The sum of two numbers x and y (where x > y ) is equal to four times their difference. Find the value of 5xy/3(x^2 - y^2) . - Quantitative Aptitude MCQ Prepp 2026 | ExamTest.live

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The sum of two numbers x and y (where x > y) is equal to four times their difference. Find the value of \frac{5 x y}{3 ( x ^{2} - y ^{2} )} .

Solution & Step-by-step Explanation

Given the condition: (x + y) = 4 (x - y) Expanding and simplifying the equation: x + y = 4 x - 4 y y + 4 y = 4 x - x 5 y = 3 x \frac{x}{y} = \frac{5}{3} Let x = 5 k and y = 3 k . Now substitute these values into the required expression: \frac{5 x y}{3 ( x ^{2} - y ^{2} )} = \frac{5 ( 5 k ) ( 3 k )}{3 (( 5 k ) ^{2} - ( 3 k ) ^{2} )} \Rightarrow \frac{75 k ^{2}}{3 ( 25 k ^{2} - 9 k ^{2} )} \Rightarrow \frac{75 k ^{2}}{3 \times 16 k ^{2}} \Rightarrow \frac{75}{48} = \frac{25}{16}

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The sum of two numbers x and y (where x > y) is equal to four times their difference. Find the value of \frac{5 x y}{3 ( x ^{2} - y ^{2} )} .

25

\frac{25}{16}

16

2 \frac{3}{16}

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