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A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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17 Feb 2016, 22:41
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73% (02:16) correct 27% (02:02) wrong based on 614 sessions
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A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15, respectively. If some of the nuts of one of the three types are removed, which of the following could be the ratio of pecans to cashews to almonds remaining in the bowl? i. 1 : 2 : 3 ii. 2 : 3 : 4 iii. 4 : 7 : 10 A. I only B. II only C. III only D. I and III only E. II and III only
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A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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17 Feb 2016, 23:02
jasper20151012 wrote: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15, respectively. If some of the nuts of one of the three types are removed, which of the following could be the ratio of pecans to cashews to almonds remaining in the bowl?
i. 1 : 2 : 3
ii. 2 : 3 : 4
iii. 4 : 7 : 10
A. I only
B. II only
C. III only
D. I and III only
E. II and III only Hi, we have ratios of three items as 6:10:15.. restriction we can remove nuts only from one bowl.. inference of restriction the ratio between the other two will remain the same solution lets see the choices.. i. 1 : 2 : 3 the ratio 10:15 is same as 2:3.. so the nuts are removed from 1st bowl.. 1:2:3 will mean 5:10:15.. so 1 nut is removed from 1st bowl... POSSIBLE ii. 2 : 3 : 4 the corresponding ratios are 6:10:15.. no two ratios are equal in earlier and present.. so not possible.. \(2:3\neq{6:10}\).. \(2:4\neq{6:15}\).. \(3:4\neq{10:15}\).. so NOT POSSIBLE.iii. 4 : 7 : 10 same reason as ii.. NOT POSSIBLE..ANS A
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Re: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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17 Feb 2016, 22:46
Official Answer on Manhattanprep. To solve this question, it’s necessary to understand that if only one kind of nut is removed, the ratio of the two remaining nuts must remain unchanged. That means our correct answers must have a ratio of two nuts that corresponds to one of the nut to nut ratios we started with. i. The ratio 2 : 3 is the same as the given ratio 10 : 15. If one pecan were removed, the new ratio would be 5 : 10 : 15, or 1 : 2 : 3. ii. None of the nuts currently have a ratio of 3 : 4. The cashews and almonds do have a ratio of 2 : 3, but there are not enough pecans in the bowl to complete the ratio. iii. The ratio 4 : 10 is the same as the given ratio 6 : 15. To see this, multiply the ratio by 3/2 . The new ratio is 6 : 10.5 : 15. Unfortunately, this means that there are fewer cashews that this ratio would require. Removing cashews won’t create the desired ratio. The correct answer is (A).
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Re: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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17 Feb 2016, 22:56
IMO : A
I don’t know the standard approach for this sum, but I just tried number plug in method 6:10:15 Which means 6x,10x,15x So probably 6,10,15 12,20,30 18,30,45 24,40,60 No we try to fit with answer 6,10 ,15 to be fitted with 1:2:3 ( 2 & 3 direct fit when we multiply 5) , reduce 6 by one number 5 5, 10 15 (1:2:3) So A works Remaining I tried it doesn’t get fit



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Re: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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28 May 2017, 16:22
i managed to get this correct by implementing a shorter method. not sure if this can be applied to other/similar problems, so lmk what you think: GIVEN 6:10:15:31 CONSTRAINT: can only remove nuts from ONE jar i. 1 : 2 : 3  1:2:3:6 (total). NOTICE YOU CAN MULTIPLY BY 5 (to get us as close to 31 as possible).  Result: 5:10:15:30. * We need to make the following adjustments: 1 (6 is missing 1, giving us "5" in this example. Therefore, only removing nut from ONE jar). 2:3 is the same proportion as 10:15. CORRECT. ii. 2 : 3 : 4  2:3:4:9 (total). NOTICE YOU CAN MULTIPLY BY 3 (to get us as close to 31 as possible).  Result: 6:9:12:27. * We need to make the following adjustments: (9 should be "10". 12 should be "15". Therefore, need to remove nuts from TWO jars). INCORRECT. iii. 4 : 7 : 10  4:7:10:21 DO NOT NEED TO MULTIPLY BY ANYTHING B/C MULTIPLYING 2X WOULD GIVE YOU SAME DIFFERENCE (+/ 10)  Result: same * We need to make the following adjustments: (4 should be "5". 7 should be "10". 10 should be "15". Therefore, need to remove nuts from THREE jars). INCORRECT. Again, please let me know your thoughts! If this was helpful, kudos please



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Re: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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25 Aug 2017, 16:07
As always, get it from answer,
Given: P:C:A = 6x:10x:15x Some nuts removed from one of three types Which of the following be the ratio?
i. 1 : 2 : 3
ii. 2 : 3 : 4
iii. 4 : 7 : 10
From the given ratio we need to get the above three answer, NOTE: all the number should be divisible by common number(dividend) Let's say some quantity removed from pecans, then we need to figure out the divisor divisible by common dividend
i) if removed 1 from 6 changes to 5 =>5:10:15 = 1:2:3 SATISFIED ii) if anything removed from cashew, we not going to get divisor that is divisible by common dividend NOT SATISFIED iii) if removed a number we can able to solve, OPTION 1: 6:10:14 = 3:5:7 whatever may the number we get reduced still we gonna get same amount of peacans and cashew. NOT SATISFIED
Only satisfied option is 1
Answer will be 'A'



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Re: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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04 Jan 2019, 09:55
cheapjasper wrote: Official Answer on Manhattanprep.
To solve this question, it’s necessary to understand that if only one kind of nut is removed, the ratio of the two remaining nuts must remain unchanged. That means our correct answers must have a ratio of two nuts that corresponds to one of the nut to nut ratios we started with.
i. The ratio 2 : 3 is the same as the given ratio 10 : 15. If one pecan were removed, the new ratio would be 5 : 10 : 15, or 1 : 2 : 3.
ii. None of the nuts currently have a ratio of 3 : 4. The cashews and almonds do have a ratio of 2 : 3, but there are not enough pecans in the bowl to complete the ratio.
iii. The ratio 4 : 10 is the same as the given ratio 6 : 15. To see this, multiply the ratio by 3/2 . The new ratio is 6 : 10.5 : 15. Unfortunately, this means that there are fewer cashews that this ratio would require. Removing cashews won’t create the desired ratio.
The correct answer is (A). Your answer is incomplete. Here is the Manhattan complete explanation: To solve this question, it’s necessary to understand that if only one kind of nut is removed, the ratio of the two remaining nuts must remain unchanged. In other words, correct answers must have a ratio of two nuts that corresponds to one of the original nut to nut ratios. Case 1: The number of pecans and cashews stay the same; almonds change. 6 : 10 : __ 3 : 5 : __ Case 2: The number of pecans and almonds stay the same; cashews change. 6 : __ : 15 2 : __ : 5 Case 3: The number of cashews and almonds stay the same; pecans change. __ : 10 : 15 __ : 2 : 3 Next, check to see whether any of the new ratios match. I. The ratio of cashews to almonds of __ : 2 : 3 is the same as the ratio in Case 3 (pecans change). The ratio 1 : 2 : 3 would be equivalent to the ratio of 5 : 10 : 15. Because the original ratio was 6 : 10 : 15, this new ratio can be found by removing one pecan. Eliminate answer choices (B), (C), and (E). Skip II. None of the remaining answer choices have roman numeral II as an option. However, note that none of the pairs of nuts currently have a ratio that matches any in this option. III. The ratio 4 : __ : 10 is a multiple of the ratio 2 : __ : 5. Because the ratio of pecans to almonds must be in the ratio of 6 : 15 and 4 : 10, find a common multiple. The ratio 24 : __ : 60 would work for both. Original New 6 : 10 : 15 4 : 7 : 10 24 : 40 : 60 24 : 42 : 60 The new ratio results from an increase in the number of cashews; eliminate choice (D). The correct answer is (A).



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Re: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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14 Jan 2019, 18:10
For case #3 4:7:10  why does that not work? If each is multiplied by 1.5 and we remove from the middle, then it would match



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Re: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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21 Jan 2019, 06:19
cchen679 wrote: For case #3 4:7:10  why does that not work? If each is multiplied by 1.5 and we remove from the middle, then it would match Hi cchen679 , In my solution (below) the reason for this impossibility will become explicit. Regards, Fabio.
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Re: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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21 Jan 2019, 06:23
cheapjasper wrote: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15, respectively. If some of the nuts of one of the three types are removed, which of the following could be the ratio of pecans to cashews to almonds remaining in the bowl?
i. 1 : 2 : 3 ii. 2 : 3 : 4 iii. 4 : 7 : 10
A. I only B. II only C. III only D. I and III only E. II and III only
\(?\,\,\,:\,\,\,p:c:a\,\,{\rm{possible}}\,\,\left( {{\rm{when}}\,\,{\rm{some}}\,\,{\rm{nuts}}\,\,{\rm{of}}\,\,{\rm{one}}\,\,{\rm{type}}\,\,{\rm{removed}}} \right)\) \(p:c:a = 6:10:15\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ \matrix{ \,p = 6k \hfill \cr \,c = 10k \hfill \cr \,a = 15k \hfill \cr} \right.\,\,\,\,\,\,\left( {k > 0\,\,{\mathop{\rm int}} \left( * \right)} \right)\) \(\left( * \right)\,\,\left\{ \matrix{ \,{\mathop{\rm int}}  {\mathop{\rm int}} = a  c = 15k  10k = 5k\,\,{\mathop{\rm int}} \hfill \cr \,{\mathop{\rm int}}  {\mathop{\rm int}} = p  5k = 6k  5k = k\,\,\,{\mathop{\rm int}} \hfill \cr} \right.\) \(\left( {\rm{I}} \right)\,\,\,p:c:a = 1:2:3\,\,\,\, \Rightarrow \,\,\,{\rm{possible}}\,\,\left( {k = 1,\,\,{\rm{take}}\,\,1\,\,{\rm{pecan}}\,\,{\rm{nut}}\,\,{\rm{out}}\,\,\,\, \Rightarrow \,\,\,\,\left( {p,c,a} \right) = \left( {5,10,15} \right)} \right)\) \(\,\,\, \Rightarrow \,\,\,\,\,{\rm{refute}}\,\,\left( {\rm{B}} \right),\left( {\rm{C}} \right),\left( {\rm{E}} \right)\) \(\left( {{\rm{III}}} \right)\,\,p:c:a = 4:7:10\,\,\,\,\mathop \Rightarrow \limits^{\left( {\text{below}} \right)} \,\,\,{\rm{impossible:}}\,\,\,\) \({\rm{some}}\,\,p\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{ \,\left( {p,c,a} \right) = \left( {6k  {\rm{some}},10k,15k} \right) \hfill \cr \,{2 \over 3} = {{10k} \over {15k}} = {c \over a} \ne {7 \over {10}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\,\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]\) \({\rm{some}}\,\,c\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{ \,\left( {p,c,a} \right) = \left( {6k,10k  {\rm{some}},15k} \right) \hfill \cr \,{4 \over 7} = {p \over c} = {{6k} \over {10k  {\rm{some}}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,40k  4 \cdot {\rm{some}} = 42k\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]\) \({\rm{some}}\,\,a\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{ \,\left( {p,c,a} \right) = \left( {6k,10k,15k  {\rm{some}}} \right) \hfill \cr \,{4 \over 7} = {p \over c} = {{6k} \over {10k}} = {3 \over 5}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]\) The correct answer is (A). (Note that (II) does not need to be evaluated!) We follow the notations and rationale taught in the GMATH method. Regards, Fabio.
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Re: A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15
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