missionGMAT Posted August 26, 2011 Share Posted August 26, 2011 In the game Cako, a player is awarded one tick for every third Alb captured, and one click for every fourth Berk captured. The total score is equal to the product of clicks and ticks. If a player has a score of 77, how many Albs did he capture? (1) The difference between Albs captured and Berks captured is 7. (2) The number of Albs captured is divisible by 4. Quote Link to comment Share on other sites More sharing options...
chandudutt Posted August 28, 2011 Share Posted August 28, 2011 the answer is B. as we can say a definite NO with that second point. the first sentence is yes/might be. hence B is the answer Quote Link to comment Share on other sites More sharing options...
Masuraha Posted September 19, 2011 Share Posted September 19, 2011 The answer is A. Please let me know if it is correct then I can provide the explanation. Quote Link to comment Share on other sites More sharing options...
Lav Posted September 23, 2011 Share Posted September 23, 2011 I must say i would have dodged this question if faced on test-day because it took me a while to get it :crazy:, here it is: The question stem tells us A/3 +B/4 = 77 (assuming, Albs and Berks captured are multiples of 3 & 4, respectively) (1) tells us A - B = 7 Solving two equations gives us A=135 and B=128. BUT, so would A=136 & B=129, AND A=137 & B=130 give us 77 as per the rules of the game. So (1) is insufficient. (2) That A/4 is an integer tells us nothing about Berks. so clearly insuff. Together: A can only be 136 for it to be divisible by 4. So C it is. Quote Link to comment Share on other sites More sharing options...
sandeep_chads Posted February 10, 2012 Share Posted February 10, 2012 Hey Lav - your approach is right but I guess there is a small mistake A/3 + B/7 != 77 it is A/3 * B/7 = 77 rest all looks good. I would have marked C and gone ahead because I would see 2 variables and 2 equations, however it turns into a quadratic equation and you always run a chance to have 2 solutions to it. Quote Link to comment Share on other sites More sharing options...
Red Bozo Posted March 13, 2012 Share Posted March 13, 2012 [TABLE=class: grid, width: 100, align: center] [TR] [TD]Clicks [/TD] [TD]Berks [/TD] [TD]Ticks [/TD] [TD]Albs [/TD] [/TR] [TR] [TD]1 [/TD] [TD]4-7 [/TD] [TD]77 [/TD] [TD]231-233 [/TD] [/TR] [TR] [TD]7 [/TD] [TD]28-31 [/TD] [TD]11 [/TD] [TD]33-35 [/TD] [/TR] [TR] [TD]11 [/TD] [TD]44-47 [/TD] [TD]7 [/TD] [TD]21-23 [/TD] [/TR] [TR] [TD]77 [/TD] [TD]308-311 [/TD] [TD]1 [/TD] [TD]3-5 [/TD] [/TR] [/TABLE] I think the answer is A. Here's the explanation: Since the score is 77, we can have the clicks and ticks taking values as shown above. Also, it isn't stated that the Berk count is a multiple of 4. Likewise, the Alb count needn't bee a multiple of 3. So, if click count = x, then berk count can be 4x, 4x+1, 4x+2 or 4x+3. Likewise, if tick count = y, then alb count can be 3y, 3y+1 or 3y+2. That's the basis for the data in the above table. Now consider statement 1: Alb - Berk = 7. So obviously it can't be row1 or row 4, because of the huge differences. Row 3 also has widely-spaced Alb and Berk values. Since Albs Now consider statement 2: No of Albs captured is divisible by 4. Thr's no such value in row2 and row 3. Row 1 and Row 2 both satisfy the condition. But we still can't zero in on one of these rows. So statement 2 alone is not sufficient. So answer is A. Quote Link to comment Share on other sites More sharing options...
krusta80 Posted March 7, 2014 Share Posted March 7, 2014 Let A denote the number of Albs Let B denote the number of Berks Let t denote the number of ticks Let c denote the number of clicks From the given portion of the question, we know the following: t = floor(A/3) c = floor(B/4) total = t*c = 77 Since t and c must be non-negative integers whose product is 77, there are four possibilities: t c 1 77 --> albs can be 3,4, or 5 / berks can be 308,309,310, or 311 77 1 --> albs can be 231,232, or 233 / berks can be 4,5,6, or 7 7 11 --> albs can be 21,22, or 23 / berks can be 44,45,46, or 47 11 7 --> albs can be 33,34, or 35 / berks can be 28,29,30, or 31 Moving on... (1) |A - B| = 7 This eliminates three of the four aforementioned possibilities for t and c, but let's take a closer look at A and B... A can be 33,34, or 35 B can be 28,29,30, or 31 Only A = 35 and B = 28 gives a difference of exactly 7. SUFFICIENT! (2) A mod 4 = 0 Without going through every possible number of Albs, both 4 and 232 are divisible evenly by 4. Not sufficient alone. ANSWER IS A Quote Link to comment Share on other sites More sharing options...
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