Very Very Basic Math Question : Quant Question Archive [LOCKED]
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# Very Very Basic Math Question

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Director
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Very Very Basic Math Question [#permalink]

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11 Jun 2007, 06:04
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Hi All,

I have a very basic math question I was hoping all could answer.

According to the ManhattanGMAT Number Properties guide the following is true:

(3^3)(5^3) = 15^3

Basically, when multiplying expressions with the same exp, multiple the bases first. Ok, so please explain this:

(r^3)(r^3)=?

That should equal r^6, but according to ManhattanGMAT rules wouldn't it be r^3 (incorrect)?

Thanks.
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11 Jun 2007, 06:19
The reasonning is correct:

o (3^3)(5^3)
= (5*3)^3
= (15)^3

o (r^3)(r^3)
= (r*r)^3
= (r^2)^3
= r^(2*3)
= r^6

Hope that helps
VP
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11 Jun 2007, 06:27
(r^3)(r^3)= r*r*r*r*r*r = r^6 -

I can't see how you can get r^3

Last edited by KillerSquirrel on 11 Jun 2007, 06:28, edited 1 time in total.
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11 Jun 2007, 06:27

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))

I first divide the fraction and then multiple getting:
m^9 * p^8 * r^3 (r^3 since doing it this way we have like r terms and exp so I keep the exp)

OA is m^9 * p^8 * r^6, because they multiplied first and then divided. Please explain why we get different answers based on how the questions is solved.
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11 Jun 2007, 06:33
jimmyjamesdonkey wrote:

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))

I first divide the fraction and then multiple getting:
m^9 * p^8 * r^3 (r^3 since doing it this way we have like r terms and exp so I keep the exp)

OA is m^9 * p^8 * r^6, because they multiplied first and then divided. Please explain why we get different answers based on how the questions is solved.

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4)) =

(m^5)(p^6)(r^3) * (p^2)(r^3)(m^4)=

(m^9)(p^8)(r^6)

since m^2/m = m*m/m = m

and since r^3*r^2 = r*r*r*r*r = r^5

Director
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11 Jun 2007, 06:38
I guess there is two rules going on at the same time!

Rule 1: When multiplying expressions with the same base, ADD the exp first.
Rule 2: When multiplying expression with the same exp, multiply bases first.

In this case r3 * r3 we have BOTH the same base and same exponent! So I guess when that happens, we add the exponents (like the rule says "first", duh)

FYI...I knew how to do this with no prob BEFORE reading the prep book and confusing me
Director
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11 Jun 2007, 06:42
To answer this question, you may use the following algebric property:

x^y/x^m = x^y . x^-m = x^(y-m)

use this for simplifying the early experession between the square brackets, and then simply multiply the variables [ by summing the exponents ] to arrive at the correct OA answer you provided. If you need me to show you the arithmatic in details, I'll more than glad to do so.
VP
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11 Jun 2007, 06:44
jimmyjamesdonkey wrote:
I guess there is two rules going on at the same time!

Rule 1: When multiplying expressions with the same base, ADD the exp first.
Rule 2: When multiplying expression with the same exp, multiply bases first.

In this case r3 * r3 we have BOTH the same base and same exponent! So I guess when that happens, we add the exponents (like the rule says "first", duh)

FYI...I knew how to do this with no prob BEFORE reading the prep book and confusing me

r^3*r^3 = r*r*r*r*r*r = r^6 meaning - rule one

r^3*m^3 = r*r*r*m*m*m = (r*m)*(r*m)*(r*m) = (r*m)^3 meaning rule 2

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11 Jun 2007, 06:45
jimmyjamesdonkey wrote:
I guess there is two rules going on at the same time!

Rule 1: When multiplying expressions with the same base, ADD the exp first.
Rule 2: When multiplying expression with the same exp, multiply bases first.

In this case r3 * r3 we have BOTH the same base and same exponent! So I guess when that happens, we add the exponents (like the rule says "first", duh)

FYI...I knew how to do this with no prob BEFORE reading the prep book and confusing me

The bold is right but u could have the different way to make it as the one used by MGMAT....

(r^2)^3 = r^(2*3) = r^(3*2) = (r^3)^2 = r^6

It's important to keep in mind that both exist ... It's sometimes tested
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11 Jun 2007, 06:50
jimmyjamesdonkey wrote:

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))

I first divide the fraction and then multiple getting:
m^9 * p^8 * r^3 (r^3 since doing it this way we have like r terms and exp so I keep the exp)

OA is m^9 * p^8 * r^6, because they multiplied first and then divided. Please explain why we get different answers based on how the questions is solved.

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))
= m^a * p^b * r^c

Where u count for a, b and c :
o plus an exponent if in the numerator
o minus an exponent if in the denominator

So,
o a = 8 - 3 + 4 = 9
o b = 7 - 1 + 2 = 8
o c = 12 - 9 + 3 = 6
VP
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11 Jun 2007, 06:52
Fig wrote:
jimmyjamesdonkey wrote:

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))

I first divide the fraction and then multiple getting:
m^9 * p^8 * r^3 (r^3 since doing it this way we have like r terms and exp so I keep the exp)

OA is m^9 * p^8 * r^6, because they multiplied first and then divided. Please explain why we get different answers based on how the questions is solved.

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))
= m^a * p^b * r^c

Where u count for a, b and c :
o plus an exponent if in the numerator
o minus an exponent if in the denominator

So,
o a = 8 - 3 + 4 = 9
o b = 7 - 1 + 2 = 8
o c = 12 - 9 + 3 = 6

good explanation - Fig

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11 Jun 2007, 06:59
KillerSquirrel wrote:
Fig wrote:
jimmyjamesdonkey wrote:

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))

I first divide the fraction and then multiple getting:
m^9 * p^8 * r^3 (r^3 since doing it this way we have like r terms and exp so I keep the exp)

OA is m^9 * p^8 * r^6, because they multiplied first and then divided. Please explain why we get different answers based on how the questions is solved.

{ ( (m^8)(p^7)(r^12) ) / ( (m^3)(r^9)(p) ) } * ((p^2)(r^3)(m^4))
= m^a * p^b * r^c

Where u count for a, b and c :
o plus an exponent if in the numerator
o minus an exponent if in the denominator

So,
o a = 8 - 3 + 4 = 9
o b = 7 - 1 + 2 = 8
o c = 12 - 9 + 3 = 6

good explanation - Fig

Thanks
11 Jun 2007, 06:59
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