(1) 4 < (x-1)² < 16 implies 2 < (x - 1) < 4 implies 3 < x < 5ruplun wrote:What is the value of integer x?
(1) 4<(x-1)*(x-1)<16
(2) 4<(x-1)*(x+1)<16
Anurag@Gurome wrote:(1) 4 < (x-1)² < 16 implies 2 < (x - 1) < 4 implies 3 < x < 5ruplun wrote:What is the value of integer x?
(1) 4<(x-1)*(x-1)<16
(2) 4<(x-1)*(x+1)<16
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.
Yes..You are right.. Answer should be C .sourabh33 wrote:Anurag@Gurome wrote:(1) 4 < (x-1)² < 16 implies 2 < (x - 1) < 4 implies 3 < x < 5ruplun wrote:What is the value of integer x?
(1) 4<(x-1)*(x-1)<16
(2) 4<(x-1)*(x+1)<16
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
1)ruplun wrote:What is the value of integer x?
(1) 4<(x-1)*(x-1)<16
(2) 4<(x-1)*(x+1)<16
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.manpsingh87 wrote:1)ruplun wrote:What is the value of integer x?
(1) 4<(x-1)*(x-1)<16
(2) 4<(x-1)*(x+1)<16
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
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);uslalas22 wrote: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.manpsingh87 wrote:1)ruplun wrote:What is the value of integer x?
(1) 4<(x-1)*(x-1)<16
(2) 4<(x-1)*(x+1)<16
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
(1) 4 < (x-1)² < 16 implies 2 < (x - 1) < 4 implies 3 < x < 5
x = 4
So, (1) is SUFFICIENT.
