In an election the supporters of two candidates
A
and
B
were taken to polling booth in two different vehicles, capable of carrying 10 and 15 voters respectively. At least 90 vehicles were required to carry a total of 1200 voters, Given that candidate
B
won, what is the maximum difference in the number of votes?
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well, lets begin by assigning variables to number of vehicles for A and B.
we then have 1 0 x + 1 5 y = 1 2 0 0 also, ⌈ x ⌉ + ⌈ y ⌉ ≥ 9 0 we use ceiling function for x and y as there is a possibility that y takes decimal values and number of vehicles can't be decimals.
so for maximum votes in favour of B, we need to have maximum value for variable y
trying out x = 3 we get y = 7 8 but x + y = 8 1 which is < 9 0
trying out x = 6 we get y = 7 6 but x + y = 8 2 which is < 9 0
from trying the two sets above, we note that for every increase in x by 3 there is overall increase in number of vehicles by 1
hence, for when x = 3 0 we get y = 6 0 and we satisfy x + y ≥ 9 0
Thus, number of supporters of A = 3 0 0 and B = 9 0 0 giving us the difference 6 0 0
Hey yo.
for total number of vehicles,
A + B = 90
A = 90 - B (1st)
for total number of voters,
10A + 15B = 1200 (2nd)
substitute (1st) into (2nd),
10(90 -B) + 15B = 1200
5B = 300
B = 60
A = 90 - 60 = 30
A has 30 vehicles while B has 60 vehicles,
A has (30 x 10) = 300 votes,
B has (60 x 15) = 900 votes,
The votes difference as B won the election = 900 - 300 = 600,
Thanks....
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A + B = 90 and 10A + 15B = 1200 solving, A = 30, B = 60
10A = 300 and 15B = 900 won by 900 – 300 = 600