I suppose you want to test whether there is significant evidence that either opinion X or Y has stronger support.
Simulated data. One model for your data, in which opinion X is very slightly more popular than opinion Y, might be that you have $N \sim \mathsf{Pois}(\lambda = 90)$
people answering each of the questions, and then that opinion X gets
$x \sim \mathsf{Binom}(N, p=.55)$ votes and opinion Y gets $y = N-x$ votes.
Simulating this model, we have data:
set.seed(2018); N = rpois(10, 90)
x = rbinom(10, N, .55); y = N-x
cbind(x,y, d=x-y)
x y d
[1,] 51 34 17
[2,] 36 45 -9
[3,] 51 41 10
[4,] 53 53 0
[5,] 51 36 15
[6,] 56 38 18
[7,] 59 34 25
[8,] 50 41 9
[9,] 49 49 0
[10,] 47 44 3
Below we show results of various tests to see which ones detect the slight
difference in preference of opinion X over opinion Y assumed in the simulation
of these data. All alternatives are two-sided.
Sign test. The simplest test is a sign test, which looks at the number of negative
signs in the last column (disregarding the 0's). It does not reject (5% level) the null
hypothesis that the two opinions are equally popular.
The p-value is 0.0703. The sign test has relative low power. It "ignores" that
there are several large positive values in the last column and only one negative value.
2*(pbinom(1, 8, .5))
[1] 0.0703125
Wilcoxon Signed Rank test. A (one-sample) Wilcoxon signed-rank test rejects with approximate p-value 0.035.
The warning messages are because of the 0s and because $\pm 9$ is considered a tie.
wilcox.test(x-y)
Wilcoxon signed rank test with continuity correction
data: x - y
V = 33.5, p-value = 0.03547
alternative hypothesis: true location is not equal to 0
Warning messages:
1: In wilcox.test.default(x - y) : cannot compute exact p-value with ties
2: In wilcox.test.default(x - y) :
cannot compute exact p-value with zeroes
One-sample t test. A one-sample t test rejects the null hypothesis. The p-value is 0.025. The data
may not be normal, but the t test is known to be robust.
t.test(x-y)
One Sample t-test
data: x - y
t = 2.695, df = 9, p-value = 0.02459
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
1.413365 16.186635
sample estimates:
mean of x
8.8
Permutation test. A permutation test, using the sum of differences as metric, rejects the
null hypothesis. When signs were randomly permuted, we observed 85 unique
sums of differences. The p value is 0.015; it is the probability that
sums of permuted differences lie outside the two vertical bars (at $\pm 88)$
in the histogram. The p-value is simulated and subject to unimportant
variation with 100,000 iterations. Three subsequent runs gave essentially the same p-value. [The Wilcoxon test is a kind of permutation test, but the
current test needs no adjustments for 0's and ties.]
d.obs = sum(x-y); d.obs
[1] 88
d.perm = replicate( 10^5, sum((x-y)*sample(c(-1,1), 10, rep=T)) )
mean(abs(d.perm) > abs(d.obs))
[1] 0.01529
length(unique(d.perm))
[1] 85

Summary. In summary, only the notoriously low-powered sign test failed to find
a significant difference in popularity between opinions X and Y. The permutation test gave the smallest p-value. On various grounds, one might wonder whether
the assumptions for the Wilcoxon and t tests are met.
Note: You could also do a chi-squared test of homogeneity using the matrix
of x
and y
values. It gives a non-significant result, which indicates
there is no significant difference in the distribution of opinions between X and Y across the ten questions. This test is not appropriate for assessing
whether opinions X and Y are equally popular.
chisq.test(cbind(x,y))
Pearson's Chi-squared test
data: cbind(x, y)
X-squared = 10.894, df = 9, p-value = 0.2831