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In a certain learning experiment, each participant had three trials an

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In a certain learning experiment, each participant had three trials an  [#permalink]

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New post 23 Jul 2017, 10:09
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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  [#permalink]

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New post 20 Aug 2017, 06:25
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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 long-ish 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 3x-1. 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  [#permalink]

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New post 23 Jul 2017, 10:35
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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  [#permalink]

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New post 23 Jul 2017, 18:24
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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 = 1

Case 2
Trial 1 = 1. Trial 2 = -1, with weighted score of -2.

1 + (-2) = -1

2. 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  [#permalink]

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New post 23 Jul 2017, 19:14
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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  [#permalink]

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New post 25 Jul 2017, 01:27
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 long-ish time to solve, albeit it is fairly simple.
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Re: In a certain learning experiment, each participant had three trials an  [#permalink]

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New post 26 Jul 2017, 15:33
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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  [#permalink]

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New post 10 Sep 2017, 02:12
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  [#permalink]

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New post 16 Oct 2017, 11:36
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  [#permalink]

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Re: In a certain learning experiment, each participant had three trials an   [#permalink] 22 Oct 2018, 04:14
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