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cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

by cans » Sat May 28, 2011 8:09 am

What is the value of x^2-1?

(1) The value of x^2 - 5x is -6

(2) The value of (x^2 - 1)/(x + 1) is 1

OA later.

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Source: Beat The GMAT — Data Sufficiency | Sat May 28, 2011 8:09 am

SoCan: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Sun May 15, 2011 9:19 amFollowed by: 1 member; Thanked: 18 times

by SoCan » Sat May 28, 2011 8:25 am

(1) factor x^2 - 5x + 6 into (x-3)(x-2). Two roots, so insufficient.
(2) factor the numerator so you have (x+1)(x-1). (x+1) in the numerator cancels out (x+1) in the denominator, leaving x-1=1, so x=2, allowing you to solve x^2-1.

B

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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

by manpsingh87 » Sat May 28, 2011 9:05 am

cans wrote:What is the value of x^2-1?

(1) The value of x^2 - 5x is -6

(2) The value of (x^2 - 1)/(x + 1) is 1

OA later.

x^2-1..??

1)x^2-5x=-6;
x^2-5x+6=0;
x^2-3x-2x+6=0;
x(x-3)-2(x-3)=0;
(x-3)*(x-2)=0;
x=3; or x=2;
when x=3 x^2-1=9-1=8;
when x=2; x^2-1= 4-1=3;

hence 1 alone is not sufficient to answer the question.

2)x^2-1/(x+1)=1;
x^2-1=x+1;
x^2-x-2=0;
(x-2)*(x+1)=0;
x=2;or x=-1;
now when x=-1; x^2-1/x+1 will be of the form 0/0=0; hence -1 is not the desired solution, hence only possible solution here is 2; so B alone is sufficient to answer the question..!!!

O Excellence... my search for you is on... you can be far.. but not beyond my reach!

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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 amFollowed by: 33 members; Thanked: 118 times; GMAT Score:710

by bblast » Sat May 28, 2011 10:44 am

Mr Singh, are u a math tutor or something ? You explanations are truly excellent !!!.

BTW I remember having read somewhere that we should not cancel out variables in equations until we know the sign of the unknown. Is this precisely the reason that in statement 2 you first solved the equation as a quadratic and then rejected the value x= -1 ?? you could have straight away cancelled x+1 as Socan did (i did the same) ?

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SoCan: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Sun May 15, 2011 9:19 amFollowed by: 1 member; Thanked: 18 times

by SoCan » Sat May 28, 2011 1:05 pm

bblast wrote:BTW I remember having read somewhere that we should not cancel out variables in equations until we know the sign of the unknown.

You're thinking of inequalities, not equations. In inequalities, dividing by a negative number flips the sign, and if you don't know the sign, you don't know whether to flip.

If you don't let yourself divide variables in equations without knowing the sign, there are going to be a number of problems that will be difficult to solve.

Whenever you have an expression like (x^2-y^2)/(x-y) or (x^2-y^2)/(x+y), factoring and canceling just takes out the root that would give an undefined answer (i.e., divide by zero).

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sourabh33: Master | Next Rank: 500 Posts; Posts: 135; Joined: Tue Mar 22, 2011 12:01 pmFollowed by: 3 members; Thanked: 21 times; GMAT Score:720

by sourabh33 » Sat May 28, 2011 10:19 pm

IMO, as Socan said, there is no point retaining roots in the equations unless all the elements of the equation have an unknown factor, unless the given equation cannot be further simplified , unless we do not want to loose equation properties, and unless we find it amusing to solve polynomials

For example in the equation below we can quickly identify the solution x=5 by simplifying the equation (and invariable it will be said that x is not equal to 3,2,1 as then the equation would be undefined)

(x-4).(x-3).(x-2).(x-1) / (x-3).(x-2).(x-1) = 1

Versus, the same could be solved as below

(x^2 - 7x + 12).(x^2 - 3x + 2) = (x^2 -5x +6 ).(x-1)
(x^4 -3x^3 +2x^2 -7x^3 +21x^2 -14x +12x^2 -36x +24) = (x^3 -x^2 -5x^2 +5x +6x -6)
(x^4 - 10x^3 +35x^2 -50x +24) = (x^3 -6x^2 +11x -6)
(x^4 -11x^3 +41x^2 -61x +30)=0
(x -1)*(x -2)*(x -3)*(x -5)=0
Now the factors are 1,2,3, & 5
By putting all the values and testing in the original equation (x-4).(x-3).(x-2).(x-1) / (x-3).(x-2).(x-1) = 1 we will find that only 5 is a possible solution

so, IMO, it is always preferable to simplify by cancelling in equation.

Situations in which cancelling should be not preferable are when all the terms of the equation have an unknown variable. Canceling under such situations may lead to loosing of vital solution when the unknown variable could be equal to 0

For example

By cancelling/simplifying we can obtain
x^3 - x^2 = x^2 - x
x(x^2 -x) = x(x -1)
x.x.(x-1)=x(x-1)
x = 1

But by solving and not cancelling we get
x^3 - 2x^2 + x = 0
x*(x -1)*(x -1) = 0
x = either 0 or 1
so we can see that there are benefits of not canceling, but only under certain situations

Even in inequalities, cancelling variables on the same side does not affect the sign, and should be preferable for quick solutions

Example

if
(x^2 - 1)/(x+1) > 1
then
(x-1)>1 (Cancelling (x-1) on both numerator and denominator, one has to be careful not to cancel on both the side in inequalities)
x > 2

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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

by manpsingh87 » Sat May 28, 2011 11:37 pm

sourabh33 wrote: For example in the equation below we can quickly identify the solution x=5 by simplifying the equation (and invariable it will be said that x is not equal to 3,2,1 as then the equation would be undefined)

(x-4).(x-3).(x-2).(x-1) / (x-3).(x-2).(x-1) = 1

Versus, the same could be solved as below

(x^2 - 7x + 12).(x^2 - 3x + 2) = (x^2 -5x +6 ).(x-1)
(x^4 -3x^3 +2x^2 -7x^3 +21x^2 -14x +12x^2 -36x +24) = (x^3 -x^2 -5x^2 +5x +6x -6)
(x^4 - 10x^3 +35x^2 -50x +24) = (x^3 -6x^2 +11x -6)
(x^4 -11x^3 +41x^2 -61x +30)=0
(x -1)*(x -2)*(x -3)*(x -5)=0
Now the factors are 1,2,3, & 5
By putting all the values and testing in the original equation (x-4).(x-3).(x-2).(x-1) / (x-3).(x-2).(x-1) = 1 we will find that only 5 is a possible solution

(x-4).(x-3).(x-2).(x-1) / (x-3).(x-2).(x-1) = 1;
no need to solve it by using the polynomial equation as suggested by you,, it can easily be solved by the following method without forming equations,,
(x-4).(x-3).(x-2).(x-1)=(x-3).(x-2).(x-1);
(x-4).(x-3).(x-2).(x-1)-(x-3).(x-2).(x-1);
taking (x-3).(x-2).(x-1) common we have;
(x-3).(x-2).(x-1).(x-4-1);
(x-3).(x-2).(x-1).(x-5)=0;
and to know why only (x-5) will be selected as a desired solution, one must have a basic understanding of mathematics, we all know that anything divided by zero is undefined,thus when we have x=3, x=2,x=1 then our denominator will be zero,, but since here these factors are also present in numerator therefore numerator will also turn into zero, hence our result becomes 0/0=0; otherwise
it would have been, undefined...!!!

yes its safe to cancel out factors in this specific case but its not advisable, because in real exam you might carry this instinct of canceling factors in those cases where canceling the factors will result into the different solution, so its better to develop good problem skill that will do the wonders for you in the real exam.
for example consider the following case;

2^18 when divided when divided by 12;
if you will cancel out the common factors you will see that problem reduces to 2^16/3; which leaves a remainder of 1; so is remainder actually 1..??? well the answer is not 1, its 4, because we've failed to incorporate the cancelling factor 4 while giving final answer..!!!

otherwise its completely up to you,,, do whatever you want to and follow whatever you have to its your wish...!!!

O Excellence... my search for you is on... you can be far.. but not beyond my reach!

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sourabh33: Master | Next Rank: 500 Posts; Posts: 135; Joined: Tue Mar 22, 2011 12:01 pmFollowed by: 3 members; Thanked: 21 times; GMAT Score:720

by sourabh33 » Sun May 29, 2011 12:41 am

Nice post manpansingh!

Before anything,

and to know why only (x-5) will be selected as a desired solution, one must have a basic understanding of mathematics, we all know that anything divided by zero is undefined,thus when we have x=3, x=2,x=1 then our denominator will be zero,, but since here these factors are also present in numerator therefore numerator will also turn into zero, hence our result becomes 0/0=0; otherwise
it would have been, undefined...!!!

0/0 = 0!!!!, IMO, even 0/0 is undefined, at least on gmat; so even if these factors are present in numerator the result will still be undefined so we can safely ignore 3,2,1

https://www.manhattangmat.com/forums/num ... t4998.html
https://en.wikipedia.org/wiki/Division_by_zero
https://www.newton.dep.anl.gov/askasci/m ... h99259.htm

Second,

(x-4).(x-3).(x-2).(x-1) / (x-3).(x-2).(x-1) = 1;
no need to solve it by using the polynomial equation as suggested by you,, it can easily be solved by the following method without forming equations,,
(x-4).(x-3).(x-2).(x-1)=(x-3).(x-2).(x-1);
(x-4).(x-3).(x-2).(x-1)-(x-3).(x-2).(x-1);
taking (x-3).(x-2).(x-1) common we have;
(x-3).(x-2).(x-1).(x-4-1);
(x-3).(x-2).(x-1).(x-5)=0;

This example was intentionally presented for the ease of explanation, calculation and understanding. You may appreciate the complexity when one has to retain the undefined roots, instead of canceling them out in the following example

(x-4).(x-3).(x-2).(x-1) / (x-3).(x-2).(x-1) - (m-24).(m^3-267).(m^1/2-233).(m^33-233) / (m^3-267).(m^1/2-233).(m^33-233)= (y-4).(y^3-67).(y^1/2-33).(y^33-33) / (y^3-67).(y^1/2-33).(y^33-33) + (z-4).(z^3-67).(z^1/2-33).(z^33-33) / (z^3-67).(z^1/2-33).(z^33-33) + 99

This could be simple reduced to

(x-4) - (m-24) = (y-4) + (z-4) + 99, but if one has to retain the roots then one unnecessarily has to waste precious time when it is certain that the other solutions will be undefined.

Third,

2^18 when divided when divided by 12; if you will cancel out the common factors you will see that problem reduces to 2^16/3; which leaves a remainder of 1; so is remainder actually 1..??? well the answer is not 1, its 4, because we've failed to incorporate the cancelling factor 4 while giving final answer..!!!

2^18/12 -> for this the remainder is 4 because this question asks to find out the remainder when divided by 4 and not the remainder when 2^16/3; However -> 2^18/12 = 2^16/3

Simplifying your example

When 8 is divided by 6 the remainder is 2, versus when 4 is divided by 3 the remainder is 1

There is no discrepancy in above because the question is asking to find a specific remainder when divided by a particular number and not their reduced form

However

Z^213 + 2^18/12 = z^213 + 2^16/3

Finally, I think your posts are one of very best of all. I was just trying to present my viewpoint, but, as you said, its completely up to you,,, do whatever you want to and follow whatever you have to its your wish...!!!

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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

by manpsingh87 » Sun May 29, 2011 7:07 am

sourabh33 wrote:Nice post manpansingh!

Before anything,
and to know why only (x-5) will be selected as a desired solution, one must have a basic understanding of mathematics, we all know that anything divided by zero is undefined,thus when we have x=3, x=2,x=1 then our denominator will be zero,, but since here these factors are also present in numerator therefore numerator will also turn into zero, hence our result becomes 0/0=0; otherwise
it would have been, undefined...!!!
0/0 = 0!!!!, IMO, even 0/0 is undefined, at least on gmat; so even if these factors are present in numerator the result will still be undefined so we can safely ignore 3,2,1

https://www.manhattangmat.com/forums/num ... t4998.html
https://en.wikipedia.org/wiki/Division_by_zero
https://www.newton.dep.anl.gov/askasci/m ... h99259.htm

Second,
(x-4).(x-3).(x-2).(x-1) / (x-3).(x-2).(x-1) = 1;
no need to solve it by using the polynomial equation as suggested by you,, it can easily be solved by the following method without forming equations,,
(x-4).(x-3).(x-2).(x-1)=(x-3).(x-2).(x-1);
(x-4).(x-3).(x-2).(x-1)-(x-3).(x-2).(x-1);
taking (x-3).(x-2).(x-1) common we have;
(x-3).(x-2).(x-1).(x-4-1);
(x-3).(x-2).(x-1).(x-5)=0;
This example was intentionally presented for the ease of explanation, calculation and understanding. You may appreciate the complexity when one has to retain the undefined roots, instead of canceling them out in the following example

(x-4).(x-3).(x-2).(x-1) / (x-3).(x-2).(x-1) - (m-24).(m^3-267).(m^1/2-233).(m^33-233) / (m^3-267).(m^1/2-233).(m^33-233)= (y-4).(y^3-67).(y^1/2-33).(y^33-33) / (y^3-67).(y^1/2-33).(y^33-33) + (z-4).(z^3-67).(z^1/2-33).(z^33-33) / (z^3-67).(z^1/2-33).(z^33-33) + 99

This could be simple reduced to

(x-4) - (m-24) = (y-4) + (z-4) + 99, but if one has to retain the roots then one unnecessarily has to waste precious time when it is certain that the other solutions will be undefined.

Third,
2^18 when divided when divided by 12; if you will cancel out the common factors you will see that problem reduces to 2^16/3; which leaves a remainder of 1; so is remainder actually 1..??? well the answer is not 1, its 4, because we've failed to incorporate the cancelling factor 4 while giving final answer..!!!
2^18/12 -> for this the remainder is 4 because this question asks to find out the remainder when divided by 4 and not the remainder when 2^16/3; However -> 2^18/12 = 2^16/3

Simplifying your example

When 8 is divided by 6 the remainder is 2, versus when 4 is divided by 3 the remainder is 1

There is no discrepancy in above because the question is asking to find a specific remainder when divided by a particular number and not their reduced form

However

Z^213 + 2^18/12 = z^213 + 2^16/3

Finally, I think your posts are one of very best of all. I was just trying to present my viewpoint, but, as you said, its completely up to you,,, do whatever you want to and follow whatever you have to its your wish...!!!

well i'll say a nice and well researched post..!!! good job....!!

The value of (x^2 - 1)/(x + 1) is 1 ;

this actually is a remainder question..!! well as we know that,, Numerator= Quotient*Denominator+Remainder;
here quotient=1 and remainder =0; that is why i intentionally, give 2^18/12 example..!!! i hope you understand the point that i'm trying to make..!!!

any ways good job and all the best,, and thanks a lot for expressing your views in a respectful tone..!!!

O Excellence... my search for you is on... you can be far.. but not beyond my reach!

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itsmebharat: Master | Next Rank: 500 Posts; Posts: 150; Joined: Tue Aug 18, 2009 7:36 amFollowed by: 2 members; Location: gurgaon; Thanked: 8 times

by itsmebharat » Mon May 30, 2011 12:49 am

IMO B

I am not an Expert, please feel free to suggest if there is an error.

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