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Three students were given three tests, the results of the first two te 10 Jul 2015, 02:39
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79% (02:31) correct 21% (02:58) wrong based on 250 sessions

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Three students were given three tests, the results of the first two tests are shown above. Student 1 and Student 3 received equal scores on Test 3. As a result, Student 1’s average (arithmetic mean) score over the three tests was 50 percent higher than Student 3’s average. What was the score Students 1 and 3 received on Test 3?

A. 33
B. 34
C. 35
D. 36
E. 39


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B
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Three students were given three tests, the results of the first two te 10 Jul 2015, 03:32
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Bunuel

Three students were given three tests, the results of the first two tests are shown above. Student 1 and Student 3 received equal scores on Test 3. As a result, Student 1’s average (arithmetic mean) score over the three tests was 50 percent higher than Student 3’s average. What was the score Students 1 and 3 received on Test 3?

A. 33
B. 34
C. 35
D. 36
E. 39


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30+50+x=(14+28+x)*3/2..
x=34
ans B
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Three students were given three tests, the results of the first two te 10 Jul 2015, 04:07
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Bunuel

Three students were given three tests, the results of the first two tests are shown above. Student 1 and Student 3 received equal scores on Test 3. As a result, Student 1’s average (arithmetic mean) score over the three tests was 50 percent higher than Student 3’s average. What was the score Students 1 and 3 received on Test 3?

A. 33
B. 34
C. 35
D. 36
E. 39


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Per the question, (30+50+x) / 3 = 1.5* (14+28+x)/3 ---> x=34. Thus B is the correct answer.
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Three students were given three tests, the results of the first two te 13 Jul 2015, 02:33
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Bunuel

Three students were given three tests, the results of the first two tests are shown above. Student 1 and Student 3 received equal scores on Test 3. As a result, Student 1’s average (arithmetic mean) score over the three tests was 50 percent higher than Student 3’s average. What was the score Students 1 and 3 received on Test 3?

A. 33
B. 34
C. 35
D. 36
E. 39


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800score Official Solution:

Let x be the score that both Student 1 and Student 3 received on Test 3.
Since Student 1's average is 50% higher than Student 3's average, Student 1's average is 1.5 times that of Student 3.
The average for Student 1 is (30 + 50 + x)/3, and the average for Student 3 is (14 + 28 + x)/3.

So we can solve for x as follows:
(30 + 50 + x)/3 = ((14 + 28 + x)/3) × 1.5
(80 + x)/3 = ((42 + x)/3) × 1.5
(80 + x)/3 = (63 + 1.5x)/3
80 + x = 63 + 1.5x
17 = 0.5x
x = 34.

The correct answer is choice (B).
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Three students were given three tests, the results of the first two te 31 May 2017, 10:22
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I got the answer by process elimination
Answer B

The average of student 1 with 34 is 38
The average of student 3 with 34 is 25 1/3

25 1/3 x 1 1/2
38

The answer has to be B. :-D
Process elimination can be a life saver sometime.
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Three students were given three tests, the results of the first two te 05 Jun 2017, 16:14
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Bunuel

Three students were given three tests, the results of the first two tests are shown above. Student 1 and Student 3 received equal scores on Test 3. As a result, Student 1’s average (arithmetic mean) score over the three tests was 50 percent higher than Student 3’s average. What was the score Students 1 and 3 received on Test 3?

A. 33
B. 34
C. 35
D. 36
E. 39
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We can let x = the score for student 1 and student 3 on test number 3. Thus, the average score for student 1 is:

(30 + 50 + x)/3 = (80 + x)/3

The average for student 3 is:

(14 + 28 + x)/3 = (42 + x)/3

Student 1’s average (arithmetic mean) score over the three tests was 50 percent higher than Student 3’s average. Using that fact, we can create the following equation to determine x:

(80 + x)/3 = 1.5[(42 + x)/3]

(80 + x) = 1.5(42 + x)

80 + x = 63 + 1.5x

17 = 0.5x

34 = x

Answer: B
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Three students were given three tests, the results of the first two te 21 Dec 2018, 16:05
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