Question

7. In an election only two candidates participate,
Candidate 'P' got 50% less votes than 'Q. Had o
got 200 votes less there would have been a tie.
What is the 8 times of the number of total votes
polled.
(a) 800 (b) 7200 (c) 3200
(d) 9600
(e) 3600​

Answer 1

Let Q have 2x votes.

Then P have x no. of votes

According to questions.

2x-200=x+200

X=400

Then total votes (P+Q)= 2x+x

= 3*400

= 1200

8 times of 1200 = 9600

Answer 2

The 8 times the number of total votes polled is (d) 9600 votes.

Given: There are only 2 candidates P and Q in an election. Candidate 'P' got 50% fewer votes than 'Q. Had 'Q' got 200 votes less there would have been a tie.

To Find: The 8 times the number of total votes polled.

Solution:

• We should keep in mind that no votes polled were wasted. So, if any candidate gets a certain amount of votes it means the other candidate has gained those votes.

Coming to the numerical,

Let the number of votes 'Q' gets be x.

So, the number of votes 'P' gets = x - ( 50x / 100 )

                                                      = x/2

According to the problem, it is said that, if candidate 'Q' got 200 votes less there would have been a tie. So, it means that the 200 votes lost by Q shall be gained by P. Framing an equation, we get;

       x - 200 = ( x/2 ) + 200

  ⇒ x - x/2 = 400

  ⇒ x = 800

So, votes received by candidate 'Q' = 800

And, votes received by candidate 'P' = 800/2 = 400

The total votes polled = 800 + 400

                                     = 1200 votes

So, 8 times the number of total votes polled = 8 × 1200 votes

                                                                               = 9600 votes.

Hence, the 8 times the number of total votes polled is (d) 9600 votes.

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