Find the relation between the given below Quantities: Quantity I: A can complete a work in ‘x – 6’ days and B can complete the same work in ‘x + 6’ days. The same work can be completed when A works for 6 days and B works for 12 days. In how many days both A and B together can complete the work? Quantity II: A can complete a work in ‘x + 3’ days and B can complete the same work in ‘x + 3’ days. The same work can be completed when A completes 75% of the work and B does the rest of the work in a total of 9 days. In how many days both A and B together can complete the work when A works with 25% higher efficiency?

ATQ, Quantity I A’s efficiency = 1/(x – 6) B’s efficiency = 1/(x + 6) Given: 6 days of A and 12 days of B complete the whole work So, 6/(x – 6) + 12/(x + 6) = 1 6(x + 6) + 12(x – 6) = (x – 6)(x + 6) 6x + 36 + 12x – 72 = x² – 36 18x – 36 = x² – 36 x² – 18x = 0 x(x – 18) = 0 x = 18 (taking positive value) So, A alone can complete in x – 6 = 18 – 6 = 12 days B alone can complete in x + 6 = 18 + 6 = 24 days LCM of 12 and 24 = 24 units (assume total work) Work done by A in 1 day = 24/12 = 2 units Work done by B in 1 day = 24/24 = 1 unit Total work per day together = 2 + 1 = 3 units Time taken together = 24 ÷ 3 = 8 days Quantity II A alone can complete in x + 3 days, B alone in x + 3 days A completes 75% and B completes 25% of the work in 9 days Time taken by A = 75% of (x + 3) = 3(x + 3)/4 Time taken by B = 25% of (x + 3) = (x + 3)/4 Total time = 3(x + 3)/4 + (x + 3)/4 = x + 3 Given total time = 9 days So, x + 3 = 9 x = 6 So, A alone can complete in x + 3 = 9 days B alone can complete in x + 3 = 9 days A’s original efficiency = 1/9 With 25% higher efficiency, A’s new efficiency = 1/9 × 125/100 = 5/36 B’s efficiency = 1/9 = 4/36 LCM for denominators 36 ⇒ treat total work as 36 units Work done by A in 1 day = 5 units Work done by B in 1 day = 4 units Total work per day together = 5 + 4 = 9 units Time taken together = 36 ÷ 9 = 4 days Thus, Quantity I = 8 days and Quantity II = 4 days So, Quantity I > Quantity II