Use one array for all data.

[MHenderson]

[{value: 1, korean: "하나", sino: "일"}, ...]

I think the find method is all I need to query such an array depending on the game mode.

https://www.w3schools.com/jsref/jsref_find.asp

[MHenderson]

const numbersArray = [
  { value: 1, korean: "하나", sino: "일" },
  { value: 2, korean: "둘", sino: "이" },
  { value: 3, korean: "셋", sino: "삼" },
  { value: 4, korean: "넷", sino: "사" },
  { value: 5, korean: "다섯", sino: "오" },
  { value: 6, korean: "여섯", sino: "육" },
  { value: 7, korean: "일곱", sino: "칠" },
  { value: 8, korean: "여덟", sino: "팔" },
  { value: 9, korean: "아홉", sino: "구" },
  { value: 10, korean: "열", sino: "십" }
];

x = numbersArray.find((x)=>x.value==6);

// Log to console
console.log(x.korean)
console.log(x.sino)

[MHenderson]

This might be preferable:

const result = numbersArray.find(({ value }) => value === 5);

console.log(result.korean)
console.log(result.sino)

[MHenderson]

An alternative might be to initialise a new Question/Answer array, depending on mode

// korean to number game
exercises = numbersArray.map(d => ({ question: d.korean, answer: d.value}) )

// sino-korean to number game
exercises = numbersArray.map(d => ({ question: d.sino, answer: d.value}) )

// number to korean guessing
exercises = numbersArray.map(d => ({ question: d.value, answer: d.korean}) )

// number to sino-korean guessing
exercises = numbersArray.map(d => ({ question: d.value, answer: d.sino}) )