Question

x% of y +?% of x= x% of (x+y)​

Answer 1

Step-by-step explanation:

$let \: question \: mar \: be \: m \\ \frac{x}{100} \times y + \frac{m}{100} \times x = \frac{x}{100} \times (x + y) \\ \frac{xy}{100} + \frac{mx}{100} = ( {x}^{2} + xy) \div 100 \\ \\ xy + mx = {x}^{2} + xy \\ \\ mx = {x}^{2} \\ \\ m = x$

Answer 2

(x% of y) + (?% of x) = x% of (x+y)

Let ? be z

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x% of y :

$\frac{x}{100} \times y$

$= \dfrac{xy}{100}$

z% of x :

$\frac{z}{100} \times x$

$= \dfrac{zx}{100}$

x% of (x + y) :

$\frac{x}{10} \times (x + y)$

$\frac{x \times (x + y)}{100}$

$\dfrac{ {x}^{2} + xy }{100}$

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Hence,

$\frac{xy}{100} + \frac{zx}{100} = \frac{ {x}^{2} + xy}{100}$

$= \dfrac{xy + zx}{100} = \frac{ {x}^{2} + xy }{100}$

[Note : When the denominators are the same for LHS (Left Hand Side) and RHS (Right Hand Side) we can actually cancel it directly, but it's usually best to cross multiply first to avoid any mistakes]

by cross multiplication,

$100(xy + zx) = 100( {x}^{2} + xy)$

$100xy + 100zx = 100 {x}^{2} + 100xy$

by bringing 100xy to the LHS,

$100xy + 100zx - 100xy = 100 {x}^{2}$

(+100xy) and (-100xy) gets cancelled.

$100zx = 100 {x}^{2}$

$100 \times z \times x = 100 \times x \times x$

$z = \dfrac{100 \times x \times x}{100 \times x}$

z = x

Therefore,

?% = x%

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VERIFICATION :

Substituting ? for x,

(x% of y) + (x% of x) = x% of (x+y)

$\dfrac{xy}{100} + \dfrac{ {x}^{2} }{100} = \dfrac{ {x}^{2} + xy }{100}$

LHS = RHS.

Hence verified

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