Let the elements of the first data set be denoted as \( x_1, x_2, \ldots, x_{13} \).
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In the second data set, each element is increased by 5: \[ y_i = x_i + 5 \quad \text{for } i = 1, 2, \ldots, 13. \] Thus, the elements of the second data set are \( y_1, y_2, \ldots, y_{13} \).
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In the third data set, each element of the second data set is multiplied by 3: \[ z_i = 3y_i = 3(x_i + 5) = 3x_i + 15 \quad \text{for } i = 1, 2, \ldots, 13. \] Thus, the elements of the third data set are \( z_1, z_2, \ldots, z_{13} \).
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We know the median of the third set is 42. Since there are 13 elements in this data set, the median is the 7th element when the elements are ordered: \[ z_7 = 42. \]
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From the expression for \( z_i \): \[ z_7 = 3x_7 + 15. \] Setting this equal to 42 gives: \[ 3x_7 + 15 = 42. \]
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To solve for \( x_7 \): \[ 3x_7 = 42 - 15, \] \[ 3x_7 = 27, \] \[ x_7 = \frac{27}{3} = 9. \]
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The median of the first data set is therefore: \[ \boxed{9}. \]