. And so your total production is going to be rate, your rate times time and then your time is going to be the production divided by the rate but I think sometimes we think too much in terms of form and you loose the intuition but I just wanted to show you that that there is a formula for it if you need one but the easiest way to think of it well I have to produce 200 volts I produce, if said, I have produce a 100 every seconds it's like it take me two seconds so if I wrap the volts eight every second I take 200 divided by eight.

So 200 divided by eight and I can even do the unit’s. 200 volts divided by eight volt per second that’s equal to 200 over eight times volts times if this isn’t the denominator you just inverse it and you got, seconds per volt, the volt came sliding. So you know you got the unit’s right. So the sound degree that should give you conviction that you got the formula right and 200 divided eight is 25, so it is 25 seconds it choice B.

Problem 83, what is the decimal equivalent of one fifth to the fifth power. One fifth to the fifth power, so let's think about this, let's try it one fifth as a decimal that’s .2, so it's .2 to the fifth power. So let's just I don’t know, let's just multiply it out, .2 times .2 is equal to what?

Two times two and we have two digit’s behind the decimals, so it's .04 and then you multiply that times .2 times .2, two times four is eight and now we have three numbers behind the decimal point that’s going to be equal to .08, sorry, .008 right, one, two, three, three dot number is going to the decimal. So this is two to the third and then .008 times .2 to the right, this is to the third power that’s equal to two times eight is 16 and then how many numbers do we have behind the decimal point?

One, two, three, four so were going to have .1234 so that’s to the fourth power, we have one left so .0016 times .2 is equal to, two times 16 is 32 and now we have one, two, three, four, five numbers behind the decimal point. So one, two, three, four, five that’s the answer and that is choice A. Another way actually the easiest way to think about it is without having even doing this, you can set it all, what's two to the fifth power. Two to the fifth power is 32, right, and had you that alone because there’s only one choice it even has a 32 behind the decimal point and you could said, “Oh!” And every time to I take to power then adding one digit behind the decimal point it's going to be 32 which going to have five number behind the decimal point, so .12345 I did it the small way.

Next question 84, 90 minus eight times 20, all right the way they did, 20 divided by four all of that over one half this is just a fast simplification problem that becomes 90 minus, what 20 divided four, 20 divided four is five, five times eight is 40 all that over one half that becomes 50 over one half which is the same thing as 50 times two over one, which is equal to 100, that’s choice C. That’s just one you just have to do quickly and not make a careless mistake.

I hope I did it quickly. I don’t know if I did a careless mistake and let's see 85 a dealer originally brought a 100 identically batteries at a total cost of q dollars. So q is equal to 100 batteries if each battery was sold at 50% above the original cost, so let b is the cost to each of the battery were sold at, so b is going to be 50% above the original cost. So it's going to be 1.5 times the original cost, right, this is 50% above the original cost for battery.

And then in terms of q for who many dollars was each battery sold? Oh, they just drew us you know, let's see a dealer, a dealer originally brought a 100 identical batteries at a total cost of q dollars, if each battery was sold—oh, I'm sorry, so this is the cost right? So q is equal to 100 times what the dealer paid for it. Right, a dealer originally bought a 100 identical at a total cost of q dollars. Right, if each battery was sold at 50% above the original cost for back battery, so the selling price, this is the original cost, it's equal to 1.5 times the cost. Times the cost, in terms of q for how many dollars was each battery sold.

So let's write q in terms of the cost, so cost is equal to q divided by a 100, it's divided both sides of that equation a hundred. And then we could just substitute C or the cost here, for here. The selling price is a 150% of this. So selling price is equal to 1.5 times c which q over a hundred. And let's see how they did, so selling price is equal to 1.5 q over a hundred I don’t expect to see that as a choice, but if we multiply, if we multiply the top and the bottom by two that equals three q over 200, three q over 200 that’s choice a. Then make sure with your variables right now.

Problem 86, in increasing sequence, in an increasing sequence of 10 consecutive integers the sum of the first five integers is 560 what is the sum of the last five integers in the sequence. Okay, 10 consecutive integers, so x plus one plus x plus, we will right this way x plus one plus x plus two. Well actually the first one we must call it x, x plus three x plus four all right, this are the first—this right here is going to be the first five consecutive integers if we started x right.

When you take the sum of those that well the sum of those is one two three four five x, five x let see one plus two is three, three plus three is six, six plus four is 10. 5x plus 10 is going to be equal to, they tell us it's equal to 560, fair enough or we could say 5x is equal to 550 I think we can use that information. Now what is the next five integers going to be, it's going to be x plus five x plus six x plus seven x plus eight and then x plus nine, right this are five. And what are those at up to be, that’s five x plus five plus six is 11 plus seven is 18 plus eight is 26 plus nine, 26 plus nine is 35, so it's 5x plus 35 and they want to know, they want to know what this is equal too right, that’s the next five. We know what 5x is it's equal to 550 from this setup.

So it's going to be equal to 550 plus 35 which is too, 35 plus 550 is 585, 585 that’s choice A. And the trick to be able to do it fast is just to recognize that you didn’t have to solve all the way for x that you could reuse this 5x when you took the sum of the next five. And I'm all out of time see you in the next video.

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