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## GMAT

### Course: GMAT > Unit 1

Lesson 1: Problem solving- GMAT: Math 1
- GMAT: Math 2
- GMAT: Math 3
- GMAT: Math 4
- GMAT: Math 5
- GMAT: Math 6
- GMAT: Math 7
- GMAT: Math 8
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- GMAT: Math 10
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- GMAT: Math 13
- GMAT: Math 14
- GMAT: Math 15
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- GMAT: Math 17
- GMAT: Math 18
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- GMAT: Math 41
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- GMAT: Math 48
- GMAT: Math 49
- GMAT: Math 50
- GMAT: Math 51
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- GMAT: Math 53
- GMAT: Math 54

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# GMAT: Math 15

81-86, pgs. 162-163. Created by Sal Khan.

## Want to join the conversation?

- On problem 85, at the8:30mark, why do we multiply 1.5q/100 times 2/2?(5 votes)
- On problem 85, I don't understand why Sal multiplied each side by 2/2? I followed all the way till he decided to multiply each side by 2.Any direction is much appreciated(2 votes)

- Does anyone understand why we multiplied 1.5q/100 x 2/2 for question 85?(2 votes)
- 1.5Q/200 was not there in the options so he multiplied by 2 to get answer(1 vote)

- The easiest way that I've learned after seeing several different methods is (n) + (n+1) + (n+2) .... + (n+s) = Sum and s= last # in series. It works for even numbered sets where finding the median may be a bit trickier. Also works well for a series of odd or even consecutive integers like 10 consecutive even integers that add up to 110. n + (n+2) + (n+4) + (n+6) + (n+8) + (n+10) + (n+12) + (n+14) + (n+16) + (n+18) . 10n+90=110 or n=2. A couple tricks to speed up the calculation is recognizing the last sum is (n +2(# of integers-1). A 2nd trick is using gauss law n(n+1)/2 and tweek it a bit. 2*[n(n+1)/2] where n is the # of integers. 2[(9)(10)/2 =90 saving a lot of time.(1 vote)
- When he says 120 bolts every 40 seconds, you don't know if 3 bolts are in one seconds because they might wait until the

Last second and it generates 120?(1 vote)

## Video transcript

We're on problem 81. They give us this equation, y
is equal to 248 minus 398x. Which of the following values of
x gives the greatest value of y in the equation above? To get as large a possible y,
y is 248 minus this thing. So what we want to do is, since
we're subtracting this from 248, we want to
minimize this. As small as this can be. And even if we can make what
we're subtracting negative, that'd be awesome, because when
you subtract a negative, then that'll get us above 248. And actually, if we look at all
of the choices, they have one negative number there. You might say if we put
a 0 here, then y is going to be 248. But what if we put a negative
number in for x, they have as choice E, negative 1. What happens when we put
negative 1 there? Then y is equal to 248 minus
negative 1 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 we're subtracting a negative. That's even better than
subtracting 0, and it's way better than subtracting a
positive, if we're trying to maximize y. So the choice is E. 82. Machine A produces bolts at a
uniform rate of 120 bolts every 40 seconds. So that's the same as-- 120
bolts every 40 seconds is the same thing as 3 bolts
per second. OK, that's A. Let me write that
down, that's A. Machine B produces bolts at a
uniform rate of, so this is B, of 120 bolts every 20 seconds. Well that's equal to
5 bolts per second. If the two machines run
simultaneously, how many seconds will it take for
them to produce a total of 200 bolts? So if you're running
simultaneously, what's your speed per second? This one's going to do 3 per
second, this one's going to do 5 per second. So if they run simultaneously,
you're going to be producing 8 per second, 3 plus 5. So 8 bolts per second is
their combined rate. And we want to know how
many seconds will it take to do 200 bolts. So we just take 200 bolts
divided by 8 seconds. Another way with distance,
distance is equal to rate times time, I guess you could
view this as production is equal to rate times time. Normally rate is how fast you're
moving per second, now it's how much you're producing
per second, so your total production is going to be your
rate times time, and then your time is going to be
the production divided by the rate. But sometimes when you think too
much in terms of formulas, you lose the intuition. But I just wanted to show you
that there is a formula for it, if you need one. But the easiest way to think
it, I have to produce 200 bolts, if I said I produce
100 every second, it'll take me 2 seconds. So if I have to produce 8
every second, I take 200 divided by 8. And I can even do the units, 200
bolts divided by 8 bolts per second. That's equal to 200 over 8 times
bolts times-- if this isn't the denominator, you just
inverse it when you go out-- seconds per bolt, the
bolts can't [UNINTELLIGIBLE]. So you know you got the units
right, so to some degree that should give you conviction
that you got the formula right. And 200 divided by 8 is 25. So it's 25 seconds. It's choice B. Problem 83. What is the decimal equivalent
of 1/5 to the fifth power? So let's think about this. Let's write 1/5 as a decimal,
that's 0.2. So it's 0.2 to the 5th power. Let's just multiply it out. 0.2 times 0.2 is
equal to what? 2 times 2, and we have
two digits behind the decimal, so it's 0.04. And you multiply
that times 0.2. 2 times 4 is 8, and now we have
three numbers behind the decimal point, that's
going to be equal to 0.008 One, two, three. Three numbers behind
the decimal. So this is 2 to the third. And then 0.008 times 0.2,
let me write this is to the third power. That's equal to 2 times 8 is 16,
and then how many numbers do we have behind the
decimal point? One, two, three, four. So we're going to have point
one, two, three, four. So that's to the 4th power. We have one left, so 0.0016
times 0.2 is equal to, 2 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. Actually the easiest way to
think about it is, without even having to do this, you
could've said, what's 2 to the fifth power? 2 to the fifth power is 32. And actually that alone, because
there's only one choice that even has a 32 behind
the decimal point. And you can say, every time I
take to a power, I'm adding one digit behind the decimal
point, so it's going to be 32, but it's going to have five
numbers behind the decimal point, so point one, two,
three, four, five. I did it the slow way. Next question, 84. 90 minus 8 times 20-- I'll write
it the way they did-- 20 divided by 4, all of
that over 1/2. This is just a fast
simplification problem. That becomes 90 minus-- what's
20 divided by 4? 20 by 4 is 5. 5 times 8 is 40, all
that over 1/2. That becomes 50 over 1/2, which
is the same thing as 50 times 2 over 1, 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 did it quickly, I don't know
if I did a careless mistake. Let's see, 85. A dealer originally bought 100
identical 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 b is the cost each battery
was 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. This is
50% above the original cost per battery. Then in terms of q, for
how many dollars was each battery sold? A dealer originally bought 100
identical batteries at a total cost of q dollars. If each battery was sold--
oh I'm sorry. So this is the cost. So q is
equal to 100 times what the dealer paid for it. A dealer originally bought 100
identical batteries at the total cost of q dollars. If each battery was sold at 50%
above the original cost per battery, this is the
original cost, is equal to 1.5 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 100. Just divided both sides of
that equation by 100. And then we could just
substitute c, or the cost here, the selling price
is 150% of this. So selling price is equal
to 1.5 times c, which is q over a 100. And let's see how they did it. So selling price is equal
to 1.5q over 100. I don't expect to see
that as a choice. But if we multiply the top and
the bottom by 2, that equals 3q over 200. That's choice A. Just need to make sure you
get your variables right on that one. Problem 86. In an increasing sequence of ten
consecutive integers, the sum of the first five
integers is 560. What is the sum of the last five
integers in the sequence? OK, ten consecutive integers. So x plus 1 plus x plus-- let me
write it this way-- x plus 1 plus x plus 2, well actually
the first one, let's call it x, x plus 3, x plus 4. This right here is going to be
the first five consecutive integers if we start at x. When you take the sum of those--
well the sum of those is 1, 2, 3, 4, 5x. Let's see, 1 plus 2 is 3, 3 plus
3 is 6, 6 plus 4 is 10. 5x plus 10 is going to
be 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 5, x
plus 6, x plus 7, x plus 8, and x plus 9. These are five, and what
do those add up to be. That's 5x plus-- 5 plus 6
is 11 plus 7 is 18 plus 8 is 26 plus 9. 26 plus 9 is 35. So that's 5x plus 35. And they want to know what
this is equal to. That's the next five. Well we know what 5x
is, it's equal 550. From this set up, so it's going
to be equal to 550 plus 35, which is equal to 35
plus 550, which is 585. That's choice A. And the trick here to being 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.