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## GMAT

### Course: GMAT > Unit 1

Lesson 1: Problem solving- GMAT: Math 1
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- GMAT: Math 47
- GMAT: Math 48
- GMAT: Math 49
- GMAT: Math 50
- GMAT: Math 51
- GMAT: Math 52
- GMAT: Math 53
- GMAT: Math 54

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# GMAT: Math 53

245-247, pg. 186. Created by Sal Khan.

## Want to join the conversation?

- can anyone explain:

If a-b=p and a+b=k, then a^2-b^2=

A) pk

B) p^2-k^2

C) p+k

D)p^2/k^2

E) K^2-p^2(0 votes)- a - b = p --- (1)

a + b = k --- (2)

so (1) + (2); 2a = p + k --> a = (p + k)/2 --> a^2 = (1/4)( p+k)^2 = (1/4)(p^2 +2pk +k^2)

and (2) - (1); 2b = k - p --> b = (k -p)/2 --> b^2 = (1/4)( k-p)^2 = (1/4)(k^2 -2pk +p^2)

finally a^2 - b^2 = (1/4)(p^2 +2pk + k^2 - k^2 + 2pk - p^2) =(1/4)( 4pk) = pk

This answer is A).(2 votes)

- how is it possible to be able to switch the equation around and yet be able to find the answer. i really dont understand............ why arent all the sides the same too?(1 vote)

## Video transcript

We're on problem 245. If x plus y is equal to a, and
x minus y is equal to b, then what is-- they want to know
what 2xy is equal to. Let's solve for x and y in terms
of a and b, and then just figure out what
this equals to. So we have two equations
with two unknowns. Let's just add them together
to solve for x. We get x plus x is 2x-- the
y's cancel out-- is equal to a plus b. Or x is equal to a
plus b over 2. Now if we had x plus
y is equal to a-- I just re-wrote this. And now if we multiply both
sides of this by negative 1, so we get minus x plus y
is equal to minus b. And now add these
two equations. So I'm essentially subtracting
this equation from that one. That cancels out, so I get 2y is
equal to a minus b, or y is equal to a minus b over 2. And now we can figure out
what 2xy is equal to. 2xy is equal to 2 times a
plus b over 2, times a minus b over 2. This 2 will cancel out with
one of these 2's. And we're left with--
and what's a plus b times a minus b? It's a squared minus
b squared over 2. Which is choice A. Problem 246. Let me do it in magenta. 246. A rectangular circuit board is
to have width w inches-- let me draw it. So let's say it has width
w inches, perimeter of p inches-- so let me just put that
at the side right here-- perimeter of p inches, and
area of k square inches. Which of the following equations
must be true? And so they want us to
relate this width to the area to the perimeter. Let me introduce another
variable. Let's call this, right here,
let's call this the height of the circuit board. So now we can do some
interesting things. If that's the height, then
that's also the height. If that's the width, then
this is also the width. So let's see what the
perimeter would be. It will be 2 times the width,
plus 2 times the height is equal to the perimeter. And then we could also
say width times height is equal to area. But if we can solve for height
in terms of the perimeter and the width, then we could use
that to get an expression that doesn't involve this variable. So let's do that. So if you divide both sides of
this by 2, you get width plus height is equal to
perimeter over 2. And then you get height is
equal to perimeter over 2 minus width. And so the area, k, will
be equal to the width times the height. Instead of writing an h there,
let's write what we just figured out. p over 2 minus w. And then that is equal to
pw/2 minus w squared. Let's see, when I look at the
solution, they don't have any fractions in it, so let's
multiple both sides of the equation by 2. We can ignore this. k is equal
to pw/2 minus w squared. Multiply both sides by 2,
you get 2k is equal to pw minus w squared. Let's add w squared
the both sides. You get w squared plus
2k is equal to pw. Let's subtract, because all of
the choices have them setting equal to 0. So then we could subtract pw
from both sides, and you get-- oh wait, I made a mistake. If we multiply both sides of
this by 2, 2 times k is 2k. 2 times pw/2 is pw. 2 times minus w squared
is minus 2w squared. So in this step we have to add
2w squared to both sides. Sorry about that. So we have 2w squared plus
2k is equal to pw. Subtract pw from both sides, you
get 2w squared minus pw, plus 2k is equal to 0. And that is choice E, right? 2w squared minus pw plus 2k. That is choice E. Next question. Problem 247. An arithmetic sequence is a
sequence in which each term after the first is equal to the
sum of the preceding term and a constant. If the list of letters shown
above is an arithmetic sequence-- so they wrote
p, r, s, t, u. So all they're saying is that,
the difference between p and r is going to be some number. And the difference between
r and s is going to be that same number. And the difference between
s and t is going to be that same number. So an example of an arithmetic
sequence, this could be 1, 2, 3, 4, 5. Because every number is one more
than the one before it. So anyway, they say which of the
following must also be an arithmetic sequence? So choice one, they write
2p, 2r, 2s, 2t, 2u. Well, let's just use
our example. If this was p, r, s, t, and u,
what is 2 times all of that? Well, then it'll be 2, 4-- no
no, sorry-- it'll be, yeah, 2, 4, 6, 8, 10. So now instead of incrementing
by one every time, we're incrementing by two
every time. But it's still an arithmetic
sequence because the difference between each
number and the number before it is a constant. It's always equal to 2. So choice number one
is definitely an arithmetic sequence. Choice two. And if you say, well, that works
for that example, how do I know it'll work for
all examples? So the other way to think
about is, whatever the difference is between p and r,
now you're going to have twice the difference between
2p and 2r. And it was the same distance
between r and s, now you're going to have twice that
distance between 2r and 2s. So here in our particular
example, we went from 1 to 2, but it could have been
something else. OK, Statement two says, p
minus 3, r minus 3-- so they're just shifting
everything; I don't even have to write them alll-- all
the way to u minus 3. So they just took everything in
this sequence and made them three less. But if r minus p is equal to
some number, r minus 3, minus p minus 3 is going to be
the exact same thing. I can even prove that
to you, right? r minus p-- sorry, r minus
3, minus p minus 3. That's equal to r minus
3, minus p plus 3. And so these cancel out. So the difference between this
and this ends up to be r minus p, the difference between
that and that. And that should make sense
intuitively, we're just shifting all the numbers
down by three. So that shouldn't change the difference between the numbers. So two is still an arithmetic
sequence. Statement three. p squared, r squared-- so we're
just squaring all the numbers-- t squared,
and u squared. So let's just use our
particular example. If p was 1, then p squared is
1, 2 squared is 4, s squared is 9-- right, 3 squared is
nine, 4 squared is 16. Now what's the difference
between the numbers? The difference here is three. The difference here is five. The difference here is seven. And this is interesting in and
of itself, that-- well, first of all, let's just answer
our question. The difference is now
not constant. We have a different difference
between each successive number, right? At the beginning, the difference
is two every time. Here it's three, then it changes
to five, then it changes to seven. So three is not an arithmetic
sequence. So the answer is
D, one and two. This is something interesting. And if you've never experimented
with it, this is something to look at. I've always been fascinated by
the distance between the perfect squares increases by
increasing odd numbers, which is just something
to think about. Anyway, next problem. Actually, I have two problems
left out of all of the problem-solving problems, so
instead of just doing one problem and going over time, let
me do two of them in the the next video. See you soon.