4 boys and 5 girls can do 1/2 work in 6 days. After 1 boy and 2 girls are added and 1/3 work is done in 3 days. How many boys must be added to complete the remaining work in 1 day?

Question

Genesis Allen · Accepted Answer

There may be a more efficient way of calculating this but I can not spot it at the moment.

See explanantion

#color(blue)("Setting up the model")# Let the work rate per day for boys be #w_b# Let the work rate per day for girls by #w_g# Let the total amount or effort (work) be #W# Let the count of days be #d# Let the added count of boys be #x# 4 boys with 5 girls for 6 days: #d(4w_b+5w_g)=W color(white)("d") -> color(white)("d")6(4w_b+5w_g)=W/2" "....Eqn(1)# Add 1 boy and 2 girls: #d(5w_b+7w_g)=W color(white)("d") -> color(white)("d")3(5w_b+7w_g)=W/3" "...Eqn(2)# 3 unknowns and 2 equations thus not solvable in this form. We need to find a way to express #w_b and w_g# in terms of #W#. Thus only have 1 unknown. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine the work rate of girls")# #24w_b+30w_g=W/2" "..................Equation(1_a)# #15w_b+21w_g=W/3" "...................Equation(2_a)# #color(brown)([8xxEqn(2_a)]-[5xxEqn(1_a) ]" eliminates "w_b)# #120w_b+168w_g=(8W)/3" "...................Equation(2_b)# #ul(120w_b+150w_g=(5W)/2)" ".................Equestion(1_b)# #color(white)("d") 0 color(white)("d")color(white)(".d")+color(white)("d")18w_g=W/6# #color(blue)(color(white)("ddddddd")bar(ul(|color(white)(.)w_g=W/108color(white)(.)|)))# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine the work rate of boys")# Substitute foe #w_g# in #Eqn(1) larr" Chosen as even numbers"# #24w_b+30w_g=W/2color(white)("d")->color(white)("d")24w_b+30(W/108)=W/2 # #color(white)("dddddddddddddddd")->color(white)("d")24w_b+color(white)("ddd")(5W)/18color(white)(".d")=W/2 # #color(white)("ddddddddd")color(blue)(ul(bar(|w_b=W/108|))# #color(brown)("The work rate for each boys and girls is "W/108" per day")# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine the count of boys for a 1 day completion")# #color(brown)("Depends on which equation you add the boys to")# I choose #Equation(2)# #d(5w_b+7w_g)=W color(white)("d") -> color(white)("d")3(5w_b+7w_g)=W/3" "...Eqn(2)# Add #x# boys, change #d=3" to "d=1# and change #W/3# to just #W# #d(5w_b+7w_g)=W color(white)("d") -> color(white)("d")1[(5+x)w_b+7w_g]=W# But #w_b=w_g=W/108# #color(white)("d")(5+x)W/108+(7W)/108 = W# Divide both sides by #W# #(5+x)/108+7/108=1# #x=108-7-5# #color(blue)(x="additional boys " = 108-12=96)#