Question Number 207610 by Davidtim last updated on 20/May/24
$${we}\:{have}\:\mathrm{100}\:{money}\:{if}\:{we}\:{want}\:{to}\:{buy}\:\mathrm{100}\:{Donkys}\:{Horses} \\ $$$${and}\:{Camels}\:{mixed}\:{while}\:\mathrm{20}\:{Donkys} \\ $$$${cost}\:\mathrm{1}\:{money},\:\mathrm{1}\:{Horse}\:{costs}\:\mathrm{1}\:{money}\:{and} \\ $$$$\mathrm{1}\:{Camel}\:{costs}\:\mathrm{5}\:{money}. \\ $$$${find}\:{total}\:{number}\:{of}\:{Donkys},\:{Horses}, \\ $$$${and}\:{Camels}. \\ $$
Commented by Davidtim last updated on 21/May/24
$${please}\:{help}\:{me}!!! \\ $$
Answered by Tinku Tara last updated on 21/May/24
$$\mathrm{If}\:\mathrm{we}\:\mathrm{buy}\:{d}\:\mathrm{donkeys},\:{h}\:\mathrm{horses}\:\mathrm{and} \\ $$$${c}\:\mathrm{camels}. \\ $$$$\mathrm{c}+\mathrm{h}+\mathrm{d}=\mathrm{100}\:\:\left(\mathrm{i}\right) \\ $$$$\mathrm{money}\:\mathrm{for}\:\mathrm{camels}\:\mathrm{5c},\:\mathrm{horses}\:{h}, \\ $$$$\mathrm{and}\:\mathrm{donkeys}\:\mathrm{d}/\mathrm{20} \\ $$$$\mathrm{5c}+\mathrm{h}+\frac{\mathrm{d}}{\mathrm{20}}=\mathrm{100}\:\:\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{ii}\right)−\left(\mathrm{i}\right) \\ $$$$\mathrm{4c}−\frac{\mathrm{19d}}{\mathrm{20}}=\mathrm{0} \\ $$$$\mathrm{19d}=\mathrm{80c} \\ $$$$\mathrm{d},\:\mathrm{c}\:\mathrm{are}\:\mathrm{integers}\:\mathrm{so}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{is} \\ $$$$\mathrm{c}=\mathrm{19} \\ $$$$\mathrm{d}=\mathrm{80} \\ $$$$\mathrm{c}+\mathrm{d}+\mathrm{h}=\mathrm{100} \\ $$$$\Rightarrow\mathrm{h}=\mathrm{1} \\ $$$$\mathrm{C}{amels}\:\mathrm{19} \\ $$$$\mathrm{Donkeys}\:\mathrm{80} \\ $$$$\mathrm{Horses}\:\mathrm{1} \\ $$
Commented by Davidtim last updated on 21/May/24
$${so}\:{much}\:{thanks} \\ $$