Math Problem Statement

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?

Solution

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

  • Let AA be the number of stickers Ally has.
  • Let BB be the number of stickers Beth has.
  • Let CC 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: AB+C=14A=14(B+C)\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: BA+C=25B=25(A+C)\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+56C = B + 56

Step 2: Solve the system of equations.

We now have three equations:

  1. A=14(B+C)A = \frac{1}{4}(B + C)
  2. B=25(A+C)B = \frac{2}{5}(A + C)
  3. C=B+56C = B + 56

Substitute equation (3) into equation (1) to eliminate CC: A=14(B+(B+56))=14(2B+56)=14(2B+56)=12B+14A = \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=12B+14(Equation 4)A = \frac{1}{2}B + 14 \quad \text{(Equation 4)}

Substitute equation (3) into equation (2) to eliminate CC: B=25(A+(B+56))=25(A+B+56)B = \frac{2}{5}(A + (B + 56)) = \frac{2}{5}(A + B + 56) Simplifying: B=25(A+B+56)=25(A+B)+25×56=25(A+B)+22.4B = \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)+1125B = 2(A + B) + 112 Simplify: 5B=2A+2B+1125B = 2A + 2B + 112 Subtract 2B2B from both sides: 3B=2A+112(Equation 5)3B = 2A + 112 \quad \text{(Equation 5)}

Step 3: Solve the system of linear equations.

We now have two equations:

  1. A=12B+14A = \frac{1}{2}B + 14 (from Equation 4)
  2. 3B=2A+1123B = 2A + 112 (from Equation 5)

Substitute A=12B+14A = \frac{1}{2}B + 14 into 3B=2A+1123B = 2A + 112: 3B=2(12B+14)+1123B = 2\left(\frac{1}{2}B + 14\right) + 112 Simplify: 3B=B+28+1123B = B + 28 + 112 3B=B+1403B = B + 140 Subtract BB from both sides: 2B=1402B = 140 Solve for BB: B=70B = 70

Step 4: Find AA and CC.

Now that we know B=70B = 70, substitute into equation A=12B+14A = \frac{1}{2}B + 14: A=12(70)+14=35+14=49A = \frac{1}{2}(70) + 14 = 35 + 14 = 49

Now, substitute B=70B = 70 into C=B+56C = B + 56: C=70+56=126C = 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=245A + 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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Ratios
Systems of Equations

Formulas

Ratio equation: A / (B + C) = 1/4
Ratio equation: B / (A + C) = 2/5
Linear equations: Substituting and solving equations

Theorems

Substitution method
Elimination method

Suitable Grade Level

Grades 7-10