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distributive property - silly question

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Gurpinder: Legendary Member; Posts: 659; Joined: Mon Dec 14, 2009 8:12 amFollowed by: 3 members; Thanked: 32 times

by Gurpinder » Thu Sep 02, 2010 10:26 am

xd-(x-w)m

are we suppose to distribute the M or do x-w first and then multiply by M?

thanks all

"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.

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Source: Beat The GMAT — Problem Solving | Thu Sep 02, 2010 10:26 am

puneetdua: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Wed Jun 16, 2010 7:57 pm; Location: Delhi,India; Thanked: 1 times

by puneetdua » Thu Sep 02, 2010 10:49 am

We will first solve the expression inside the Bracket and then multiply my M.

Please post the Question , if you have any doubt in that ..it might help us here to understand one more new funda

Thanks
Puneet

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Gurpinder: Legendary Member; Posts: 659; Joined: Mon Dec 14, 2009 8:12 amFollowed by: 3 members; Thanked: 32 times

by Gurpinder » Thu Sep 02, 2010 10:52 am

puneetdua wrote:We will first solve the expression inside the Bracket and then multiply my M.

Please post the Question , if you have any doubt in that ..it might help us here to understand one more new funda

In MGMAT book, they went from xd-(x-w)m --> xd - xm + wm

how?

"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.

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niksworth: Legendary Member; Posts: 520; Joined: Mon Jun 14, 2010 10:44 amFollowed by: 6 members; Thanked: 70 times

by niksworth » Thu Sep 02, 2010 10:53 am

Both will result in the same value! However, always follow the BODMAS rule.

E.g. x=5, d=4, w=3, m=2

Distribute m first

xd-(x-w)m
=5*4 - (5-3)*2
=20 - (5*2 - 3*2)
=20 - (10 - 6)
=20 - 4
=16

Subtract first
xd-(x-w)m
=5*4 - (5-3)*2
=20 - 2*2
=20 - 4
=16

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niksworth: Legendary Member; Posts: 520; Joined: Mon Jun 14, 2010 10:44 amFollowed by: 6 members; Thanked: 70 times

by niksworth » Thu Sep 02, 2010 10:55 am

Gurpinder wrote:
puneetdua wrote:We will first solve the expression inside the Bracket and then multiply my M.

Please post the Question , if you have any doubt in that ..it might help us here to understand one more new funda
In MGMAT book, they went from xd-(x-w)m --> xd - xm + wm

how?

xd - (x-w)m
=xd - (xm - wm)
=xd - xm + wm (sign changes on opening brackets if preceded by a negative sign)

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Gurpinder: Legendary Member; Posts: 659; Joined: Mon Dec 14, 2009 8:12 amFollowed by: 3 members; Thanked: 32 times

by Gurpinder » Thu Sep 02, 2010 11:05 am

niksworth wrote:
Gurpinder wrote:
puneetdua wrote:We will first solve the expression inside the Bracket and then multiply my M.

Please post the Question , if you have any doubt in that ..it might help us here to understand one more new funda
In MGMAT book, they went from xd-(x-w)m --> xd - xm + wm

how?
xd - (x-w)m
=xd - (xm - wm)
=xd - xm + wm (sign changes on opening brackets if preceded by a negative sign)

alright!

so I guess the rule is:
x(a+b) = xa+xb
(a+b)x = xa+xb

x can be in front or at the back!

Correct?

"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.

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niksworth: Legendary Member; Posts: 520; Joined: Mon Jun 14, 2010 10:44 amFollowed by: 6 members; Thanked: 70 times

by niksworth » Thu Sep 02, 2010 11:31 am

Gurpinder wrote:
niksworth wrote:
Gurpinder wrote:
puneetdua wrote:We will first solve the expression inside the Bracket and then multiply my M.

Please post the Question , if you have any doubt in that ..it might help us here to understand one more new funda
In MGMAT book, they went from xd-(x-w)m --> xd - xm + wm

how?
xd - (x-w)m
=xd - (xm - wm)
=xd - xm + wm (sign changes on opening brackets if preceded by a negative sign)
alright!

so I guess the rule is:
x(a+b) = xa+xb
(a+b)x = xa+xb

x can be in front or at the back!

Correct?

I think you are getting unnecessarily confused.

x(a+b) = xa+xb = ax + bx = ax + xb = xa +bx
(a+b)x = xa+xb = ax + bx = ax + xb = xa +bx

Read about the following on the net and your concepts will be clear -
1) Commutative law
2) Associate law
3) Distributive law

Check the following link.
https://www.mathsisfun.com/associative-c ... utive.html

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Gurpinder: Legendary Member; Posts: 659; Joined: Mon Dec 14, 2009 8:12 amFollowed by: 3 members; Thanked: 32 times

by Gurpinder » Thu Sep 02, 2010 11:35 am

Umm...I think I might have not been clear.

x(a+b) = xa+xb
(a+b)x = xa+xb

i meant there. regardless of whether the x is in front of the bracket or behind it, you are to distribute the x?

clear?

"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.

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niksworth: Legendary Member; Posts: 520; Joined: Mon Jun 14, 2010 10:44 amFollowed by: 6 members; Thanked: 70 times

by niksworth » Thu Sep 02, 2010 11:46 am

Gurpinder wrote:Umm...I think I might have not been clear.

x(a+b) = xa+xb
(a+b)x = xa+xb

i meant there. regardless of whether the x is in front of the bracket or behind it, you are to distribute the x?

clear?

Yeah. Absolutely!

E.g.
5*(3+2) = (5*3 + 5*2) = (3*5+2*5) = 15 + 10 = 25
(3+2)*5 = (3*5 + 2*5) = (5*3 + 5*2) = 15 + 10 = 25

Order of multiplication does not matter [By commutative law]

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Gurpinder: Legendary Member; Posts: 659; Joined: Mon Dec 14, 2009 8:12 amFollowed by: 3 members; Thanked: 32 times

by Gurpinder » Thu Sep 02, 2010 11:48 am

niksworth wrote:
Gurpinder wrote:Umm...I think I might have not been clear.

x(a+b) = xa+xb
(a+b)x = xa+xb

i meant there. regardless of whether the x is in front of the bracket or behind it, you are to distribute the x?

clear?
Yeah. Absolutely!

E.g.
5*(3+2) = (5*3 + 5*2) = (3*5+2*5) = 15 + 10 = 25
(3+2)*5 = (3*5 + 2*5) = (5*3 + 5*2) = 15 + 10 = 25

Order of multiplication does not matter [By commutative law]

Thanks!

"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.

Like this post

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