Question

Represent the following situations mathematically.
i) The Sum of two numbers is 40. We need to find the number if the sum of reciprocal is 2/5

​ii) We need to find two consecutive odd integers ,the sum of whose squares is 202​

Answer 1

Solution:-

i) Let one of the two number be x

•According to question :The sum of two numbers is 40

Then the other number be =40-x

•Their reciprocals are 1/x and 1/(40-x)

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Therefore, the sum of their reciprocal

$\: \: \sf \implies \dfrac{1}{x} + \dfrac{1}{40 - x}$

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Now, According to problem situation

$\: \sf \: \: \implies \dfrac{1}{x} + \dfrac{1}{40 - x} = \dfrac{2}{5}$

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$\: \: \: \: \sf \implies \dfrac{40 - x + x}{x(40 - x)} = \dfrac{2}{5}$

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$\sf \: \: \: \: \implies \dfrac{40}{x(40 - x)} = \dfrac{2}{5}$

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Dividing by 2 on both side

$\sf \: \: \: \: \implies \dfrac{20}{x(40 - x)} = \dfrac{1}{5}$

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$\sf \: \: \: \: \implies x(40 - x) = (20)(5)$

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$\sf \: \: \: \: \implies40 - {x}^{2} = 100$

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$\sf \: \: \: \: \implies {x}^{2} - 40x + 100 = 0$

Therefore, one of the two numbers satisfies the quadratic equation

⠀⠀⠀⠀x²-40x+100=0

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ii) Let the smaller odd integer be x

•According to question:Two consecutive odd integer differ by 2

Then the consecutive greater odd integer =x+2

Therefore, the sum of their squares

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$\: \implies \sf {x}^{2} + {(x + 2)}^{2}$

According to the problem situation

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$\: \implies \sf {x}^{2} + {(x + 2)}^{2} = 202$

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$\: \implies \sf {x}^{2} + {x }^{2} + 4x + 4 = 202$

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$\: \implies \sf2 {x}^{2} + 4x - 198 = 0$

Dividing by 2 in both side

$\: \implies \sf {x}^{2} + 2x - 99 = 0$

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Therefore, the smaller of the two consecutive odd integers satisfies the quadratic equation

⠀⠀⠀⠀x²+2x-99=0

Important note:-

We have to represent the following situations mathematically. So we don't need to find the exact solution of this equation