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.
What was the total number of schools having exactly three of the four facilities?
- 200
- 50
- 80
- 64
The Correct Option is A
Approach Solution - 1
The correct answer is (A): \(200\)
Given,
\(F_2 = (a+x+40+x)+(30+26+x+45) = 313\).
It is also given that \(F_1\) and \(F_2\) = \(a+x+40+x = 162\).
Hence, \(30+26+x+45 = 313-162 = 151\)
Hence, \(x = 151-(30+26+45) = 50\)
The number of schools that have exactly three facilities = \(4x = 200\)
Approach Solution -2
Let the no. of schools with three of the facilities was the same irrespective of which three were be x.
Number of schools without facilities be 'n' from 1, n=80.
Number of schools where only F1 and F2 be 'b'
Number of schools where only F1 and F3 be 'c'
Number of schools where only F1 and F4 be 'd'
From the 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)
equation (ii)-(i) gives x=50 and 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.
Now the table looks will be :
Therefore, The total number of schools with exactly three of the four facilities will be 4x=200.
What was the number of schools having facilities \(F_2\) and \(F_4\)?
- 185
- 45
- 95
- 85
The Correct Option is A
Approach Solution - 1
The correct answer is (A): \(185\)
The number of schools having facilities \(F_2\) and \(F_4 = 40+x+45+x = 185\)
Approach Solution -2
Let the no. of schools with three of the facilities was the same irrespective of which three were be x.
Number of schools without facilities be 'n' from 1, n=80.
Number of schools where only F1 and F2 be 'b'
Number of schools where only F1 and F3 be 'c'
Number of schools where only F1 and F4 be 'd'
From the 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)
equation (ii)-(i) gives x=50 and 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.
Now the table looks will be :
The total number of schools having facilities F4 and F2= 45+50+50+40=185
Hence, Option (A): 185 is the correct answer.
What was the number of schools having only facilities F1 and F3?
Approach Solution - 1
Only \(F_1\) and \(F_3 = b\)
Given \(F_1\) = \(F_4\)
\(25+b+xc+a+x+40+x\)
= \(24+20+x+45+40+x+x+c\)
Hence, \(a+b = 64\)
It is given that \(a+x+40+x = 162\)
As \(x = 50, a = 22\)
Hence only \(F_1\) and \(F_3\) = \(b = 64-22 = 42.\)
Approach Solution -2
Let the no. of schools with three of the facilities was the same irrespective of which three were be x.
Number of schools without facilities be 'n' from 1, n=80.
Number of schools where only F1 and F2 be 'b'
Number of schools where only F1 and F3 be 'c'
Number of schools where only F1 and F4 be 'd'
From the 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)
equation (ii)-(i) gives x=50 and 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.
Now the table looks will be :
Therefore, The total number of schools having only F1 and F3 ⇒ c=42
What was the number of schools having only facilities F1 and F4?
Approach Solution - 1
Only \(F_1\) and \(F_4\) = \(c\)
Exactly 1 + Exactly 2 + Exactly 3 + Exactly 4 = \(600-80=520\)
\((25+30+26+20)\)+ Exactly \(2 + 200 + 40 = 520\)
Hence, Exactly \(2 = 179 = a+24+b+c+26+45\)
As \(a = 22\) and \(b = 42\), c = only \(F_1\) and \(F_4 = 20\)
This table helps to figure out that vials A & B, viable C & D, Vials E & F, Vials G & H cannot be negative simultaneously. As each group consists exclusive set of patients.
Approach Solution -2
Let the no. of schools with three of the facilities was the same irrespective of which three were be x.
Number of schools without facilities be 'n' from 1, n=80.
Number of schools where only F1 and F2 be 'b'
Number of schools where only F1 and F3 be 'c'
Number of schools where only F1 and F4 be 'd'
From the 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)
equation (ii)-(i) gives x=50 and 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.
Now the table looks will be :
Top Questions on Data Analysis
- The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km) between two adjacent intersections. Six ATMs are placed at six of the nine road intersections. Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.
The following additional information is known.
The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
The road distance between the ATM with the second highest cash requirement and the ATM located at the intersection of R-C and V3 is 12 km.
- The two plots below give the following information about six firms A, B, C, D, E, and F for 2019 and 2023.
PAT: The firm’s profits after taxes in Rs. crores,
ES: The firm’s employee strength, that is the number of employees in the firm, and PRD: The percentage of the firm’s PAT that they spend on Research and Development (R&D).
In the plots, the horizontal and vertical coordinates of point representing each firm gives their ES and PAT values respectively. The PRD values of each firm are proportional to the areas around the points representing each firm. The areas are comparable between the two plots, i.e., equal areas in the two plots represent the same PRD values for the two years.
- The numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 are placed in ten slots of the following grid based on the conditions below
1. Numbers in any row appear in an increasing order from left to right.
2. Numbers in any column appear in a decreasing order from top to bottom.
3. 1 is placed either in the same row or in the same column as 10.
4. Neither 2 nor 3 is placed in the same row or in the same column as 10.
5. Neither 7 nor 8 is placed in the same row or in the same column as 9.
6. 4 and 6 are placed in the same row. - Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches - Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even numbered players, while Zara trained only odd numbered players. After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7, with 1 being the lowest rating and 7 the highest.
The following additional information is known.
1. Xena trained more players than Yuki.
2. Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3 and Player-5 were all different.
3. Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.
4. The average of the ratings of all the players was 4.
5. Player-2 got the highest rating.
6. The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.
7. Player-4's rating was double of Player-8's and less than Player-5's. - The chart below shows the price data for seven shares – A, B, C, D, E, F, and G as a candlestick plot for a particular day. The vertical axis shows the price of the share in rupees. A share whose closing price (price at the end of the day) is more than its opening price (price at the start of the day) is called a bullish share; otherwise, it is called a bearish share. All bullish and bearish shares are shown in green and red colour respectively.
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