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. $3 \frac{1}{4}$ $3 1/4$ hours
2. $4 \frac{1}{2}$ $4 1/2$ hours
3. $4$ $4$ hours
4. $3$ $3$ 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
- 2. $1$ $1$ hour $30$ $30$ minutes
- 3. $30$ $30$ minutes
- 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. $12$ $12$ pm
2. $2$ $2$ pm
3. $11$ $11$ am
4. $4$ $4$ pm

Question 5 of 6

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

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

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

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 $560$ 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: $x$

Base: $2x$

Body: $4x$

Translate the problem into an equation based on the diagram

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

Solve for $x$

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

$7x$ $=$ $560$

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

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

Solve for $2x$

$2x$

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

$=$ $160$ cm height of the base

Solve for $4x$

$4x$

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

$=$ $320$ cm height of the body

Head: $80$ cm

Base: $160$ cm

Body: $320$ 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
$100$ $100$	$=$ $=$	$20 x$ $20x$
$20x$	$=$	$100$
$20x$ $-:20$	$=$	$100$ $-:20$	Divide both sides by $20$
$x$	$=$	$5$ hours

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

Measurement Word Problems

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Quiz summary

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1. Question

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Fantastic!

2. Question

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Keep Going!

3. Question

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Correct!

4. Question

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Great Work!

5. Question

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Exceptional!

6. 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

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

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

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

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

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