Question

Two numbers are such that square of one is 224 less than 8 times the square of the other. if two numbers be in the ratio of 4:3 the numbers are?

Answer 1

hope it helps u......

Answer 2

$Let m, n \:are \: two \: numbers$

$m : n = 4 : 3\: (given)$

$Let \: m = 4x \:and \\n = 3x$

/* According to the problem given*/

Two numbers are such that square of one is 224 less than 8 times the square of the other.

$m^{2} + 224 = 8n^{2}$

$\implies (4x)^{2} + 224 = 8\times (3x)^{2}$

$\implies 16x^{2} + 224 = 8 \times 9x^{2}$

$\implies 16x^{2} + 224 = 72x^{2}$

$\implies 224 = 72x^{2} - 16x^{2}$

$\implies 224 = 56x^{2}$

$\implies x^{2} = \frac{224}{56}$

$\implies x^{2} = 4$

$\implies x = 2$

Therefore.,

$\red { One \: number } = 4x \\= 4 \times 2 \\ \green {= 8 }$

$\red { Second \: number } = 3x \\= 3 \times 2 \\ \green {= 6 }$

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