Two trucks leave the centre of the city at the same time. One truck travels east and the other travels west. The eastbound truck travels at 6565km/h and the westbound truck at 5858km/h. When will the trucks be 5531255312km apart?
A word problem can be drawn as a diagram and then translated into an equation for easier solving.
First, draw a diagram of the problem to understand it more
Translate the problem into an equation based on the diagram
time taken in hours: tt
distance of westbound truck: 58t58t
distance of eastbound truck: 65t65t
Distance West ++ Distance East
==
5531255312
58t+65t58t+65t
==
5531255312
Since S=dtS=dt, it means that d=S×td=S×t.
Solve for tt
58t+65t58t+65t
==
5531255312
123t123t
==
5531255312
123t123t÷123÷123
==
5531255312÷123÷123
Divide both sides by 123123
tt
==
412412 hours
412412 hours
Question 2 of 6
2. Question
A lioness is 128128 metres from a gazelle. The lioness starts to sprint towards the gazelle at 2222m/s. At the same time, the gazelle starts to sprint at 1818m/s. When will the lioness catch the gazelle?
A word problem can be drawn as a diagram and then translated into an equation for easier solving.
First, draw a diagram of the problem to understand it more
Translate the problem into an equation based on the diagram
Note that the gazelle and lioness would meet when their distance becomes equal
time taken in seconds: tt
distance of the gazelle: 128+18t128+18t
distance of the lioness: 22t22t
speed of the gazelle: 18m/s18m/s
speed of the lioness: 22m/s22m/s
distance of the gazelledistance of the gazelle
==
distance of the lionessdistance of the lioness
128+18t128+18t
==
22t22t
Solve for tt
128+18t128+18t
==
22t22t
128+18t128+18t-18t−18t
==
22t22t-18t−18t
Subtract 18t18t from both sides
128128
==
4t4t
4t4t
==
128128
4t4t÷4÷4
==
128128÷4÷4
Divide both sides by 44
tt
==
3232 seconds
3232 seconds
Question 3 of 6
3. Question
Two submarines 5555km apart aimed and fired their torpedoes toward each other. Torpedo A averages 5050km/h and Torpedo B averages a speed of 6060km/h. When and where do they collide and impact each other?
A word problem can be drawn as a diagram and then translated into an equation for easier solving.
First, draw a diagram of the problem to understand it more
Translate the problem into an equation based on the diagram
time taken in hours: tt
distance of Torpedo A: 50t50t
distance of Torpedo B: 60t60t
Distance A + Distance B: 55km55km
50t+60t50t+60t
==
5555
Since S=dtS=dt, it means that d=S×td=S×t.
Solve for tt
50t+60t50t+60t
==
5555
110t110t
==
5555
100t100t÷110÷110
==
5555÷110÷110
Divide both sides by 110110
tt
==
1212 hour or 3030 minutes
3030 minutes
Question 4 of 6
4. Question
Two cars left Town A for Town B. The first car left at 88am and averaged 100100km/h. The second car left at 99am and averaged 120120km/h. At what time did they meet?
A word problem can be drawn as a diagram and then translated into an equation for easier solving.
First, draw a diagram of the problem to understand it more
Translate the problem into an equation based on the diagram
Note that the two cars would meet when their distance becomes equal
hours that it will take for them to meet: xx
distance of first car: 100+100x100+100x (this car left an hour earlier, because of its speed, it has a 100100km advantage)
distance of second car: 120x120x
distance of first cardistance of first car
==
distance of second cardistance of second car
100+100x100+100x
==
120x120x
Since S=dtS=dt, it means that d=S×td=S×t. Let time t=xt=x so we can have d=S×xd=S×x.
Solve for xx
100+100x100+100x
==
120x120x
100+100x100+100x-100x−100x
==
120x120x-100x−100x
Subtract 100x100x from both sides
100100
==
20x20x
20x20x
==
100100
20x20x÷20÷20
==
100100÷20÷20
Divide both sides by 2020
xx
==
55 hours
55 hours added to 99am is 22pm.
Therefore, the two cars met at 22 pm
22pm
Question 5 of 6
5. Question
A pipe is 4.84.8 metres long. It is cut into three pieces. One piece is twice the length of the shortest piece. The other piece is 6060cm longer than the shortest piece. Calculate the length of each piece in centimetres.
A word problem can be drawn as a diagram and then translated into an equation for easier solving.
First, draw a diagram of the problem to understand it more
shortest piece: xx
twice the shortest: 2x2x
6060cm longer than the shortest: x+60x+60
Translate the problem into an equation based on the diagram, then make sure all units are in centimetres
xx++2x2x+(+(x+60x+60cm))
==
4.84.8m
xx++2x2x+(+(x+60x+60cm))
==
480480cm
Solve for xx
xx++2x2x+(x+60cm)
=
480cm
4x+60
=
480
4x+60-60
=
480-60
Subtract 60 from both sides
4x
=
420
4x÷4
=
420÷4
Divide both sides by 4
x
=
105
Solve for 2x
2x
=
2(105)
Substitute x
=
210
Solve for x+60
x+60
=
105+60
Substitute x
=
165
105,165,210
Question 6 of 6
6. Question
The body of a lamp post is twice as long as its base. Its base is twice as long as its head. Altogether, the lamp post is 560cm tall. How tall is each section?