Let's analyze and solve the problem given in the image.

Problem Statement:

Classes 5A and 5B have the same number of pupils. The following information is provided:

1. The ratio of the number of boys in 5A to the number of boys in 5B is 2:3.
2. The ratio of the number of girls in 5A to the number of girls in 5B is 7:4.

(a) Find the ratio of the number of boys to the number of girls in 5A.

(b) If there are 15 fewer girls than boys in 5B, how many pupils are there in 5B?

Solution:

(a) Find the ratio of the number of boys to the number of girls in 5A.

Let:

• The number of boys in 5A = $2x$
• The number of boys in 5B = $3x$
• The number of girls in 5A = $7y$
• The number of girls in 5B = $4y$

Since the total number of students in both classes is the same:

$\text{Number of students in 5A} = 2x + 7y$ $\text{Number of students in 5B} = 3x + 4y$

So, $2x + 7y = 3x + 4y$.

Subtract $2x + 4y$ from both sides:

$3y = x$

Therefore, the number of boys in 5A is $2x = 2(3y) = 6y$.

The ratio of the number of boys to the number of girls in 5A is:

$\text{Ratio} = \frac{6y}{7y} = \frac{6}{7}$

So, the ratio is $\boxed{6:7}$.

(b) If there are 15 fewer girls than boys in 5B, how many pupils are there in 5B?

From the problem statement: $3x - 4y = 15$

We know $x = 3y$, so substituting in the equation:

$3(3y) - 4y = 15$

Simplifying:

$9y - 4y = 15$

$5y = 15$

$y = 3$

Now, substituting $y = 3$ back into $x = 3y$:

$x = 9$

So, the number of students in 5B is:

$3x + 4y = 3(9) + 4(3) = 27 + 12 = 39$

Thus, there are $\boxed{39}$ pupils in 5B.

Would you like further details or have any questions?

Follow-Up Questions:

1. What is the total number of students in class 5A?
2. If 10 more girls join 5A, what will be the new ratio of boys to girls in 5A?
3. If 5 boys leave 5B, how will that affect the total number of pupils in 5B?
4. What is the total number of girls in both classes combined?
5. How would the problem change if the ratio of boys in 5A to 5B was reversed?

Tip:

When dealing with ratio problems, express all variables in terms of one unknown whenever possible to simplify solving the equations.