tomtom

1/2 > 2/5

Hi, GAD is totally right. 25 is not 35% otherwise it would mean that the two halves just consumed the same amount of water or it is stated that one half are water savers and the other is not so by definition water savers use 35% less water than non water savers. Saying that X represents the amount of water consumed by the water savers half and Y the amount of water consumed by the non water savers half: X + Y = 50 and X = Y-35Y/100 Y-35Y/100 + Y = 50 2Y - 35Y/100 = 50 165Y = 5000 Y ~ 30 By deduction X ~ 20 and we saved here 10BG. However this answer is not listed. Anyone on this one ? Thanks

Hi, I don't agree with the statement saying that 2 should be excluded. The question talks about prime and even numbers indeed, and would have had to mention "no even prime number" in case it had to be excluded. As long as the exclusion isn't mentioned explicitely, I guess we should not assume it. Also having two dices with the number 2 still gives us one dice with a prime number and the other with an even number so why not !? However my answer does not match the official answer ! Here is how I'm doing it: We have 3 prime numbers from 1 to 6, and 3 even numbers from 1 to 6 so the probability of having a prime number on a dice is 3/6 and the probability of having an even number on a dice is 3/6. Also we have two ways to make the combination, the prime number first and the even number next or the even number first and the prime next. We end up with: 3/6*3/6*2 = 1/2 A=1/2 We have 3 even numbers from 1 to 6 so having an even number on both dices results in having: 3/6*3/6 = 1/4 B = 1/4 The official answer seems to say that B is greatest than A though. Does anyone have any thought on this one ? Thanks

Hi, 0 is not part of the 100th otherwise it is not a 3-digit integer anymore i.e 011 = 11 is a 2-digit integer. Tom

Hi, The answer is C. x-2*sq.root(x*y)+y = (sq.root(x)-sq.root(y))^2 Hence sq.root(x-2*sq.root(x*y)+y) = sq.root((sq.root(x)-sq.root(y))^2) = sq.root(x)-sq.root(y) In the example above: case 2: x = 1 and y = 3 Here, quantity A = 1-sq.root(3), and quantity B = 1-sq.root(3) So, the two quantities are equal Tom

Hi, Here is what I would do. Question 1 S1 = sq. root (((1-M1)2 + (2-M1)2 + (X-M1)2)/3) S2 = sq. root ((1-M2)2 + (2-M2)2 + (1-M2)2)/3) So compare S1 and S2 is the same as comparing (X-M1)2 and (1-M2)2 what finally leads us to compare |X-M1| and |1-M2|. We can easily calculate M2 = 4/3. We can also see that X So we end up comparing |X-M1| and 4/3. There is no way to determine which could be greater than the other only knowing that X For instance, if X = 0 |X-M1| = |0+1| = 1 and 1 4/3. I'd say answer D. Question 2 m m=1, n=2, p=3 => median = 2 and mean = 3 m=-1, n=0, p=1 => median = 0 and mean = 0 Hence we cannot determine. Tom

Hi, This is how I would do it. |x2-5x| = 1 x2-5x = 1 ==> x2-5x-1 = 0 or x2-5x = -1 ==> x2-5x+1 = 0 We now have quadratic equations of the form: ax2 + bx + c = 0 Let's call the d the discriminant. By definition d = b2-4ac. The rule is : if d if d = 0 there is a unique root, if d > 0 there is two real roots In this case we can calculate two discriminants: d1 = (-5)2 - 4*1*(-1) = 29 d2 = (-5)2 - 4*1*1 = 21 Following the previous rule, d1 > 0 then there is two real roots, and d2 > 0 then we have another two real roots. There is 4 solutions for this equation. Tom