LCM of 32, 40, 48 and 60 = 480
The number divisible by 480 between 4000 and 6000 are 4320, 4800, 5280 and 5760.

Correct Option: C

LCM of 32, 40, 48 and 60 = 480
The number divisible by 480 between 4000 and 6000 are 4320, 4800, 5280 and 5760.

Hence, required number of numbers are 4.

Let the numbers be 66a and 66b, where a and b are co-primes.

According to the question,
66a + 66b = 1056
⇒ 66(a + b) = 1056
⇒ (a + b) = 1056/66 = 16
∴ Possible values of a and b are
(a = 1, b = 15), (a = 3, b = 13).
(a = 5, b = 11 ), (a = 7, b = 9)

Correct Option: C

Let the numbers be 66a and 66b, where a and b are co-primes.

According to the question,
66a + 66b = 1056
⇒ 66(a + b) = 1056
⇒ (a + b) = 1056/66 = 16
∴ Possible values of a and b are
(a = 1, b = 15), (a = 3, b = 13).
(a = 5, b = 11 ), (a = 7, b = 9)
∴ Numbers are
(66 x 1, 66 x 15), (66 x 3, 66 x 13),
(66 x 5, 66 x 11), (66 x 7, 66 x 9).
∴ Possible number of pairs = 4

Second number = (LCM x HCF) / first number

Correct Option: A

Second number = (LCM x HCF) / first number
= (72 x 12)/ 24 = 36
Difference between the two numbers
= 36 - 24 = 12

According to the question,
1st number = 3M, 2nd number = 4M
where, M = HCF

But given, M = 4
We know that,
LCM = Product of two numbers / HCF

Correct Option: D

According to the question,
1st number = 3M, 2nd number = 4M
where, M = HCF

But given, M = 4
We know that,
LCM = Product of two numbers / HCF
= (3M x 4M) / M
LCM = 12M = 12 x 4 = 48

HCF of (22n + 7, 33n + 10) is always 1.
lllustraction
For n = 1, HCF(29, 43) ⇒ HCF = 1
For n = 2, HCF(51, 76) ⇒ HCF = 1
For n = 3, HCF(73, 109) ⇒ HCF = 1

Correct Option: B

HCF of (22n + 7, 33n + 10) is always 1.
lllustraction
For n = 1, HCF(29, 43) ⇒ HCF = 1
For n = 2, HCF(51, 76) ⇒ HCF = 1
For n = 3, HCF(73, 109) ⇒ HCF = 1