Logical Reasoning

Arrangements

Medium

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:

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

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

	F	R	P	I
A-one	A	A	A	B
Best Ed	B	C	D	D
Cosmopolitan	B	D	D	C
Dominance	D	D	B	C
Education Aid	A	A	B	A
Fancy	A	A	B	B
Global	C	F	D	D
High Q	C	D	D	B

It is further known that in terms of overall scores:

High Q is better than Best Ed;

Best Ed is better than Cosmopolitan; and

Education Aid is better than A-one.

How many colleges receive the accreditation of AAA?

Entered answer:

✅ Correct Answer: 3

Slide 1/11

Understanding the set :-

This is a weighted ranking deduction puzzle.

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.

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)

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.

College	F	R	P	I
A-one	50	50	50	40
Best Ed	40	30	20	20
Cosmopolitan	40	20	20	30
Dominance	20	20	20	30
Education Aid	50	50	40	50
Fancy	50	50	40	40
Global	30	0	20	20
High Q	30	20	20	40

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.

College	F	R	P	I
A-one	50	50	50	40
Best Ed	40	30	20	20
Cosmopolitan	40	20	20	30
Dominance	20	20	20	30
Education Aid	50	50	40	50
Fancy	50	50	40	40
Global	30	0	20	20
High Q	30	20	20	40

Clue 1, 2 and 3 says,

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

Therefore, We can say that High Q > Cosmopolitan

College	F	R	P	I
A-one	50	50	50	40
Best Ed	40	30	20	20
Cosmopolitan	40	20	20	30
Dominance	20	20	20	30
Education Aid	50	50	40	50
Fancy	50	50	40	40
Global	30	0	20	20
High Q	30	20	20	40

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.

College	F	R	P	I
A-one	50	50	50	40
Best Ed	40	30	20	20
Cosmopolitan	40	20	20	30
Dominance	20	20	20	30
Education Aid	50	50	40	50
Fancy	50	50	40	40
Global	30	0	20	20
High Q	30	20	20	40

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.

College	F	R	P	I
A-one	50	50	50	40
Best Ed	40	30	20	20
Cosmopolitan	40	20	20	30
Dominance	20	20	20	30
Education Aid	50	50	40	50
Fancy	50	50	40	40
Global	30	0	20	20
High Q	30	20	20	40

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.

College	F	R	P	I
A-one	50	50	50	40
Best Ed	40	30	20	20
Cosmopolitan	40	20	20	30
Dominance	20	20	20	30
Education Aid	50	50	40	50
Fancy	50	50	40	40
Global	30	0	20	20
High Q	30	20	20	40

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

R > I > P > F

R > I > F > P

College	F	R	P	I
A-one	50	50	50	40
Best Ed	40	30	20	20
Cosmopolitan	40	20	20	30
Dominance	20	20	20	30
Education Aid	50	50	40	50
Fancy	50	50	40	40
Global	30	0	20	20
High Q	30	20	20	40

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

= $0.1*40+0.4*30+0.2*20+0.3*20 = 26$

In this case overall score received by High Q

= $0.1*30+0.4*20+0.2*20+0.3*40 = 27$

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

Hence, this is a possible case.

College	F	R	P	I
A-one	50	50	50	40
Best Ed	40	30	20	20
Cosmopolitan	40	20	20	30
Dominance	20	20	20	30
Education Aid	50	50	40	50
Fancy	50	50	40	40
Global	30	0	20	20
High Q	30	20	20	40

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

= $0.2*40+0.4*30+0.1*20+0.3*20 = 28$

In this case overall score received by High Q

= $0.2*30+0.4*20+0.1*20+0.3*40 = 28$

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

College	F (0.1)	R (0.4)	P (0.2)	I (0.3)	Overall score	Accreditation
A-one	50	50	50	40	47	AAA
Best Ed	40	30	20	20	26	BBA
Cosmopolitan	40	20	20	30	25	BBA
Dominance	20	20	20	30	27	BBA
Education Aid	50	50	40	50	48	AAA
Fancy	50	50	40	40	45	AAA
Global	30	0	20	20	13	Junk
High Q	30	20	20	40	27	BBA

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

A-one - $50*0.1 + 50 * 0.4 + 50 * 0.2 + 40 * 0.4 = 47$

Therefore, accreditation - AAA

Best Ed - $40 * 0.1 + 30* 0.4 + 20*0.2+ 20*0.3 = 26$

Therefore, accreditation- BBA

Cosmopolitan - $40*0.1+20*0.4+20*0.2+30*0.3 = 25$

Therefore, accreditation - BBA

Dominance - $20*0.1+20*0.4+20*0.2+30*0.3 = 27$

Therefore, accreditation - BBA

Education Aid - $50*0.1+50*0.4+40*0.2+50*0.3 = 48$

Therefore, accreditation - AAA

Fancy - $50*0.1+50*0.4+40*0.2+40*0.3 = 45$

Therefore, accreditation - AAA

Global - $30*0.1 + 0*0.4+20*0.2+20*0.3 = 13$

Therefore, accreditation - JUNK

High Q - $30*0.1+20*0.4+20*0.2+40*0.3 = 27$

Therefore, accreditation - BBA

College	F (0.1)	R (0.4)	P (0.2)	I (0.3)	Overall score	Accreditation
A-one	50	50	50	40	47	AAA
Best Ed	40	30	20	20	26	BBA
Cosmopolitan	40	20	20	30	25	BBA
Dominance	20	20	20	30	27	BBA
Education Aid	50	50	40	50	48	AAA
Fancy	50	50	40	40	45	AAA
Global	30	0	20	20	13	Junk
High Q	30	20	20	40	27	BBA

Three colleges received AAA rating.

Question 17

What is the weight of the faculty quality parameter?

Question 19

What is the highest overall score among the eight colleges?

Question 20

How many colleges have overall scores between 31 and 40, both inclusive?

How many colleges receive the accreditation of AAA?

Solution

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