Quantitative Aptitude Problem Solving - 8 Video Lecture

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.