Question 433560
Let {{{a}}} = tons of 20% mixture to be used
Let {{{b}}} = tons of 80% mixture to be used
given:
(1) {{{ a + b = 10 }}} tons
Tons of concrete in 20% mixture = {{{.2a}}}
Tons of concrete in 80% mixture = {{{.8b}}}
(2) {{{ (.2a + .8b) / 10 = .62 }}}
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(2) {{{ .2a + .8b = 6.2 }}}
(2) {{{ 2a + 8b = 62 }}}
Multiply both sides of (1) by {{{2}}}  and
subtract (1) from (2)
(2) {{{ 2a + 8b = 62 }}}
(1) {{{ -2a - 2b = 20 }}}
{{{ 6b = 42 }}}
{{{ b = 7 }}}
and, since
{{{ a+ b = 10 }}}
{{{ a = 3 }}}
3 tons of 20% mixture are to be used
7 tons of 80% mixture are to be used
check:
(2) {{{ (.2a + .8b) / 10 = .62 }}}
(2) {{{ (.2*3 + .8*7) / 10 = .62 }}}
(2) {{{ (.6 + 5.6) / 10 = .62 }}}
{{{ 6.2/10 = .62 }}}
{{{ 6.2 = 6.2 }}}
OK