diff --git a/README.md b/README.md
index a2518a6..642960b 100644
--- a/README.md
+++ b/README.md
@@ -49,6 +49,8 @@ TODO: question14.c
 - [Quick Sort](/arrays/question5.c)
 - [Bubble Sort](/arrays/question6.c)
 - [Counting Sort](/arrays/question7.c)
+- [find a pair in an array whose sum is X](/arrays/question8.c)
+- [Moore Voting algorithm to find majority element in an array](/arrays/question9.c)
 
 
 # Topic1: Introduction
diff --git a/arrays/question8.c b/arrays/question8.c
index 49f6525..64aa86e 100644
--- a/arrays/question8.c
+++ b/arrays/question8.c
@@ -69,4 +69,51 @@ int main(){
 	findPairs(a,length, sum);
 
 }
+/*
+METHOD3: using quick sort technique
+but finding if a pair exists or not
+*/
+#include <stdio.h>
+#include <stdlib.h>
 
+int cmpfunc(const void*a, const void*b){
+	return (*(int*)a-*(int*)b);
+}
+
+int findPairs(int arr[], int size, int sum){
+	int left, right;
+
+	qsort(arr,size,sizeof(int),cmpfunc);
+
+	left = 0;
+	right = size - 1;
+	while(left<right){
+		if(arr[left]+arr[right]==sum){
+			return 1;
+		}else if(arr[left]+arr[right]<sum){
+			left++;
+		}else{
+			right--;
+		}
+	}
+}
+
+int main(){
+	int size, index, sum, *arr;
+	printf("enter number of elements in the array\n");
+	scanf("%d",&size);
+	//allocate memory for array
+	arr = (int *)malloc(sizeof(int)*size);
+	printf("enter the elements of the array\n");
+	for(index=0; index<size; index++){
+		scanf("%d",&arr[index]);
+	}
+	printf("enter the sum value\n");
+	scanf("%d",&sum);
+	if(findPairs(arr,size,sum)){
+		printf("Pairs found\n");
+	}else{
+		printf("Pairs not found\n");
+	}
+
+}
diff --git a/arrays/question9.c b/arrays/question9.c
new file mode 100644
index 0000000..502accf
--- /dev/null
+++ b/arrays/question9.c
@@ -0,0 +1,55 @@
+/*
+Given an array A of size n, find an element that occurs more than n/2 times
+
+MOORE VOTING ALGORITHM
+
+In this algo every element votes for itself. If a different element in encountered then the vote collected
+for previous element is used to cancel out the vote for this element.
+When votes is zero and new element is found voter changes
+The number of votes have to be greater than or equal to 1 to satisfy the condition given in the problem
+always
+If it is greater than or equal to one, we loop the array again to find how many times is the voter occuring.
+
+Time complexity:
+O(n)
+
+*/
+#include <stdio.h>
+#include <stdlib.h>
+
+int main(){
+
+	int index,voter, votes=0, counter=0;
+	int a[] = {2,2,5,6,2,2};
+	int length = sizeof(a)/sizeof(a[0]);
+	//checking voter and votes
+	for(index = 0; index<length; index++){
+		if(votes == 0){
+			voter = a[index];
+			votes++;
+		}else{
+			if(voter == a[index]){
+				votes++;
+			}else{
+				votes--;
+			}
+		}
+	}
+	//checking if that element is really repeating more than n/2 times or not
+	if(votes >= 1){
+
+		for(index=0; index<length; index++){
+			if(a[index]==voter){
+				counter++;
+			}
+		}
+
+		if(counter > length/2){
+			printf("the element that occurs more than n/2 times is %d\n", voter);
+		}else{
+			printf("no element found\n");
+		}
+
+	}
+
+}
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