3. Rudi, Sam and Teh had 162 stickers altogether. Rudi gave \( \frac{3}{10} \) of his stickers to Sam and \( \frac{1}{4} \) of his stickers to Ted. After that, all three boys had the same number of stickers Who had more stickers at first, Sam or Ted, and how many more?

Let's solve the problem step by step. ### **Step 1: Define the Variables** Let: - \( R \) = Number of stickers Rudi initially had - \( S \) = Number of stickers Sam initially had - \( T \) = Number of stickers Ted initially had Given: \[ R + S + T = 162 \] ### **Step 2: Express Transfers** Rudi gives: - \( \frac{3}{10}R \) stickers to Sam - \( \frac{1}{4}R \) stickers to Ted After the transfers: - Rudi has: \[ R - \left( \frac{3}{10}R + \frac{1}{4}R \right) = R - \frac{11}{20}R = \frac{9}{20}R \] - Sam has: \[ S + \frac{3}{10}R \] - Ted has: \[ T + \frac{1}{4}R \] ### **Step 3: Set Final Sticker Counts Equal** After the transfers, all three have the same number of stickers: \[ \frac{9}{20}R = S + \frac{3}{10}R = T + \frac{1}{4}R \] Solving for \( S \) and \( T \) in terms of \( R \): - For Sam: \[ S + \frac{3}{10}R = \frac{9}{20}R \] \[ S = \frac{9}{20}R - \frac{3}{10}R = \frac{3}{20}R \] - For Ted: \[ T + \frac{1}{4}R = \frac{9}{20}R \] \[ T = \frac{9}{20}R - \frac{1}{4}R = \frac{1}{5}R \] ### **Step 4: Substitute Back into the Total** Now substitute \( S \) and \( T \) back into the total: \[ R + \frac{3}{20}R + \frac{1}{5}R = 162 \] \[ R + \frac{3}{20}R + \frac{4}{20}R = 162 \] \[ \frac{27}{20}R = 162 \] \[ R = 162 \times \frac{20}{27} = 120 \] Thus: - \( R = 120 \) - \( S = \frac{3}{20} \times 120 = 18 \) - \( T = \frac{1}{5} \times 120 = 24 \) ### **Step 5: Compare Sam and Ted** - **Sam initially had 18 stickers.** - **Ted initially had 24 stickers.** **Conclusion:** **Ted had more stickers at first, having 6 more stickers than Sam.** ### **Final Answer** Ted initially had six more stickers than Sam.