Correct Option: A

(A + B)’s 1 day’s work =	1
	15

(A + B + C)’s 1 day’s work =	1
	8

∴ C’s 1 day’s work

=	1	-	1	=	15 - 8	=	7
	8		15		120		120

Let (B + C) can complete the work in x days.
∴ (A + C) can complete the work in (x + 2) days.

∴ (B + C)’s 1 day’s work =	1
	x

(A + C)’s 1 day’s work =	1
	x + 2

∴ B’s 1 day’s work

=	1	-	7	=	120 - 7x
	x		120		120x

and, A’s 1 day’s work

=	1	-	7	=	120 - 7(x + 2)
	x + 2		120		120(x + 2)

=	106 - 7x
	120(x + 2)

Now, A’s 1 day’s work + B’s 1 day’s work = (A + B)’s 1 day’s work

⇒	106 - 7x	+	120 - 7x	=	1
	120(x + 2)		120x		15

⇒	106x - 7x2 + 120x + 240 - 7x2 - 14x	=	1
	120x(x + 2)		15

⇒ – 14x² + 212x + 240 = 8x² + 16x
⇒ 22x² – 196x – 240 = 0
⇒ 11x² – 98x – 120 = 0
⇒ 11x² – 110x + 12x –120 = 0
⇒ 11x (x – 10) + 12 (x – 10)= 0
⇒ (x – 10) (11x + 12) = 0

⇒ x =10, and -	12
	11

But no. of days cannot be negative
∴ x = 10
∴ A’s 1 day’s work

=	1	-	7
	10 + 2		120

=	1	-	7
	12		120

=	10 - 7	=	3	=	1
	120		120		40

∴ A alone can complete the work in 40 days.