WORD PROBLEMS ON PARTIAL AND JOINT VARIATION

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$1150n-700n=3600450n=3600$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.