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manikant

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· 1 Months ago

Mani (Question: 127796.0 )

why is not the extra 5mems added to the total students? Question asked is no of students appeared....as given in the question, 5 more students are added and these should be included in the final list. Plz explain.

Question:

In an examination, the number of those who passed and the number of those who failed were in the ratio 25 : 4. If five more had appeared and the number of failures was 2 less than earlier, the ratio of passers to failures would have been 22 : 3. The number of students who appeared at the examination is:

Options:
A)

145

B)

150

C)

155 

D)

180

E)

None of these

Solution:

Let the number of passed student and that of failed students be 25x and 4x respectively.

According to the question, if 5 more had appeared i.e. 25x + 4x + 5

= 29x + 5, number of failure was 2

less i.e. 4x - 2, then  passedfailures   =  223   

Passed= appeared - failed

(29x+5)−(4x−2)4x−2    =  223   

⇒  29x−4x+5+24x−2    =  223   

⇒  25x+74x−2  =  223   

⇒ 75x + 21 = 88x – 44

⇒ 21 + 44 = 88x – 75x 

⇒ 13x = 65

x = 5

No. of Students who appeared = 29x 

= 29 × 5

= 145 

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