"> Word Problems 9 | Job Exam Rare Mathematics (JEMrare)

Word Problems 9

Problem 9: In an election one candidate secures 43% of total votes and is rejected by his competitor who takes a lead of 420 votes. What is the total number of votes ?

a) 1200                 b) 1300                       c) 1710               d) 3000



Solution:

Lets suppose,

Total Votes = x

Further we suppose winning candidate is B and rejected candidate is A. So as per statement of question, A gets 43% of total votes i-e,

A= \frac{43}{100}x

and B takes lead of 420 votes, i-e he takes 420 votes more than A, so it simply means that B first takes as much votes as taken by A and additionally he takes more 420 votes. so we can write;

B= \frac{43}{100}x + 420

But total votes were x, which were divided between A & B, so sum of their votes should be equal to x i-e,

x= A+B

Putting values of A and B, we get;

x= \frac{43}{100}x+ \frac{43}{100}x+420

=2( \frac{43}{100}x)+420

shift ‘x’ terms to one side of ‘=’ sign;

= \frac{43}{50}x+420

x- \frac{43}{50}x=420

we get

\frac{7x}{50}=420

then,

x= 420\times\frac{50}{7}

x= 60\times50=3000

So total number of votes was  3000

 

 

 

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