Sure, let's go through the questions one by one with detailed explanations.
### Question 34
The price of milk has risen by 75%. Let's find the percentage reduction in consumption needed to maintain the same expenditure.
- Original Price: [tex]$P$[/tex]
- New Price: [tex]$P + 0.75P = 1.75P$[/tex]
Let the original consumption be [tex]$C$[/tex] units. The original expenditure is [tex]$P \cdot C$[/tex].
To keep the expenditure the same, at the new price, the new consumption should be [tex]$C'$[/tex] units. Therefore:
[tex]\[ \text{New Expenditure} = \text{Original Expenditure} \][/tex]
[tex]\[ 1.75P \cdot C' = P \cdot C \][/tex]
[tex]\[ C' = \frac{P \cdot C}{1.75P} \][/tex]
[tex]\[ C' = \frac{C}{1.75} \][/tex]
Let's simplify [tex]$\frac{C}{1.75}$[/tex]:
[tex]\[ C' = \frac{C}{\frac{7}{4}} = C \cdot \frac{4}{7} \][/tex]
The percentage reduction in consumption is:
[tex]\[ \text{Reduction} = \left(1 - \frac{C'}{C}\right) \times 100 = \left(1 - \frac{4}{7}\right) \times 100 \][/tex]
[tex]\[ \text{Reduction} = \left(\frac{7-4}{7}\right) \times 100 \][/tex]
[tex]\[ \text{Reduction} = \frac{3}{7} \times 100 \][/tex]
[tex]\[ \text{Reduction} = 42 \frac{6}{7} \% \][/tex]
So the correct answer is:
c) [tex]$42 \frac{6}{7} \%$[/tex]
### Question 35
The diagonal of a cube can be found using the Pythagorean theorem in 3D space.
A cube has sides of length [tex]$l$[/tex]. The body diagonal (longest diagonal passing through the cube and connecting opposite corners) can be calculated as:
[tex]\[ d = \sqrt{l^2 + l^2 + l^2} \][/tex]
[tex]\[ d = \sqrt{3l^2} \][/tex]
[tex]\[ d = \sqrt{3} l \][/tex]
So the correct answer is:
a) [tex]$\sqrt{3} l$[/tex]
### Question 36
We need to determine which option cannot be a multiple of either [tex]$x$[/tex] or [tex]$y$[/tex], given that [tex]$x > y > 1$[/tex] and both [tex]$x$[/tex] and [tex]$y$[/tex] are integers.
Let's break it down:
- For an expression to not be a multiple of either [tex]$x$[/tex] or [tex]$y$[/tex], it must not be divisible by [tex]$x$[/tex] or [tex]$y$[/tex].
For each option:
a) [tex]\( x - 1 \)[/tex]:
- [tex]$x$[/tex] and [tex]$x - 1$[/tex] are consecutive integers, so [tex]$x - 1$[/tex] is definitely not a multiple of [tex]$x$[/tex]. But there's no guarantee that [tex]$y$[/tex] won't divide [tex]$x - 1$[/tex].
b) [tex]\( y + 1 \)[/tex]:
- Similarly, [tex]$y$[/tex] and [tex]$y + 1$[/tex] are consecutive integers, so [tex]$y + 1$[/tex] cannot be a multiple of [tex]$y$[/tex]. But [tex]$x$[/tex] might divide [tex]$y + 1$[/tex].
c) [tex]\( y - 1 \)[/tex]:
- [tex]$y$[/tex] and [tex]$y - 1$[/tex] are consecutive integers, so [tex]$y - 1$[/tex] is not divisible by [tex]$y$[/tex]. But again, we can't ensure [tex]$x$[/tex] does not divide [tex]$y - 1$[/tex].
d) [tex]\( x + y \)[/tex]:
- Depending on the values of [tex]$x$[/tex] and [tex]$y$[/tex], [tex]$x + y$[/tex] could be a multiple of both, either, or neither.
The key point is to find the option which, under all circumstances, wouldn't satisfy being a multiple of either [tex]$x$[/tex] or [tex]$y$[/tex]. Thus, the correct approach is evaluating possible integers:
- [tex]\( x - 1 \)[/tex] is universally not a multiple of [tex]$x$[/tex], and by its definition, [tex]$x - 1$[/tex] won't align synchronously with other multiples of [tex]$y$[/tex] either because [tex]$y$[/tex] is distinct from [tex]$x$[/tex].
Thus, [tex]\( x - 1 \)[/tex] reliably stands out as the number that cannot be a multiple of either [tex]$x$[/tex] or [tex]$y$[/tex] consistently.
So, the correct answer is:
a) [tex]\( x - 1 \)[/tex]
Hence:
- Question 34: c) [tex]$42 \frac{6}{7} \%$[/tex]
- Question 35: a) [tex]$\sqrt{3} l$[/tex]
- Question 36: a) [tex]$x-1$[/tex]
