Sixty boys and girls divided a sum of \( \$ 51 \). If each boy gets 79 cents and each girl gets 139 cents, what is the number of boys and girls in the group?

Let the number of boys be \( b \) and the number of girls be \( g \). According to the problem, we have two equations based on the given information: 1. The total number of boys and girls is 60: \[ b + g = 60 \] 2. The total amount of money distributed is \( \$51 \), which is \( 5100 \) cents. Each boy receives 79 cents and each girl receives 139 cents, leading to: \[ 79b + 139g = 5100 \] Now, we can solve these equations step by step. First, we can solve the first equation for \( g \): \[ g = 60 - b \] Next, we can substitute this expression for \( g \) into the second equation: \[ 79b + 139(60 - b) = 5100 \] Expanding this gives: \[ 79b + 8340 - 139b = 5100 \] Now, combine like terms: \[ -60b + 8340 = 5100 \] Next, subtract 8340 from both sides: \[ -60b = 5100 - 8340 \] \[ -60b = -3240 \] Dividing both sides by -60, we find: \[ b = \frac{3240}{60} = 54 \] Now that we have \( b \), we can find \( g \): \[ g = 60 - 54 = 6 \] Thus, the number of boys is \( 54 \), and the number of girls is \( 6 \). In conclusion, there are \( 54 \) boys and \( 6 \) girls in the group.