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

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. $45$ $45$ minutes
- 3. $1$ $1$ hour
- 4. $30$ $30$ 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. $4$ $4$ pm
2. $11$ $11$ am
3. $12$ $12$ pm
4. $2$ $2$ 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. $110, 170, 220$ $110,170,220$
2. $100, 160, 200$ $100,160,200$
3. $105, 170, 200$ $105,170,200$
4. $105, 165, 210$ $105,165,210$

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

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

$128 + 18 t$ $128+18t$	$=$ $=$	$22 t$ $22t$
$128 + 18 t$ $128+18t$ $- 18 t$ $-18t$	$=$ $=$	$22 t$ $22t$ $- 18 t$ $-18t$	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

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

$100 + 100 x$ $100+100x$	$=$ $=$	$120 x$ $120x$
$100 + 100 x$ $100+100x$ $- 100 x$ $-100x$	$=$ $=$	$120 x$ $120x$ $- 100 x$ $-100x$	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

$x$ $x$ $+$ $+$ $2 x$ $2x$ $+ ($ $+($ $x + 60$ $x+60$ 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 $4$ $4$
$x$ $x$	$=$ $=$	$105$ $105$

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3. Question

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Quizzes

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

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

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

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

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

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