A can do a work in 15 days and B in 20 days. They work together for 6 days, then A leaves. In how many more days will B finish the remaining work? MCQ with Answer and Explanation

Explanation:
Difference in percentage = 32% - 20% = 12%. Difference in marks = 42 + 30 = 72. 12% of Total = 72. Total = 600.

Explanation:
Equivalent discount = a + b - (ab / 100). 10 + 20 - (10 * 20 / 100) = 30 - 2 = 28%.

Explanation:
Let B=x, A=1.2x. (1.2x+10) = 1.1(x+10) → 1.2x+10 = 1.1x+11 → 0.1x=1 → x=10. B=10, A=12. That's not in options. I need larger. Let's set ratio: A = 1.25B. After 10 years, A+10 = 1.2(B+10)? Not enough. I'll craft a clean one: A is 20% more than B. After 5 years, A will be 10% more than B. Then 1.2x+5 = 1.1(x+5) → 1.2x+5=1.1x+5.5 → 0.1x=0.5 → x=5, A=6. Not. So I need ages in decades. Let A=1.2B, after 10 years A+10 = 1.1(B+10) gives 1.2B+10=1.1B+11 → 0.1B=1 → B=10, A=12. Too small. So maybe after 20 years: 1.2B+20 = 1.1(B+20) → 1.2B+20=1.1B+22 → 0.1B=2 → B=20, A=24. Not 50. To get 50, need bigger multipliers. Let A=1.5B, after 10 years A+10 = 1.25(B+10) → 1.5B+10=1.25B+12.5 → 0.25B=2.5 → B=10, A=15. Not. I'll set A=2B, after 10 years A+10=1.5(B+10) → 2B+10=1.5B+15 → 0.5B=5 → B=10, A=20. Not. So to get A=50, let B=40, A=48 (20% more is 48). After 10 years, A=58, B=50, A is 16% more. Not 10%. This is tricky. I'll just use a standard age problem with percentages: Father is 25% older than mother, sum is 90, find father's age. 1.25M + M = 90 → M=40, F=50. That's easy. For tough: after some years, ratio changes. I'll do: A is 25% older than B. 10 years ago, A was 50% older than B. Find present age of A. Let B=x, A=1.25x. 1.25x-10 = 1.5(x-10) → 1.25x-10=1.5x-15 → 0.25x=5 → x=20, A=25. Not 50. So I'll just skip age and add another percentage. I'll replace this question with a different one.