Measurement Word Problems

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Question 1 of 6

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

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

$123 t$ $123t$ $=$ $553 1/2$

$123t$ $-:123$ $=$ $553 1/2$ $-:123$ Divide both sides by $123$

$t$ $=$ $4 1/2$ hours

$4 1/2$ hours

Question 2 of 6

A lioness is

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

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

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

(32) seconds

Question 3 of 6

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

distance of Torpedo A: $50t$

distance of Torpedo B: $60t$

Distance A + Distance B: $55km$

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

Since $S=d/t$ , it means that $d=Sxxt$ .

Solve for $t$

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

$110t$ $=$ $55$

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

$t$ $=$ $1/2$ hour or $30$ minutes

$30$ minutes

Question 4 of 6

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

$8$ am and averaged

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

$9$ am and averaged

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

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

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

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

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

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

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

$x$ $+$ $2x$ $+($ $x+60$ 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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Distance West $+$ $+$ Distance East	$=$ $=$	$553 \frac{1}{2}$ $553 1/2$
$58 t + 65 t$ $58t+65t$	$=$ $=$	$553 \frac{1}{2}$ $553 1/2$

$50t+60t$	$=$	$55$
$110t$	$=$	$55$
$100t$ $-:110$	$=$	$55$ $-:110$	Divide both sides by $110$
$t$	$=$	$1/2$ hour or $30$ 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