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In a certain learning experiment, each participant had three trials an
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23 Jul 2017, 11:09
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In a certain learning experiment, each participant had three trials and was assigned, for each trial, a score of either 2, 1, 0, 1, or 2. The participant's final score consisted of the sum of the first trial score, 2 times the second trial score, and 3 times the third trial score. If Anne received scores of 1 and 1 for her first two trials, not necessarily in that order, which of the following could NOT be her final score? A. 4 B. 2 C. 1 D. 5 E. 6
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In a certain learning experiment, each participant had three trials an
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20 Aug 2017, 07:25
bfistein wrote: Took me 03:17 to solve, did it the same way as above. Seems like one of those questions that is just going to take a longish time to solve, albeit it is fairly simple. Not really. There's no need to calculate each individual sums. 2 possible scores for the first two trials 1,1. Let 3rd trial score = x. Total score can be 3x+1 or 3x1. Can never be a multiple of 3. So, 6 (or in that case even 6,3,0,3,6) can't come.




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Re: In a certain learning experiment, each participant had three trials an
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23 Jul 2017, 11:35
Score in first trial = x Score in 2nd trial = y Score in 3rd trial = z allowed values for x , y and z are 2, 1, 0, 1, 2 final score = x + 2y + 3z given 2 scores are 1 and 1 so x = 1 or y = 1 or z = 1 and x = 1 or y = 1 or z = 1 sum can be ==> 1 + (2 * 1) + 3z ==> 3z  1 ==> equation 1 ==> 1 + 2y + ( 3* 1) ==> 2y  2 ==> equation 2 ==> 1 + (2 * 1) + 3z ==> 3z + 1 ===> equation 3 for z = 2 equation1 = 6  1= 7 and equation 3 = 6 + 1 = 5 for z = 1 equation1 = 3  1= 4 and equation 3 = 3 + 1 = 2 for z = 0 equation1 = 0  1 = 1 and equation 3 = 0 + 1 = 1 for z = 1 equation1 = 3  1 = 2 and equation 3 = 3 + 1 = 4 for z = 2 equation1 = 6  1 = 5 and equation 3 = 6 + 1 = 7 for y = 2 equation 2 = 2y  2 = 4  2 = 6 for y = 1 equation 2 = 2y  2 = 2  2 = 4 for y = 0 equation 2 = 2y  2 =  2 = 2 for y = 1 equation 2 = 2y  2 = 2  2 = 0 for y = 2 equation 2 = 2y  2 = 4  2 = 2 Among the answer choice all except 6 ( choice E ) can be the final score
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In a certain learning experiment, each participant had three trials an
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23 Jul 2017, 19:24
carcass wrote: In a certain learning experiment, each participant had three trials and was assigned, for each trial, a score of either 2, 1, 0, 1, or 2. The participant's final score consisted of the sum of the first trial score, 2 times the second trial score, and 3 times the third trial score. If Anne received scores of 1 and 1 for her first two trials, not necessarily in that order, which of the following could NOT be her final score?
A. 4
B. 2
C. 1
D. 5
E. 6 1. There are two cases for the total value of first two trials' scores of 1 and 1. Case 1 Trial 1 = 1. Trial 2 = 1, with weighted score of 2. 1 + 2 = 1Case 2 Trial 1 = 1. Trial 2 = 1, with weighted score of 2. 1 + (2) = 12. Add Trial 3 weighted score, 3C, to 1 or 1 Possible values, C, for Trial 3 are 2, 1, 0, 1, or 2 Which answer choice can we not achieve by addition or subtraction of 3C in combination with either 1 or 1? A. 4. Possible. C is 1, 3C is 3. 1  3 = 4. Reject B. 2. Possible. C is 1, 3C is 3. 1  3 = 2. Reject C. 1. Possible. C is 0, 3C is 0. 1 + 0 = 1. Reject D. 5. Possible. C is 2, 3C is 6. 1 + 6 = 5. Reject E. 6. Impossible by default. And there is no way to get +5 (to add to Case 1) or +7 (to add to Case 2) from 3 or a multiple of 3. Answer E
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Re: In a certain learning experiment, each participant had three trials an
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23 Jul 2017, 20:14
There are 2 possibilities Case 1: Trial score 1(x) : 1, Trial score 2(y) : 1 (Sum = 1*1 + 2*1 = 1) Case 2: Trial score 1(x) : 1, Trial score 2(y) : 1 (Sum = 1*1 + 2*1 = 1) As for the third score, there are 5 possibilities each in case 1 and case 2 2,1,0,1,2 which has to multiplied by 3 and added to the previous sum(from both cases) The numbers which has to be added to attain the possible sums are 6,3,0,3,6. Possible sums in case 1 : 7,4,1,2,5 Possible sums in case 2 : 5,2,1,4,7 The only possible sum which is not part of this list is 6(Option E)
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Re: In a certain learning experiment, each participant had three trials an
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25 Jul 2017, 02:27
Took me 03:17 to solve, did it the same way as above. Seems like one of those questions that is just going to take a longish time to solve, albeit it is fairly simple.



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Re: In a certain learning experiment, each participant had three trials an
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26 Jul 2017, 16:33
carcass wrote: In a certain learning experiment, each participant had three trials and was assigned, for each trial, a score of either 2, 1, 0, 1, or 2. The participant's final score consisted of the sum of the first trial score, 2 times the second trial score, and 3 times the third trial score. If Anne received scores of 1 and 1 for her first two trials, not necessarily in that order, which of the following could NOT be her final score?
A. 4
B. 2
C. 1
D. 5
E. 6 If Anne received 1 for her first trial and 1 for her second, then the sum of her first two trials is 1 + 2(1) = 1 + 2 = 1. If Anne received 1 for her first trial and 1 for her second, then the sum of her first two trials is 1 + 2(1) = 1 + 2 = 1. Let’s analyze each answer choice: A) 4 If the sum of the first two trials is 1 and the third trial is 1, then the sum can be 1 + 3(1) = 4. B) 2 If the sum of the first two trials is 1 and the third trial is 1, then the sum can be 1 + 3(1) = 2. C) 1 If the sum of the first two trials is 1 and the third trial is 0, then the sum can be 1 + 3(0) = 1. D) 5 If the sum of the first two trials is 1 and the third trial is 2, then the sum can be 1 + 3(2) = 5. E) 6 We see there is no way to get a sum of 6. Answer: E
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Re: In a certain learning experiment, each participant had three trials an
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10 Sep 2017, 03:12
carcass wrote: In a certain learning experiment, each participant had three trials and was assigned, for each trial, a score of either 2, 1, 0, 1, or 2. The participant's final score consisted of the sum of the first trial score, 2 times the second trial score, and 3 times the third trial score. If Anne received scores of 1 and 1 for her first two trials, not necessarily in that order, which of the following could NOT be her final score?
A. 4
B. 2
C. 1
D. 5
E. 6 Final Score = (S1) + (S2 x 2) + (S3 x 3) given s1 and s2 = 1 or 1 1 + (2) + 6 = 5 so eliminate 5 1 + (2) + (3) = 4 so eliminate 4 1 + (2) + (3) = 2 1 + 2 = 0 = 1 eliminate 1 we are left with 6 Answer E



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Re: In a certain learning experiment, each participant had three trials an
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16 Oct 2017, 12:36
For this question, i believe the simplest way is to test the answer choices. From the stem, it flows that her score, before the last trial, will be 1 or 1. To get 4 as final score she must have 4 or 1. This is possible if she gets 1 in the last trial. Keep on testing, we see that the answer choices A through D work. But E does not work: to have 6 as final score, she must get a score of 5 or 7 in the last trial. both value are not multiple of 3.
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Re: In a certain learning experiment, each participant had three trials an
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18 Apr 2019, 11:27
How do I do this in 2mins? ScottTargetTestPrep wrote: carcass wrote: In a certain learning experiment, each participant had three trials and was assigned, for each trial, a score of either 2, 1, 0, 1, or 2. The participant's final score consisted of the sum of the first trial score, 2 times the second trial score, and 3 times the third trial score. If Anne received scores of 1 and 1 for her first two trials, not necessarily in that order, which of the following could NOT be her final score?
A. 4
B. 2
C. 1
D. 5
E. 6 If Anne received 1 for her first trial and 1 for her second, then the sum of her first two trials is 1 + 2(1) = 1 + 2 = 1. If Anne received 1 for her first trial and 1 for her second, then the sum of her first two trials is 1 + 2(1) = 1 + 2 = 1. Let’s analyze each answer choice: A) 4 If the sum of the first two trials is 1 and the third trial is 1, then the sum can be 1 + 3(1) = 4. B) 2 If the sum of the first two trials is 1 and the third trial is 1, then the sum can be 1 + 3(1) = 2. C) 1 If the sum of the first two trials is 1 and the third trial is 0, then the sum can be 1 + 3(0) = 1. D) 5 If the sum of the first two trials is 1 and the third trial is 2, then the sum can be 1 + 3(2) = 5. E) 6 We see there is no way to get a sum of 6. Answer: E




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