Latest Wipro Aptitude Question SOLUTION: There are two sections in a question paper each contain five questions. A student has to answer 6 questions. Maximum no. of questions that can be answered from any section is 4. Ho

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There are two sections in a question paper each contain five questions. A student has to answer 6 questions. Maximum no. of questions that can be answered from any section is 4. How many ways he can attempt the paper?

(a) 50
(b) 100
(c) 120
(d) 200

Asked In Wipro K BASHA (12 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

Pargat Pal Singh  (12 years ago)

If the max number from each section is 4 and a total of 6 questions must be answered then the possible ways to answer are:

4 from A and 2 from B

or 2 from A and 4 from B

or 3 from both A and B

Number of ways of choosing 4 from part A AND 2 from part B = 5C4 * 5C2 = 50 ... (AND means multiply)

Number of ways of choosing 2 from part A AND 4 from part B = 5C2 * 5C4 = 50

Number of ways of choosing 3 from part A AND 3 from part B = 5C3 * 5C3 = 100


So total number of ways to answer the paper = 50 + 50 + 100 = 200
........ option (d)

Is this solution Helpfull? Yes (38) | No   

Reck  (12 years ago)

5C4*5C2=50
5C3*5C3=100
5C2*5C4=50
50+100+50=200

Is this solution Helpfull? Yes (7) | No   

Sulekha  (12 years ago)

(d) 200 should be correct.

Is this solution Helpfull? Yes (3) | No   

IQBAL MEHDI  (12 years ago)

(d)200 ans.
sol:-
Student can choose from the two sections as:
s1(5) s2(5)
----- -----
2 4
3 3
4 2
hence,p(E)=5C2*5C4+5C3*5C3+5C4*5C2=50+100+50=200 ANS.

Is this solution Helpfull? Yes (3) | No   

magesh  (12 years ago)

5c4+5c2=65.
5c3+5c3=40.
5c2+5c4=65.
65+40+65=170.
5c1=120.
therefore the prob. of attempting is 170-120=50.

Is this solution Helpfull? Yes | No (16)  

RAKESH KUMAR  (9 years ago)

4 from A and 2 from B

or 2 from A and 4 from B

or 3 from both A and B

Number of ways of choosing 4 from part A AND 2 from part B = 5C4 * 5C2 = 50 ... (AND means multiply)

Number of ways of choosing 2 from part A AND 4 from part B = 5C2 * 5C4 = 50

Number of ways of choosing 3 from part A AND 3 from part B = 5C3 * 5C3 = 100


So total number of ways to answer the paper = 50 + 50 + 100 = 200 → choice (d)

Is this solution Helpfull? Yes | No (1)  

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