Jack, Tom, Beth, and Ruth each have a number of marbles. Jack has $10$ marbles, the least, while Tom has $27$ marbles, the highest. Which of the following could be the average of the number of marbles present with all of them, given that no two of them has the same number of marbles? $\newline$ $A)\ \$13$ $\newline$ $B)\ \$17$ $\newline$ $C)\ \$23$ $\newline$ $D)\ \$26$

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Q. Jack, Tom, Beth, and Ruth each have a number of marbles. Jack has

10

marbles, the least, while Tom has

27

marbles, the highest. Which of the following could be the average of the number of marbles present with all of them, given that no two of them has the same number of marbles?

\newline

A)\ \$13

\newline

B)\ \$17

\newline

C)\ \$23

\newline

D)\ \$26

Denote Marbles: Let's denote the number of marbles Jack, Tom, Beth, and Ruth have as $J$ , $T$ , $B$ , and $R$ respectively. We know that $J = 10$ and $T = 27$ . Since no two of them have the same number of marbles, $B$ and $R$ must have a number of marbles that is between $10$ and $27$ , but not equal to $10$ or $27$ .
Calculate Average: The average number of marbles $A$ can be calculated by adding the number of marbles each person has and dividing by $4$ , since there are $4$ people. So, $A = (J + T + B + R) / 4$ .
Check Option (A): We know that $J + T = 10 + 27 = 37$ . Now, we need to find possible values for $B$ and $R$ such that when added to $37$ and divided by $4$ , the result is one of the given options (A, B, C, D).
Check Option (B): Let's check each option to see if it could be the average: $\newline$ For option (A) $13$ : $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 13$ . Multiplying both sides by $4$ gives us $37 + B + R = 52$ . This means $B + R = 52 - 37 = 15$ . Since $B$ and $R$ must be different and between $10$ and $27$ , they cannot add up to $15$ . This option is not possible.
Check Option (C): For option (B) $17$ : $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 17$ . Multiplying both sides by $4$ gives us $37 + B + R = 68$ . This means $B + R = 68 - 37 = 31$ . Since $B$ and $R$ must be different and between $10$ and $27$ , they could potentially add up to $31$ with $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 17$ $0$ and $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 17$ $1$ , for example. This option is possible.
Check Option (D): For option (C) $23$ : $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 23$ . Multiplying both sides by $4$ gives us $37 + B + R = 92$ . This means $B + R = 92 - 37 = 55$ . Since the maximum number of marbles $R$ can have is $26$ (one less than Tom), and the minimum for $B$ is $11$ (one more than Jack), the maximum sum of $B$ and $R$ is $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 23$ $1$ , which is less than $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 23$ $2$ . This option is not possible.
Check Option (D): For option (C) $23$ : $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 23$ . Multiplying both sides by $4$ gives us $37 + B + R = 92$ . This means $B + R = 92 - 37 = 55$ . Since the maximum number of marbles $R$ can have is $26$ (one less than Tom), and the minimum for $B$ is $11$ (one more than Jack), the maximum sum of $B$ and $R$ is $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 23$ $1$ , which is less than $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 23$ $2$ . This option is not possible.For option (D) $26$ : $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 23$ $4$ . Multiplying both sides by $4$ gives us $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 23$ $6$ . This means $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 23$ $7$ . Since the maximum number of marbles $R$ can have is $26$ and the minimum for $B$ is $11$ , the maximum sum of $B$ and $R$ is $A = (J + T + B + R) / 4 = (37 + B + R) / 4 = 23$ $1$ , which is less than $4$ $5$ . This option is not possible.