Two positive numbers differ by 2880. When the greater number is divided by the smaller number, the quotient is 6 and the remainder is 120. What is the HCF of the greater of the two given numbers and 3916?​

Correct option is B

Given:

Two positive numbers differ by 2880;

Greater number ÷ smaller number gives quotient = 6 and remainder = 120;

Required: HCF of the greater number and 3916.

Formula Used:

Dividend = Divisor × Quotient + Remainder

Solution:

Let smaller number = x

Then greater number = 6x + 120

Now, 

(6x + 120) - x = 2880

5x + 120 = 2880

5x = 2760

x = 552

Greater number = 6x + 120 = 6 × 552 + 120

= 3312 + 120 = 3432

Now, HCF of 3432 and 3916 

3432 = $$2^3 \times 3^1 \times 11^1 \times 13^1$$​​

3916 = $$2^2 \times 11^1 \times 89^1$$​​

HCF = $$2^2 \times 11 = 4 \times 11$$​ = 44