Number Patterns & Puzzles MCQ Test — Class 8 CBSE

Solution:

In a two-digit number, the digit in the tens place represents its value multiplied by 10.

The digit in the units place represents its value multiplied by 1.

So, if 'a' is in the tens place and 'b' is in the units place, the number is (a × 10) + (b × 1). — Number = 10a + b

Solution:

The original number in generalized form is 100x + 10y + z.

When the digits are reversed, 'z' moves to the hundreds place, 'y' remains in the tens place, and 'x' moves to the units place.

Therefore, the new number will be (z × 100) + (y × 10) + (x × 1). — New number = 100z + 10y + x

Solution:

Let the tens digit be 'a' and the units digit be 'b'. The original number is 10a + b.

The number obtained by reversing its digits is 10b + a.

Given that the units digit b = 4. So, the original number is 10a + 4, and the reversed number is 10(4) + a = 40 + a.

The sum of the two numbers is (10a + 4) + (40 + a) = 11a + 44.

We are given that the sum is 132. So, 11a + 44 = 132.

Subtract 44 from both sides: 11a = 132 - 44 = 88.

Divide by 11: a = 88 / 11 = 8. Thus, the tens digit of the original number is 8.

Solution:

From the units column, we have 8 + B. The result's units digit is 1. Since B is a single digit (0-9), 8 + B must be 11 (as 8+B cannot be 1 or 21).

So, B = 11 - 8 = 3.

This also means there is a carry-over of 1 to the tens column.

From the tens column, we have A + 2 + (carry-over 1) = 7. So, A + 3 = 7. This implies A = 4.

Thus, the value of B is 3.

Solution:

The divisibility rule for 10 states that a number is divisible by 10 if and only if its units digit is 0.

Let's check the units digit of each option:

A) 105 (units digit is 5) - Not divisible by 10.

B) 210 (units digit is 0) - Divisible by 10.

C) 345 (units digit is 5) - Not divisible by 10.

D) 501 (units digit is 1) - Not divisible by 10.

Solution:

The divisibility rule for 5 states that a number is divisible by 5 if its units digit is either 0 or 5.

For example, 25, 130, 785 are all divisible by 5 because their units digits are 5 or 0.

Solution:

The divisibility rule for 2 states that a number is divisible by 2 if its units digit is 0, 2, 4, 6, or 8.

In the number 34_, the underscore represents the units digit.

We need the units digit to be an even number from the given options.

A) 1 is odd. B) 3 is odd. C) 5 is odd. D) 6 is even.

Therefore, replacing the underscore with 6 (making the number 346) will make it divisible by 2.

Solution:

According to the divisibility rule for 3, the sum of the digits of a number must be divisible by 3.

For the number 5X2, the sum of the digits is 5 + X + 2 = 7 + X.

We need 7 + X to be a multiple of 3. Since X must be a single non-zero digit (1-9), let's test values for X.

If X = 1, sum = 7 + 1 = 8 (not divisible by 3).

If X = 2, sum = 7 + 2 = 9 (divisible by 3). This is the smallest non-zero digit that works.

If X = 3, sum = 7 + 3 = 10 (not divisible by 3).

If X = 4, sum = 7 + 4 = 11 (not divisible by 3).

If X = 5, sum = 7 + 5 = 12 (divisible by 3).

Therefore, the smallest possible non-zero digit for X is 2.

Solution:

Let's analyze each statement:

A) All numbers divisible by 3 are also divisible by 9. (False. For example, 6 is divisible by 3 but not by 9).

B) All numbers divisible by 9 are also divisible by 3. (True. If a number is a multiple of 9, it can be written as 9k. Since 9k = 3 × (3k), it is also a multiple of 3).

C) A number is divisible by 3 if its units digit is 3, 6, or 9. (False. The rule for 3 involves the sum of digits, not just the units digit. For example, 13 is not divisible by 3).

D) A number is divisible by 9 if the sum of its digits is 3. (False. For divisibility by 9, the sum of digits must be a multiple of 9, not 3. For example, 12 has a sum of digits 3, but is not divisible by 9).

Solution:

The number 6 is a composite number, and its prime factors are 2 and 3 (6 = 2 × 3).

For a number to be divisible by 6, it must be divisible by both of its prime factors, 2 and 3, simultaneously.

For example, 12 is divisible by both 2 and 3, and thus by 6. However, 4 is divisible by 2 but not 3 (not by 6), and 9 is divisible by 3 but not 2 (not by 6).