# Real Number (Class X )

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Questions: 12 | Attempts: 686

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• 1.

### For some integer ‘m’ every odd integer is of form:

• A.

M + 1

• B.

2 m

• C.

• D.

2 m + 1

D. 2 m + 1
Explanation
The given answer, 2m + 1, is the correct form for representing every odd integer. In this form, m is an integer, and when it is multiplied by 2 and added to 1, the result is always an odd number. This can be observed by substituting different values of m into the formula. Therefore, 2m + 1 is the correct expression to represent odd integers.

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• 2.

### H. C. F. of 96 and 404 is

• A.

8

• B.

4696

• C.

96

• D.

4

D. 4
Explanation
The H.C.F. (Highest Common Factor) is the highest number that divides both 96 and 404 without leaving a remainder. In this case, the only number that satisfies this condition is 4. Therefore, 4 is the correct answer.

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• 3.

### Write the HCF of the smallest composite number and the smallest even number

• A.

4

• B.

0

• C.

1

• D.

2

D. 2
Explanation
The HCF (Highest Common Factor) is the largest number that divides two or more numbers without leaving a remainder. In this case, the smallest composite number is 4, which can be divided by 1, 2, and 4. The smallest even number is 2, which can be divided by 1 and 2. The highest number that is common to both 4 and 2 is 2, as it is the largest number that divides both numbers without leaving a remainder. Therefore, the HCF of the smallest composite number and the smallest even number is 2.

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• 4.

### HCF of the smallest composite number and the smallest prime number is

• A.

0

• B.

2

• C.

4

• D.

1

B. 2
Explanation
The HCF (Highest Common Factor) of any two numbers is the largest number that divides both of them without leaving a remainder. In this case, the smallest composite number is 4 (as it is the smallest number that is divisible by numbers other than 1 and itself) and the smallest prime number is 2. The only number that divides both 4 and 2 without leaving a remainder is 2, making it the HCF of the two numbers. Thus, the correct answer is 2.

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• 5.

### If n is a positive integer , then n2 – n is always

• A.

Multiple of 2 and 4

• B.

Prime number

• C.

Odd number

• D.

Even number

D. Even number
Explanation
When we expand the expression n^2 - n, we get n(n-1). Since n is a positive integer, both n and n-1 will be positive integers. Now, consider two consecutive positive integers. One of them will always be even and the other will always be odd. Thus, when we multiply an even number with an odd number, the result will always be an even number. Therefore, n^2 - n will always be an even number.

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• 6.

### What is the HCF of 235 and 395?

• A.

25

• B.

5

• C.

35

• D.

15

B. 5
Explanation
The HCF (Highest Common Factor) is the largest number that can evenly divide both 235 and 395. In this case, the only number that can divide both 235 and 395 is 5. Therefore, the HCF of 235 and 395 is 5.

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• 7.

### H.C.F. of two consecutive even numbers is

• A.

2

• B.

0

• C.

4

• D.

1

A. 2
Explanation
The H.C.F. (Highest Common Factor) of two consecutive even numbers is always 2. This is because every even number is divisible by 2. Since the two consecutive even numbers have no other common factors besides 2, the H.C.F. is 2.

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• 8.

### 4052 = 420 x q + 272 then value of q is

• A.

12

• B.

7

• C.

8

• D.

9

D. 9
Explanation
The given equation states that 4052 is equal to 420 times q plus 272. To find the value of q, we need to isolate it on one side of the equation. We can do this by subtracting 272 from both sides of the equation, which gives us 3780 = 420q. Next, we can divide both sides of the equation by 420, resulting in q = 9. Therefore, the value of q is 9.

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• 9.

### For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, 0 ≤ r < b, if b = 4 then which is not the value of r?

• A.

2

• B.

1

• C.

3

• D.

4

D. 4
Explanation
For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, 0 ≤ r < b. In this case, b = 4. The possible values for r when b = 4 are 0, 1, 2, and 3. The only value that is not possible for r is 4.

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• 10.

### For a=n and b=3, which of the following represents the correct mathematical form of Euclid’s division lemma?

• A.

N = q+ 3

• B.

N = 3q - r

• C.

N = 3q +r

• D.

N = 3q  + n

C. N = 3q +r
Explanation
Euclid's division lemma states that for any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. In this case, the correct mathematical form of Euclid's division lemma is n = 3q + r. This equation represents the division of n by 3, where q is the quotient and r is the remainder.

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• 11.

### Euclid’s division lemma states that if a and b are two positive integers, then there exist unique integers q and r such that:

• A.

a = bq + r, 0 ≤ r ≤ b

• B.

A = bq + r, 0 < r < b

• C.

A = bq + r, 0 < b ≤ r

• D.

A = bq + r, 0 ≤ r < b

D. A = bq + r, 0 ≤ r < b
Explanation
The correct answer is a = bq + r, 0 ≤ r < b. This is because Euclid's division lemma states that when dividing a positive integer a by a positive integer b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. This means that the remainder (r) when dividing a by b is always greater than or equal to 0 and less than b.

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• 12.

### Complete the statement : Any positive odd integer is of the form 6q + 1, or 6q + 3, or ___________, where q is some positive interger.

• A.

6q + 5

• B.

6q + 4

• C.

6q + 2

• D.

6q

A. 6q + 5
Explanation
Any positive odd integer can be represented as 6q + 1, where q is a positive integer. The given statement suggests that the positive odd integer can also be expressed as 6q + 3 or 6q + 5. This is because when q is a positive integer, multiplying it by 6 and adding 3 or 5 will always result in an odd number. Therefore, the correct answer is 6q + 5.

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• Current Version
• Mar 07, 2023
Quiz Edited by
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• Apr 08, 2020
Quiz Created by
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