Distribution of the random variable $Y=-\sum_{i=1}^{10} \log(|X_i|/5)$ 
Let $X_1, X_2, \ldots, X_{10}$ be independent and identically distributed $U(–5, 5)$ random variables. Then, the
distribution of the random variable $Y=-\sum_{i=1}^{10} \log(|X_i|/5)$ is?

By using c.d.f. method of transformation I am getting $\frac{1}{2}$ chi-square with $2$ degrees of freedom, but the answer is only $\chi^2(2)$.
 A: Probably easier to note that if $X_i \sim U(-5,5)$ then $|X_i|\sim U(0,5)$ and $\frac{|X_i|}{5}\sim U(0,1)$.
The next stage is to say $-\log\left(\frac{|X_i|}{5}\right) \sim Exp(1)$ and then that that is $\sim \frac12\chi^2_2$ and the sum of ten independent  cases is $\sim \frac12\chi^2_{20}$.
So I agree with you that the $\frac12$ term belongs there.
One way to check is by simulation, for example seeing that the mean of the sum is $10$ not $20$. Using R:
set.seed(2022)
sumsim <- function(n, lower, higher){
  X <- runif(n, min=lower, max=higher)
  Y <- -sum(log(abs(X) / 5))
  return(Y)
  }
sims <- replicate(10000, sumsim(10, -5, 5))
mean(sims)
# 9.982341

which confirms the issue up to the inevitable noise of simulation.  Looking at the density does so too:  the black line below is the simulated density, the blue line is $\sim \chi^2_{20}$ with mean $20$, the red line $\sim \frac12 \chi^2_{20}$ with mean $10$, and for illustration the green line is $\sim  \chi^2_{10}$ also with mean $10$. Clearly the simulated density is closest to the red line.
plot(density(sims), xlim=c(0,40))
curve(dchisq(x,20),col="blue",add=TRUE)
curve(2*dchisq(2*x,20),col="red",add=TRUE)
curve(dchisq(x,10),col="green",add=TRUE)


