The average weight of a political party is decreased by 1, when some new politician joined the party whose strength is (\frac{1}{4}) of the existing politicians and the total weight of the new politicians is 209 kg. What is the average weight the politicians joined if is known that in any case the number of politicians always must be greater than 50 but less than 100.

Let x be original politicians, so x/4 new politicians joined. Total weight of new politicians is 209 kg, so their average is 209 ÷ (x/4) = 836/x. The average decreases by 1: (total weight + 209)/(x + x/4) = original average - 1. If original average is A, then (xA + 209)/(5x/4) = A - 1. Solving gives A - 836/x = 1, so A = (836/x) + 1. New politician average = 836/x = A - 1. The politicians joined reduces average, so their average < original average. This means 836/x < (836/x) + 1, which is always true. Given 50 < x + x/4 < 100 (must be > 50 and < 100), x must be divisible by 4. Options: 40 (fails), 44 (fails), 48 (55 total, works), 52 (65 total, works), 56 (70 total, works), 60 (75 total, works), 64 (80 total, works), 68 (85 total, works), 72 (90 total, works), 76 (95 total, works), 80 (100 total, fails). Checking new average = 836/x: For x=48, avg=17.42; for x=52, avg=16.08; for x=56, avg=14.93; for x=60, avg=13.93; for x=64, avg=13.06; for x=68, avg=12.29; for x=72, avg=11.61; for x=76, avg=11. So x=76 gives average = 11, which matches option A.