$1$ . Aileen had $\frac{3}{5}$ as many beads as Betty. When Aileen gave Betty some beads, the number of beads Aileen had to the number of beads Betty had become $1: 3$ . Betty then gave $\frac{2}{3}$ of her beads away and had $30$ beads left. How many beads did Aileen have at first?

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1

. Aileen had

\frac{3}{5}

as many beads as Betty. When Aileen gave Betty some beads, the number of beads Aileen had to the number of beads Betty had become

1: 3

. Betty then gave

\frac{2}{3}

of her beads away and had

30

beads left. How many beads did Aileen have at first?

Set Equations: Let's denote the number of beads Aileen had at first as $A$ and the number of beads Betty had at first as $B$ . According to the problem, Aileen had $(3/5)$ as many beads as Betty, so we can write this as: $\newline$ $A = (3/5) \times B$
Ratio Calculation: After Aileen gave some beads to Betty, the ratio of Aileen's beads to Betty's beads became $1:3$ . Let's denote the number of beads Aileen gave to Betty as $x$ . So, the new number of beads Aileen and Betty have are $A - x$ and $B + x$ , respectively. The ratio can be written as: $\newline$ $(A - x) / (B + x) = 1/3$
Betty's Beads: Betty then gave away $(\frac{2}{3})$ of her beads and had $30$ beads left. This means that the number of beads Betty had after receiving beads from Aileen and before giving away $(\frac{2}{3})$ of them was $3$ times $30$ , because she gave away $2$ parts of the total $3$ parts she had. So we can write: $\newline$ $B + x = 3 \times 30$ $\newline$ $B + x = 90$
Substitute and Solve: Now we have two equations: $\newline$ $1$ ) $A = \frac{3}{5} \cdot B$ $\newline$ $2$ ) $B + x = 90$ $\newline$ We can substitute the value of $A$ from equation $1$ into the ratio equation to find the value of $x$ : $\newline$ $\frac{(\frac{3}{5} \cdot B) - x}{B + x} = \frac{1}{3}$
Final Calculation: Multiplying both sides of the equation by $3(B + x)$ to get rid of the fraction, we get: $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $\newline$ $\frac{9B}{5} - 3x = B + x$
Final Calculation: Multiplying both sides of the equation by $3(B + x)$ to get rid of the fraction, we get: $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $\newline$ $\frac{9B}{5} - 3x = B + x$ Now, let's solve for $x$ by moving all terms involving $x$ to one side and all terms involving $B$ to the other side: $\newline$ $\frac{9B}{5} - B = 3x + x$ $\newline$ $\frac{9B - 5B}{5} = 4x$ $\newline$ $\frac{4B}{5} = 4x$
Final Calculation: Multiplying both sides of the equation by $3(B + x)$ to get rid of the fraction, we get: $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $\newline$ $\frac{9B}{5} - 3x = B + x$ Now, let's solve for $x$ by moving all terms involving $x$ to one side and all terms involving $B$ to the other side: $\newline$ $\frac{9B}{5} - B = 3x + x$ $\newline$ $\frac{9B - 5B}{5} = 4x$ $\newline$ $\frac{4B}{5} = 4x$ Divide both sides by $4$ to solve for $x$ : $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $1$ $\newline$ Now we can substitute this value of $x$ back into the equation $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $3$ to find $B$ : $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $5$
Final Calculation: Multiplying both sides of the equation by $3(B + x)$ to get rid of the fraction, we get: $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $\newline$ $\frac{9B}{5} - 3x = B + x$ Now, let's solve for $x$ by moving all terms involving $x$ to one side and all terms involving $B$ to the other side: $\newline$ $\frac{9B}{5} - B = 3x + x$ $\newline$ $\frac{9B - 5B}{5} = 4x$ $\newline$ $\frac{4B}{5} = 4x$ Divide both sides by $4$ to solve for $x$ : $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $1$ $\newline$ Now we can substitute this value of $x$ back into the equation $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $3$ to find $B$ : $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $5$ Multiplying all terms by $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $6$ to get rid of the fraction, we get: $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $7$ $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $8$
Final Calculation: Multiplying both sides of the equation by $3(B + x)$ to get rid of the fraction, we get: $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $\newline$ $\frac{9B}{5} - 3x = B + x$ Now, let's solve for $x$ by moving all terms involving $x$ to one side and all terms involving $B$ to the other side: $\newline$ $\frac{9B}{5} - B = 3x + x$ $\newline$ $\frac{9B - 5B}{5} = 4x$ $\newline$ $\frac{4B}{5} = 4x$ Divide both sides by $4$ to solve for $x$ : $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $1$ $\newline$ Now we can substitute this value of $x$ back into the equation $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $3$ to find $B$ : $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $5$ Multiplying all terms by $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $6$ to get rid of the fraction, we get: $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $7$ $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $8$ Divide both sides by $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $9$ to solve for $B$ : $\newline$ $\frac{9B}{5} - 3x = B + x$ $1$ $\newline$ $\frac{9B}{5} - 3x = B + x$ $2$
Final Calculation: Multiplying both sides of the equation by $3(B + x)$ to get rid of the fraction, we get: $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $\newline$ $\frac{9B}{5} - 3x = B + x$ Now, let's solve for $x$ by moving all terms involving $x$ to one side and all terms involving $B$ to the other side: $\newline$ $\frac{9B}{5} - B = 3x + x$ $\newline$ $\frac{9B - 5B}{5} = 4x$ $\newline$ $\frac{4B}{5} = 4x$ Divide both sides by $4$ to solve for $x$ : $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $1$ $\newline$ Now we can substitute this value of $x$ back into the equation $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $3$ to find $B$ : $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $5$ Multiplying all terms by $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $6$ to get rid of the fraction, we get: $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $7$ $\newline$ $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $8$ Divide both sides by $3 \times \left(\frac{3}{5} \times B - x\right) = B + x$ $9$ to solve for $B$ : $\newline$ $\frac{9B}{5} - 3x = B + x$ $1$ $\newline$ $\frac{9B}{5} - 3x = B + x$ $2$ Now that we have the value of $B$ , we can find the initial number of beads Aileen had ( $\frac{9B}{5} - 3x = B + x$ $4$ ) using the first equation $\frac{9B}{5} - 3x = B + x$ $5$ : $\newline$ $\frac{9B}{5} - 3x = B + x$ $6$ $\newline$ $\frac{9B}{5} - 3x = B + x$ $7$