At first, we find out the HCF of 1134 and 1215

∴ HCF of 1134 and 1215 is 81.
∴ Required HCF = HCF of 513 and 81.

Correct Option: A

At first, we find out the HCF of 1134 and 1215

∴ HCF of 1134 and 1215 is 81.
∴ Required HCF = HCF of 513 and 81.

∴ HCF of given numbers = 27

Here , Last divisor = 18 and quotient = 3
∴ Dividend = 18 × 3 = 54
Now, divisor = 54, quotient = 7 and remainder = 18
∴ Dividend = 7 × 54 + 18 = 378 + 18 = 396
Now, divisor = 396, quotient = 2 and remainder = 54
∴ Dividend = 2 × 396 + 54 = 792 + 54 = 846

Correct Option: C

Here , Last divisor = 18 and quotient = 3
∴ Dividend = 18 × 3 = 54
Now, divisor = 54, quotient = 7 and remainder = 18
∴ Dividend = 7 × 54 + 18 = 378 + 18 = 396
Now, divisor = 396, quotient = 2 and remainder = 54
∴ Dividend = 2 × 396 + 54 = 792 + 54 = 846
Hence, the required numbers are 396 and 846.

Here, the last divisor = 49 and quotient = 2
or, Symbolically the process of finding the HCF by division method can be shown in this way.

the dividend (c) = 49 × 2 = 98
Now, divisor = 98,
quotient = 3 and remainder = 49
∴ Dividend (a) = 98 × 3 + 49 = 294 + 49 = 343
Again, divisor = 343, quotient = 17 and remainder = 98
∴ Dividend (b) = 343 × 17 + 98 = 5831+ 98 = 5929
Thus, two numbers are 343 and 5929.
Second method to solve this question :

Correct Option: B

Here, the last divisor = 49 and quotient = 2
or, Symbolically the process of finding the HCF by division method can be shown in this way.

the dividend (c) = 49 × 2 = 98
Now, divisor = 98,
quotient = 3 and remainder = 49
∴ Dividend (a) = 98 × 3 + 49 = 294 + 49 = 343
Again, divisor = 343, quotient = 17 and remainder = 98
∴ Dividend (b) = 343 × 17 + 98 = 5831+ 98 = 5929
Thus, two numbers are 343 and 5929.
Second method to solve this question :

Obviously, the greatest number will divide completely the numbers (2930 – 7) and (3246 – 7),
i.e., 2923 and 3239.
Hence, the greatest number will be the HCF of 2923 and 3239.

Correct Option: A

Obviously, the greatest number will divide completely the numbers (2930 – 7) and (3246 – 7),
i.e., 2923 and 3239.
Hence, the greatest number will be the HCF of 2923 and 3239.

∴ HCF = 79
Hence, the required number = 79

Let the remainder in each case be t.
Then, (11284 – t) and (7655 – t) are exactly divisible by that three digit number.
Hence, their difference is [(11284 – t ) – (7655 – t ] = 3629 will also be exactly divisible by that three digit number. In other words that divisor will be a factor of 3629.
Now, 3629 = 19 ×191

Correct Option: D

Let the remainder in each case be t.
Then, (11284 – t) and (7655 – t) are exactly divisible by that three digit number.
Hence, their difference is [(11284 – t ) – (7655 – t ] = 3629 will also be exactly divisible by that three digit number. In other words that divisor will be a factor of 3629.
Now, 3629 = 19 ×191
Since both 19 and 191 are prime numbers, the three digit number is 191.
Hence, the required number = 191