PH = Numbers of doors on Pang’s Honda
PT = Numbers of doors on Pang’s Toyota
PM = Numbers of doors on Pang’s Mazda
FH = Numbers of doors on Farid’s Honda
FT = Numbers of doors on Farid’s Toyota
FM = Numbers of doors on Farid’s Mazda
HH = Numbers of doors on Harjit’s Honda
HT = Numbers of doors on Harjit’s Toyota
HM = Numbers of doors on Harjit’s Mazda
Cars of each owner has together 11 doors:
one two-door + one four-door + one five-door = 2 + 4 + 5 = 11 doors
So you have system of equations:
FH + FT + FM = 11
PH + PT + PM = 11
HH + HT + HM = 11
Farid’s Honda has the same number of doors as Pang’s Toyota mean:
FH = PT
Harjit’s Honda has the same number of doors as Farid’s Toyota mean:
HH = FT
Farid’s Mazda is a two-door mean:
FM = 2
Pang’s Mazda is a four-door mean:
PM = 4
Put this values in the initial system:
PT + FT + 2 = 11
PH + PT + 4 = 11
FT + HT + HM = 11
Your system become:
PT + FT = 9
PH + PT = 7
FT + HT + HM = 11
First equation:
PT + FT = 9
PT = 9 - FT
One car has 5 doors, other has 4 doors.
PT = 5
FT = 4
The solution PT = 4 , FT = 9 - PT = 9 - 4 = 5 is not possible because Pang’s Mazda has four doors ( PM = 4 ) so in that case Pang’s Toyota also will bee have four-door cars.
In that case, Pang would have two four-door cars.
Second equation:
PH + PT = 7
One car has 5 doors, other has 2 doors.
PH = 2
PT = 5
The solution PH = 5 , PT = 2 is not possible because
Pang still has Toyota with five doors ( PT = 5 )
In that case, Pang would have two five-door cars.
In the initial system of equations:
FH + FT + FM = 11
FH + 4 + 2 = 11
FH + 6 = 11
FH = 5
Third equation:
FT + HT + HM = 11
4 + HT + HM = 11
Subtract 4 to both sides
HT + HM = 7
One car has two doors and the other five doors.
HT = 2
HM = 5
The solution HT = 5 , HM = 2 is not possible because
Pang’s Toyota is five-door car ( PT = 5 )
None of the same car brand has the same number of doors.
Therefore Harjit’s Toyota can’t have five doors.
In the initial system of equations:
HH + HT + HM = 11
HH + 2 + 5 = 11
HH + 7 = 11
HH = 4
Answers:
a. Harjit , because HM = 5
b. Pang, because PT = 5
c. Harjit, because HT = 2
d. Harjit , because HH = 4
e. Farid, because FH = 5
f. Pang, because PH = 2
Three neighbors – Farid, Pang and Harjit – each own 3 cars; one two-door, one
four-door and one five-door. None of the same car brand has the same number of
doors. They each have a Honda, Toyota and Mazda. Farid’s Honda has the same
number of doors as Pang’s Toyota. Harjit’s Honda has the same number of doors
as Farid’s Toyota. Farid’s Mazda is a two-door and Pang’s Mazda is a four-door.
a. Who has a five-door Mazda?
b. Who has a five-door Toyota?
c. Who has a two-door Toyota?
d. Who has a four-door Honda?
e. Who has a five-door Honda?
f. Who has a two-door Honda?
g. Please show how you get the answers.
1 answer
1 answer