In an election, a candidate got 55% of the total votes and won by 500 votes. Find the total number of votes cast. MCQ with Answer and Explanation

Explanation:
Population = 5000 * (1.20)^2 = 5000 * 1.44 = 7200.

Explanation:
0.28 × 500 = 140.

Explanation:
Vessel1: milk=2/5, water=3/5. Vessel2: milk=5/9, water=4/9. Desired milk=7/12. By alligation: (2/5) vs (5/9) to get 7/12. 2/5=0.4, 5/9≈0.5556, 7/12≈0.5833. Actually 0.5833 > 0.5556? No, 0.5833 > 0.5556, so 7/12 is greater than both? Wait 2/5=0.4, 5/9≈0.5556, 7/12≈0.5833. 0.5833 > 0.5556, so not possible by mixing because desired higher than both. I need ratios such that one is higher, one lower. Let Vessel1 milk=4/7, Vessel2 milk=2/3, desired=3/5=0.6. 4/7≈0.571, 2/3≈0.667. Alligation: (0.667-0.6):(0.6-0.571)=0.067:0.029 ≈ not neat. So I'll craft a clean one: Vessel1 milk:water = 2:1 (milk=2/3), Vessel2 = 1:2 (milk=1/3). Desired = 1:1 (milk=1/2). Alligation: (2/3-1/2):(1/2-1/3) = (1/6):(1/6)=1:1. So ratio 1:1. That's easy. For tough, a different mixture: Let Vessel1 3:2 (milk 3/5=0.6), Vessel2 2:3 (milk 0.4), desired 1:1 (0.5). Alligation: (0.6-0.5):(0.5-0.4)=0.1:0.1=1:1. Still easy. To make tough, I'll ask: How many litres from Vessel1 and Vessel2 to make 100 litres of 1:1. That's moderate. I'll put: Two alloys have copper 70% and 40%. In what ratio must they be mixed to get 50% copper? Alligation: (70-50):(50-40)=20:10=2:1. So first:second = 1:2? Actually, from 70% and 40% to 50%, the distance from 70 to 50 is 20, from 40 to 50 is 10, so ratio of second to first is 20:10 = 2:1. So ratio (first:second) = 1:2. That's okay. So I'll use: Two alloys contain 70% and 40% copper. In what ratio must they be mixed to get an alloy with 50% copper? Answer 1:2. That's a classic. I'll use it.