manish8109

Refer to Q3: http://www.www.urch.com/forums/gmat-problem-solving/51123-challenge-gmat-help-manish8109-ctg1983-chix475ntu-arjmen.html

Statement 1: k=4a+3 => 5k = 20a+15 => 5k/6 = 20a/6 + 15/6 it will always give different remainder. Ex: If a=1 then k=7 5k=35 => 35/6 = rem=5 If a=2 then k=11 5k=55 => 55/6 rem= 1 Insufficient Statement 2: k=12a+3 => 5k = 60a+15 => 5k/6 = 10a + 15/6 15/6 will always give remainder 3 -------Sufficient Ans(B)

Kth term = (-1)^(k+1) * 1/2^k Ist term = 1/2 , 2nd term= -1/4 IIIrd term = 1/8 T = 1/2 -1/4 + 1/8 -1/16 + 1/32 - 1/64 + 1/128 -1/256 +1/512 -1/1024 = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 Ans is (D) b/w 1/4 & 1/2

(x/3)/(y/4) = 5/6 => 4x/3y = 5/6 => x/y = 5*3/6*4 => x/y = 15/24 => x/y= 3*5/3*8 => x/y= 5/8-------(Ans)

Q1. Refer to http://www.www.urch.com/forums/gmat-data-sufficiency/48203-gmatprep.html

Let the no. of employees be x Avg. salary = 2000 = total salary/x Total salary=2000x Avg. salary(150 people) = 4000 Salary for 150worker = 4000*150 = 600000 2000x = 600000 + (x-150)*1250 => 2000x = 600000 + 1250x -187500 => 750x = 412500 => x=550 worker

1. Please Refer to http://www.www.urch.com/forums/gmat-problem-solving/82963-can-someone-show-how-solve-answer.html 3. X/2 at 10:00 am so, X/2/4 i.e. x/8 at 9:00 am-----Ans(A) 5. Correct question A certain basketball team that has played 2/3 of its games has a record of 17 wins and 3 losses. What is the greatest number of the remaining games that the team can lose and still win at least 3/4 of all of its games? (A) 7 (B) 6 © 5 (D) 4 (E) 3

Refer to http://www.www.urch.com/forums/gmat-problem-solving/48771-2-gmatprep-questions.html

Let the no. of hardback copies sold be x after issuing the first paperback copy 9x + (x+36)= 441 => 10x+36 =441 => 10x=405 => x=40.5 No. of paperback copies sold = 9*40.5 = 364.5-----Ans©

Sum of temperatures=5x Sum of 3 greatest > Sum of the remaining temp. A) Not possible as 6x>5x B) 4x Could be possible C) 5x/3 => 5x-5x/3 = 10x/3 > 5x/3 ----It means 5x/3 is not the sum of 3 greatest temperature. D) 3x/2 => 5x-3x/2 = 7x/2> 3x/2 -----Same reason as above not possible So Maximum sum possible for sum of 3 greatest of these temperatures could be 4x-----ANS(B)

Let the first solution be 400mL i.e. 300 alcohol & 100 ml water Let the Second solution be 100mL i.e. 40 Alcohol & 60 ml water Combining both 340/500 = 34/50=17/25-----ANS

Congratulations.....................

IIL, 3= sqrt(3) * sqrt(3) so, 3*3/sqrt(3) = 3*sqrt(3) *sqrt(3) /sqrt(3) cancel sqrt(3) from both numerator & denominator = 3*sqrt(3) Regards manish

Zuckerman, I concur with you. I only used it to establish that x-y=1/2 where 1/2 is a positive number it means x > y but when if y 1. So, both x & y has to be positive to satisfy both the conditions. It is always better to keep the fractional nature of ratios intact. Regards Manish