Avg q3

[Saumyojit]

There are 3 classes having 20, 25 and 30 students respectively having average marks in
an examination as 20,25 and 30 respectively. The three classes are represented by A, B and C and you have the following information about the three classes.
a. In class A highest score is 22 and lowest score is 18
b. In class B highest score is 31 and lowest score is 23
c. In class C highest score is 33 and lowest score is 26


If five students are transferred from A to B, what can be said about the max possible average score of class B ?


No of students B has = 30 where 5 is from class A

Sum of the marks of each student of class B = 25 original students of class B * marks₁ + 5 students from class A * marks₂


marks2 will be 22 as the five students have already inherited the highest score of class A and then came to class B.

As highest score can be inherited by only one person but bcoz 22 is the highest score of class A not of B that why multiple students which in this case is 5 are obtaining multiple 22' s.


but what will be marks₁ ? i guess 25 marks as 25 students of class B cannot obtain class B highest score i.e 31 .

 

[Saumyojit]

we can find the average then by dividing sum of marks of class B by 30


Highest can be obtained by atmost one .
So when 5 students have been transferred to class B , it does not matter to class B whether multiple students are having highest score of class A i.e 22 as it is not highest score of class B .

so , (25 * 25 + 5 * 22 ) / 30 = 24.5 (AVG)

 

[Saumyojit]

@Dr.Peterson @JeffM

 

[Dr.Peterson]

we can find the average then by dividing sum of marks of class B by 30


Highest can be obtained by atmost one .
So when 5 students have been transferred to class B , it does not matter to class B whether multiple students are having highest score of class A i.e 22 as it is not highest score of class B .

so , (25 * 25 + 5 * 22 ) / 30 = 24.5 (AVG)

You've given an answer; are you saying you are not sure it's right, and you need someone else to convince you?

Your explanation could be clearer, but it is correct. (I don't know why you made the comment about multiple students, or why you think that at most one can have the highest score.)

There are 3 classes having 20, 25 and 30 students respectively having average marks in
an examination as 20,25 and 30 respectively. The three classes are represented by A, B and C and you have the following information about the three classes.
a. In class A highest score is 22 and lowest score is 18
b. In class B highest score is 31 and lowest score is 23
c. In class C highest score is 33 and lowest score is 26

If five students are transferred from A to B, what can be said about the max possible average score of class B ?

(Since there is more information here than you need, I assume there are other questions about the same situation.)

The highest possible scores for each of the 5 transferred from A is 22; this will result in the highest possible new average for B. The current total in B is 25*25 = 625; the new total will be 25*25 + 5*22 = 735. There will be 30 students, so the new average is 735/30 = 24.5.

Now, it is conceivable that in a question like this there couldn't actually be 5 students with the maximum score, while having the indicated average; for instance, this would happen if there had been only 6 in that class.

We can check this by finding the lowest possible average if 5 have the maximum 22 and the rest are no more than 18: if the other 15 all got 18, then the average would be (15*18 + 5*22)/20 = 19. That implies it is possible to get an average of 20 -- unless the scores can only be integers, and we can't choose them to get exactly 20!

To check that issue, consider that we need the total to be 20*20 = 400, so the other 15 must make a total of 400 - 5*22 = 290. This means they average 19 1/3; we can let 2/3 of them have 19, and the other 1/3 have 20. (This is a nice thing to know about averages.) That is, (10*19 + 5*20 + 5*22)/20 = 400/20 = 20. So it is in fact possible to have 5 with 22 and the rest between 18 and 22 inclusive.

That's probably more work than you were expected to do ...

 

[Saumyojit]

consider that we need the total to be 20*20 = 400, so the other 15 must make a total of 400 - 5*22 = 290. This means they average 19 1/3; we can let 2/3 of them have 19, and the other 1/3 have 20. ( This is a nice thing to know about averages) That is,

(10*19 + 5*20 + 5*22) /20 = 400/20 = 20.

we can let 2/3 of them have 19, and the other 1/3 have 20.
How did you came up with this from 19 1/3 ?

I understand that 2/3 rd of 15 is 10 and 1/3rd is 5 .

(10*19 + 5*20 + 5*22) / 20 = 400/20 = 20. So it is in fact possible to have 5 with 22

Yeah that means those EACH 5 students out of 20 student have actually literally got 22 and so the avg is 22 .

But highest and lowest means there can be one of each at most right .

Just an eg , if out of 4 workers 2 workers A and B got rs 500 and other two Rs 300 and Rs 100 each .

Who has the highest earning ? Will it be "cannot be determined " or A and B ?

What is the highest amount ? Will it be " cannot be determined " or 500 ?

 

[Dr.Peterson]

But highest and lowest means there can be one of each at most right .

Just an eg , if out of 4 workers 2 workers A and B got rs 500 and other two Rs 300 and Rs 100 each .

Who has the highest earning ? Will it be "cannot be determined " or A and B ?

What is the highest amount ? Will it be " cannot be determined " or 500 ?

I'm not sure whether you are misunderstanding the English, or the mathematical concept.

There will always be a highest score, regardless of how many have it.

The problem says "highest score is 22", not "the single highest-scoring person got 22". There is nothing at all here to suggest that only one person has that score.

In your example, obviously the highest amount is 500; and both A and B earn it. There is no one highest-earning person, but we could still say they share that role. But the problem doesn't say anything like that.

we can let 2/3 of them have 19, and the other 1/3 have 20.
How did you came up with this from 19 1/3 ?

I understand that 2/3 rd of 15 is 10 and 1/3rd is 5 .

This is the "nice thing to know": Weighted averages work just like balancing objects. If A has weight a and B has weight b, then the distances to the average (balance point) will have the reverse ratio as the weights, b:a. In this case, we have a 1:2 ratio of distances, (19 1/3 - 19) : (20-19 1/3) = 1/3 : 2/3 = 1:2; so the number with each score will be 2:1. I think of this as, "twice as many people at 19 pull the average twice as hard as the people at 20, so that the average will be half as far from 19 as from 20". I just reversed that thinking.

As another example, if your current grade is 70 for 4 exams, and your next exam has a grade of 90, then since the weights have a ratio of 4:1, the distances to the average will have the ratio 1:4, so it will be 1/5 of the way from 70 to 90. Alternatively, since the new grade is worth only 1/5 of the new total, it will pull the average only 1/5 of the way from 70 to 90, namely up 4 points to 74. Check it out: (4*70 + 1*90)/5 = 74. There are many kinds of problems for which this provides a convenient shortcut.

Of course, you don't have to do any of this; I just wanted to do a quick check, so I didn't use a long algebraic solution or trial and error.

 

[Saumyojit]

In your example, obviously the highest amount is 500; and both A and B earn it. There is no one highest-earning person, but we could still say they share that role. But the problem doesn't say anything like tha

http://cetking.com/wp-content/uploa...book-PDF-Ck-Quant-for-CAT-CET-other-exams.pdf

Please see q8 of Lod iii

There in the book it is given The highest earning Cannot be determined .





obviously the highest amount is 500 and

Who has the highest earning ? it "cannot be determined .

 

[Saumyojit]

(19 1/3 - 19) : (20-19 1/3)

: and ( written closely will print as Smily

 

[Dr.Peterson]

http://cetking.com/wp-content/uploa...book-PDF-Ck-Quant-for-CAT-CET-other-exams.pdf

Please see q8 of Lod iii

There in the book it is given The highest earning Cannot be determined .

obviously the highest amount is 500 and

Who has the highest earning ? it "cannot be determined .

I have no desire to try to figure out what this problem is saying, and why, if you're right, the answer is "cannot be determined". But (a) they do seem to allow answers like "Both B and D", contrary to your idea that there can't be more than one person with the highest value, and (b) this is not what the problem you were asking about asks, so it is irrelevant. There can always be more than one person with the highest score or earning, even if you can't say who.

Can you show a better reason for your belief? And are you willing to admit that you might be wrong? (Otherwise, why should I keep trying?)

 

[Saumyojit]

(25 * 25 + 5 * 22 ) / 30 = 24.5 (AVG)

Can you say to me why

Using highest score of class B for those 25 students and then evaluating
max possible average score of class B = ( 31 * 25 + 5 * 22 ) / 30 is Wrong ?

 

[Dr.Peterson]

Can you say to me why

Using highest score of class B for those 25 students and then evaluating
max possible average score of class B = ( 31 * 25 + 5 * 22 ) / 30 is Wrong ?

Are you saying you think that's right? WHY? You have the burden of proof.

What does 31 * 25 give you? What does the whole calculation mean?

 

[Saumyojit]

Yeah I understood where was I going wrong.
I cannot change the sum of marks of class B which is 625 for 25 students which is a fact.

 

[Saumyojit]

If A has weight a and B has weight b, then the distances to the average (balance point) will have the reverse ratio as the weights, b:a.

Quite fascinating .
When did you learn this?

Because the average looks nice when it is a integer so you subtracted from both ends .

 

[Dr.Peterson]

Quite fascinating .
When did you learn this?

I presumably learned it in physics, but probably several other times before I realized how many places it can be useful. That may have come from a colleague who used it in solving "mixture problems" without algebra; and it probably has a name, now that I think about it. Such problems used to be called "alligation", and maybe that name can be used for the method.

 

[Saumyojit]

(Since there is more information here than you need, I assume there are other questions about the same situation.)

yeah .



Follow up from OP.

Stuck in q12 .

A -> B = 5 people , B -> A = 5 people

A = 20 students original then 5 left A= 25 students

Sum of marks of A = 400

Sum of marks of B = 625

B = 25 students then 5 joined B = 30 students

Sum of marks of B after 5 joining = 25 * 25 + 20 * 5 = 725


As 5 people left from class A there will be a new Average of class A with 15 students which is represented by NewAvg

Sum of marks of A after 5 person left = ( NewAvg * 15 ) / 15


As 5 people left from class B there will be a new Average of class B with 20 students which is represented by Avgmarks

Sum of marks of B after independent set of 5 people left= Avgmarks * 20 + 20 * 5

Sum of marks of A after independent set of 5 person joined= ( Avgmarks * 5 + NewAvg * 15 )




then?

 

[Saumyojit]

yeah .

View attachment 30063

Follow up from OP.

Stuck in q12 .

A -> B = 5 people , B -> A = 5 people

A = 20 students original then 5 left A= 25 students

Sum of marks of A = 400

Sum of marks of B = 625

B = 25 students then 5 joined B = 30 students

Sum of marks of B after 5 joining = 25 * 25 + 20 * 5 = 725


As 5 people left from class A there will be a new Average of class A with 15 students which is represented by NewAvg

Sum of marks of A after 5 person left = ( NewAvg * 15 ) / 15


As 5 people left from class B there will be a new Average of class B with 20 students which is represented by Avgmarks

Sum of marks of B after independent set of 5 people left= Avgmarks * 20 + 20 * 5

Sum of marks of A after independent set of 5 person joined= ( Avgmarks * 5 + NewAvg * 15 )




then?

many things I have written are wrong although the approach is same .

 

[Saumyojit]

A -> B = 5 people , B -> A = 5 people

A = 20 students original then 5 left

A= 15 students

Sum of marks of A = 400

Sum of marks of B = 625

B = 25 students then 5 joined B = 30 students

Avg is the average marks of 5 transferred students from class A.


Sum of marks of 30 students of B after 5 joining = 25 * 25 + Avg * 5 =


As 5 people left from class A there will be a new Average of class A with 15 students which is represented by NewAvg

Sum of marks of A after 5 person left = NewAvg * 15


Sum of marks of 30 students in class B =
( 25*25) + ( Avg * 5)


Avg2 = Average marks per student out of 25 students in B class after 5 students left.

Sum of marks of B after independent set of 5 people left= Avg2 * 25


Avg3 is the average marks per student of 20 original students of class B

Avg2 * 25 = Avg3 * 20 + Avg * 5



A1 is the average marks of transferred 5 students from B

Sum of marks of class A after independent set of 5 person joined=
( A1* 5 + NewAvg * 15 )


Now this is rectified and corrected




then?

 

[Dr.Peterson]

I don't have time now to look through any details, but if I were you, I would first try to find what it would take for A to increase and B decrease, and so on, to see if any combinations seem impossible. I would not be introducing lots of variables.

 

[Saumyojit]

Nothing is coming

 

[Saumyojit]

@Dr.Peterson

Testing with two cases

Case 1 : When highest score of Transferred class is Taken

A's Average after 5 gone = 19 1/3

When 5 students each with highest score of 22 has transferred from class A

B's Average becomes (30students) = 24.5


When 5 new students are transferred from B to A with high score of 31 of Class b

Then B's average becomes 29

Avg * 20 + 31 * 5 = 735

Avg = 29


And Average marks of class A=
(155 + 290) / 20 = 17.8

So , Average of B will decrease when avg of class A decreases
But B's average increases from previous after second operation and then A's average decreases.

Case 2: When lowest score of Transferred class is Taken i.e 23 of B and 18 of A

Same above process is done

A's avg increases then B's avg decreases .
B's average increases from previous then A's average increases from previous