Measurement Word Problems

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Question 1 of 6

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

65

$65$ km/h and the westbound truck at

58

$58$ km/h. When will the trucks be

553 \frac{1}{2}

$553 1/2$ km apart?

1. $4$ $4$ hours
2. $3 \frac{1}{4}$ $3 1/4$ hours
3. $3$ $3$ hours
4. $4 \frac{1}{2}$ $4 1/2$ hours

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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

$t$

distance of westbound truck:

58 t

$58t$

distance of eastbound truck:

65 t

$65t$

Distance West $+$ $+$ Distance East	$=$ $=$	$553 \frac{1}{2}$ $553 1/2$
$58 t + 65 t$ $58t+65t$	$=$ $=$	$553 \frac{1}{2}$ $553 1/2$

Since

S = \frac{d}{t}

$S=d/t$ , it means that

d = S \times t

$d=Sxxt$ .

Solve for

t

$t$

$58 t + 65 t$ $58t+65t$	$=$ $=$	$553 \frac{1}{2}$ $553 1/2$

$123 t$ $123t$	$=$ $=$	$553 \frac{1}{2}$ $553 1/2$

$123 t$ $123t$ $\div 123$ $-:123$	$=$ $=$	$553 \frac{1}{2}$ $553 1/2$ $\div 123$ $-:123$	Divide both sides by $123$ $123$

$t$ $t$	$=$ $=$	$4 \frac{1}{2}$ $4 1/2$ hours

4 \frac{1}{2}

$4 1/2$ hours

Question 2 of 6

A lioness is

128

$128$ metres from a gazelle. The lioness starts to sprint towards the gazelle at

22

$22$ m/s. At the same time, the gazelle starts to sprint at

18

$18$ m/s. When will the lioness catch the gazelle?

(32) seconds

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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

$t$

distance of the gazelle:

128 + 18 t

$128+18t$

distance of the lioness:

22 t

$22t$

speed of the gazelle:

18 m/s

$18\text(m/s)$

speed of the lioness:

22 m/s

$22\text(m/s)$

$distance of the gazelle$ $\text(distance of the gazelle)$	$=$ $=$	$distance of the lioness$ $\text(distance of the lioness)$
$128 + 18 t$ $128+18t$	$=$ $=$	$22 t$ $22t$

Solve for

t

$t$

$128 + 18 t$ $128+18t$	$=$ $=$	$22 t$ $22t$
$128 + 18 t$ $128+18t$ $- 18 t$ $-18t$	$=$ $=$	$22 t$ $22t$ $- 18 t$ $-18t$	Subtract $18 t$ $18t$ from both sides
$128$ $128$	$=$ $=$	$4 t$ $4t$
$4 t$ $4t$	$=$ $=$	$128$ $128$
$4 t$ $4t$ $\div 4$ $-:4$	$=$ $=$	$128$ $128$ $\div 4$ $-:4$	Divide both sides by $4$ $4$
$t$ $t$	$=$ $=$	$32$ $32$ seconds

32

$32$ seconds

Question 3 of 6

3. Question
Two submarines $55$ $55$ km apart aimed and fired their torpedoes toward each other. Torpedo A averages $50$ $50$ km/h and Torpedo B averages a speed of $60$ $60$ km/h. When and where do they collide and impact each other?
- 1. $1$ $1$ hour $30$ $30$ minutes
- 2. $30$ $30$ minutes
- 3. $1$ $1$ hour
- 4. $45$ $45$ minutes
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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: $t$ $t$

distance of Torpedo A: $50 t$ $50t$

distance of Torpedo B: $60 t$ $60t$

Distance A + Distance B: $55 k m$ $55km$

$50 t + 60 t$ $50t+60t$ $=$ $=$ $55$ $55$

Since $S = \frac{d}{t}$ $S=d/t$ , it means that $d = S \times t$ $d=Sxxt$ .

Solve for $t$ $t$

$50 t + 60 t$ $50t+60t$ $=$ $=$ $55$ $55$

$110 t$ $110t$ $=$ $=$ $55$ $55$

$100 t$ $100t$ $\div 110$ $-:110$ $=$ $=$ $55$ $55$ $\div 110$ $-:110$ Divide both sides by $110$ $110$

$t$ $t$ $=$ $=$ $\frac{1}{2}$ $1/2$ hour or $30$ $30$ minutes

$30$ $30$ minutes

Question 4 of 6

Two cars left Town A for Town B. The first car left at

8

$8$ am and averaged

100

$100$ km/h. The second car left at

9

$9$ am and averaged

120

$120$ km/h. At what time did they meet?

1. $2$ $2$ pm
2. $4$ $4$ pm
3. $12$ $12$ pm
4. $11$ $11$ am

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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

$x$

distance of first car:

100 + 100 x

$100+100x$ (this car left an hour earlier, because of its speed, it has a

100

$100$ km advantage)

distance of second car:

120 x

$120x$

$distance of first car$ $\text(distance of first car)$	$=$ $=$	$distance of second car$ $\text(distance of second car)$
$100 + 100 x$ $100+100x$	$=$ $=$	$120 x$ $120x$

Since

S = \frac{d}{t}

$S=d/t$ , it means that

d = S \times t

$d=Sxxt$ . Let time

t = x

$t=x$ so we can have

d = S \times x

$d=Sxxx$ .

Solve for

x

$x$

$100 + 100 x$ $100+100x$	$=$ $=$	$120 x$ $120x$
$100 + 100 x$ $100+100x$ $- 100 x$ $-100x$	$=$ $=$	$120 x$ $120x$ $- 100 x$ $-100x$	Subtract $100 x$ $100x$ from both sides
$100$ $100$	$=$ $=$	$20 x$ $20x$
$20 x$ $20x$	$=$ $=$	$100$ $100$
$20 x$ $20x$ $\div 20$ $-:20$	$=$ $=$	$100$ $100$ $\div 20$ $-:20$	Divide both sides by $20$ $20$
$x$ $x$	$=$ $=$	$5$ $5$ hours

5

$5$ hours added to

9

$9$ am is

2

$2$ pm.

Therefore, the two cars met at

2

$2$ pm

2

$2$ pm

Question 5 of 6

A pipe is

4.8

$4.8$ metres long. It is cut into three pieces. One piece is twice the length of the shortest piece. The other piece is

60

$60$ cm longer than the shortest piece. Calculate the length of each piece in centimetres.

1. $105, 170, 200$ $105,170,200$
2. $100, 160, 200$ $100,160,200$
3. $110, 170, 220$ $110,170,220$
4. $105, 165, 210$ $105,165,210$

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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: $x$ $x$

twice the shortest: $2 x$ $2x$

60

$60$ cm longer than the shortest: $x + 60$ $x+60$

Translate the problem into an equation based on the diagram, then make sure all units are in centimetres

$x$ $x$ $+$ $+$ $2 x$ $2x$ $+ ($ $+($ $x + 60$ $x+60$ cm $)$ $)$	$=$ $=$	$4.8$ $4.8$ m
$x$ $x$ $+$ $+$ $2 x$ $2x$ $+ ($ $+($ $x + 60$ $x+60$ cm $)$ $)$	$=$ $=$	$480$ $480$ cm

Solve for

x

$x$

$x$ $x$ $+$ $+$ $2 x$ $2x$ $+ ($ $+($ $x + 60$ $x+60$ cm $)$ $)$	$=$ $=$	$480$ $480$ cm
$4 x + 60$ $4x+60$	$=$ $=$	$480$ $480$
$4 x + 60$ $4x+60$ $- 60$ $-60$	$=$ $=$	$480$ $480$ $- 60$ $-60$	Subtract $60$ $60$ from both sides
$4 x$ $4x$	$=$ $=$	$420$ $420$
$4 x$ $4x$ $\div 4$ $-:4$	$=$ $=$	$420$ $420$ $\div 4$ $-:4$	Divide both sides by $4$ $4$
$x$ $x$	$=$ $=$	$105$ $105$

Solve for

2 x

$2x$

	$2 x$ $2x$
$=$ $=$	$2 (105)$ $2(105)$	Substitute $x$ $x$
$=$ $=$	$210$ $210$

Solve for

x + 60

$x+60$

	$x + 60$ $x+60$
$=$ $=$	$105 + 60$ $105+60$	Substitute $x$ $x$
$=$ $=$	$165$ $165$

105, 165, 210

$105,165,210$

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 $560$ $560$ cm tall. How tall is each section?
- Head: (80)cm
  
  Base: (160)cm
  
  Body: (320)cm
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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

Head: $x$ $x$

Base: $2 x$ $2x$

Body: $4 x$ $4x$

Translate the problem into an equation based on the diagram

$x + 2 x + 4 x$ $x+2x+4x$ $=$ $=$ $560$ $560$ cm

Solve for $x$ $x$

$x + 2 x + 4 x$ $x+2x+4x$ $=$ $=$ $560$ $560$ cm

$7 x$ $7x$ $=$ $=$ $560$ $560$

$7 x$ $7x$ $\div 7$ $-:7$ $=$ $=$ $560$ $560$ $\div 7$ $-:7$ Divide both sides by $7$ $7$

$x$ $x$ $=$ $=$ $80$ $80$ cm height of the head

Solve for $2 x$ $2x$

$2 x$ $2x$

$=$ $=$ $2 (80)$ $2(80)$ Substitute $x$ $x$

$=$ $=$ $160$ $160$ cm height of the base

Solve for $4 x$ $4x$

$4 x$ $4x$

$=$ $=$ $4 (80)$ $4(80)$ Substitute $x$ $x$

$=$ $=$ $320$ $320$ cm height of the body

Head: $80$ $80$ cm

Base: $160$ $160$ cm

Body: $320$ $320$ cm

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Quiz summary

Information

1. Question

Hint

Fantastic!

2. Question

Hint

Keep Going!

3. Question

Hint

Correct!

4. Question

Hint

Great Work!

5. Question

Hint

Exceptional!

6. Question

Hint

Fantastic!

Quizzes

$50 t + 60 t$ $50t+60t$	$=$ $=$	$55$ $55$
$110 t$ $110t$	$=$ $=$	$55$ $55$
$100 t$ $100t$ $\div 110$ $-:110$	$=$ $=$	$55$ $55$ $\div 110$ $-:110$	Divide both sides by $110$ $110$
$t$ $t$	$=$ $=$	$\frac{1}{2}$ $1/2$ hour or $30$ $30$ minutes

$x + 2 x + 4 x$ $x+2x+4x$	$=$ $=$	$560$ $560$ cm
$7 x$ $7x$	$=$ $=$	$560$ $560$
$7 x$ $7x$ $\div 7$ $-:7$	$=$ $=$	$560$ $560$ $\div 7$ $-:7$	Divide both sides by $7$ $7$
$x$ $x$	$=$ $=$	$80$ $80$ cm	height of the head

	$2 x$ $2x$
$=$ $=$	$2 (80)$ $2(80)$	Substitute $x$ $x$
$=$ $=$	$160$ $160$ cm	height of the base

	$4 x$ $4x$
$=$ $=$	$4 (80)$ $4(80)$	Substitute $x$ $x$
$=$ $=$	$320$ $320$ cm	height of the body