Question

In an election,there were two candidates A and B. Out of all the voters in the constituency, only 80% were polled and out of the votes polled. A got 56% votes and own the election by 21120 votes.Find:
(i)the total number of voters in the constituency
(ii)the number of votes polled
(iii)the number of votes cast favour of A
(iv)the number of votes cast in favour of B

Answer 1

Let the total number of votes be 'x'.

then polled votes = 80% of x = 80x/100.

Number of votes A got = 56% of polled votes = 56% of 80x/100

Number of votes B got = (100-56)% of polled votes = 44% of 80x/100

thus A won by (56-44)% of 80x/100 = 12% of 80x/100

and it is given in answer that

12*80*x/100*100 = 21120

from here we get1.) x = 220000= total votes.

2.) number of votes polled = 80x/100 = 176000

3.) A got = 56× 17600/100 = 98560 votes

4.) B got = 44× 17600/100 = 77440 votes.

p.s- there might be calculation mistakes