All questions of Playing with Numbers for Class 6 Exam

When we divide 650 by 2 and 25 we get whole number, it is not possible with other options. so option d is correct.

Solution:
The largest number that will divide 76, 113 and 186 leaving remainder 4, 5, 6 respectively is also known as the highest common factor (HCF) of (76-4), (113-5) and (186-6).

Step-by-step solution:
• Subtract the remainder from each number to get the actual number.
• So, the actual numbers are 72, 108, and 180.
• Now, find the common factors of these numbers.
• The common factors of 72, 108, and 180 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
• Out of these common factors, select the highest one which leaves the same remainder when divided by 4, 5, and 6 respectively.
• The highest common factor (HCF) of (76-4), (113-5) and (186-6) is 36.
• Hence, the correct option is (c) 36.

Therefore, the largest number that will divide 76, 113 and 186 leaving remainder 4, 5, 6 respectively is 36.

To determine the two numbers nearest to 1870 that are each divisible by 22, we can use the concept of divisibility and apply it to the given options.

Divisibility Rule of 22: A number is divisible by 22 if it is divisible by both 2 and 11.

Let's analyze the given options one by one to see if they satisfy the divisibility rule of 22.

a) 1848, 1892:
- 1848: This number is divisible by 2 because the last digit is even (8). However, to determine if it is divisible by 11, we can use the divisibility rule of 11, which states that the difference between the sum of the digits in the odd place and the sum of the digits in the even place should be either 0 or a multiple of 11.
- In this case, 1 - 8 + 4 - 8 = -11, which is a multiple of 11. Therefore, 1848 is divisible by 11 and consequently divisible by 22.
- 1892: This number is divisible by 2 because the last digit is even (2). However, to determine if it is divisible by 11, we can use the divisibility rule of 11.
- In this case, 1 - 8 + 9 - 2 = 0, which is a multiple of 11. Therefore, 1892 is divisible by 11 and consequently divisible by 22.

Since both 1848 and 1892 satisfy the divisibility rule of 22, option A is correct.

b) 1893, 1914:
- 1893: This number is not divisible by 2 because the last digit is odd (3). Therefore, it is not divisible by 22.
- 1914: This number is divisible by 2 because the last digit is even (4). However, it is not divisible by 11 since 1 - 9 + 1 - 4 = -11, which is not a multiple of 11. Therefore, it is not divisible by 22.

Option B is incorrect.

c) 1826, 1914:
- 1826: This number is divisible by 2 because the last digit is even (6). However, it is not divisible by 11 since 1 - 8 + 2 - 6 = -11, which is not a multiple of 11. Therefore, it is not divisible by 22.
- 1914: This number is divisible by 2 because the last digit is even (4). However, it is not divisible by 11 since 1 - 9 + 1 - 4 = -11, which is not a multiple of 11. Therefore, it is not divisible by 22.

Option C is incorrect.

d) None of these:
Since we have found that option A is correct, option D is incorrect.

Therefore, the two numbers nearest to 1870 that are each divisible by 22 are 1848 and 1892, as stated in option A.

To find the other number, we can use the formula:

LCM × HCF = Product of the two numbers

Given that the HCF is 23 and the LCM is 1449, we can substitute these values into the formula:

1449 × 23 = Product of the two numbers

Therefore, the product of the two numbers is 33327.

We are also given that one of the numbers is 161. Let's call the other number x.

So, the equation becomes:

161 × x = 33327

To find the value of x, we can divide both sides of the equation by 161:

x = 33327 ÷ 161

Performing the division, we get:

x ≈ 207

Therefore, the other number is approximately 207.

Hence, the correct answer is option D) 207.