diff --git a/Challenge/yuzunsang/everyArray/README.md b/Challenge/yuzunsang/everyArray/README.md
new file mode 100644
index 00000000..442558f6
--- /dev/null
+++ b/Challenge/yuzunsang/everyArray/README.md
@@ -0,0 +1,9 @@
+# 문제제목
+
+## 설명
+
+every를 이용해서 모든 원소가 짝수인지 아닌지를 판별하세요
+
+## Expected Output 
+
+true
\ No newline at end of file
diff --git a/Challenge/yuzunsang/everyArray/solve.js b/Challenge/yuzunsang/everyArray/solve.js
new file mode 100644
index 00000000..b08e33b5
--- /dev/null
+++ b/Challenge/yuzunsang/everyArray/solve.js
@@ -0,0 +1,7 @@
+function solution(inputArray) {
+    const result = inputArray.every((el) => el % 2 === 0);
+
+    return result;
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/everyArray/solve.test.js b/Challenge/yuzunsang/everyArray/solve.test.js
new file mode 100644
index 00000000..7dac26a4
--- /dev/null
+++ b/Challenge/yuzunsang/everyArray/solve.test.js
@@ -0,0 +1,21 @@
+const { solution } = require("./solve");
+
+const test1 = {
+  input: [2, 4, 6, 8, 10],
+  answer: true,
+};
+
+const test2 = {
+  input: [2, 3, 6, 8, 10],
+  answer: false,
+};
+
+describe("everyArray", () => {
+  test("모두 짝수면 true여야 한다.", () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+
+  test("홀수가 있으면 false여야 한다.", () => {
+    expect(solution(test2.input)).toEqual(test2.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/expDivOdd/README.md b/Challenge/yuzunsang/expDivOdd/README.md
new file mode 100644
index 00000000..c07ab10b
--- /dev/null
+++ b/Challenge/yuzunsang/expDivOdd/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+제곱한 후 3으로 나눈 나머지가 홀수인 것 을 뽑은 배열의 총 합을 구하세요.
diff --git a/Challenge/yuzunsang/expDivOdd/solve.js b/Challenge/yuzunsang/expDivOdd/solve.js
new file mode 100644
index 00000000..0cba14ac
--- /dev/null
+++ b/Challenge/yuzunsang/expDivOdd/solve.js
@@ -0,0 +1,13 @@
+function solution (inputArray) {
+    let ansArray = [];
+
+    for (let i = 0; i < inputArray.length; i++) {
+        inputArray[i] **= 2;
+
+        if ((inputArray[i] % 3) % 2 === 1) ansArray.push(inputArray[i]);
+    }
+
+    return ansArray.reduce((a, b) => a + b);
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/expDivOdd/solve.test.js b/Challenge/yuzunsang/expDivOdd/solve.test.js
new file mode 100644
index 00000000..ca37fcc3
--- /dev/null
+++ b/Challenge/yuzunsang/expDivOdd/solve.test.js
@@ -0,0 +1,21 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [1, 7, 3, 4, 6],
+  answer: 66,
+};
+
+const test2 = {
+  input: [2, 3, 6, 8, 10],
+  answer: 168,
+};
+
+describe('everyArray', () => {
+  test('1, 7, 3, 4, 6', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+
+  test('2, 3, 6, 8, 10', () => {
+    expect(solution(test2.input)).toEqual(test2.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/filterAge/README.md b/Challenge/yuzunsang/filterAge/README.md
new file mode 100644
index 00000000..873a15c1
--- /dev/null
+++ b/Challenge/yuzunsang/filterAge/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+배열 원소의 age가 30이상 50미만인 사람만 있는 객체의 배열을 만드세요
diff --git a/Challenge/yuzunsang/filterAge/solve.js b/Challenge/yuzunsang/filterAge/solve.js
new file mode 100644
index 00000000..4d68fd9a
--- /dev/null
+++ b/Challenge/yuzunsang/filterAge/solve.js
@@ -0,0 +1,7 @@
+// write your codes
+function solution(inputArray) {
+    const result = inputArray.filter((item) => item.age >= 30 && item.age < 50);
+    return result;
+}
+  
+exports.solution = solution;
\ No newline at end of file
diff --git a/Challenge/yuzunsang/filterAge/solve.test.js b/Challenge/yuzunsang/filterAge/solve.test.js
new file mode 100644
index 00000000..b9081601
--- /dev/null
+++ b/Challenge/yuzunsang/filterAge/solve.test.js
@@ -0,0 +1,32 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [
+    {
+      name: '영미',
+      age: 25,
+    },
+    {
+      name: '일미',
+      age: 35,
+    },
+    {
+      name: '이미',
+      age: 45,
+    },
+    {
+      name: '삼미',
+      age: 55,
+    },
+  ],
+  answer: [
+    { name: '일미', age: 35 },
+    { name: '이미', age: 45 },
+  ],
+};
+
+describe('filterAge', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/filterIntersection/README.md b/Challenge/yuzunsang/filterIntersection/README.md
new file mode 100644
index 00000000..d7c9ad39
--- /dev/null
+++ b/Challenge/yuzunsang/filterIntersection/README.md
@@ -0,0 +1,5 @@
+# 문제제목
+
+## 설명
+
+두 배열의 교집합을 출력하세요!
diff --git a/Challenge/yuzunsang/filterIntersection/solve.js b/Challenge/yuzunsang/filterIntersection/solve.js
new file mode 100644
index 00000000..20bdb691
--- /dev/null
+++ b/Challenge/yuzunsang/filterIntersection/solve.js
@@ -0,0 +1,7 @@
+function solution (inputArray1, inputArray2) {
+    // filter는 배열에서 조건에 맞는 원소만을 남겨둘 때 사용
+    // includes는 어떠한 원소가 배열에 포함되어 있는 지를 boolean으로 반환
+    return inputArray1.filter((el) => inputArray2.includes(el));
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/filterIntersection/solve.test.js b/Challenge/yuzunsang/filterIntersection/solve.test.js
new file mode 100644
index 00000000..576cbc86
--- /dev/null
+++ b/Challenge/yuzunsang/filterIntersection/solve.test.js
@@ -0,0 +1,15 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: {
+    A: [1, 2, 3, 4, 5],
+    B: [3, 4, 5, 6, 7],
+  },
+  answer: [3, 4, 5],
+};
+
+describe('filterIntersection', () => {
+  test('test1', () => {
+    expect(solution(test1.input.A, test1.input.B)).toEqual(test1.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/filterOdd/README.md b/Challenge/yuzunsang/filterOdd/README.md
new file mode 100644
index 00000000..cf9eda79
--- /dev/null
+++ b/Challenge/yuzunsang/filterOdd/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+홀수만 뽑아 배열로 만드세요
diff --git a/Challenge/yuzunsang/filterOdd/solve.js b/Challenge/yuzunsang/filterOdd/solve.js
new file mode 100644
index 00000000..602d3415
--- /dev/null
+++ b/Challenge/yuzunsang/filterOdd/solve.js
@@ -0,0 +1,8 @@
+// write your codes
+
+function solution(inputArray) {
+    // filter는 배열에서 조건에 맞는 원소만 남겨두는 것
+    return inputArray.filter((v) => v % 2 === 1);
+}
+  
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/filterOdd/solve.test.js b/Challenge/yuzunsang/filterOdd/solve.test.js
new file mode 100644
index 00000000..8cfd7f5a
--- /dev/null
+++ b/Challenge/yuzunsang/filterOdd/solve.test.js
@@ -0,0 +1,28 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [4, 2, 5, 1, 3],
+  answer: [5, 1, 3],
+};
+
+const test2 = {
+  input: [4, 2, 6, 8, 50, 16],
+  answer: [],
+};
+
+const test3 = {
+  input: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
+  answer: [1, 3, 5, 7, 9, 11],
+};
+
+describe('filterOdd', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+  test('test2', () => {
+    expect(solution(test2.input)).toEqual(test2.answer);
+  });
+  test('test3', () => {
+    expect(solution(test3.input)).toEqual(test3.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/findWord/README.md b/Challenge/yuzunsang/findWord/README.md
new file mode 100644
index 00000000..8ef123f4
--- /dev/null
+++ b/Challenge/yuzunsang/findWord/README.md
@@ -0,0 +1,9 @@
+# 문제제목
+
+## 설명
+
+용가리라는 단어가 있으면 true 없으면 false를 출력
+
+## Expected Output 
+
+true
diff --git a/Challenge/yuzunsang/findWord/solve.js b/Challenge/yuzunsang/findWord/solve.js
new file mode 100644
index 00000000..238bcc8a
--- /dev/null
+++ b/Challenge/yuzunsang/findWord/solve.js
@@ -0,0 +1,6 @@
+// write your codes
+function solution(inputArray) {
+    return inputArray.includes('용가리');
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/findWord/solve.test.js b/Challenge/yuzunsang/findWord/solve.test.js
new file mode 100644
index 00000000..804114f6
--- /dev/null
+++ b/Challenge/yuzunsang/findWord/solve.test.js
@@ -0,0 +1,26 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: ['잠', '자', '고', '싶', '다', '용가리'],
+  answer: true,
+};
+const test2 = {
+  input: ['맛있는', '용가리치킨'],
+  answer: false,
+};
+const test3 = {
+  input: ['고질라', '용가리 ', '울트라맨'],
+  answer: false,
+};
+
+describe('findWord', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+  test('test2', () => {
+    expect(solution(test2.input)).toEqual(test2.answer);
+  });
+  test('test3', () => {
+    expect(solution(test3.input)).toEqual(test3.answer);
+  });
+});
\ No newline at end of file
diff --git a/Challenge/yuzunsang/forEachFilter/README.md b/Challenge/yuzunsang/forEachFilter/README.md
new file mode 100644
index 00000000..6dcbc7c2
--- /dev/null
+++ b/Challenge/yuzunsang/forEachFilter/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+배열 원소중 40 이상인 수만 뽑아 배열을 만드세요.
diff --git a/Challenge/yuzunsang/forEachFilter/solve.js b/Challenge/yuzunsang/forEachFilter/solve.js
new file mode 100644
index 00000000..efb28746
--- /dev/null
+++ b/Challenge/yuzunsang/forEachFilter/solve.js
@@ -0,0 +1,6 @@
+// write your codes
+function solution(inputArray) {
+    return inputArray.filter((el) => el >= 40);
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/forEachFilter/solve.test.js b/Challenge/yuzunsang/forEachFilter/solve.test.js
new file mode 100644
index 00000000..3d73b25c
--- /dev/null
+++ b/Challenge/yuzunsang/forEachFilter/solve.test.js
@@ -0,0 +1,12 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [100, 10, 20, 40],
+  answer: [100, 40],
+};
+
+describe('forEachFilter', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/forEachFilterIsNaN/README.md b/Challenge/yuzunsang/forEachFilterIsNaN/README.md
new file mode 100644
index 00000000..570c030e
--- /dev/null
+++ b/Challenge/yuzunsang/forEachFilterIsNaN/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+배열 원소중 숫자인 원소만 뽑아 배열을 만드세요.
diff --git a/Challenge/yuzunsang/forEachFilterIsNaN/solve.js b/Challenge/yuzunsang/forEachFilterIsNaN/solve.js
new file mode 100644
index 00000000..0b204fe9
--- /dev/null
+++ b/Challenge/yuzunsang/forEachFilterIsNaN/solve.js
@@ -0,0 +1,6 @@
+// write your codes
+function solution(inputArray) {
+    return inputArray.filter((el) => typeof el === 'number');
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/forEachFilterIsNaN/solve.test.js b/Challenge/yuzunsang/forEachFilterIsNaN/solve.test.js
new file mode 100644
index 00000000..d6e118bb
--- /dev/null
+++ b/Challenge/yuzunsang/forEachFilterIsNaN/solve.test.js
@@ -0,0 +1,28 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [1, 40, '라매', '개발자', 51.5, 'a', 88],
+  answer: [1, 40, 51.5, 88],
+};
+
+const test2 = {
+  input: [1, 2, 3, '4', 5, '6'],
+  answer: [1, 2, 3, 5],
+};
+
+const test3 = {
+  input: [-3, -2, -1, 0, 1, 2, 3],
+  answer: [-3, -2, -1, 0, 1, 2, 3],
+};
+
+describe('forEachFilterIsNaN', () => {
+  test('test1: 숫자,문자 판별', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+  test('test2: 문자로된 숫자 판별', () => {
+    expect(solution(test2.input)).toEqual(test2.answer);
+  });
+  test('test3: 음수 양수 포함', () => {
+    expect(solution(test3.input)).toEqual(test3.answer);
+  });
+});
\ No newline at end of file
diff --git a/Challenge/yuzunsang/forEachMap/README.md b/Challenge/yuzunsang/forEachMap/README.md
new file mode 100644
index 00000000..1e8e03c8
--- /dev/null
+++ b/Challenge/yuzunsang/forEachMap/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+forEach 메소드를 사용해서 배열의 각 원소 끝에 '%'를 붙인 문자열 배열을 출력하세요
diff --git a/Challenge/yuzunsang/forEachMap/solve.js b/Challenge/yuzunsang/forEachMap/solve.js
new file mode 100644
index 00000000..4d934950
--- /dev/null
+++ b/Challenge/yuzunsang/forEachMap/solve.js
@@ -0,0 +1,10 @@
+// write your codes
+function solution(inputArray) {
+    const answer = [];
+    
+    inputArray.forEach((item) => answer.push(`${item}%`));
+
+    return answer;
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/forEachMap/solve.test.js b/Challenge/yuzunsang/forEachMap/solve.test.js
new file mode 100644
index 00000000..de4868c6
--- /dev/null
+++ b/Challenge/yuzunsang/forEachMap/solve.test.js
@@ -0,0 +1,12 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [100, 10, 20, 40],
+  answer: ['100%', '10%', '20%', '40%'],
+};
+
+describe('forEachMap', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/forEachReduce/README.md b/Challenge/yuzunsang/forEachReduce/README.md
new file mode 100644
index 00000000..a984df2d
--- /dev/null
+++ b/Challenge/yuzunsang/forEachReduce/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+forEach 메소드를 사용해서 배열의 총 합을 출력하는 코드를 작성하세요
diff --git a/Challenge/yuzunsang/forEachReduce/solve.js b/Challenge/yuzunsang/forEachReduce/solve.js
new file mode 100644
index 00000000..48e350fe
--- /dev/null
+++ b/Challenge/yuzunsang/forEachReduce/solve.js
@@ -0,0 +1,10 @@
+// write your codes
+function solution(inputArray) {
+    let sum = 0;
+
+    inputArray.forEach((item) => sum += item);
+
+    return sum;
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/forEachReduce/solve.test.js b/Challenge/yuzunsang/forEachReduce/solve.test.js
new file mode 100644
index 00000000..397e4b50
--- /dev/null
+++ b/Challenge/yuzunsang/forEachReduce/solve.test.js
@@ -0,0 +1,28 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [100, 10, 20, 40],
+  answer: 170,
+};
+
+const test2 = {
+  input: [120, -20, -30, 0, 15],
+  answer: 85,
+};
+
+const test3 = {
+  input: [-10, -20, -30],
+  answer: -60,
+};
+
+describe('forEachReduce', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+  test('test2: 음수가 포함된 계산', () => {
+    expect(solution(test2.input)).toEqual(test2.answer);
+  });
+  test('test3: 음수만 존재하는 계산', () => {
+    expect(solution(test3.input)).toEqual(test3.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/mapAddPercent/README.md b/Challenge/yuzunsang/mapAddPercent/README.md
new file mode 100644
index 00000000..75e73042
--- /dev/null
+++ b/Challenge/yuzunsang/mapAddPercent/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+map 메소드를 사용해 배열 각각 숫자 뒤에 %를 붙인 문자열을 만드세요
diff --git a/Challenge/yuzunsang/mapAddPercent/solve.js b/Challenge/yuzunsang/mapAddPercent/solve.js
new file mode 100644
index 00000000..f9832889
--- /dev/null
+++ b/Challenge/yuzunsang/mapAddPercent/solve.js
@@ -0,0 +1,6 @@
+// write your codes
+function solution(inputArray) {
+    return inputArray.map((item) => `${item}%`);
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/mapAddPercent/solve.test.js b/Challenge/yuzunsang/mapAddPercent/solve.test.js
new file mode 100644
index 00000000..dac1b3cc
--- /dev/null
+++ b/Challenge/yuzunsang/mapAddPercent/solve.test.js
@@ -0,0 +1,12 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [100, 10, 20, 40],
+  answer: ['100%', '10%', '20%', '40%'],
+};
+
+describe('mapAddPercent', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/mapAppendOrder/README.md b/Challenge/yuzunsang/mapAppendOrder/README.md
new file mode 100644
index 00000000..1435623a
--- /dev/null
+++ b/Challenge/yuzunsang/mapAppendOrder/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+배열의 값을 name 프로퍼티에 넣고 몇번째 원소인지를 order에 넣은 객체의 배열을 출력하세요
diff --git a/Challenge/yuzunsang/mapAppendOrder/solve.js b/Challenge/yuzunsang/mapAppendOrder/solve.js
new file mode 100644
index 00000000..e7f8b51c
--- /dev/null
+++ b/Challenge/yuzunsang/mapAppendOrder/solve.js
@@ -0,0 +1,10 @@
+// write your codes
+function solution(inputArray) {
+    let answer = [];
+
+    inputArray.map((el, idx) => answer.push({ name : el, order : idx + 1 }));
+
+    return answer;
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/mapAppendOrder/solve.test.js b/Challenge/yuzunsang/mapAppendOrder/solve.test.js
new file mode 100644
index 00000000..e08db10c
--- /dev/null
+++ b/Challenge/yuzunsang/mapAppendOrder/solve.test.js
@@ -0,0 +1,16 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: ['홍길동', '둘리', '루피'],
+  answer: [
+    { name: '홍길동', order: 1 },
+    { name: '둘리', order: 2 },
+    { name: '루피', order: 3 },
+  ],
+};
+
+describe('mapAppendOrder', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/reduceMaxValueNIndex/README.md b/Challenge/yuzunsang/reduceMaxValueNIndex/README.md
new file mode 100644
index 00000000..965bdec6
--- /dev/null
+++ b/Challenge/yuzunsang/reduceMaxValueNIndex/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+reduce 메소드를 사용해 최댓값의 값을 maxValue에, 해당 값의 index를 idx에 넣은 객체를 출력하세요
diff --git a/Challenge/yuzunsang/reduceMaxValueNIndex/practice.js b/Challenge/yuzunsang/reduceMaxValueNIndex/practice.js
new file mode 100644
index 00000000..e817c378
--- /dev/null
+++ b/Challenge/yuzunsang/reduceMaxValueNIndex/practice.js
@@ -0,0 +1,95 @@
+// 1. 숫자 배열의 합 구하기
+// const numbers = [1, 2, 3, 4, 5];
+
+// const sum = numbers.reduce((acc, cur) => {
+//     return acc + cur;
+// }, 0);
+
+// console.log(sum);
+
+// 2. 객체 배열에서 나이의 총합 구하기
+// const users = [
+//     { name: "Kim", age: 16 },
+//     { name: "Lee", age: 25 },
+//     { name: "Park", age: 30 },
+//     { name: "Choi", age: 14 },
+//     { name: "Jeon", age: 19 },
+//   ];
+
+//   const totalAge = users.reduce((acc, cur) => {
+//     return acc + cur.age;
+//   }, 0);
+
+//   console.log(totalAge);
+
+// 3. 문자열 배열을 하나의 문자열로 합치기
+// const words = ["Hello", " ", "World", "!"];
+
+// const sentence = words.reduce((acc, cur) => {
+//     acc += cur;
+//     return acc;
+// }, "");
+
+// console.log(sentence);
+
+// 4. 배열의 최댓값 찾기
+// const numbers = [10, 3, 20, 15, 7];
+
+// const findMax = numbers.reduce((acc, cur) => {
+//     return Math.max(acc, cur);
+// }, numbers[0]);
+
+// console.log(findMax);
+
+// 5. 각 문자열의 등장 횟수 세기
+// const words = ["apple", "banana", "apple", "orange", "banana", "orange", "apple"];
+
+// const wordCounter = words.reduce((acc, cur) => {
+//     // acc[cur]에서 cur가 key 가 되고, acc[cur]는 value가 됨.
+//     acc[cur] = (acc[cur] || 0) + 1;
+
+//     return acc;
+// }
+// , {});
+
+// console.log(wordCounter);
+
+// 6. 객체 배열에서 각 카테고리별 요소 개수 세기
+const items = [
+    { name: "A", category: "X" },
+    { name: "B", category: "Y" },
+    { name: "C", category: "X" },
+    { name: "D", category: "Z" },
+    { name: "E", category: "Y" },
+    { name: "F", category: "X" },
+  ];
+
+  const cntItemsByCategories = items.reduce((acc, cur) => {
+    acc[cur.category] = (acc[cur.category] || 0) + 1;
+
+    return acc;
+  }, {});
+
+  console.log(cntItemsByCategories);
+
+//   console.log(cntItemsByCategories);
+
+// 7. 중첩 배열을 단일 배열로 평탄화하기
+// const nestedArrays = [[1, 2], [3, 4], [5, 6], [7, 8]];
+
+// const normalizeToOne = nestedArrays.reduce((acc, cur) => {
+//     return acc.concat(cur);
+// }, []);
+
+// console.log(normalizeToOne);
+
+// 8. initialValue 를 사용하는 예제: 숫자 배열에서 각 숫자의 제곱을 담은 새 배열 생성하기
+
+// const numbers = [1, 2, 3, 4, 5];
+
+// const result = numbers.reduce((acc, cur) => {
+//     acc.push(cur ** 2);
+//     return acc;
+// }, []);
+
+// console.log(result);
\ No newline at end of file
diff --git a/Challenge/yuzunsang/reduceMaxValueNIndex/solve.js b/Challenge/yuzunsang/reduceMaxValueNIndex/solve.js
new file mode 100644
index 00000000..857c03e6
--- /dev/null
+++ b/Challenge/yuzunsang/reduceMaxValueNIndex/solve.js
@@ -0,0 +1,11 @@
+// write your codes
+function solution(inputArray) {
+    return result = inputArray.reduce((acc, cur, index) => {
+        if (acc.maxValue < cur) {
+            return { maxValue : cur, idx : index };
+        }
+        return acc;
+    }
+    , { maxValue: -Infinity, idx: -1 })};
+
+exports.solution = solution;
\ No newline at end of file
diff --git a/Challenge/yuzunsang/reduceMaxValueNIndex/solve.test.js b/Challenge/yuzunsang/reduceMaxValueNIndex/solve.test.js
new file mode 100644
index 00000000..0daef5b8
--- /dev/null
+++ b/Challenge/yuzunsang/reduceMaxValueNIndex/solve.test.js
@@ -0,0 +1,28 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [3, 29, 38, 12, 57, 74, 40, 85, 61],
+  answer: { maxValue: 85, idx: 7 },
+};
+
+const test2 = {
+  input: [-24, -2, -13, -49, -999999, -17],
+  answer: { maxValue: -2, idx: 1 },
+};
+//최댓값이 중복일 때에는 먼저 나온 최댓값의 인덱스를 유지하도록 설정하였습니다.
+const test3 = {
+  input: [2, -20, 21, -874, 99, -16, -29, 99],
+  answer: { maxValue: 99, idx: 4 },
+};
+
+describe('reduceMaxValueNIndex', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+  test('test2 : 음수 테스트', () => {
+    expect(solution(test2.input)).toEqual(test2.answer);
+  });
+  test('test3 : 중복된 최대값 테스트', () => {
+    expect(solution(test3.input)).toEqual(test3.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/reduceNameNickname/README.md b/Challenge/yuzunsang/reduceNameNickname/README.md
new file mode 100644
index 00000000..b6920e05
--- /dev/null
+++ b/Challenge/yuzunsang/reduceNameNickname/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+입력받은 객채배열의 nickname을 key, name을 value로 하는 객체를 출력하세요
diff --git a/Challenge/yuzunsang/reduceNameNickname/solve.js b/Challenge/yuzunsang/reduceNameNickname/solve.js
new file mode 100644
index 00000000..82025f58
--- /dev/null
+++ b/Challenge/yuzunsang/reduceNameNickname/solve.js
@@ -0,0 +1,11 @@
+// write your codes
+
+function solution(inputArray) {
+    return result = inputArray.reduce((acc, cur) => {
+        acc[cur.nickname] = cur.name;
+        // cur.nickname -> key, acc[cur.nickname] -> value 
+        return acc;
+    }, {});
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/reduceNameNickname/solve.test.js b/Challenge/yuzunsang/reduceNameNickname/solve.test.js
new file mode 100644
index 00000000..67b74529
--- /dev/null
+++ b/Challenge/yuzunsang/reduceNameNickname/solve.test.js
@@ -0,0 +1,25 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [
+    {
+      name: '홍길동',
+      nickname: 'hong',
+    },
+    {
+      name: '둘리',
+      nickname: '2li',
+    },
+    {
+      name: '오스트랄로피테쿠스',
+      nickname: '1Cin',
+    },
+  ],
+  answer: { hong: '홍길동', '2li': '둘리', '1Cin': '오스트랄로피테쿠스' },
+};
+
+describe('reduceNameNickname', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/reduceSum/README.md b/Challenge/yuzunsang/reduceSum/README.md
new file mode 100644
index 00000000..8c607eb2
--- /dev/null
+++ b/Challenge/yuzunsang/reduceSum/README.md
@@ -0,0 +1,9 @@
+# 문제제목
+
+## 설명
+
+reduce 메소드를 사용해서 배열의 모든 수의 합을 구하세요.
+
+## Expected Output 
+
+106
diff --git a/Challenge/yuzunsang/reduceSum/solve.js b/Challenge/yuzunsang/reduceSum/solve.js
new file mode 100644
index 00000000..17da581e
--- /dev/null
+++ b/Challenge/yuzunsang/reduceSum/solve.js
@@ -0,0 +1,6 @@
+// write your codes
+function solution(inputArray) {
+    return inputArray.reduce((acc, cur) => acc + cur, 0);
+}
+
+exports.solution = solution;
\ No newline at end of file
diff --git a/Challenge/yuzunsang/reduceSum/solve.test.js b/Challenge/yuzunsang/reduceSum/solve.test.js
new file mode 100644
index 00000000..046c40cb
--- /dev/null
+++ b/Challenge/yuzunsang/reduceSum/solve.test.js
@@ -0,0 +1,12 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [10, 3, 20, 5, 8, 60],
+  answer: 106,
+};
+
+describe('reduceSum', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/sortByPrice/README.md b/Challenge/yuzunsang/sortByPrice/README.md
new file mode 100644
index 00000000..b1ccf70e
--- /dev/null
+++ b/Challenge/yuzunsang/sortByPrice/README.md
@@ -0,0 +1,14 @@
+# 문제제목
+
+## 설명
+
+배열안의 객체를 price를 기준으로 오름차순 정렬한 배열을 출력하세요
+
+## Expected Output 
+
+[
+  { name: '사과', price: 1000 },
+  { name: '당근', price: 2000 },
+  { name: '수박', price: 5000 },
+  { name: '참외', price: 10000 }
+]
diff --git a/Challenge/yuzunsang/sortByPrice/practice.js b/Challenge/yuzunsang/sortByPrice/practice.js
new file mode 100644
index 00000000..550d0f58
--- /dev/null
+++ b/Challenge/yuzunsang/sortByPrice/practice.js
@@ -0,0 +1,46 @@
+const account = [
+    {id : 'jsh', addr : '서울'},
+    {id : 'root2', addr : '인천'},
+    {id : 'ghd', addr : '광주'},
+    {id : 'root3', addr : '경기'},
+    {id : 'root5', addr : '경북'},
+    {id : 'rnv', addr : '세종'},
+    {id : 'ekz', addr : '전북'},
+    {id : 'gae', addr : '전남'},
+    {id : 'root4', addr : '전남fds'},
+    {id : 'root1', addr : '전남fds'}
+];
+
+const compare = function(a, b) {
+
+    if ((b.id.indexOf('root') === 0 && (a.id.indexOf('root') === 0))) {
+        let newA = a.id.split('root');
+        let newB = b.id.split('root');
+
+        return newA[1] - newB[1];
+        // if (newA[1] > newB[1]) {
+        //     return 1;
+        // } else if (newA[1] < newB[1]) {
+        //     return -1;
+        // } else {
+        //     return 0;
+        // }
+    }
+    else if ((a.id.indexOf('root') === 0)) {
+        return -1;
+    }
+    else if ((b.id.indexOf('root') === 0)) {
+        return 1;
+    }
+    else {
+        if (a.id > b.id) {
+            return 1;
+        } else if (a.id < b.id) {
+            return -1;
+        } else {
+            return 0;
+        }
+    }
+};
+
+console.log(account.sort(compare));
\ No newline at end of file
diff --git a/Challenge/yuzunsang/sortByPrice/solve.js b/Challenge/yuzunsang/sortByPrice/solve.js
new file mode 100644
index 00000000..3da8cbfd
--- /dev/null
+++ b/Challenge/yuzunsang/sortByPrice/solve.js
@@ -0,0 +1,6 @@
+// write your codes
+function solution(inputArray) {
+    return inputArray.sort((a, b) => a.price - b.price);
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/sortByPrice/solve.test.js b/Challenge/yuzunsang/sortByPrice/solve.test.js
new file mode 100644
index 00000000..3d6a3431
--- /dev/null
+++ b/Challenge/yuzunsang/sortByPrice/solve.test.js
@@ -0,0 +1,34 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [
+    {
+      name: '사과',
+      price: 1000,
+    },
+    {
+      name: '수박',
+      price: 5000,
+    },
+    {
+      name: '당근',
+      price: 2000,
+    },
+    {
+      name: '참외',
+      price: 10000,
+    },
+  ],
+  answer: [
+    { name: '사과', price: 1000 },
+    { name: '당근', price: 2000 },
+    { name: '수박', price: 5000 },
+    { name: '참외', price: 10000 },
+  ],
+};
+
+describe('sortByPrice', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/sortByPriceAndQuantity/README.md b/Challenge/yuzunsang/sortByPriceAndQuantity/README.md
new file mode 100644
index 00000000..41cd5c95
--- /dev/null
+++ b/Challenge/yuzunsang/sortByPriceAndQuantity/README.md
@@ -0,0 +1,16 @@
+# 문제제목
+
+## 설명
+
+배열안의 객체를 price를 기준으로 오름차순 정렬한 배열을 출력하세요
+만약 price가 같다면 quantity기준으로 오름차순 정렬하세요
+
+## Expected Output 
+
+[
+  { name: '사과', price: 1000, quantity: 2 },
+  { name: '오이', price: 2000, quantity: 49 },
+  { name: '당근', price: 2000, quantity: 50 },
+  { name: '참외', price: 5000, quantity: 10 },
+  { name: '수박', price: 5000, quantity: 20 }
+]
diff --git a/Challenge/yuzunsang/sortByPriceAndQuantity/solve.js b/Challenge/yuzunsang/sortByPriceAndQuantity/solve.js
new file mode 100644
index 00000000..01a8cf42
--- /dev/null
+++ b/Challenge/yuzunsang/sortByPriceAndQuantity/solve.js
@@ -0,0 +1,13 @@
+// write your codes
+function solution(inputArray) {
+    return inputArray.sort((a, b) => {
+        if (a.price !== b.price) {
+            return a.price - b.price;
+        }
+        else {
+            return a.quantity - b.quantity;
+        }
+    });
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/sortByPriceAndQuantity/solve.test.js b/Challenge/yuzunsang/sortByPriceAndQuantity/solve.test.js
new file mode 100644
index 00000000..424f42b7
--- /dev/null
+++ b/Challenge/yuzunsang/sortByPriceAndQuantity/solve.test.js
@@ -0,0 +1,44 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [
+    {
+      name: '사과',
+      price: 1000,
+      quantity: 2,
+    },
+    {
+      name: '수박',
+      price: 5000,
+      quantity: 20,
+    },
+    {
+      name: '당근',
+      price: 2000,
+      quantity: 50,
+    },
+    {
+      name: '참외',
+      price: 5000,
+      quantity: 10,
+    },
+    {
+      name: '오이',
+      price: 2000,
+      quantity: 49,
+    },
+  ],
+  answer: [
+    { name: '사과', price: 1000, quantity: 2 },
+    { name: '오이', price: 2000, quantity: 49 },
+    { name: '당근', price: 2000, quantity: 50 },
+    { name: '참외', price: 5000, quantity: 10 },
+    { name: '수박', price: 5000, quantity: 20 },
+  ],
+};
+
+describe('sortByPriceAndQuantity', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+});
diff --git a/Challenge/yuzunsang/spreadOperatorMaxValue/README.md b/Challenge/yuzunsang/spreadOperatorMaxValue/README.md
new file mode 100644
index 00000000..164b68a1
--- /dev/null
+++ b/Challenge/yuzunsang/spreadOperatorMaxValue/README.md
@@ -0,0 +1,3 @@
+## 설명
+
+Spread Operator를 이용해서 값의 최대, 최소값을 구하세요
diff --git a/Challenge/yuzunsang/spreadOperatorMaxValue/solve.js b/Challenge/yuzunsang/spreadOperatorMaxValue/solve.js
new file mode 100644
index 00000000..de517bc6
--- /dev/null
+++ b/Challenge/yuzunsang/spreadOperatorMaxValue/solve.js
@@ -0,0 +1,6 @@
+// write your codes
+function solution(inputArray) {
+    return `max : ${Math.max(...inputArray)}, min : ${Math.min(...inputArray)}`;
+}
+
+exports.solution = solution;
diff --git a/Challenge/yuzunsang/spreadOperatorMaxValue/solve.test.js b/Challenge/yuzunsang/spreadOperatorMaxValue/solve.test.js
new file mode 100644
index 00000000..26f2abf9
--- /dev/null
+++ b/Challenge/yuzunsang/spreadOperatorMaxValue/solve.test.js
@@ -0,0 +1,12 @@
+const { solution } = require('./solve');
+
+const test1 = {
+  input: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
+  answer: 'max : 10, min : 1',
+};
+
+describe('spreadOperatorMaxValue', () => {
+  test('test1', () => {
+    expect(solution(test1.input)).toEqual(test1.answer);
+  });
+});