Question

the ratio of red apples to green apples in a basket is 4:6 if there is 40 apples in the basket how many are red?

Answer 1

`\bold{Method\ 1:}\\\\\mathrm{Let\ }4x\ \mathrm{and\ }6x\ \mathrm{be\ the\ number\ of\ red\ and\ green\ apples\ in\ the\ basket}\\\mathrm{respectively.}\\\mathrm{Given,}\\\mathrm{Total\ number\ of\ apples=40}\\\mathrm{or,\ }4x+6x=40\\\mathrm{or,\ }10x=40\\\mathrm{or,\ }x=4\\\mathrm{Number\ of\ red\ apples=}4x=4(4)=16\\`

`\bold{Method\ 2:}\\\\\mathrm{Let\ }x\ \mathrm{and}\ 40-x\ \mathrm{be\ the\ number\ of\ red\ and\ green\ apples\ respectively.}\\\mathrm{Given:}\\\mathrm{Number\ of\ red\ apples:\ Number\ of\ green\ apples=4:6}\\\mathrm{or,\ }(x)/(40-x)=(4)/(6)\\\\\mathrm{or,\ }6x=160-4x\\\mathrm{or,\ }10x=160\\\mathrm{or,\ }x=16\\\mathrm{So\ the\ number\ of\ red\ apples=16.}`

`\bold{Method\ 3:}\\\\\mathrm{Let}\ x\ \mathrm{and}\ y\ \mathrm{be\ the\ number\ of\ red\ and\ green\ apples\ in\ the\ basket\ respectively.}\\\mathrm{First\ case:}\\\mathrm{}x:y=4:6\\\mathrm{or,\ }(x)/(y)=(4)/(6)\\\\\mathrm{or,\ }6x=4y......(1)`

`\mathrm{Second\ case:}\\x+y=40\\\mathrm{or,\ }4(x+y)=4(40)\\\mathrm{or,\ }4x+4y=160\\\mathrm{From\ equation(1),\ substitute}\ 4y\ \mathrm{with}\ 6x.\\\mathrm{or,\ }4x+6x=160\\\mathrm{or,\ }10x=160\\\mathrm{or,\ }x=16\\\mathrm{So\ the\ number\ of\ red\ apples=16}`