CAT 2020 Slot 3 DILR Question & Solution

Logical ReasoningEasy

Data Set

A survey of 600 schools in India was conducted to gather information about their online teaching learning processes (OTLP). The following four facilities were studied.

F1: Own software for OTLP
F2: Trained teachers for OTLP
F3: Training materials for OTLP
F4: All students having Laptops

The following observations were summarized from the survey.

1. 80 schools did not have any of the four facilities - F1, F2, F3, F4.
2. 40 schools had all four facilities.
3. The number of schools with only F1, only F2, only F3, and only F4 was 25, 30, 26 and 20 respectively.
4. The number of schools with exactly three of the facilities was the same irrespective of which three were considered.
5. 313 schools had F2.
6. 26 schools had only F2 and F3 (but neither F1 nor F4).
7. Among the schools having F4, 24 had only F3, and 45 had only F2.
8. 162 schools had both F1 and F2.
9. The number of schools having F1 was the same as the number of schools having F4.

Question 1

What was the total number of schools having exactly three of the four facilities?

200

Solution:

Let the number of schools with exactly three of the facilities was the same irrespective of which three were considered be x.

Number of schools with none of the facilities be 'n' from 1, n=80.

Number of schools with only F1 and F2 be 'b'

Number of schools with only F1 and F3 be 'c'

Number of schools with only F1 and F4 be 'd'

From the information given in the question we will get the following Venn diagram.

From 5, b+141+3x=313 => b+3x=172....(i)

From 8, b+x+40+x=162 => b+2x=122....(ii)

(ii)-(i) gives x=50 => b=22

From 9, 237+3x+c+d=279+3x=d => c=42

Total number of schools =600 => 313+25+c+x+d+26+24+20+80=600 => d=20.

The final table looks like:

The total number of schools with exactly three of the four facilities= 4x=200.

Question 2

What was the number of schools having facilities F2 and F4?

185

Solution:

Let the number of schools with exactly three of the facilities was the same irrespective of which three were considered be x.

Number of schools with none of the facilities be 'n' from 1, n=80.

Number of schools with only F1 and F2 be 'b'

Number of schools with only F1 and F3 be 'c'

Number of schools with only F1 and F4 be 'd'

From the information given in the question we will get the following Venn diagram.

From 5, b+141+3x=313 => b+3x=172....(i)

From 8, b+x+40+x=162 => b+2x=122....(ii)

(ii)-(i) gives x=50 => b=22

From 9, 237+3x+c+d=279+3x=d => c=42

Total number of schools =600 => 313+25+c+x+d+26+24+20+80=600 => d=20.

The final table looks like:

The total number of schools having facilities F4 and F2= 45+50+50+40=185.

Question 3

What was the number of schools having only facilities F1 and F3?

Solution:

Let the number of schools with exactly three of the facilities was the same irrespective of which three were considered be x.

Number of schools with none of the facilities be 'n' from 1, n=80.

Number of schools with only F1 and F2 be 'b'

Number of schools with only F1 and F3 be 'c'

Number of schools with only F1 and F4 be 'd'

From the information given in the question we will get the following Venn diagram.

From 5, b+141+3x=313 => b+3x=172....(i)

From 8, b+x+40+x=162 => b+2x=122....(ii)

(ii)-(i) gives x=50 => b=22

From 9, 237+3x+c+d=279+3x=d => c=42

Total number of schools =600 => 313+25+c+x+d+26+24+20+80=600 => d=20.

The final table looks like:

The total number of schools having only F1 and F3= c=42

Question 4

What was the number of schools having only facilities F1 and F4?

Solution:

Let the number of schools with exactly three of the facilities was the same irrespective of which three were considered be x.

Number of schools with none of the facilities be 'n' from 1, n=80.

Number of schools with only F1 and F2 be 'b'

Number of schools with only F1 and F3 be 'c'

Number of schools with only F1 and F4 be 'd'

From the information given in the question we will get the following Venn diagram.

From 5, b+141+3x=313 => b+3x=172....(i)

From 8, b+x+40+x=162 => b+2x=122....(ii)

(ii)-(i) gives x=50 => b=22

From 9, 237+3x+c+d=279+3x=d => c=42

Total number of schools =600 => 313+25+c+x+d+26+24+20+80=600 => d=20.

The final table looks like:

The total number of schools having only F1 and F4=20.