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Source: — Data Sufficiency |

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by Anurag@Gurome » Thu May 26, 2011 4:16 am
ruplun wrote:What is the value of integer x?

(1) 4<(x-1)*(x-1)<16

(2) 4<(x-1)*(x+1)<16
(1) 4 < (x-1)² < 16 implies 2 < (x - 1) < 4 implies 3 < x < 5
x = 4
So, (1) is SUFFICIENT.

(2) 4 < (x-1)*(x+1) < 16 implies 4 < (x² - 1) < 16 implies x = 3, 4
We don't get a definite answer.
So, (2) is NOT sufficient.

The correct answer is A.
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by sourabh33 » Thu May 26, 2011 4:37 am
Anurag@Gurome wrote:
ruplun wrote:What is the value of integer x?

(1) 4<(x-1)*(x-1)<16

(2) 4<(x-1)*(x+1)<16
(1) 4 < (x-1)² < 16 implies 2 < (x - 1) < 4 implies 3 < x < 5
x = 4
So, (1) is SUFFICIENT.

(2) 4 < (x-1)*(x+1) < 16 implies 4 < (x² - 1) < 16 implies x = 3, 4
We don't get a definite answer.
So, (2) is NOT sufficient.

The correct answer is A.

Hi Anurag

A small doubt

For Equation 1

4 < (x-1)² < 16

if x = -2, will it not satisfy the equation above
And if yes, then Statement 1 should be insufficient

Possible values of x - -2, 4

For Equation 2

The possible values of x could be -3,-4,3,4

Therefore this is insufficient

Solving For Equation 1 & 2 together

x = 4, therefore sufficient
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by Frankenstein » Thu May 26, 2011 8:33 am
sourabh33 wrote:
Anurag@Gurome wrote:
ruplun wrote:What is the value of integer x?

(1) 4<(x-1)*(x-1)<16

(2) 4<(x-1)*(x+1)<16
(1) 4 < (x-1)² < 16 implies 2 < (x - 1) < 4 implies 3 < x < 5
x = 4
So, (1) is SUFFICIENT.

(2) 4 < (x-1)*(x+1) < 16 implies 4 < (x² - 1) < 16 implies x = 3, 4
We don't get a definite answer.
So, (2) is NOT sufficient.

The correct answer is A.

Hi Anurag

A small doubt

For Equation 1

4 < (x-1)² < 16

if x = -2, will it not satisfy the equation above
And if yes, then Statement 1 should be insufficient

Possible values of x - -2, 4

For Equation 2

The possible values of x could be -3,-4,3,4

Therefore this is insufficient

Solving For Equation 1 & 2 together

x = 4, therefore sufficient
Yes..You are right.. Answer should be C .

Cheers!
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by manpsingh87 » Thu May 26, 2011 8:53 am
ruplun wrote:What is the value of integer x?

(1) 4<(x-1)*(x-1)<16

(2) 4<(x-1)*(x+1)<16
1)
4<(x-1)^2<16;
(x-1)^2>4;
(x-1)^2-4>0;
now using the formula a^2-b^2=(a-b)*(b-a); we have;
(x-1-2)*(x-1+2)>0;
(x-3)(x+1)>0;
i.e. x-3 and x+3 are of same sign; and this inequality will hold true for x<-1 and x>3;
now the only integer for x<-1; which will satisfy the inequality 4<(x-1)^2<16; is x=-2;
and the only integer for x>3; which will satisfy the inequality 4<(x-1)^2<16 is x=3;
as here two different values are possible for x; hence 1) alone is not sufficient to answer the question...!!!

2) also 4<(x^2-1)<16;
now this inequality will hold true for x=-3,3,-4,4; hence 2 alone is also not sufficient to answer the question..!!!

combining 1 and 2 we get common solution x=4; hence it should be C
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by uslalas22 » Thu May 26, 2011 7:48 pm
manpsingh87 wrote:
ruplun wrote:What is the value of integer x?

(1) 4<(x-1)*(x-1)<16

(2) 4<(x-1)*(x+1)<16
1)
4<(x-1)^2<16;
(x-1)^2>4;
(x-1)^2-4>0;
now using the formula a^2-b^2=(a-b)*(b-a); we have;
(x-1-2)*(x-1+2)>0;
(x-3)(x+1)>0;
i.e. x-3 and x+3 are of same sign; and this inequality will hold true for x<-1 and x>3;
now the only integer for x<-1; which will satisfy the inequality 4<(x-1)^2<16; is x=-2;
and the only integer for x>3; which will satisfy the inequality 4<(x-1)^2<16 is x=3;
as here two different values are possible for x; hence 1) alone is not sufficient to answer the question...!!!

2) also 4<(x^2-1)<16;
now this inequality will hold true for x=-3,3,-4,4; hence 2 alone is also not sufficient to answer the question..!!!

combining 1 and 2 we get common solution x=4; hence it should be C
Manpsingh I think you have your formulas mixed up. a^2-b^2 does not equal (a-b)*(b-a), in fact it equals (a-b)(a+b) which would be relevant to the second statement, but since its "1" as noted in the solution provided, easier just to recognize x2-1.
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by manpsingh87 » Thu May 26, 2011 10:09 pm
uslalas22 wrote:
manpsingh87 wrote:
ruplun wrote:What is the value of integer x?

(1) 4<(x-1)*(x-1)<16

(2) 4<(x-1)*(x+1)<16
1)
4<(x-1)^2<16;
(x-1)^2>4;
(x-1)^2-4>0;
now using the formula a^2-b^2=(a-b)*(b-a); we have;
(x-1-2)*(x-1+2)>0;
(x-3)(x+1)>0;
i.e. x-3 and x+3 are of same sign; and this inequality will hold true for x<-1 and x>3;
now the only integer for x<-1; which will satisfy the inequality 4<(x-1)^2<16; is x=-2;
and the only integer for x>3; which will satisfy the inequality 4<(x-1)^2<16 is x=3;
as here two different values are possible for x; hence 1) alone is not sufficient to answer the question...!!!

2) also 4<(x^2-1)<16;
now this inequality will hold true for x=-3,3,-4,4; hence 2 alone is also not sufficient to answer the question..!!!

combining 1 and 2 we get common solution x=4; hence it should be C
Manpsingh I think you have your formulas mixed up. a^2-b^2 does not equal (a-b)*(b-a), in fact it equals (a-b)(a+b) which would be relevant to the second statement, but since its "1" as noted in the solution provided, easier just to recognize x2-1.
if you didn't able to comprehend it than it doesn't mean its wrong,, a^2-b^2 indeed is equal to (a-b)*(a+b);
lets apply this to the equation (x-1)^2-2^2; here a=x-1 and b=2;
now a-b= ((x-1)-2); (x-3);
a+b= ((x-1)+2); (x+1);

which is exactly what i've written,,,
Manpsingh I think you have your formulas mixed up
;
i think i'm correct as i have proven myself above.!!!
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by kumadil2011 » Tue Nov 22, 2011 11:35 am
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by shekhar.kataria » Mon Dec 12, 2011 5:52 pm
Hi Anurag

The Answer to the Quoted statement would be not sufficient to answer because both -2 and 4 can satisfy the Equation. This is contary to your explanation.

Original Answer Please.

(1) 4 < (x-1)² < 16 implies 2 < (x - 1) < 4 implies 3 < x < 5
x = 4
So, (1) is SUFFICIENT.
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