Question

1. in a class, there are 21 boys. \\(\frac{3}{5}\\) of the girls wear spectacles. if \\(\frac{1}{4}\\) of the class are girls who wear spectacles, how many pupils are there in the class?

Explanation:

Step 1: Define variables

Let the total number of pupils in the class be \( x \), and the number of girls be \( g \). We know the number of boys is 21, so \( x = 21 + g \), which means \( g = x - 21 \).

Step 2: Translate the spectacle - wearing information into an equation

We are given that \( \frac{3}{5} \) of the girls wear spectacles, and \( \frac{1}{4} \) of the class are girls who wear spectacles. So, \( \frac{3}{5}g=\frac{1}{4}x \).

Step 3: Substitute \( g \) into the equation

Substitute \( g = x - 21 \) into \( \frac{3}{5}g=\frac{1}{4}x \). We get \( \frac{3}{5}(x - 21)=\frac{1}{4}x \).

Step 4: Expand the left - hand side

Using the distributive property \( a(b - c)=ab - ac \), where \( a=\frac{3}{5} \), \( b = x \), and \( c = 21 \), we have \( \frac{3}{5}x-\frac{3\times21}{5}=\frac{1}{4}x \), which simplifies to \( \frac{3}{5}x-\frac{63}{5}=\frac{1}{4}x \).

Step 5: Move the \( x \) terms to one side

Subtract \( \frac{1}{4}x \) from both sides and add \( \frac{63}{5} \) to both sides:
\( \frac{3}{5}x-\frac{1}{4}x=\frac{63}{5} \)
Find a common denominator for the left - hand side. The common denominator of 5 and 4 is 20. So, \( \frac{3\times4}{5\times4}x-\frac{1\times5}{4\times5}x=\frac{63}{5} \), which is \( \frac{12}{20}x-\frac{5}{20}x=\frac{63}{5} \).
Simplifying the left - hand side gives \( \frac{7}{20}x=\frac{63}{5} \).

Step 6: Solve for \( x \)

Multiply both sides by the reciprocal of \( \frac{7}{20} \), which is \( \frac{20}{7} \):
\( x=\frac{63}{5}\times\frac{20}{7} \)
Simplify the right - hand side: \( \frac{63\times20}{5\times7}=\frac{63\times4}{7}=9\times4 = 36 \).

Answer:

The total number of pupils in the class is 36.