BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

distributive property - silly question

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 659
Joined: Mon Dec 14, 2009 8:12 am
Thanked: 32 times
Followed by:3 members

distributive property - silly question

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.
Top
Source: — Problem Solving |
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
Top
User avatar
Legendary Member
Posts: 659
Joined: Mon Dec 14, 2009 8:12 am
Thanked: 32 times
Followed by:3 members

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.
Top
Legendary Member
Posts: 520
Joined: Mon Jun 14, 2010 10:44 am
Thanked: 70 times
Followed by:6 members

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
Top
Legendary Member
Posts: 520
Joined: Mon Jun 14, 2010 10:44 am
Thanked: 70 times
Followed by:6 members

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)
Top
User avatar
Legendary Member
Posts: 659
Joined: Mon Dec 14, 2009 8:12 am
Thanked: 32 times
Followed by:3 members

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.
Top
Legendary Member
Posts: 520
Joined: Mon Jun 14, 2010 10:44 am
Thanked: 70 times
Followed by:6 members

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
Top
User avatar
Legendary Member
Posts: 659
Joined: Mon Dec 14, 2009 8:12 am
Thanked: 32 times
Followed by:3 members

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.
Top
Legendary Member
Posts: 520
Joined: Mon Jun 14, 2010 10:44 am
Thanked: 70 times
Followed by:6 members

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]
Top
User avatar
Legendary Member
Posts: 659
Joined: Mon Dec 14, 2009 8:12 am
Thanked: 32 times
Followed by:3 members

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.
Top
Post new topic Post Reply
• Page 1 of 1