Question

Question on Average

Answer 1

hi

Answer 2

There are 3 classes A ,B and C having 20, 25 and 30 students
respectively having avg marks 20 ,25 and 30.

A : highest score 22, lowest 18

B: highest 31, lowest 23

C: highest 33, lowest 26

5 students are are transferred from C to B,
then 5 students are transferred from B to A,
then 5 students are transferred from from A to B,
and last 5 students are transferred from B to C

if we set the highest possible average of class C
as the primary objective and the secondary objective
is to achieve minimum possible average of class B.

then the minimum possible average of class B is ?

a.) 22.2
b.) 23
c.) 22.6
d.) 22

Answer 3

My answer


\(\large\tt \begin{align} \color{black}{B_{avg(min)}=\dfrac{625-31\times 5+26\times 5 -31\times 5+18\times 5}{25}\\~\\
~~~~~~~~~~~~~~~ =21.4\\
\text{but thats not given in the options}}\end{align}\)

Answer 4

D

Answer 5

@amistre64

Answer 6

@phi

Answer 7

the primary objective is to maximize C's avg. What did you get for C ?

Answer 8

for c_max=(900-26*5+31*5)/30=30.83

Answer 9

I am thinking that we have to keep track of B's average at each step.
C to B,
B to A, what is B's ave after this step ?
A to B,
B to C

Answer 10

b_(after cmax)=(625-31*5+26*5)/25=24

Answer 11

B cannot have more than 6 entries with the value of 31
otherwise it's not possible to get an average of 25: (7*31+ 18*23)/25 > 25

we have to "save" 5 of the 31's in B for the last step (moving them to C to maximize C's average)

that means we can move 1 31 and 4 others to A in step 2
I am not sure what the 4 others are allowed to be ?

Answer 12

oh ok

Answer 13

but wolfram shows their can be 12 values of 31

http://www.wolframalpha.com/input/?i=solve+31x%2B23y%3D625%2Cx%5Cge+0%2Cy%5Cge+0+over+integers

Answer 14

but that adds up to 11+12 = 23
we need 25, so add another constraint x+y= 25 (students)

Answer 15

oh ,sry u r right

Answer 16

so I think (if we want to maximize C) we need B to have
5 31's , 15 23's to balance and 5 more values that average to 25

Answer 17

there can 5*25 more

Answer 18

B has 5 31's , 15 23's , 5 that average to 25. I think we should use 5 25's
OK, here is what I came up with
C-> B 5 smallest from C into B +5*26 added to B
B-> A "second" biggest set of 5 to A. We are saving the 5 31s for C
-5*26 from B (all of the other values in B are 25 or smaller)
A->B move 5 smallest from A to B. add 5*18
B->C move 5*31 from B to C -5*31 from B

so we have for B
625 +5*26 - 5*26 + 5*18 - 5*31
625 - 5*13
625 - 65
560

average is 560/25 = 22.4

Answer 19

which does not match your choices. So close!

Answer 20

book gives 22.6

Answer 21

there can also be 6*31,23*7,24*2 for B

http://www.wolframalpha.com/input/?i=solve+31x%2B23y%2B24z%3D625%2Cx%2By%2Bz%3D25%2Cx%5Cgeq+0%2Cy%5Cgeq+0%2Cz%5Cgeq+0+over+integers

Answer 22

typo*23 *17

Answer 23

can we do a backword method as we know answer is 22.6

Answer 24

*backward

Answer 25

I was looking at B:
6 31's 1 25 18 23's
average 25, total number 25

Answer 26

C->B + 5*26 to B
B->A -31 + -26*4 (leaves 5 31's for C in the last step)
A->B +5*18
B->C -5*31

average is (625 +5*26 -31 - 4*26 +5*18 -5*31)/25

Answer 27

oh yess!!

that gives 22.6!

(600-31-25-23*3+18*5)/25=22.6

Answer 28

I got 22.2

Answer 29

http://www.wolframalpha.com/input/?i=%28600-31-25-23*3%2B18*5%29%2F25

Answer 30

but C->B is the first step, so you have some 26's in B that you can move out in the second step.