Let's denote the number of cards Ruchard initially had as \( r \). Then, Esther had \( 5r \) cards since she had five times as many cards as Ruchard.
-
Initially, Esther has \( 5r \) cards and Ruchard has \( r \) cards.
-
Esther gives \( \frac{1}{4} \) of her cards to Ruchard: \[ \frac{1}{4} \times 5r = \frac{5r}{4} \] After this transaction, Esther has: \[ 5r - \frac{5r}{4} = \frac{20r}{4} - \frac{5r}{4} = \frac{15r}{4} \] Ruchard receives these cards, so now he has: \[ r + \frac{5r}{4} = \frac{4r}{4} + \frac{5r}{4} = \frac{9r}{4} \]
-
Next, Ruchard gives \( \frac{1}{6} \) of his cards to Esther: \[ \frac{1}{6} \times \frac{9r}{4} = \frac{9r}{24} = \frac{3r}{8} \] After this transaction, Ruchard has: \[ \frac{9r}{4} - \frac{3r}{8} \] To subtract these fractions, we convert \( \frac{9r}{4} \) into eighths: \[ \frac{9r}{4} = \frac{18r}{8} \] So, \[ \frac{18r}{8} - \frac{3r}{8} = \frac{15r}{8} \]
Esther receives these cards, so now she has: \[ \frac{15r}{4} + \frac{3r}{8} \] Again, convert \( \frac{15r}{4} \) into eighths: \[ \frac{15r}{4} = \frac{30r}{8} \] Therefore, \[ \frac{30r}{8} + \frac{3r}{8} = \frac{33r}{8} \]
-
Finally, we know that in the end, Esther has 90 cards more than Ruchard: \[ \frac{33r}{8} = \frac{15r}{8} + 90 \]
To solve for \( r \), we first simplify the equation: \[ \frac{33r}{8} - \frac{15r}{8} = 90 \] This gives: \[ \frac{18r}{8} = 90 \] Simplifying, we multiply both sides by 8: \[ 18r = 720 \] Now, dividing both sides by 18: \[ r = 40 \]
-
Finally, we can find the number of cards Esther initially had: \[ 5r = 5 \times 40 = 200 \]
Therefore, Esther had initially 200 cards.
2 answers