IB myp 4-5 MATHEMATICS – Practice Questions- All Topics
Topic :Function-System of Equations/Simultaneous Equation
Topic :Function- Weightage : 21 %
All Questions for Topic : Mappings,Function notation,Linear functions,y=mx+c,Parallel and Perpendicular lines,System of equations/simultaneous equations,Quadratic functions,Algorithms
Question (6 Marks)
Find the value of $X$ for the following expression.
▶️Answer/Explanation
Ans:
To find the values of $x$ and $y$ in the given system of equations:
Equation 1: $x + y = 22$
Equation 2: $x – y = 12$
We can solve this system of equations using the method of elimination or substitution.
Using the method of elimination, we can add both equations to eliminate the variable $y$:
$(x + y) + (x – y) = 22 + 12$
$2x = 34$
Dividing both sides of the equation by 2, we get:
$x = 17$
Substituting the value of $x$ into either equation, we can find the value of $y$:
$x + y = 22$
$17 + y = 22$
$y = 22 – 17$
$y = 5$
Therefore, the solution to the system of equations is $x = 17$ and $y = 5$.
Now , $\rm{X}=2x+4y\Rightarrow 2\times (17)=4\times (5) =54$
Question (5 Marks)
The basic scales as shown in the image use specific items to weigh fruit and vegetables at a market. The handler uses batteries and metal weights.
As shown in the diagram, the combined weight of two metal weights and three batteries is 305 grams $(\mathrm{g})$.
One metal weight measures $100 \mathrm{~g}$.
Question (a) : 3 marks
Using the information from the diagram, find the weight of one battery.
▶️Answer/Explanation
Ans:
Let’s assume the weight of one battery is \(x\) grams.
According to the information given, the combined weight of two metal weights and three batteries is 305 grams.
We can express this information in the form of an equation:
\[2 \times 100 \, \text{g} + 3 \times x \, \text{g} = 305 \, \text{g}\]
Now, let’s solve for the weight of one battery (\(x\)) by isolating it on one side of the equation:
\[3 \times x \, \text{g} = 305 \, \text{g} – 2 \times 100 \, \text{g}\]
Simplifying:
\[3 \times x \, \text{g} = 305 \, \text{g} – 200 \, \text{g}\]
\[3 \times x \, \text{g} = 105 \, \text{g}\]
Dividing both sides of the equation by 3:
\[x \, \text{g} = \frac{105 \, \text{g}}{3}\]
\[x \, \text{g} = 35 \, \text{g}\]
Therefore, the weight of one battery is 35 grams.