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29 Apr 2025, 21:31
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Sum of Large: 19
Small: 5
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
84% (02:19) correct
16%
(02:42)
wrong
based on 383
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In a certain game, 50 large marbles are numbered 1 through 50 and placed in a jar, while 10 small marbles are numbered 1 through 10 and placed in a separate jar. To play the game, a person chooses 3 large marbles and 1 small marble, and is awarded a quantity of points equal to the average of the numbers on all 4 of the marbles she chose. During one game, one of the marbles that Alex chose had the number 48 on it, and he received 18 points.
In the first column, identify a number that could be the sum of the other two large marbles chosen. In the second column, identify the corresponding value of the small marble chosen. The two choices must be consistent with each other. Make only two selections, one in each column.
In the first column, identify a number that could be the sum of the other two large marbles chosen. In the second column, identify the corresponding value of the small marble chosen. The two choices must be consistent with each other. Make only two selections, one in each column.
| Sum of Large | Small | Value |
| 5 | ||
| 8 | ||
| 10 | ||
| 11 | ||
| 13 | ||
| 19 |
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Official Answer
Sum of Large: 19
Small: 5
poojaarora1818
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01 May 2025, 04:11
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Solution:
Given in the question:
1. 50 large marbles are numbered 1 through 50.
2. 10 small marbles are numbered 1 through 10.
3. A person chooses 3 large marbles and 1 small marble.
4. The quantity of points awarded is equal to the average of the numbers on all 4 marbles.
5. One of the marbles that Alex chose had the number 48 on it, and he received 18 points.
As mentioned in the question itself, the points awarded are equal to the average of the numbers on all 4 marbles. Let's assume the sum of 2 large marbles is X +Y and the one small marble is Z. So, our equation would be \(\frac{48 + X + Y + Z}{4 }\)= 18. So, X + Y + Z = 24. Now, we have to look into the options available to us that match up with the values for the sum of 2 large and one small marble, and their sum should be 24.
Hence, the only option that fits for a sum of two large marbles would be 19, and one small marble is 5.
Given in the question:
1. 50 large marbles are numbered 1 through 50.
2. 10 small marbles are numbered 1 through 10.
3. A person chooses 3 large marbles and 1 small marble.
4. The quantity of points awarded is equal to the average of the numbers on all 4 marbles.
5. One of the marbles that Alex chose had the number 48 on it, and he received 18 points.
As mentioned in the question itself, the points awarded are equal to the average of the numbers on all 4 marbles. Let's assume the sum of 2 large marbles is X +Y and the one small marble is Z. So, our equation would be \(\frac{48 + X + Y + Z}{4 }\)= 18. So, X + Y + Z = 24. Now, we have to look into the options available to us that match up with the values for the sum of 2 large and one small marble, and their sum should be 24.
Hence, the only option that fits for a sum of two large marbles would be 19, and one small marble is 5.
Bismuth83
01 May 2025, 08:38
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Bismuth83
Given that the average of the 4 marbels = 18 -> 18*4 = 72 = 48 + L + L + S.
We need to find L + L (I will treat it as a single term called "Large") and S.
24 = Large + S
- S can be any number from 1 - 10 included.
If S = 10 -> Large = 14 -> Not possible
If S = 5 -> Large = 19 -> Possible and thus the correct solution
L + L = 19 and S = 5
