tarkumar

assuming x2y = x^2*y -12y = 4y + 9y => y=0 ..Suff 18y = 9y+9y ...infinite solutions..Insuff

Detailed solution x^2 > |x| x > 0 then x^2 > x x(x-1) > 0 x 1 but for positive x x>1 ......... equ 1 x -x x(x+1) > 0 x 0 but for negative x x From 1 and 2 x^2 > 1

v^2-vw = 0 v(v-w) =0 either v =0 or v-w =0 or both but v and w are different integers so v = 0 Suff v is not 2 nothing else can be said about v. Insuff

(x+1)^2=x^2 x^2+2x+1 = x^2 which simplifies to statement 1 as 2x+1 = 0 Hope this helps

37 > 36 (6^2) and 576 24^2 so required numbers even integers from 8 to 24 (both inclusive)

50! (51 - 1 ) / 49! * (50 - 1) = 50* 49! * 50 / (49! * 49) = 2500/49

Let D be total distance between station A and B Speed of train leaving A Va = D/4 Speed of train leaving B Vb = D/3.5 By 7 am. train A would have covered half the distance so remaining distance = D/2 Relative speed of train A wrt B = Va - (-Vb) = Va + Vb = D/4 + D/3.5 using relative distance = relative speed * time D/2 = (D/4 + D/3.5) * t t = 14/15 hour = 56 minutes Hope this helps

You can take test once in a seven days period.

12C2 is the answer so 12*11/2 is right

Lets speed be P km/h and Q kmph respectively Using relative distance = relative velocity * time relative distance = 120 km (P +Q) * 1 = 120 ................ (i) (P - Q ) * 6 = 120 ................ (ii) solving i and ii P = 70, Q = 50

Where are options :hmm:??

Alternate way number of codes when second digit is zero + number of codes when second digit is one = 8 * 1 * 9 (second and third digits can not be 0 at same time) + 8*1*10 = 72 + 80 = 152

Total number of codes with constraint (second and third digits cannot be 0 at the same time) = 8*2*10 = 160 Number of codes where second and third digits are zero = 8*1*1 = 8 Required codes = 160 - 8 = 152

A x^2 > 9 means x>3 or x given x is -ve x Suff