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The ratio of the number of cards that Wilbur had to the number of cards Xavier has was 5:9 at first. After Wilbur gave \( {2\over 5}\frac{}{}\)of his cards to his cousin and Xavier lost 84 of his cards. Xavier had twice as many cards as Wilbur. How many cards did both of them have at first?

 Aug 18, 2021
 #1
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Let X be Xavier at first, and W be Wilbur at first. The problem states that \(W \over X\)\(=\) \( 5 \over 9\) ==> 5X = 9W ==> X = (9/5)W

The amounts Wilbur and Xavier have now are \(3 \over 5 \)W and X - 84. The problem also states that X - 84 = 2(\(3 \over 5\)W) ==> X - 84 = \(6 \over 5\)W ==> X = \(6 \over 5\)W + 84

(9/5)W = (6/5)W + 84 =====> 9W = 6W + 420 =====> 3W = 420 =====> W = 140, so X = 252

 Aug 19, 2021
 #2
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W : X = 5 : 9

(5x - 2/5(5x)) / 9x - 84 = 1/2

x = 28

5(28) + 9(28) = 392

 Aug 19, 2021

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