supersuj

this question becomes much easier if you plug in any number for a1 Take a1=2 a2=2 a3=4 a4=8 let a4=an=S=8 a5=a(n+1)=16=2S a6=a(n+2)=32=4S a7=a(n+3)=64=8S

http://www.www.urch.com/forums/gmat-data-sufficiency/105167-number-seconds.html

agree with genius. IMO, E

46 in verbal..thats terrific ! Congratulations!

S(n)= 1 + 2 + 3 + 4 + ....n S(2n) = 1 + 2 + 3 + 4 + ....2n Hope this helps!

Agree with E. I believe the question should have been " is n negative ? " In that case, answer is C

questions asks whether sqrt{(x-3)^2} = 3-x we know that sqrt {(x-3)^2} = mod (x-3) and thus equals (x-3) if x>3 or equals (3-x) if x so we need to know whether x>3 or x statement 1--> we get x is not equal to 3 but that does not tell us whether x is greater than or less than 3 ..INSUFFICIENT statement 2--> -x*mod(x) > 0 since mod(x) is ALWAYS greater than 0 , x must be NEGATIVE for the above inequality to hold true hence x Hope this helps!

IMO, C I1 = P1*x I2= (60,000-P1)*y statement 1---> x=3y/4 I1+I2= 4080 = P1*x + (60,000-P1)*4/3*x 1 equation, 2 unknowns ( P1 and x) ...INSUFFICIENT statement 2--> I1/I2 = 3/2 = (P1*x)/(60,000-P1)*y 1 equation, 2 unknowns( P1 and y) ..INSUFFICIENT Together, we get y=4/3*x and 2 equations with 2 unknowns..SUFFICIENT

IMO, A statement 1--> SUFFICIENT..a line with a negative slope will always intersect quad II statement 2--> INSUFFICIENT..dont know the slope of the line ( a line with a +ve slope will not intersect quad II)

IMO, B statement 1--> m/y=x/r ..INSUFFICIENT ( if m=x, then true...if m not equal to x..then not true) statement 2--> (m+x)/(r+y) = x/y then, my+xy=rx+xy so my=rx and m/r = x/y...SUFFICIENT

IMO, C lcm(x,y) * gcf(x*y) = x*y xy= 10*180 = 1800

Maybe, statement 2 should have been " the 15th root of w is 4" then statement 2 would be w^(1/15)=4 w= 4^15 = 2^30 w^(1/3)= 2^10

I dont think this is a GMAT question the value of w is different in options 1 and 2 Anyways, I get D statement 1--> w^(1/5) = 64 w= 64^5 = 2^30 therefore, w^(1/3) = 2^10...SUFFICIENT statement 2--> w^(1/14) = 4 w=4^14 = 2^28 w^(1/3) = 2^(28/3)..SUFFICIENT

good question! i get D since a/b= even , a=even*b and thus a=even a-b= even, so b MUST be even as well a=even*b ( a MUST be even and MUST be divisible by 4) (a+2)/2 will always be ODD since a must be divisible by 4 Answer is D

congratulations! 710 is a very competitive score for the top schools. Good Luck!

John, We are asked is |x| is now, from statement 2, we know x is between 0 and 1 So, we are 100% sure that value of x exists between -1 and 1, which is enough to answer the question that |x| you can think of this in another way since we need to know if |x| taking any value of x between 0 and 1 , we get |x|

If for a given statement, the value of x lies between (-1,1)..then that statement is sufficient

8/36 = 2/9 ..agree with john.3030

agree with rajtitan

IMO, D Is |x| which means we need to show if x lies in (-1,1) statement 1--> x^2 so x-1 and it lies in (-1,1)..SUFFICIENT statement 2--> |x| which means that x can be only between (0,1)..SUFFICIENT

agree with A y-intercept = -6

IMO, C perimeter of circle C= 2*pi*r perimeter of square S = 4s 4s=2*pi*r s= 0.5*pi*r area of S/area of C = s^2/(pi*r^2) = (0.5^2*pi^2^*r^2)/(pi*r^2) = 0.25*pi =1/4*22/7 = 22/28 = 11/14 closest option is C, 3/4

IMO, E 165-125= 40 for each increment of 10, there is a tenfold increase therefore for an increment of 40, there is 10*10*10*10 increase = 10,000

Excellent score! I am contemplating a retake as well after scoring a 680 last month but i am too busy working on my applications now. All the best for yours!

Since we know the x-co-ordinate of the vertex= 4 the smallest value of g(x) will be at x=4 At x=4, we get g(x)=-4 now we know this parabola opens UPWARD..and g(x) will go from -4 all the way to infinity.. if you can draw a graph, you will understand it better. Hope this helps!