abhigup01

I wonder why TwinnSplitter's awesome post is not sticky anymore

D is better than E because even if you don't put 'that' to maintain parallelism, it is understood. Moreover, if at all you would want to have 'that' in this sentence, you would want to have it immediately after 'and' and just before ', despite the recent illnesses'. What do you think Erin?

Revenue from this particular kind of product in 1994 = (2/100) * 8 billion = 0.16 billion Revenue from this particular kind of product in 1995 = (2.3/100)*10 billion = 0.23 billion Revenue increase % = ((0.23 – 0.16)/0.16)*100 = 700/16

Solution: Assuming that by ‘ratio between the circles they ran’ you mean ratio between the angles formed at the center by the arc created due to running, Distance covered = R * (angle 1 in radians) = r * (angle 2 in radians) So (angle 1 in radians)/ (angle 2 in radians) = (r/R) = 3/7

Same as DKumar's solution

(10P5)/(10P4 ) = 6 to 1

Area of Equilateral triangle with side a is A = ½ * base * height = ½ * a * (a Cos 30) = (a square)(Root 3)/4 ð 9(root 3) = (a square)(Root 3)/4 ð a = 6 If in-radius is r, the a = 2r (cos30) = r (root 3) ð r = 6/(root 3) Area of circle = Pi (r square) = Pi (36/3) = 12 Pi

The question is not completely unambiguous as it’s not mentioned that lengths of strings are natural numbers. Lengths of strings are usually continuous rather than discrete in nature so it should have been mentioned in the question that lengths are natural numbers. Anyways, going by this assumption, and supposing that A,B,C,D,E,F, and G are those lengths in ascending order, (A+B+C+D+E+F+G) = 68 * 7 = 476 -- (Eqn 1) Median D = 84. Now, for G to be as high as possible, E and F should be as close to median D as possible. That requires E = F = 84. Replacing these values and also G = 4A + 14, in Eqn 1 ð A+B+C+84+84+84+4A+14 = 476 ð 5A + B + C = 210 --- (Eqn 2) Now, since G is directly proportional to A, the higher the A the higher the G. For A to be as high as possible, B and C in Eqn 2 should be as low as possible. Lowest possible value of B as well as C is A. So 5A + A + A = 210 ð A = 30 ð G = 4 * 30 + 14 = 134

Let length be x and width be w As per the first scenario, (b+2) (x+2) = M => bx + 2b + 2x + 4 = M ---- (1) As per the second scenario, (b+4) (x+4) = M + 52 => bx + 4b + 4x + 16 = M + 52 ---- (2) On subtracting Eqn.(1) from Eqn.(2), i.e. Eqn(2) – Eqn(1) ð 2b + 2x + 12 = 52 Perimeter P = 2 (b+x) = 52 - 12 = 40

Since (n/4)+(r/8) =(s/8)+(t/6) ==> (2n +r)/ 8 = (3s+4t)/24 ==> 2n +r = s + (4t/3) Unless we know the values of both r and s, we can't determine whether (s + (4t/3)) is bigger or (2s+t) is.