We have to find such numbers which are divisible by 12 (LCM of 4 and 6).
Number of numbers divisible by 12 and lying between 1 to 600

Correct Option: C

We have to find such numbers which are divisible by 12 (LCM of 4 and 6).
Number of numbers divisible by 12 and lying between 1 to 600

(x – 2) is a factor of polynomial P (x) = x³ + x² – 5x + λ.
∴  P(2) = 0 (on putting x = 2)

Correct Option: B

(x – 2) is a factor of polynomial P (x) = x³ + x² – 5x + λ.
∴  P(2) = 0 (on putting x = 2)
⇒  2³ + 2² – 5 × 2 + λ = 0
⇒  8 + 4 – 10 + λ = 0
⇒  λ + 2 = 0
∴  λ = - 2

According to question ,
Required Number = 100p + 10q + r
∵  10q + r = 6m
∴  Number = 100x + 6m, where m is a positive integer.

Correct Option: C

According to question ,
Required Number = 100p + 10q + r
∵  10q + r = 6m
∴  Number = 100x + 6m, where m is a positive integer.
Number = 2 (50p + 3m)
Hence required answer is 2 .

LCM of 4, 5 and 6 = 60
Quotient on dividing 800 by 60 = 13
Quotient on dividing 400 by 60 = 6
∴  Required answer = Quotient on dividing 800 by 60 – Quotient on dividing 400 by 60

Correct Option: A

LCM of 4, 5 and 6 = 60
Quotient on dividing 800 by 60 = 13
Quotient on dividing 400 by 60 = 6
∴  Required answer = Quotient on dividing 800 by 60 – Quotient on dividing 400 by 60
∴  Required answer = 13 – 6 = 7
2nd Method to solve this question :
First number greater than 400
that is divisible by 60 = 420
Smaller number than 800 that
is divisible by 60 = 780
It is an Arithmetic Progression with common difference = 60
By t_n = a + (n – 1)d
780 = 420 + (n – 1) × 60
⇒ (n – 1) × 60 = 780 – 420 = 360
⇒ (n – 1) = 360 ÷ 60 = 6
⇒ n = 6 + 1 = 7

Here, the first divisor (289) is a multiple of second divisor (17).
∴ Required remainder = Remainder obtained on dividing 18 by 17

Correct Option: D

Here, the first divisor (289) is a multiple of second divisor (17).
∴ Required remainder = Remainder obtained on dividing 18 by 17 = 1
Therefore required answer is 1.