Skip to main contentSkip to solution

An agency entrusted to accredit colleges looks at four parameters: faculty quality (F), reputation (R), placement quality (P), and infrastructure (I). The four parameters are used to arrive at an overall score, which the agency uses to give an accreditation to the colleges. In each parameter, there are five possible letter grades given, each carrying certain points: A (50 points), B (40 points), C (30 points), D (20 points), and F (0 points). The overall score for a college is the weighted sum of the points scored in the four parameters. The weights of the parameters are 0.1, 0.2, 0.3 and 0.4 in some order, but the order is not disclosed.

Accreditation is awarded based on the following scheme:

RangeAccreditation
Overall score ≥ 45AAA
35 ≤ Overall score < 45BAA
25 ≤ Overall score < 35BBA
15 ≤ Overall score < 25BBB
Overall score < 15Junk

Eight colleges apply for accreditation, and receive the following grades in the four parameters (F, R, P, and I):

FRPI
A-oneAAAB
Best EdBCDD
CosmopolitanBDDC
DominanceDDBC
Education AidAABA
FancyAABB
GlobalCFDD
High QCDDB

It is further known that in terms of overall scores:

  1. High Q is better than Best Ed;
  2. Best Ed is better than Cosmopolitan; and
  3. Education Aid is better than A-one.

What is the highest overall score among the eight colleges?

Entered answer:

Solution

✅ Correct Answer: 48
Slide 1/11

Understanding the set :-

This is a weighted ranking deduction puzzle.

  1. We are told:

a) The grades convert to points.

b) Weights are fixed but in unknown order.

c) The ranking conditions tell us relative total scores.

  1. Parameters and their weights

Parameters: F, R, P, I.

Each parameter has grades:

A = 50 points

B = 40 points

C = 30 points

D = 20 points

F = 0 points

Weights (one for each parameter, but unknown order): 0.1, 0.2, 0.3, 0.4

The overall score =

(points in F × weight_F) + (points in R × weight_R) + (points in P × weight_P) + (points in I × weight_I)

  1. Also, :

a) Data table 1 = scoring scheme (points per grade & weights).

b) Data table 2 = input data (college grades).

c) Extra statements = constraints that help you deduce the missing weight order and rankings.

d) Output = deduced ranking + accreditation categories.

Now, lets try solving the set using the following information.

CollegeFRPIOverall scoreAccreditation
A-one50505040
Best Ed40302020
Cosmopolitan40202030
Dominance20202030
Education Aid50504050
Fancy50504040
Global3002020
High Q30202040

We can make the following table from the set.

Also, We know, A (50 points), B (40 points), C (30 points), D (20 points), and F (0 points).

we can directly fill this information from the table given in the set.

CollegeFRPIOverall scoreAccreditation
A-one50505040
Best Ed40302020
Cosmopolitan40202030
Dominance20202030
Education Aid50504050
Fancy50504040
Global3002020
High Q30202040

Clue 1, 2 and 3 says,

High Q > Best Ed > Cosmopolitan and Education Aid > A-one

Therefore, We can say that High Q > Cosmopolitan

CollegeFRPIOverall scoreAccreditation
A-one50505040
Best Ed40302020
Cosmopolitan40202030
Dominance20202030
Education Aid50504050
Fancy50504040
Global3002020
High Q30202040

From the table we can see that,

both High Q and Cosmopolitan got same points in reputation (R) and placement quality (P).

Also, High Q received more points in infrastructure (I) than Cosmopolitan

whereas Cosmopolitan received more points in faculty Quality (F) than High Q.

Hence, we can say that Infrastructure's weight should be greater than Faculty quality. i.e. I > F.

CollegeFRPIOverall scoreAccreditation
A-one50505040
Best Ed40302020
Cosmopolitan40202030
Dominance20202030
Education Aid50504050
Fancy50504040
Global3002020
High Q30202040

Similarly,

We can see that both Best Ed and Cosmopolitan got same points in faculty Quality (F) and placement quality (P).

Best Ed received more points in reputation (R) than Cosmopolitan

whereas Cosmopolitan received more points in infrastructure (I) than Best Ed.

Hence, we can say that reputation's weight should be greater than infrastructure. i.e. R > I.

CollegeFRPIOverall scoreAccreditation
A-one50505040
Best Ed40302020
Cosmopolitan40202030
Dominance20202030
Education Aid50504050
Fancy50504040
Global3002020
High Q30202040

Similarly,

We can see that both Education Aid and A-one got same points in faculty Quality (F) and reputation (R).

Education Aid received more points in infrastructure (I) than A-one

whereas A-one received more points in placement quality (P) than Education Aid.

Hence, we can say that reputation's weight should be greater than infrastructure. i.e. I > P.

CollegeFRPIOverall scoreAccreditation
A-one50505040
Best Ed40302020
Cosmopolitan40202030
Dominance20202030
Education Aid50504050
Fancy50504040
Global3002020
High Q30202040

Now, from the previous slides we can now form two cases :-

  1. R > I > P > F
  1. R > I > F > P
CollegeFRPIOverall scoreAccreditation
A-one50505040
Best Ed40302020
Cosmopolitan40202030
Dominance20202030
Education Aid50504050
Fancy50504040
Global3002020
High Q30202040

CASE 1 :- Order of weights assigned = R > I > P > F

=> R = 0.4, I = 0.3. P = 0.2, F = 0.1

In this case overall score received by Best Ed

= 0.1∗40+0.4∗30+0.2∗20+0.3∗20=260.1*40+0.4*30+0.2*20+0.3*20 = 26

In this case overall score received by High Q

= 0.1∗30+0.4∗20+0.2∗20+0.3∗40=270.1*30+0.4*20+0.2*20+0.3*40 = 27

We can see that High Q's overall score is higher than Best Ed.

Hence, this is a possible case.

CollegeFRPIOverall scoreAccreditation
A-one50505040
Best Ed40302020
Cosmopolitan40202030
Dominance20202030
Education Aid50504050
Fancy50504040
Global3002020
High Q30202040

Case 2: Order of weights assigned = R > I > F > P

=> R = 0.4, I = 0.3. P = 0.1, F = 0.2

In this case overall score received by Best Ed

= 0.2∗40+0.4∗30+0.1∗20+0.3∗20=280.2*40+0.4*30+0.1*20+0.3*20 = 28

In this case overall score received by High Q

= 0.2∗30+0.4∗20+0.1∗20+0.3∗40=280.2*30+0.4*20+0.1*20+0.3*40 = 28

We can see that High Q's overall score is not greater than the overall score received Best Ed.

Hence, this case is not possible.

Therefore, orders of weights assigned = R > I > P > F

R = 0.4, I = 0.3. P = 0.2, F = 0.1

CollegeF (0.1)R (0.4)P (0.2)I (0.3)Overall scoreAccreditation
A-one5050504047AAA
Best Ed4030202026BBA
Cosmopolitan4020203025BBA
Dominance2020203027BBA
Education Aid5050405048AAA
Fancy5050404045AAA
Global300202013Junk
High Q3020204027BBA

Now, lets calculate total score of all the college :-

  1. A-one - 50∗0.1+50∗0.4+50∗0.2+40∗0.4=4750*0.1 + 50 * 0.4 + 50 * 0.2 + 40 * 0.4 = 47

Therefore, accreditation - AAA

  1. Best Ed - 40∗0.1+30∗0.4+20∗0.2+20∗0.3=2640 * 0.1 + 30* 0.4 + 20*0.2+ 20*0.3 = 26

Therefore, accreditation- BBA

  1. Cosmopolitan - 40∗0.1+20∗0.4+20∗0.2+30∗0.3=2540*0.1+20*0.4+20*0.2+30*0.3 = 25

Therefore, accreditation - BBA

  1. Dominance - 20∗0.1+20∗0.4+20∗0.2+30∗0.3=2720*0.1+20*0.4+20*0.2+30*0.3 = 27

Therefore, accreditation - BBA

  1. Education Aid - 50∗0.1+50∗0.4+40∗0.2+50∗0.3=4850*0.1+50*0.4+40*0.2+50*0.3 = 48

Therefore, accreditation - AAA

  1. Fancy - 50∗0.1+50∗0.4+40∗0.2+40∗0.3=4550*0.1+50*0.4+40*0.2+40*0.3 = 45

Therefore, accreditation - AAA

  1. Global - 30∗0.1+0∗0.4+20∗0.2+20∗0.3=1330*0.1 + 0*0.4+20*0.2+20*0.3 = 13

Therefore, accreditation - JUNK

  1. High Q - 30∗0.1+20∗0.4+20∗0.2+40∗0.3=2730*0.1+20*0.4+20*0.2+40*0.3 = 27

Therefore, accreditation - BBA

CollegeF (0.1)R (0.4)P (0.2)I (0.3)Overall scoreAccreditation
A-one5050504047AAA
Best Ed4030202026BBA
Cosmopolitan4020203025BBA
Dominance2020203027BBA
Education Aid5050405048AAA
Fancy5050404045AAA
Global300202013Junk
High Q3020204027BBA

Highest overall score among the eight colleges is 48.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question