Question 1130453
Let {{{ n }}}= the original number of calculators
{{{ 300/n }}} = the original cost/calculator
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{{{ ( 300/n - 1 )*( n + 10 ) = 300 }}}
{{{ 300 - n + 3000/n - 10 = 300 }}}
{{{ 3000/n - n = 10  }}}
{{{ 3000 - n^2 = 10n }}}
{{{ -n^2 - 10n + 3000 = 0 }}}
{{{ ( n + 60 )*( -n + 50 ) = 0 }}} ( by inspection )
{{{ n = 50 }}} ( can't use the negative solution )
50 calculators can be bought at original price
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check:
{{{ ( 300/50 - 1 )*( 50 + 10 ) = 300 }}}
{{{ ( 6 - 1 )*60 = 300 }}}
{{{ 5*60 = 300 }}}
{{{ 300 = 300 }}}
OK