Latest L&t Aptitude Question SOLUTION: Q. Find the least number of candidates in an examination so that the percentage of successful candidates should be 76.8%:

• Placement Papers
• /
• L&t
• /
• Numerical Ability
• /
• Percentage

L&t Question Papers: L&T Company Profile

(#M40032132) L&T Numerical Ability question Keep an EYE

Q. Find the least number of candidates in an examination so that the percentage of successful candidates should be 76.8%:

Asked In L&t K.Shyam Prasad Rao (10 years ago)

Unsolved Read Solution (8)

Is this Puzzle helpful?   (88)   (32) Submit Your Solution Percentage

Read Solution (8) :

Nagendra Singh Parmar  (10 years ago)

suppose no. of total candidates are 'y' and successful are 'x'.
(x/y)*(100)=76.8 or x/y=0.768
or y=x/0.768
here we need minimum value of y, which is obtained by minimum value of x.
if put x=768 then y=1000
now we will try decrease value of x than we will find less value of y. so factorized 768 by 2 further.
if put x=384 then y=500
if put x=192 then y=250
if put x=96 then y=125(answer)
if put x=48 then y=62.5(not integer)
so ans is 125(minimum)

Is this solution Helpfull? Yes (30) | No (1)  

saranya sankar  (10 years ago)

answer is 125

bcoz...
No. of successful candidates = 76.8% of x
x = total students =(768/(10*100))*x = (96/125)x
Which must be a whole no.
The reqd least no. = 125

Is this solution Helpfull? Yes (25) | No (12)  

umang agarwal  (10 years ago)

125.
Let x be the total students.
76.8x/100 = integer
It should be integer.
(384/5)x/100 = 96x/125
From this we get x should be min value 125.

Is this solution Helpfull? Yes (4) | No (9)  

ankit thakur  (8 years ago)

76.8----------->100(suppose out of 100 students 76.8 students are passed )
so multiply both by 5 to make it a whole number
384------------>500
and now divide the both no. by 2
so we get 96----------->125

Is this solution Helpfull? Yes (4) | No   

varsha  (10 years ago)

let total candidate is x
no of successful candidate 76.8%
pass candidate is y
(x/y)*(100)=76.8 or x/y=0.768
or y=x/0.768
here we need minimum value of y, which is obtained by minimum value of x.
if put x=768 then y=1000
now we will try decrease value of x than we will find less value of y. so factorized 768 by 2 further.
put x=384 then y=500
put x=192 then y=250
put x=96 then y=125(answer)
put x=48 then y=62.5
62.5 is not integer
so answer is 125

Is this solution Helpfull? Yes (1) | No   

Bharat Jain  (10 years ago)

100 will be

Is this solution Helpfull? Yes | No (6)  

Chirag Ranjan Lahiri  (10 years ago)

125.
Let x be the total students.
76.8x/100 = integer
It should be integer.
(384/5)x/100 = 96x/125
From this we get x should be min value 125.

Is this solution Helpfull? Yes | No   

sriraghavan  (9 years ago)

success percentage is 76.8,so multiply in by 5 to get whole number then the least number of candidates in 384

Is this solution Helpfull? Yes | No (3)  

Please Login to Read Solution.