Question 890566
Let {{{ g[A] }}} = number of guavas in crate A
Let {{{ p[A] }}}= number of papayas in crate A
Let {{{ g[B] }}} = number of guavas in crate B
Let {{{ p[B] }}}= number of papayas in crate B
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(1) {{{ g[A] / p[A] = 5/3 }}}
(2) {{{ g[B] = (4/9)*p[B] }}}
(2) {{{ g[B] / p[B] = 4/9 }}}
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After transfer:
{{{ g[A] / 2 }}} guavas in crate A
{{{ g[A] / 2 + g[B] }}} guavas in crate B
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{{{ 308 }}} fruits in crate A
(3) {{{ g[A] / 2 + p[A] = 308 }}}
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(4) {{{ ( g[A] / 2 + g[B] ) / p[B] = 5/6 }}}
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(3) {{{ g[A] / 2 + p[A] = 308 }}}
and, from (1):
(1) {{{ g[A] / p[A] = 5/3 }}}
(1) {{{ g[A] = (5/3)*p[A] }}}
(1) {{{ p[A] = ( 3/5 )*g[A] }}}
By substitution:
(3) {{{ g[A] / 2 + ( 3/5 )*g[A] = 308 }}}
(3) {{{ g[A] *( 1/2 + 3/5 ) = 308 }}}
(3) {{{ g[A]*( 11/10 ) = 308 }}}
(3) {{{ g[A] = ( 10/11 )*308 }}}
(3) {{{ g[A] = 280 }}}
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(2) {{{ g[B] / p[B] = 4/9 }}}
(2) {{{ p[B] = (9/4)*g[B] }}}
And, after transfer:
(4) {{{ (( g[A] / 2 + g[B] )) / p[B] = 5/6 }}}
(4) {{{ (( 280 / 2 + g[B] )) / p[B] = 5/6 }}}
(4) {{{ (( 140 + g[B] )) / p[B] = 5/6 }}}
(4) {{{ 140 + g[B] = (5/6)*p[B] }}}
Substitute (2) into (4)
(4) {{{ 140 + g[B] = (5/6)*(9/4)*g[B] }}}
(4) {{{ 140 + g[B] = ( 15/8 )*g[B] }}}
(4) {{{ ( 7/8 )*g[B] = 140 }}}
(4) {{{ g[B] = (8/7)*140 }}}
(4) {{{ g[B] = 160 }}}
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{{{ g[A] / 2 = 280/2 }}}
{{{ g[A] / 2 = 140 }}} 
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(a) 140 guavas were transferred to crate B
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(4) {{{ ( g[A] / 2 + g[B] ) / p[B] = 5/6 }}}
(4) {{{ ( 280 / 2 + 160 ) / p[B] = 5/6 }}}
(4) {{{ ( 140 + 160 ) / p[B] = 5/6 }}}
(4) {{{ p[B] = 300*( 6/5 ) }}}
(4) {{{ p[B] = 360 }}}
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After the transfer, 
{{{ g[A] / 2 + g[B] + p[B] = 140 + 160 + 360 }}}
{{{ g[A] / 2 + g[B] + p[B] = 660 }}}
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(b) There are 660 fruits in crate B
check:
(1) {{{ g[A] / p[A] = 5/3 }}}
(1) {{{ p[A] = (3/5)*g[A] }}}
(1) {{{ p[A] = (3/5)*280 }}}
(1) {{{ p[A] = 168 }}}
and
(3) {{{ g[A] / 2 + p[A] = 308 }}}
(3) {{{ 140 + 168 = 308 }}}
(3) {{{ 308 = 308 }}}
OK