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 65km/h and the westbound truck at 58km/h. When will the trucks be 55312km 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: t
distance of westbound truck: 58t
distance of eastbound truck: 65t
Distance West + Distance East
=
55312
58t+65t
=
55312
Since S=dt, it means that d=S×t.
Solve for t
58t+65t
=
55312
123t
=
55312
123t÷123
=
55312÷123
Divide both sides by 123
t
=
412 hours
412 hours
Question 2 of 6
2. Question
A lioness is 128 metres from a gazelle. The lioness starts to sprint towards the gazelle at 22m/s. At the same time, the gazelle starts to sprint at 18m/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: t
distance of the gazelle: 128+18t
distance of the lioness: 22t
speed of the gazelle: 18m/s
speed of the lioness: 22m/s
distance of the gazelle
=
distance of the lioness
128+18t
=
22t
Solve for t
128+18t
=
22t
128+18t−18t
=
22t−18t
Subtract 18t from both sides
128
=
4t
4t
=
128
4t÷4
=
128÷4
Divide both sides by 4
t
=
32 seconds
32 seconds
Question 3 of 6
3. Question
Two submarines 55km apart aimed and fired their torpedoes toward each other. Torpedo A averages 50km/h and Torpedo B averages a speed of 60km/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: t
distance of Torpedo A: 50t
distance of Torpedo B: 60t
Distance A + Distance B: 55km
50t+60t
=
55
Since S=dt, it means that d=S×t.
Solve for t
50t+60t
=
55
110t
=
55
100t÷110
=
55÷110
Divide both sides by 110
t
=
12 hour or 30 minutes
30 minutes
Question 4 of 6
4. Question
Two cars left Town A for Town B. The first car left at 8am and averaged 100km/h. The second car left at 9am and averaged 120km/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: x
distance of first car: 100+100x (this car left an hour earlier, because of its speed, it has a 100km advantage)
distance of second car: 120x
distance of first car
=
distance of second car
100+100x
=
120x
Since S=dt, it means that d=S×t. Let time t=x so we can have d=S×x.
Solve for x
100+100x
=
120x
100+100x−100x
=
120x−100x
Subtract 100x from both sides
100
=
20x
20x
=
100
20x÷20
=
100÷20
Divide both sides by 20
x
=
5 hours
5 hours added to 9am is 2pm.
Therefore, the two cars met at 2 pm
2pm
Question 5 of 6
5. Question
A pipe is 4.8 metres long. It is cut into three pieces. One piece is twice the length of the shortest piece. The other piece is 60cm 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: x
twice the shortest: 2x
60cm longer than the shortest: x+60
Translate the problem into an equation based on the diagram, then make sure all units are in centimetres
x+2x+(x+60cm)
=
4.8m
x+2x+(x+60cm)
=
480cm
Solve for x
x+2x+(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?