WORD PROBLEMS ON PARTIAL AND JOINT VARIATION

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As the diligent mathematician that you are, your mind would have been ingrained in partial and joint variations.

So, today we are going to be trying more real-life applications of these variations.


Example 1
The time taken for throw-ins in a match varies directly as the number of throw-ins and inversely as the number of free-kick taken. In a match where 5 free kicks were taken, it took 10 minutes to have 20 throw-ins. Find
1. The law of variation 
2. Calculate the time it will take to have 50 throw-ins in a match when 2 free-kicks were taken.

Solution:
Let the time taken to have throw-in be $x$
Let the number of throw-ins be $y$.
Let the number of free-kicks be $z$

From the first sentence
$x\alpha\frac{y}{z}$
$x=\frac{ky}{z}$

$20$ throw-ins and $5$ free-kickswere taken in $10$ minutes
$10=\frac{k(20)}{5}$
$10=\frac{20k}{5}$
$10=4k$

Divide both side by $4$
$\frac{10}{4}=\frac{4k}{4}$
$k=\frac{5}{2}$

Therefore the law of variation is 
$x=\frac{5y}{2z}$

Let calculate how long it will take to have 50 throw-ins and 2 free-kicks in the match
$x=\frac{5(50)}{2(2)}$
$x=\frac{250}{4}$
$x=62.5$

So, therefore, it will take 62.5 minutes to have 50 throw-ins in a match where there are 2 free-kicks.

Example 2
The mass of a balloon varies jointly as its specific gravity and the cube of its diameter. The mass of the balloon was 850g when the diameter is $6$cm and the specific gravity is $75$. Find the mass of a balloon of specific gravity $10.5$ and a diameter of $8$cm.

Solution: 
Let the mass of the balloon by $x$
Let the specific gravity be $y$
Let the diameter of the balloon be $z$
$x\alpha yz^3$
$x=kyz^3$
$850=k7.5(6^3)$
$850=k7.5(216)$
$850=k1620$

Divide both side by $1620$
$\frac{850}{1620}=\frac{1620k}{1620}$
$\frac{850}{1620}=k$
$k=\frac{850}{1620}$
$k=\frac{85}{162}$

So $x=\frac{85yz^3$}{162}$

Let calculate the mass of a ballon with a specific gravity of $10.5$ and diameter of $8$cm
$x=\frac{85(10.5)(8)^3$}{162}$
$x=\frac{85(10.5)(512)}{162}$
$x=2820.7$

The mass of the balloon with a specific gravity of $10.5$ and diameter of $82$cm is approximately $2821g$

Example 3
The charges for electricity consumption is partly constant and partly varies directly as the number of units used. The charge for 180 units is €2700 while the charge for 300 units is €3100. Calculate the charge that will be paid for 90 units.

Solution:
Let the charge for electricity consumption be $c$
Let the number of units be $n$

From the first sentence
$c=a+kn$  where $a$ and $k$ are constants

Charges for 180 units is €2700
$2700=a+k(180)$
$2700=a+180k$......eqn 1

Charges for 300 units is €3100
$3100=a+k(300)$
$3100=a+300k$......eqn 2

To derive the value of $a$ and $k$, we solve this equation simultaneously,

Substract eqn 1 from eqn 2
$3100=a+300k$
$-(2700=a+180k$)
$400=120k$

Divide both sides by $120$
$\frac{400}{120}=\frac{120k}{120}$
$k=\frac{400}{120}$
$k=\frac{10}{3}$

To solve for $a$, we will substitute the value of $k$ in eqn 1
$2700=a+\frac{180(10)}{3}$
$2700=a+\frac{1800}{3}$
$2700=a+600$

Collect like terms
$2700-600=a$
$2100=a$
$a=2100$

$c=2100+\frac{10n}{3}$

Let calculate the charges for 90 units
$c=2100+\frac{10(90)}{3}$
$c=2100+\frac{900}{3}$
$c=2100+300$
$c=2400$

So, €2400 will be charged if 90 units were used

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Example 4
The cost of producing a book in a factory is partly constant and partly varies inversely with the number of books produced per day. The cost of producing 4 books per day is €1600 and the cost of producing 5 books per day is €1420. Find the number of books that will be produced per day if the factory decided to bring down the cost to €1150.

Solution:
Let the cost of the book be $b$
Let the number of books per day be $n$

From the first sentence
$b=a+\frac{k}{n}$ where $a$ and $k$ are constants

€1600 is the cost of 4 books
$1600=a+\frac{k}{4}$

Taking the L.C.M
$1600=\frac{4a+k}{4}$

Cross multiply
$1600\times4=4a+k$
$6400=4a+k$.......eqn 1

The cost of 5 books is €1420
$1420=a+\frac{k}{5}$

Taking the L.C.M
$1420=\frac{5a+k}{5}$

Cross multiply
$1420\times5=5a+k$
$7100=5a+k$.......eqn 2

Substract eqn 1 from eqn 2
$7100=5a+k$
$-(6400=4a+k)$
$700=a$
$a=700$

To obtain $k$, Substitute $700$ for $a$ in eqn 1
$6400=4(700)+k$
$6400=2800+k$
$6400-2800=k$
$k=3600$

$b=700+\frac{3600}{n}$

Now, let's determine the number of books the factory can produce if it decides to reduce the cost to $1150$
$1150=700+\frac{3600}{n}$

Taking the L.C.M
$1150=\frac{700n+3600}{n}$

Cross multiply
$1150n=700n+3600
$1150n-700n=3600$
$450n=3600$

To determine the value of $n$, divide both sides by $450$
$\frac{450n}{450}=\frac{3600}{450}$
$n=8$

Therefore, the company will have to produce $8$ cars to bring down the cost to $€1150.

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