Numbers Questions

The numbers questions and answers provided on this page might help children grasp the subject quickly. In practically all competitive and board examinations, many questions are asked from the concept called “Numbers”. Students can use these questions to acquire a quick overview of the topics and practice them so that they understand the concept better. Look over the complete explanations for each question to double-check your answers. To learn more about Numbers, click here.

Numbers questions and answers are provided below for you to learn and practice.

Numbers Questions with Solutions

1. Is 91 a prime number?

Solution:

No, 91 is not a prime number. The number 91 is divisible by 1, 7, 13, and 91. Since it has 4 factors, the number 91 does not satisfy the condition for prime numbers.

2. If one-third of one-fourth of a number is 15, then three-tenth of that number is equal to ___.

Solution:

Let the unknown number be “p”.

(⅓) of (¼) of p = 15

Hence, p = 15 × 3 × 4 = 180

Thus, the required number is,

= (3/10) × 180

= 54

Hence, three-tenth of that number is equal to 54.

3. The difference between a two-digit number and a number formed by swapping its digits’ locations is 36. What is the difference between those numbers in two digits?

Solution:

Assume that ten’s digit is “a” and unit’s digit is “b”.

Then, (10a + b) – (10b + a) = 36

10 a + b – 10b – a = 36

9a – 9b = 36

9(a – b) = 36

a – b = 36/9

a – b = 4

Hence, the difference between those two digits of that number is 4.

4. Is the number 415624 divisible by 11?

Solution:

Yes, the number 415624 is divisible by 11

Explanation:

The divisibility by 11 rule says that if the difference between the sum of the digits in odd places and the sum of the digits in even places is 0 or divisible by 11, the provided number is also divisible by 11.

Given number: 415624

The sum of digits in odd places = 4 + 5 + 2 = 11

The sum of digits in even places = 1 + 6 + 4 = 11

Thus, difference = 11 – 11 = 0

Hence, the number 415624 is divisible by 11.

5. If a number is divided by 56, we get the remainder 29. If the same number is divided by 8, then find the remainder.

Solution:

We know that, Dividend/Divisor = Quotient + Remainder

Thus,

Dividend = [(Divisor × Quotient)] + Remainder

Let the unknown number (dividend) be x and quotient be y.

Thus, x = [56×y] + 29

x = 56y + 29

x = (8 × 7y) + (8 × 3) + 5

Hence, the required remainder is 5.

6. A three-digit number has a sum equal to 10. The middle digit is equal to the sum of the other two, and reversing the digits increases the number by 99. Find the number.

Solution:

Suppose the middle digit is k.

Thus, 2k=10 or k=5 applies. So, one of the numbers is 253 or 352, and the other pair is 154 or 451.

The hundreds digit is less than the units digits because the number increases when the digits are reversed.

Because the difference between 154 and 451 is not 99, the required number is 253.

7. Find the sum of all even natural numbers between 1 and 31.

Solution:

To find: Sum of all even natural numbers between 1 and 31. I.e., 2 + 4 + 6 + … + 30

The given sequence of numbers is an Arithmetic Progression (AP) because the difference between two consecutive terms is 2.

I.e.,d = 2

Also, a = 2 and the last term, l = 30.

Let the number of terms be “n”.

Thus we can write

a + (n – 1)d = 30

2 + (n-1)2 = 30

2n – 2 = 30 – 2

2n = 28 +2

2n = 30

n = 15.

Hence, the sum of all even natural numbers between 1 and 31 is:

S₁₅ = (15/2)(2+30)

S₁₅ = (15/2)(32)

S₁₅ = 15 (16)

S₁₅ = 240.

8. If a number is divided by 357, we get the remainder 39. If the same number is divided by 17, then find the remainder.

Solution:

As we know,

Dividend = [(Divisor × Quotient)] + Remainder

Let the unknown numbers (dividend) be p and quotient be q

Thus, p = [357×q] + 39

p = 357y + 39

p = (17 × 21y) + (17 × 2) + 5

Therefore, the remainder is 5.

9. Find the sum of three numbers, if the sum of squares of three numbers is equal to 138 and the sum of their products taken two at a time is equal to 131.

Solution:

Suppose the three numbers are p, q and r.

Given that, the sum of squares of three numbers = 138

Similarly, the sum of their products taken two at a time = 131

From the given information, we can write,

p² + q² + r² = 138

pq + qr + pr =131

We know the formula (a + b + c)² = a² + b² + c² + 2 (ab + bc + ca)

By substituting the values in the formula, we get

(p + q + r)² = p² + q² + r² + 2 (pq + qr + pr)

(p + q + r)² = 138 + 2(131)

(p + q + r)² = 400

Hence, p + q + r = 20

Therefore, the sum of three numbers is 20.

10. A fraction’s numerator and denominator add up to 30. 2/3 is obtained by adding 2 to the numerator and subtracting 2 from the denominator. Determine the fraction.

Solution:

Let the fraction be x/y.

Given that the sum of numerator and denominator is 30

I.e., x + y = 30 …(1)

Another condition given here is 2 is added to the numerator and 2 is subtracted from the denominator. Thus, the fraction becomes

(x+2)/(y-2) = 2/3

3(x+2) = 2(y-2)

Now, solve the equation, we get

3x + 6 = 2y – 4

3x – 2y = -10 …(2)

Now, solve the equation, we get

x = 10 and y = 20

Therefore, the fraction x/y = 10/20 = 1/2.

Practice Questions

1. Solve the expression: -84 × 29 + 365.
2. How many terms in the series 6 + 12 + 18 + 24 +… add up to 1800?
3. What is the largest four-digit number that can be divided by 88?
4. Determine a positive number that is equal to 60 times its reciprocal when increased by 17.

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