A student of eighth grade has to answer three sections, A, B and C, i

Question

A student of eighth grade has to answer three sections, A, B and C, in the final examination. There are a total of 20 questions and the maximum marks a student can score is 30. The following is the additional information known -

Section A has at least 50% of the total questions and each question has weightage of one marks. 
 Section B has at least 30% of the total questions and each question has weightage of two marks. 
 Section C has at least 5% of the total questions and each question has weightage of three marks.

Select for Maximum contribution the maximum score that any section can contribute to total score, and select for Minimum A the minimum possible number of questions that section A can have.

Aryaa03 · Answer

IMO - Maximum Contibution - 16 
         Minimum A = 11

A + B + C = 20
1) Section A has at least 50% of the total questions and each question has weightage of one marks
    ==> A is greater than or equal to 10 
2) Section B has at least 30% of the total questions and each question has weightage of two marks.
    ==> B is greater than or equal to 6
3) Section C has at least 5% of the total questions and each question has weightage of three marks.
    ==> C is greater than or equal to 1

ii) Minimum 
A*1 + B*2 + C*3 = 30 
Say A = 10 (min. as per stem);
2B = 12 , 14, 16 .. for B=6, 7, 8..
3C = 3, 6 ..for C  = 1, 2..
10*1 + 7*2 + 2*3 = 30 (Sum condition is met) but A+B+C = 10+7+2 = 19 (Invalid Case)

If A=11, B=8 and C = 1
11*1 + 8*2 + 1*3 = 30 (Sum condition is met) and also, A+B+C = 20 - Valid

i) Maximum Contibution:   If B = 8, the maximum contibution is 16.

Catman · Answer

A+B+C=20

10A+2B+3C<=30

Min value of A (atleast 50% of total questions ) = 10
Min value of B (atleast  30% of total questions)= 6  
Min value of C ( atleast 5% of total questions)= 1

Starting from Minimum value of A, A=10,  B>=6 and C>=1
(10,9,1) Not possible as Total Score >30
(10,8,2) Not possible as Total Score >30
(10,7,3) Not possible as Total Score >30
(10,6,4) Not possible as Total Score >30

Minimum Value of A can't be 10.

Possible case: (A,B,C)= (11,8,1)

When (A,B,C)= (11,8,1)  (Minimum A=11) 
Contribution by A = 11 *1=11
 Contribution by B (Maximum ) = 8*2= 16 
Contribution by C = 3*1 = 3

Iamriya · Answer

How do we know that it Min A is not a separate subquestion and part of same question where total marks from one section has to be maximised? Can anyone help me understand this?

ShubhadeepB · Answer

Algebraic approach works the best for such questions. If x, y, and z are the number of questions for sections A, B, and C, respectively, then the constraint on the total number of questions can be represented by the equation: x + y + z = 20 .......... (1)

The constraint on the total number of marks can be represented by the equation: x + 2y + 3z = 30 ............ (2).

Subtracting (1) from (2) implies y + 2z = 10 .......... (3).

There could be two possible cases satisfying the values of x, y, and z, given the constraints.

Case 1: y = 6, z = 2, and x = 12
Case 2: y = 8, z = 1, and x = 11

So, the minimum number of questions for section A will be 11 in case 2. The maximum contribution of a section to the score also arises in case 2 where section B contributes 16 out of the total 30.

arushi118 · Answer

we know:
equation 1: a+2b+3c=30 and
equation 2: a+b+c=20

Restrictions:
a>=10
b>=6
c>=1

Subtracting Equation 1 and 2: 
b+2c=10
6,2 -> a=12
8,1 -> a=11

So max contribution will be from b when a=11, b=8, and c=1. This contribution will be 16.

And minimum value that "a" can take is 11 as can be seen.

Dereno · Answer

The three sections are given as A,B, and C respectively.

There are a total of 20 questions and the maximum marks a student can score is 30.

The following is the additional information known -

Section A has at least 50% of the total questions and each question has weightage of one marks. 
 Section B has at least 30% of the total questions and each question has weightage of two marks. 
 Section C has at least 5% of the total questions and each question has weightage of three marks.

Based on the above info,

Section A = at least 10 questions , with a weightage of 1 mark.

Section B = Atleast 6 questions , with weightage of 2 marks.

Section C = Atleast 1 question , with a weightage of 3 marks.

What does this minimum condition provide :

Total questions = 10+6+1 = 17 questions. 
 
Marks. (10*1) + (6*2) + (1*3) = 25 marks.

After we have satisfied the basic conditions, What’s Available at hand = 3 Questions and 5 marks.

How can we get 5 marks from three questions.

case 1: (2,0,3) wrt questions (2,0,1) 
Case 2: (1,4,0) wrt questions ( 1,2,0)

Maximum score contribution by a section   can be :

Sec A = 10 questions = 10 marks

Sec B = 6 questions = 12 marks

Sec C = 1 question = 3 marks.

If we adopt the case 2, we get two more questions in sec B,  8 questions contributing 16 marks.

Minimum question section A   can have :

case 2:  11 questions for sec A , 8 questions for sec B, and 1 question for sec C.

= (11*1)+(8*2)+(1*3)

= 11+16+3

= 30 marks.

11 questions for section A .

A student of eighth grade has to answer three sections, A, B and C, i

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A student of eighth grade has to answer three sections, A, B and C, i