Answer and solution
Correct answer: D
Solution
1. Rohit sells 100 items over two identical days, hence 50 items per day. If he visits h households each day, every salesman visits 2h households in total.
2. Let Vikram’s visits be a and a+30. Then 2a+30=2h, so a=h−15 and his second-day visits are h+15.
3. Rohit’s rate is 50/h. Vikram sells 100(h−15)/h on Day 1 and 37.5(h+15)/h on Day 2.
4. These sales sum to 100: 100h−1500+37.5h+562.5=100h. Thus 37.5h=937.5 and h=25.
5. Vikram: visits 10,40 and sells 40,60. Rohit: visits 25,25 and sells 50,50. Arjun meets 1 household on Day 2 due to the complaint, so visits 49,1 and sells 100,0.
6. Check every salesman’s totals: 50 households and 100 items. Vikram’s rates 4 and 1.5 are 2 and 0.75 times Rohit’s rate 2.
7. Rohit meets 25 on Day 2, between 21 and 29. Answer D.
Why the rate model is valid:
Success rate means items per household, and a household can buy several items. Thus a rate of 4 is permitted; it is not a probability capped at 1. Multiplying the rate by households gives items sold. Both the daily split and the two-day totals must be satisfied together; the completed table checks all these conditions.
G Strategy shortcut
Method: Normalize Rohit’s daily success rate to r. Vikram’s rates are 2r and 0.75r, and his combined average must be r. Alligation gives household weights (1−0.75):(2−1)=1:4. Their 30-household difference is 3 parts, so Vikram visits 10 and 40. Rohit visits 25 each day.Verification: 7. Rohit meets 25 on Day 2, between 21 and 29. Answer D.
G Strategy: Method: Normalize Rohit’s daily success rate to r. Vikram’s rates are 2r and 0.75r, and his combined average must be r. Alligation gives household weights (1−0.75):(2−1)=1:4. Their 30-household difference is 3 parts, so Vikram visits 10 and 40. Rohit visits 25 each day. Verification: 7. Rohit meets 25 on Day 2, between 21 and 29. Answer D.