d Word Problem 2 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs
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Word Problem 2

Problem 2: At an election a candidate secures 40% of the votes but is defeated by the other candidate by a majority of 289 votes. Find the total number of votes recorded.

(word problems on job exam mathematics.)

Solution:

Let total votes = ‘x’

Let there be two candidates ‘A’ and ‘B’.

‘A’ gets 40% of total votes;

\Rightarrow A=\frac{40}{100}x

But ‘B’ defeated ‘A’ by 289 votes, so votes secured by ‘B’ should be equal to ‘A’s votes + 289

\Rightarrow B=\frac{40}{100}x+289

Now both ‘A’ and ‘B’ votes together should be equal to total recorded votes;

Total votes = A's Votes+ B's Votes

\Rightarrow x = \frac{40}{100}xs+ \frac{40}{100}x+289

\Rightarrow x = 2 \times \frac{40}{100}x+289

\Rightarrow x =  \frac{40}{50}x+289

\Rightarrow x -\frac{40}{50}x =  289

\Rightarrow \frac{50x-40x}{50} =  289

\Rightarrow \frac{10x}{50} =  289

\Rightarrow x=289 \times \frac{50}{10} \Rightarrow x=1445

\Rightarrow x=1445

 

 

           

Problem: In an examination75% of the candidates passed in English and 65% in Mathematics, while 15% failed in both subjects. If 495 candidates passed in both subjects, find the total number of candidates who took the exam.    

a. 8000            b. 850                   c. 800                d. 750

                                                                                                                                                                            

Solution:

 

Let there be  x=100% Candidates,   candidates,

\Rightarrow EngFail=100-75=25

and FailMaths=100-65=35

and Both Fail=15

Using formula

Total Fail=Fail Eng+Fail Math-Both

Total Fail=25+35-15=45

So,

Pass candidates = x-45 = 100-45=55% of ‘x’

But Pass candidates (as given) = 495

\Rightarrow 55% of x=495

\Rightarrow \frac{55}{100}\times x = 495

\Rightarrow x= 495 \times \frac{100}{55}

\Rightarrow x= 900

l black’, sans-serif;”>So, there were 900 candidates

 

 

 

 

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