Ally, Beth and Candy had a number of stickers. The ratio of the number of stickers Ally had to the total number of stickers Beth and Candy had was 1: 4. The ratio of the number of stickers Beth had to the total number of stickers Ally and Candy had was
2: 5. Candy had 56 more stickers than Beth. How many stickers did the 3 girls have altogether?

Let's define variables for the number of stickers each person has:

- Let \( A \) be the number of stickers Ally has.
- Let \( B \) be the number of stickers Beth has.
- Let \( C \) be the number of stickers Candy has.

### Step 1: Set up equations based on the given ratios and conditions.

1. The ratio of Ally's stickers to the total number of stickers Beth and Candy have is 1:4. This gives us the equation:
   \[
   \frac{A}{B + C} = \frac{1}{4} \quad \Rightarrow \quad A = \frac{1}{4}(B + C)
   \]

2. The ratio of Beth's stickers to the total number of stickers Ally and Candy have is 2:5. This gives us the equation:
   \[
   \frac{B}{A + C} = \frac{2}{5} \quad \Rightarrow \quad B = \frac{2}{5}(A + C)
   \]

3. Candy had 56 more stickers than Beth, which gives us:
   \[
   C = B + 56
   \]

### Step 2: Solve the system of equations.

We now have three equations:
1. \( A = \frac{1}{4}(B + C) \)
2. \( B = \frac{2}{5}(A + C) \)
3. \( C = B + 56 \)

Substitute equation (3) into equation (1) to eliminate \( C \):
\[
A = \frac{1}{4}(B + (B + 56)) = \frac{1}{4}(2B + 56) = \frac{1}{4}(2B + 56) = \frac{1}{2}B + 14
\]
So, we have:
\[
A = \frac{1}{2}B + 14 \quad \text{(Equation 4)}
\]

Substitute equation (3) into equation (2) to eliminate \( C \):
\[
B = \frac{2}{5}(A + (B + 56)) = \frac{2}{5}(A + B + 56)
\]
Simplifying:
\[
B = \frac{2}{5}(A + B + 56) = \frac{2}{5}(A + B) + \frac{2}{5} \times 56 = \frac{2}{5}(A + B) + 22.4
\]
Multiply both sides by 5 to clear the fraction:
\[
5B = 2(A + B) + 112
\]
Simplify:
\[
5B = 2A + 2B + 112
\]
Subtract \( 2B \) from both sides:
\[
3B = 2A + 112 \quad \text{(Equation 5)}
\]

### Step 3: Solve the system of linear equations.

We now have two equations:
1. \( A = \frac{1}{2}B + 14 \) (from Equation 4)
2. \( 3B = 2A + 112 \) (from Equation 5)

Substitute \( A = \frac{1}{2}B + 14 \) into \( 3B = 2A + 112 \):
\[
3B = 2\left(\frac{1}{2}B + 14\right) + 112
\]
Simplify:
\[
3B = B + 28 + 112
\]
\[
3B = B + 140
\]
Subtract \( B \) from both sides:
\[
2B = 140
\]
Solve for \( B \):
\[
B = 70
\]

### Step 4: Find \( A \) and \( C \).

Now that we know \( B = 70 \), substitute into equation \( A = \frac{1}{2}B + 14 \):
\[
A = \frac{1}{2}(70) + 14 = 35 + 14 = 49
\]

Now, substitute \( B = 70 \) into \( C = B + 56 \):
\[
C = 70 + 56 = 126
\]

### Step 5: Calculate the total number of stickers.

The total number of stickers the three girls have is:
\[
A + B + C = 49 + 70 + 126 = 245
\]

Thus, the total number of stickers the 3 girls have altogether is **245**.

Would you like more details or have any questions?

### Related Questions:
1. How can you find the total number of stickers if the ratio for Beth had been different?
2. What if Candy had 30 fewer stickers than Beth instead of 56 more?
3. How do the ratios help in solving the problem when more people are involved?
4. What changes if the total number of stickers for Ally and Beth is known instead of the ratios?
5. Can you generalize the method for solving problems with more complex ratios?

### Tip:
Always start by translating ratios into algebraic equations, then use substitution or elimination to solve them efficiently.

Ally, Beth, and Candy had a number of stickers. The ratio of the number of stickers Ally had to the total number of stickers Beth and Candy had was 1:4. The ratio of the number of stickers Beth had to the total number of stickers Ally and Candy had was 2:5. Candy had 56 more stickers than Beth. How many stickers did the 3 girls have altogether?

Learn how to solve a complex ratio problem involving three people using algebraic methods. This problem explores key concepts like setting up equations from ratios and solving systems of linear equations. Ideal for students in Grades 7-10.

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