Question

375375÷455+13.3%of 8600-15.7%of9240=40% of x. Find x =?

Answer 1

Answer:

$\:\boxed{\bf \: x = 1295.30 \: }$

Step-by-step explanation:

Given expression is

$\sf \: 375375 \div 455+13.3\% \: of \: 8600-15.7\% \: of \: 9240=40\% \: of \: x$

can be rewritten as

$\sf \: 375375 \times \dfrac{1}{455} +\dfrac{13.3}{100} \times 8600-\dfrac{15.7}{100} \times 9240=\dfrac{40}{100} \times x$

$\sf \: 825 +13.3 \times 86-\dfrac{157}{100} \times 924=\dfrac{40}{100} \times x$

$\sf \: 825 +1143.8-\dfrac{145068}{100} =\dfrac{2}{5} \times x$

$\sf \: 1968.8-1450.68 =\dfrac{2}{5} \times x$

$\sf \: 518.12 =\dfrac{2}{5} \times x$

$\sf \:x = 518.12 \times \dfrac{5}{2}$

$\sf \:x = 259.06 \times 5$

$\implies\sf \: x = 1295.30$

Hence,

$\implies\sf \:\boxed{\bf \: x = 1295.30 \: }$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$