The H.c.f. And L.c.m. Of Two Numbers Are 11 And 385 Respectively. If

1.	The H.c.f. And L.c.m. Of Two Numbers Are 11 And 385 Respectively. If One Number Lies Between 75 And 125, Then That Number Is?
Answer» Product of numbers = 11 x 385 = 4235. Let the numbers be 11a and 11b. Then 11a x 11b = 4235 ab = 35 Now, co-primes with product 35 are (1, 35) and (5, 7). So, the numbers are (11 x 1, 11 x 35) and (11 x 5, 11 x 7). Since one NUMBER lies between 75 and 125, the suitable pair is (55, 77) Hence, required number = 77. Product of numbers = 11 x 385 = 4235. Let the numbers be 11a and 11b. Then 11a x 11b = 4235 ab = 35 Now, co-primes with product 35 are (1, 35) and (5, 7). So, the numbers are (11 x 1, 11 x 35) and (11 x 5, 11 x 7). Since one number lies between 75 and 125, the suitable pair is (55, 77) Hence, required number = 77.

The H.c.f. And L.c.m. Of Two Numbers Are 11 And 385 Respectively. If One Number Lies Between 75 And 125, Then That Number Is?

Answer»

Product of numbers = 11 x 385 = 4235.

Let the numbers be 11a and 11b. Then 11a x 11b = 4235 ab = 35

Now, co-primes with product 35 are (1, 35) and (5, 7).

So, the numbers are (11 x 1, 11 x 35) and (11 x 5, 11 x 7).

Since one NUMBER lies between 75 and 125, the suitable pair is (55, 77)

Hence, required number = 77.

Product of numbers = 11 x 385 = 4235.

Let the numbers be 11a and 11b. Then 11a x 11b = 4235 ab = 35

Now, co-primes with product 35 are (1, 35) and (5, 7).

So, the numbers are (11 x 1, 11 x 35) and (11 x 5, 11 x 7).

Since one number lies between 75 and 125, the suitable pair is (55, 77)

Hence, required number = 77.