Math, asked by eswarchaitanya1591, 1 year ago

Bhamsingh and sukhalal contested the assembly election. bhimsingh won the election by 2000 votes,securing 55% of the total votes. 10% of the total voters did not votes. 26. how maney votes were cast in favour of bhimsing?

Answers

Answered by santy2
4

Let the total votes cast be x.

The votes to Bahmsingh = 0.55x

The votes to Sukhalal = 0.45x

0.55x = 0.45x + 2000

0.55x - 0.45x = 2000

0.1x = 2000

x = 20000

The total who cast their votes were 20000

=20000

= 20000 people

Those who voted for Bahmsingh are :

0.55 × 20000 = 11000

= 11000 voters.

Answered by shewanipradhan11
0

Answer:

Step-by-step explanation

Let the total number of votes cast be X

Votes Bhimsingh secured =55%X

Votes Sukhalal secured= 35%X

(10% did not vote so )

55%X=35%X+2000

0.55X=0.35X+2000

0.20X=2000

X=10000

VOTES IN FAVOUR OF BHIMSINGH= 55%X

=55%*10000

=5500

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