Latest TCS Aptitude Question SOLUTION: in an exam the maximum mark for each of 3 subject is 50 and the maximum mark for the forth subject is 100 .find the number of ways in which a student can score 60% marks in aggreg

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in an exam the maximum mark for each of 3 subject is 50 and the maximum mark for the forth subject is 100 .find the number of ways in which a student can score 60% marks in aggregate.

Asked In TCS aditi kumari (9 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

sagar  (9 years ago)

3C3 or (1C1 and 3C1)
=3C3+1C1*3C1
=1+3
=4 ways

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

Richa  (9 years ago)

We need to distribute 60% marks ie. 150marks in 4 sections namely A,B,C,D.
A+B+C+D=150

Let 'a' be the marks that he loses in section A.. ie. (Marks obtained)+(Marks lost)=50 in section A
Implies.. a+A=50
a=50-A
Similarly, b=50-B
c=50-C
d=100-D
Now, A+B+C+D=150
»(50-a)+(50-b)+(50-c)+(100-d)=150
»a+b+c+d=100

We need to distribute 100identical marks among 4 variables..
(100+4-1)C(4-1)= 103!/100! 3!

But this figure involves the cases when 'a' attains values like 51,52,53...up to 100.
But this is not possible as section A can have maximum 50 marks.

So, subtracting cases in which a=51,52,53....100

Let a=51+ã, where ã can be any value starting from zero. So that we will get a=51+0 or a=51+1 or a=51+2... And so on.. All the values of 'a' will be covered.

We know that, a+b+c+d=100
»(51+ã)+b+c+d=100
ȋ+b+c+d=49

Distribution of 49 among 4 variables= (49+4-1)C(4-1)= 52!/49! 3!

Here we have considered 'a' to be greater than 50.. Same goes when 'b' & 'c' will be greater than 50.
Therefore multiplying above by 3.
= 3x (52!/49! 3!)

Final ans.=103!/100! 3! - 3x(52!/49!3!)
=176851- 66300 = 110551

Is this solution Helpfull? Yes (2) | No (6)  

Shivam Gupta  (9 years ago)

LET THE SCORE BE
a/50
b/50
c/50
d/100
max marks=250
60% 0f 250=150
ie. a+b+c+d =150
ie 153C3 is the answer

Is this solution Helpfull? Yes (1) | No (17)  

Aguru Srinivas  (5 years ago)

please provide correct solution

Is this solution Helpfull? Yes | No   

Pavani  (3 years ago)

Correct option is
B
110551
The candidate must score 150 marks.
∴ Required number
=coefficient of x
150
in (1+x+...+x
50
)
3
(1+x+...+x
100
)
= coefficient of x
150
in (
1−x
1−x
51

​
)
3

1−x
1−x
101

​

=coefficient of x
150
in (1−x
51
)
3
(1−x
101
)(1−x)
−4

=coefficient of x
150
in (1−3x
51
+3x
102
−x
153
)(1−x
101
)(1−x)
−4

[leaving terms containing powers of x greater than 150]
=coefficient of x
150
in (1−x)
−4
−3.
=coefficient of x
99
in (1−x)
−4
+3coefficient of x
48
in (1−x)
−4
coefficient of x
49
in(1−x)
−4

=
153
C
150
​
−3.
102
​
C
99
​
+3.
51
C
48
​
−
52
C
49
​

=
6
153.152.151
​
−3.
6
102.101.100
​
+3.
6
51.50.49
​
−
6
52.51.50
​

=110551

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