Latest TCS Aptitude Question SOLUTION: Total 100 members are writing exam. In the 48 members are writing first exam. 45 members are writing second exam. 38 members are writing third exam. 5 members are writing all the t

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Total 100 members are writing exam. In the 48 members are writing first exam. 45 members are writing second exam. 38 members are writing third exam. 5 members are writing all the three exams. How many members are writing 2 exams?

Asked In TCS Sneha Singh (8 years ago)

Unsolved Read Solution (18)

Is this Puzzle helpful?   (23)   (6) Submit Your Solution Permutation and Combination

Read Solution (18) :

Yashi Das  (8 years ago)

Total number of exams written by 100 students = 48 + 45 + 38 = 131
Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.
Therefore, x + 2y + 3z = 131 also x + y + z = 100.
Given that z = 5. So x + 2y = 116 and x + y = 95.
Solving we get y = 21.
So 21 members are writing exactly 2 exams.

Is this solution Helpfull? Yes (55) | No (5)  

komal kumari  (8 years ago)

n(1)=48
n(2)=45
n(3)=38
n(1U2U3)=100
total appered in the exam= 48+45+38=131
and 5 is given all three exam so,
131-5=126
100+x=126
=>x=26
so, 26 is answer

Is this solution Helpfull? Yes (9) | No (5)  

Ayanendu Banerjee  (8 years ago)

For a group of three sets:
1. n(A cup B cup C) = n(A) + n(B) + n(C) - n(A cap B) - n(A cap C) - n(B cap C) + n(A cap B cap C)
now the required answer for this problem lies in the n(A cap B) - n(A cap C) - n(B cap C) part
the answer coming out is 36
but it is taken thrice
so the answer is 12

Is this solution Helpfull? Yes (5) | No (4)  

srimanth  (8 years ago)

correct solution is
Total number of exams written by 100 students = 48 + 45 + 38 = 131
Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.
Therefore, x + 2y + 3z = 131 also x + y + z = 100.
Given that z = 5. So x + 2y = 116 and x + y = 95.
Solving we get y = 21.
So 21 members are writing exactly 2 exams.

Is this solution Helpfull? Yes (4) | No (3)  

Deepika Jain  (8 years ago)

total members writing exam=100
total exams written=131+5 (who r writing all three exams)
members writing two exams=131-100=31
ans=31

Is this solution Helpfull? Yes (2) | No (7)  

RENUKA MADARI  (8 years ago)

100=n(f)+n(s)+n(t)-n(f inters)-n(s inter t)-n(t inter f)+n(f inter s inter t)
=48+45+38-3(inter)+5
3(inter)=136-100=36
inter=12

Is this solution Helpfull? Yes (2) | No (2)  

Sunny Grover  (8 years ago)

Answer is 16

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

s.sambasivan  (8 years ago)

I+II+III exam = 48 + 45 + 38 = 131. All the three exams written by 5. That means one or two written by 131 - (5*3) = 116. Total members 100. So members writing two exams = 116-100 =16.

Is this solution Helpfull? Yes (1) | No (3)  

navi chinnu  (7 years ago)

21 is the correct ans

Is this solution Helpfull? Yes (1) | No   

mohit  (8 years ago)

plz give me proper solution

Is this solution Helpfull? Yes | No (1)  

spoorthi  (8 years ago)

plz gie corrent ans

Is this solution Helpfull? Yes | No (1)  

vinit kishor singh  (8 years ago)

48+45+38+5-100=35

Is this solution Helpfull? Yes | No (2)  

Arnab Bose  (8 years ago)

For a group of three sets:
1. n(A cup B cup C) = n(A) + n(B) + n(C) - n(A cap B) - n(A cap C) - n(B cap C) + n(A cap B cap C)
now the required answer for this problem lies in the n(A cap B) - n(A cap C) - n(B cap C) part
the answer coming out is 36
but it is taken THRICE
so the answer is 36/3=12

Is this solution Helpfull? Yes | No (2)  

varsha jaiswal  (8 years ago)

total no. of exams written=48+45+38=131
member are writing all three exam
therefore, total exam written by five members=5*3=15
left exams=131-15= 16 exams
16 people are writing 2 exams

Is this solution Helpfull? Yes | No (4)  

VARMA  (8 years ago)

100 = 48+38+45 - (our solution) + 5
our solution = 136-100 = 36.

Is this solution Helpfull? Yes | No (3)  

Tejaswini  (8 years ago)

let X be number of persons writing first & second exam,
Y be the no. of persons writing second & third exam,
Z be no. of persons writing third & first exam
Given 5 members are writing all the 3 exams.
so, members who are writing first exam can be:
X+Y+5=48 --> (1)
likewise,
for 2nd & 3rd:
X+Z+5=45 -->(2)
Y+Z+5=38 -->(3)
solving 1,2,3 eq's, we get
X=25, Y=18 , Z=15
Total members who r writing 2 exams are:

X+Y+Z=58

Is this solution Helpfull? Yes | No (3)  

vishakha roman  (3 years ago)

21 is correct answer @yashi das good explanitaion

Is this solution Helpfull? Yes | No   

Vamsi  (9 Months ago)

Total number of exams written by 100 students = 48 + 45 + 38 = 131
Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.
Therefore, x + 2y + 3z = 131 also x + y + z = 100.
Given that z = 5. So x + 2y = 116 and x + y = 95.
Solving we get y = 21.
So 21 members are writing exactly 2 exams.

Is this solution Helpfull? Yes | No   

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