WORD PROBLEMS ON DIRECT AND INDIRECT VARIATION

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In our previous post, we discussed direct variation. We also discussed indirect variation.

we will be looking at some real-life examples today.


Example 1
When exchanging pounds for euros, the number of euros varies directly with the number of pounds. $1300$ euros is exchanged for $5$ pounds. Find the formula connecting euros and pounds. Hence, find the number of euros that is exchanged for $28$ pounds.

Let the number of euros be $e$
Let the number of pounds be $p$
$e\alpha p$
$e=kp$

$1300$ euros for 5$ pounds.
$1300=k(5)$

Divide both side by $5$
$\frac{1300}{5}=\frac{5k}{5}$
$260=k$
$k=260$

The formula connecting euros and pounds is
$e=260p$

Now, let's find the number of euros that is exchanged for $28$ pounds
$e=260(28)$
$e=7280$

Therefore, 7280 euros will be exchanged for $28$ pounds.

Example 2
The distance covered by an object is directly proportional to the square of the time taken. If the object covers 1200m in 20 seconds. Find
1. The law of variation
2. Determine the time it will take for the object to covered 300m.

Let the distance covered by the object or $d$
Let the time taken by the object be $t$

$d\alpha t^2$
$d=kt^2$
$1200=k(20^2)$
$1200=k(400)$

Divide both side by $400$
$\frac{1200}{400}=\frac{400k}{400}$
$3=k$
$k=3$

The law of variation is:
$d=3t^2$

Ley's find the time taken by the object to covered 300m
$300=3t^2$

Divide both side by $3$
$\frac{300}{3}=\frac{3t^2}{3}$
$100=t^2$

To get rid of square, we add square root to both side.
$\sqrt{100}=(\sqrt{t})^2$
$10=t$
$t=10$

So, it will take 10 seconds for the object to covered 300m.

Example 3
The ticket fee for a football match varies inversely with the number of spectators. When there are $80$ spectators, the ticket fee is $€490$. What is the ticket fee when there are $56$ spectators?

Let the ticket fee by $t$
Let the number of spectators be $s$

From the first sentence
$t\alpha\frac{1}{s}$
$t=\frac{k}{s}$

From the wording of the second sentence,
$490=\frac{k}{80}$

Cross multiply
$490\times80=k$
$k=39200$


$t=\frac{39200}{s}$

Now, let's find the ticket fee when there are $56$ spectators
$t=\frac{39200}{56}$
$t=700$

Therefore, the ticket when there are 56 spectators will be €700.

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Example 4
The fixed cost of a book company is €300,000. The variable costs, which are directly proportional to the sales, are €400,000 when the sales are €1,000,000. Calculate the total cost and the profit when the sales are €900,000.

Let the variable cost be $v$
Let sales be $s$.

From the second sentence
$v=\alpha s$
$v=ks$

The variable cost was €400,000 when the sales were €1,000,000.
$400000=k(1000000)$

Divide both side by $1000000$
$\frac{400000}{1000000}=\frac{1000000k}{1000000}$
$\frac{2}{5}=k$
$k=\frac{2}{5}$

$v=\frac{2s}{5}

When sales was €900,000
$v=\frac{2(900000)}{5}$
$v=\frac{1800000}{5}$
$v=360000$

So variable cost when sales are $€900000$ is $€360000$.

Total cost is variable cost plus fixed cost. Hence
The total cost is  $€360000+€300000=€660000$

You know that profit is total revenue less total cost hence, 
Profit will be €900000-€660000=€240,000

There you have it. In this post, we would be concluding our series on variations when we explore word problems on partial and joint variation.

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