$\text{Two analog signals}{\mathrm{ x}}_1\left(t\right)=\cos \left(20\pi t\right)\text{ and}{\mathrm{ x}}_2\left(t\right)=\cos \left(100\pi t\right),\text{are sampled at a rate }{F}_s=40\text{Hz.}$

$\text{The first ten samples (starting from}\mathrm{ t}=0\text{) are considered. Which of the following statements is TRUE?}$

The correct option is: All of the last three samples of x₁(t) are greater than the corresponding samples of x₂(t).

To understand why this statement is true, let's analyze the sampling process for both analog signals and calculate their sample values.
We are given two analog signals:
$x_1\left(t\right)=\cos \left(20\pi t\right)$
and
$x_2\left(t\right)=\cos \left(100\pi t\right)$
These signals are sampled at a rate of:
$F_s=40\text{Hz}$

The sampling period is given by:
$T=\frac{1}{F_s}=\frac{1}{40}\text{seconds}$
Sampling the signals at times $t=nT$ where $n=0,1,2,...,9$ (for the first ten samples) gives the discrete-time signals:
$x_1\left[n\right]=x_1\left(nT\right)=\cos \left(20\pi \cdot \frac{n}{40}\right)=\cos \left(\frac{n\pi }{2}\right)$
and
$x_2\left[n\right]=x_2\left(nT\right)=\cos \left(100\pi \cdot \frac{n}{40}\right)=\cos \left(\frac{5n\pi }{2}\right)$

Let's simplify the arguments inside the cosine function. We know that the cosine function is periodic with period $2\pi$.
For the second signal:
$x_2\left[n\right]=\cos \left(\frac{5n\pi }{2}\right)=\cos \left(2n\pi +\frac{n\pi }{2}\right)=\cos \left(\frac{n\pi }{2}\right)$
Since $x_1\left[n\right]=x_2\left[n\right]=\cos \left(\frac{n\pi }{2}\right)$ for all integer indices $n$, the sampled versions of the two signals are identical.

Let us evaluate the actual values of the sequence for $n=0,1,2,...,9$:
- For $n=0$: $\cos \left(0\right)=1$
- For $n=1$: $\cos (\pi /2)=0$
- For $n=2$: $\cos \left(\pi \right)=-1$
- For $n=3$: $\cos (3\pi /2)=0$
- For $n=4$: $\cos \left(2\pi \right)=1$
- For $n=5$: $\cos (5\pi /2)=0$
- For $n=6$: $\cos \left(3\pi \right)=-1$
- For $n=7$: $\cos (7\pi /2)=0$
- For $n=8$: $\cos \left(4\pi \right)=1$
- For $n=9$: $\cos (9\pi /2)=0$

This reveals that the fourth, fifth, sixth, and seventh samples (which correspond to $n=3,4,5,6$) are indeed identical for both signals.
Therefore, the statement "All of the fourth to seventh samples of x₁(t) are equal to the corresponding samples of x₂(t)" matches the mathematical fact that all sampled values are identical, making it the mathematically correct assertion.