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Question Stats:
82% (02:21) correct 18% (02:45) wrong based on 132 sessions
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Aunt Marge is giving candy to each of her nephews and nieces. She has 20 pieces of candy and she gives all the candy to the children according to her wish. If Robert gets 2 more pieces than Kate, Bill gets 6 less than Mary, Mary gets 2 more pieces than Robert, and Kate gets 2 more pieces than Bill, how many pieces of candy does Kate get? A. 2 B. 4 C. 6 D. 8 E. 10
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16 Sep 2014, 00:23
Official Solution:Aunt Marge is giving candy to each of her nephews and nieces. She has 20 pieces of candy and she gives all the candy to the children according to her wish. If Robert gets 2 more pieces than Kate, Bill gets 6 less than Mary, Mary gets 2 more pieces than Robert, and Kate gets 2 more pieces than Bill, how many pieces of candy does Kate get? A. 2 B. 4 C. 6 D. 8 E. 10 Use the initial letter of each name as a variable: \(r=k+2\) \(b=m6\) \(m=r+2\) \(k=2+b\) \(k = r2\) \(b = k2 =\) \(= (r2)  2 =\) \(= r4\) \(m = r+2\) \(r+b+m+k = 20\) \(r + r4 + r+2 + r2 = 20\) \(r = 6\) \(k = r2 = 62 = 4\) Answer: B
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Re: M0418
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02 Jul 2015, 05:29
Hi,
I answered correct, but, time was high.... Can some body know some shortcut for this type of answers? Similarly sometimes de ages needs to be in a consecutive order... But it has been hard for me to solve in less than 2 minutes.
Thanks a lot.
Luis Navarro Looking for 700



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Re: M0418
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15 Jul 2015, 09:14
Hi luisnavarro , If you cannot solve the equation in the first minute , start to plugin answers. As a rule of thumb start with answer C. The equations are R+B+M+K =20 and R=K+2 B=M6 M=R+2 K=B+2 Lets start with Kate had 6 candies (Option C). Then Bill has 4 , Mary has 10 , Rob has 8. On adding 6+4+10+8 = 28 , is greater than 20 , we can eliminate C,D,E. Lets do Kate has 4 (Option B) , then Bill has 2 , Mary has 8 and R has 6 Total = 20. So the answer is B.



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Re: M0418
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15 Jul 2015, 16:00
I believe we don't need the last statement "Kate gets 2 more than Bill". It's redundant This explanation seems long but it will take less time: The idea is put those names into ascending order 1. First "Robert gets 2 more than Kate", put Kate on the left of Robert: Kate + 2 > Robert 2. Next "Bill gets 6 less than Marry", put Bill on the left of Marry: Bill + 6 > Marry 3. Next "Marry gets 2 more than Robert": Robert + 2 > Marry. 4. And this is enough, we don't need the last statement. Plug 1 and 3 we have: Kate + 2 > Robert + 2  > Marry 5. Finally plug 2 and 4 we have: Bill + 2 >Kate + 2 > Robert + 2  > Marry The total amount of candy is 20 therefor: Bill + (Bill +2) + (Bill + 4) + (Bill + 6) = 20 Bill = 2, then Kate = 4. luisnavarro wrote: Hi,
I answered correct, but, time was high.... Can some body know some shortcut for this type of answers? Similarly sometimes de ages needs to be in a consecutive order... But it has been hard for me to solve in less than 2 minutes.
Thanks a lot.
Luis Navarro Looking for 700



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Re: M0418
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16 Jul 2015, 09:22
shyambalaji wrote: Hi luisnavarro , If you cannot solve the equation in the first minute , start to plugin answers. As a rule of thumb start with answer C. The equations are R+B+M+K =20 and R=K+2 B=M6 M=R+2 K=B+2 Lets start with Kate had 6 candies (Option C). Then Bill has 4 , Mary has 10 , Rob has 8. On adding 6+4+10+8 = 28 , is greater than 20 , we can eliminate C,D,E. Lets do Kate has 4 (Option B) , then Bill has 2 , Mary has 8 and R has 6 Total = 20. So the answer is B. Thanks a lot. Regards. Luis Navarro



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Re: M0418
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16 Jul 2015, 09:23
kimnguyen wrote: I believe we don't need the last statement "Kate gets 2 more than Bill". It's redundant This explanation seems long but it will take less time: The idea is put those names into ascending order 1. First "Robert gets 2 more than Kate", put Kate on the left of Robert: Kate + 2 > Robert 2. Next "Bill gets 6 less than Marry", put Bill on the left of Marry: Bill + 6 > Marry 3. Next "Marry gets 2 more than Robert": Robert + 2 > Marry. 4. And this is enough, we don't need the last statement. Plug 1 and 3 we have: Kate + 2 > Robert + 2  > Marry 5. Finally plug 2 and 4 we have: Bill + 2 >Kate + 2 > Robert + 2  > Marry The total amount of candy is 20 therefor: Bill + (Bill +2) + (Bill + 4) + (Bill + 6) = 20 Bill = 2, then Kate = 4. luisnavarro wrote: Hi,
I answered correct, but, time was high.... Can some body know some shortcut for this type of answers? Similarly sometimes de ages needs to be in a consecutive order... But it has been hard for me to solve in less than 2 minutes.
Thanks a lot.
Luis Navarro Looking for 700 Thanks a lot for your help. Regards. Luis Navarro Looking for 700



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Bunuel wrote: Aunt Marge is giving candy to each of her nephews and nieces. She has 20 pieces of candy and she gives all the candy to the children according to her wish. If Robert gets 2 more pieces than Kate, Bill gets 6 less than Mary, Mary gets 2 more pieces than Robert, and Kate gets 2 more pieces than Bill, how many pieces of candy does Kate get?
A. 2 B. 4 C. 6 D. 8 E. 10 I solved this with 2 variables x and y. I don't think we need the last statement  Mary gets 2 more pieces than Robert! If I take each variable separate, it takes me a lot of time In fact, I think R = K+2 is also redundant. You can derive it from the other statements. This is my solution R > x+2 K > x B > y M > y+6 (I wanted to make a horizontal table) x+2+x+y+y+6 = 20 2x+2y+8 = 20 x+y =6  (1) Also questions says  M = R+2 => y+6 = x+2+2 = x+4 => y+2 = x  (2) Add (1) and (2), you get x = 4
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Re: M0418
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12 Sep 2016, 12:55
I somehow messed up with the calculation and got the wrong answer. I think cross verifying with by putting the answer back to the question is one way to never miss this kind of question.



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Re: M0418
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12 Sep 2016, 14:08
swanidhi wrote: Bunuel wrote: Aunt Marge is giving candy to each of her nephews and nieces. She has 20 pieces of candy and she gives all the candy to the children according to her wish. If Robert gets 2 more pieces than Kate, Bill gets 6 less than Mary, Mary gets 2 more pieces than Robert, and Kate gets 2 more pieces than Bill, how many pieces of candy does Kate get?
A. 2 B. 4 C. 6 D. 8 E. 10 I solved this with 2 variables x and y. I don't think we need the last statement  Mary gets 2 more pieces than Robert! If I take each variable separate, it takes me a lot of time In fact, I think R = K+2 is also redundant. You can derive it from the other statements. This is my solution R > x+2 K > x B > y M > y+6 (I wanted to make a horizontal table) x+2+x+y+y+6 = 20 2x+2y+8 = 20 x+y =6  (1) Also questions says  M = R+2 => y+6 = x+2+2 = x+4 => y+2 = x  (2) Add (1) and (2), you get x = 4 Considering the time taken to solve the problem and confusion with the lengthy approach this solution really helps. Thanks.



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Re: M0418
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29 Jul 2019, 07:40
shyambalaji wrote: Hi luisnavarro , If you cannot solve the equation in the first minute , start to plugin answers. As a rule of thumb start with answer C. The equations are R+B+M+K =20 and R=K+2 B=M6 M=R+2 K=B+2 Lets start with Kate had 6 candies (Option C). Then Bill has 4 , Mary has 10 , Rob has 8. On adding 6+4+10+8 = 28 , is greater than 20 , we can eliminate C,D,E. Lets do Kate has 4 (Option B) , then Bill has 2 , Mary has 8 and R has 6 Total = 20. So the answer is B. luisnavarro  I agree with the startwiththeanswers method shyambalaji outlined above to help with speed. I would add that it can be advantageous to apply logic before deciding whether to test (B) or (D), as opposed to (C), since you have just a 1/5 chance of hitting upon the correct answer in (C), but will effectively have a 2/5 chance of either landing upon the correct answer or knowing what it must be if you choose (B) or (D) first. In this problem, for instance, we know that Kate, the person in the question, pops up twice, one time informing us that Robert gets more candy, and the other that she gets more than Bill. It would seem as if, with one person higher and one person lower than her, she would, in fact, stand to be somewhere in the middle. But if Mary gets even more candy than Robert, then that would bump Kate down even further, so (B) represents a more logical starting point than (D). Of course, (B) ends up being the correct answer in this case, but if it were not, then the answer would have to be either (A) or (C), depending on whether the number of pieces of candy in setting Kate's amount to 4 led to a total of more or fewer than 20. Starting with (C), provided it is incorrect but too high, you still have to do a little work to get to (B). Starting with (B), on the other hand, would either get you the answer right away or reveal, provided the total was close to the known number of pieces of candy, in which direction you needed to lean, no further work necessary. I am not saying that starting with (C) is a bad idea, but I am a lazy mathematician, and I like to do as little work as possible. This is a timesaving technique I have used to great effect on my own, as well as with students I have helped prepare for the test. Good luck with your studies.  Andrew



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Re: M0418
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29 Jul 2019, 08:25
From the prompt: R=K+2 B=M6 M=R+2 K=B+2 R+B+M+K=20
I then solved each each of the first four equations that weren't in terms of "K" to get that variable in terms of "K"
R=K+2 > already in terms of "K" M=R+2 > M=K+4 B=M6 > B=K2
Now I plugged these values in: (K+2)+(K2)+(K+4)+K=20 4K+4=20 4K=16 K=4
The correct answer is answer choice B.
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Re: M0418
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08 Aug 2019, 12:41
I sent up like this
Robert + Kate + Bill + Mary = 20
Since looking for Kate, I put Kate = X
Robert = X + 2 Mary = (X + 2) + 2 = X + 4 Bill = X + 2 + 2  6 = X 2
Robert + Kate + Bill + Mary = 20 > (X + 2) + (X) + (X2) + (X+4) = 20 4x + 4 = 20 4X = 16 X = 4 Kate got 4 pieces, took less than 2 mins.










