Home / IB MYP Year 4-5: Exntended Mathematics : Unit 3: Function -Weighted networks MYP Style Questions

IB MYP Year 4-5: Exntended Mathematics : Unit 3: Function -Weighted networks MYP Style Questions

IB myp 4-5 MATHEMATICS – Practice Questions- All Topics

Topic :Functions-Weighted Networks

Topic :Function- Weightage : 21 % 

All Questions for Topic : Representation and shape of more complex functions,Transformation of quadratic functions,Rational functions,Graphing trigonometric functions,Linear programming, including inequalities,Networks-edges and arcs, nodes/ vertices, paths,Calculating network pathways,Weighted networks,Domain and range

Question

In this question, you will make calculations for changes that individuals can make to save water on a daily basis.

Question (a) 

Write down the missing percentage for drinking and cooking on the pie chart.

▶️Answer/Explanation

Ans:

Sum of all percentage is equal to $100$

$(100-40-9-12)\%$=remaining

$=39 \%$

Question (b)

It is estimated that the daily water usage per person is 120 litres.

A regular shower has a flow rate of 8 Litres (L) per minute.
Using your answer from part (b), determine the duration of a regular shower.

▶️Answer/Explanation

Ans:

To determine the amount of water used for showering, we need to calculate $40\%$ of the daily water usage per person, which is $\rm{120~ litres}$.

To find $40\%$ of a quantity, we can multiply that quantity by $0.40$ (or $40\%$).

Water used for showering $= 120 \text{ liters} \times 0.40 = 48 \text{ liters}$.

Therefore, the amount of water used for showering is 48 liters.

Now, to determine the duration of a regular shower, we can divide the amount of water used by the flow rate:

\[\text{Duration of shower} = \frac{\text{Amount of water used}}{\text{Flow rate}}\]

Plugging in the values:

\[\text{Duration of shower} = \frac{48 \text{ liters}}{8 \text{ L/min}}\]

Simplifying the expression:

\[\text{Duration of shower} = 6 \text{ minutes}\]

Therefore, the duration of a regular shower is 6 minutes.

Question (c)

In a water-saving condition the flow rate of a shower can be reduced to be $5 \mathrm{~L}$ per minute. Given that the duration of the shower does not change:
Determine the amount of water used for showering in a water-saving condition.

▶️Answer/Explanation

Ans:

To determine the amount of water used for showering in a water-saving condition, we can use the new flow rate of 5 L per minute and the duration of the shower.

Let’s denote the amount of water used for showering in the water-saving condition as $x$ liters.

Using the equation:

\[\text{Amount of water used} = \text{Flow rate} \times \text{Duration}\]

In the original condition, we found that the duration of the shower is 6 minutes. So, in the water-saving condition, the duration of the shower remains the same.

Substituting the values into the equation:

\[x = 5 \text{ L/min} \times 6 \text{ min}\]

Simplifying the expression:

\[x = 30 \text{ liters}\]

Therefore, in a water-saving condition, the amount of water used for showering is 30 liters.

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