8 325 children attended a concert at a school auditorium. \( \frac{4}{5} \) of the boys \( \frac{1}{4} \) and \( \frac{3}{4} \) of the girls left the auditorium at the end of the concert. 29 more boys than girls remained in the auditorium. How many girls attended the concert?

Let's solve the problem step-by-step. ### **Step 1: Define Variables** - Let **\( b \)** represent the number of boys. - Let **\( g \)** represent the number of girls. Given: - The total number of children is \( b + g = 325 \). ### **Step 2: Determine the Number of Children Remaining** - **Boys Remaining:** - \( \frac{4}{5} \) of the boys **left**, so \( \frac{1}{5} \) of the boys remained. - Number of boys remaining: \( \frac{1}{5}b \). - **Girls Remaining:** - \( \frac{3}{4} \) of the girls **left**, so \( \frac{1}{4} \) of the girls remained. - Number of girls remaining: \( \frac{1}{4}g \). ### **Step 3: Setup the Equation Based on the Given Condition** It's given that **29 more boys than girls remained** in the auditorium: \[ \frac{1}{5}b = \frac{1}{4}g + 29 \] ### **Step 4: Solve the System of Equations** We have two equations: 1. \( b + g = 325 \) 2. \( \frac{1}{5}b = \frac{1}{4}g + 29 \) **From the first equation:** \[ b = 325 - g \] **Substitute \( b \) into the second equation:** \[ \frac{1}{5}(325 - g) = \frac{1}{4}g + 29 \] **Multiply every term by 20 to eliminate the fractions:** \[ 4(325 - g) = 5g + 580 \] \[ 1300 - 4g = 5g + 580 \] **Combine like terms:** \[ 1300 - 580 = 9g \] \[ 720 = 9g \] \[ g = 80 \] ### **Step 5: Conclusion** **80 girls attended the concert.** ### **Verification** - **Number of boys:** \( 325 - 80 = 245 \) - **Boys remaining:** \( \frac{1}{5} \times 245 = 49 \) - **Girls remaining:** \( \frac{1}{4} \times 80 = 20 \) - **Check:** \( 49 = 20 + 29 \) ✔️ **Answer:** 80