MCQ Test: Radar Graph Caselet - 2 - Banking Exams MCQ

Total marked price of 3 juicers = 2000 × 3 = 6000

The total money paid by them = 90x + 72x + 54x = 216x = 4320

Hence, option B is correct.
Common explanation : 
Let the marked price of juicer = 100x 
Then, The amount Chand paid = (100 – 10)% of 100x = 90% of 100x = 90x 
The amount Chandni will pay = 90% of 80% of 100x = 72x 
The amount Chanchal will pay = 90% of 80% of 75% of 100x = 54x 
According to the question, 72x – 54x = 18x = 360 
x = 20 

The amount paid by Chand = 90x

the amount paid by Chanchal = 54x

Hence, option D is correct.
Common explanation:
Let the marked price of juicer = 100x 
Then, The amount Chand paid = (100 – 10)% of 100x = 90% of 100x = 90x 
The amount Chandni will pay = 90% of 80% of 100x = 72x 
The amount Chanchal will pay = 90% of 80% of 75% of 100x = 54x 
According to the question, 72x – 54x = 18x = 360 
x = 20 

MP = 100X = 100 × 20 = 2000 
Hence, option C is correct.
Common explanation: 
Let the marked price of juicer = 100x 
Then, The amount Chand paid = (100 – 10)% of 100x = 90% of 100x = 90x 
The amount Chandni will pay = 90% of 80% of 100x = 72x 
The amount Chanchal will pay = 90% of 80% of 75% of 100x = 54x 
According to the question, 72x – 54x = 18x = 360 
x = 20 

The amount paid by Chand = 90x 
the amount paid by Chanchal = 54x 
The required ratio = 90x : 54x = 5 : 3 
Hence, option D is correct.
Common explanation: 
Let the marked price of juicer = 100x 
Then, The amount Chand paid = (100 – 10)% of 100x = 90% of 100x = 90x 
The amount Chandni will pay = 90% of 80% of 100x = 72x 
The amount Chanchal will pay = 90% of 80% of 75% of 100x = 54x 
According to the question, 72x – 54x = 18x = 360 
x = 20 

The amount Chand paid = (100 – 10)% of 100x = 90% of 100x = 90x = 90 × 20 = 1800 
The money left with him = 2500 – 1800 = 700 
Hence, option C is correct.
Common explanation: 
Let the marked price of juicer = 100x 
Then, The amount Chand paid = (100 – 10)% of 100x = 90% of 100x = 90x 
The amount Chandni will pay = 90% of 80% of 100x = 72x 
The amount Chanchal will pay = 90% of 80% of 75% of 100x = 54x 
According to the question, 72x – 54x = 18x = 360 
x = 20 

Following the common explanation, we get
Total number of 4 star hotels from Site A and C together are 60
Hence, option C is correct.
Common explanation:
It is given that total number of hotels is 720. 
4 star hotels = 2 × (3 star hotels)
5 star hotels = 3 × (4 star hotels)
Let, 3 star hotels M, 4 star hotels N and 5 star hotels P. 
N = 2M
P = 3N
M + 2M + 6M = 720
9M = 720
M = 80
N = 160
P = 480
Total 200 hotels listed on sites A. Out of which, 30% are 3 star.
60 hotels are there in 3 star category on Site A. 
Total 5 star hotels are 480. 

Ratio of the hotels on site A, B and C is 1:1:2. 
4x = 480
x = 120
Number of 5 star hotels on B website is 20% more than number of 4 star hotels on the same website.
Number of 5 Star Hotels on site B = 120 
So, number of 4 star hotels on site B would be 100. 
Total 4 star hotels are 160.
So 4 star hotels on site C will be 160 – 20 – 100 = 40
Number of 3 star hotels on website B and C are same. 

Total 80
So 3 star and 4 star hotels on site B and C should be 10. 

Following the common explanation, we get
3 star hotels on site D is 50% more than number of 4 star hotels on site A
3 star hotels on Site D = 150% of 20 = 30
Total Hotels on site D = 500
Out of Which 50% are 4 star = 250
Number of 4 star hotels on site D = 500 – 250 – 30 = 220
Hence, option B is correct.
Common explanation:
It is given that total number of hotels is 720. 
4 star hotels = 2 × (3 star hotels)
5 star hotels = 3 × (4 star hotels)
Let, 3 star hotels M, 4 star hotels N and 5 star hotels P. 
N = 2M
P = 3N
M + 2M + 6M = 720
9M = 720
M = 80
N = 160
P = 480
Total 200 hotels listed on sites A. Out of which, 30% are 3 star.
60 hotels are there in 3 star category on Site A. 
Total 5 star hotels are 480. 

Ratio of the hotels on site A, B and C is 1:1:2. 
4x = 480
x = 120
Number of 5 star hotels on B website is 20% more than number of 4 star hotels on the same website.
Number of 5 Star Hotels on site B = 120 
So, number of 4 star hotels on site B would be 100. 
Total 4 star hotels are 160.
So 4 star hotels on site C will be 160 – 20 – 100 = 40
Number of 3 star hotels on website B and C are same. 

Total 80
So 3 star and 4 star hotels on site B and C should be 10. 

Following the common explanation, we get
3 star Hotels on site A = 60

4 star Hotels on Site C = 40

Difference = 20

Hence, option A is correct.
Common explanation:
It is given that total number of hotels is 720. 
4 star hotels = 2 × (3 star hotels)
5 star hotels = 3 × (4 star hotels)
Let, 3 star hotels M, 4 star hotels N and 5 star hotels P. 
N = 2M
P = 3N
M + 2M + 6M = 720
9M = 720
M = 80
N = 160
P = 480
Total 200 hotels listed on sites A. Out of which, 30% are 3 star.
60 hotels are there in 3 star category on Site A. 
Total 5 star hotels are 480. 

Ratio of the hotels on site A, B and C is 1:1:2. 
4x = 480
x = 120
Number of 5 star hotels on B website is 20% more than number of 4 star hotels on the same website.
Number of 5 Star Hotels on site B = 120 
So, number of 4 star hotels on site B would be 100. 
Total 4 star hotels are 160.
So 4 star hotels on site C will be 160 – 20 – 100 = 40
Number of 3 star hotels on website B and C are same. 

Total 80
So 3 star and 4 star hotels on site B and C should be 10. 

Following the common explanation, we get

4 Star Hotels on Site B = 100

Total Hotels on Site A = 200

Hence, option E is correct.
Common explanation:
It is given that total number of hotels is 720. 
4 star hotels = 2 × (3 star hotels)
5 star hotels = 3 × (4 star hotels)
Let, 3 star hotels M, 4 star hotels N and 5 star hotels P. 
N = 2M
P = 3N
M + 2M + 6M = 720
9M = 720
M = 80
N = 160
P = 480
Total 200 hotels listed on sites A. Out of which, 30% are 3 star.
60 hotels are there in 3 star category on Site A. 
Total 5 star hotels are 480. 

Ratio of the hotels on site A, B and C is 1:1:2. 
4x = 480
x = 120
Number of 5 star hotels on B website is 20% more than number of 4 star hotels on the same website.
Number of 5 Star Hotels on site B = 120 
So, number of 4 star hotels on site B would be 100. 
Total 4 star hotels are 160.
So 4 star hotels on site C will be 160 – 20 – 100 = 40
Number of 3 star hotels on website B and C are same. 

Total 80
So 3 star and 4 star hotels on site B and C should be 10. 

Following the common explanation, we get

Total number of Hotels listed on site C is 290. 

Hence, option A is correct.
Common explanation:
It is given that total number of hotels is 720. 
4 star hotels = 2 × (3 star hotels)
5 star hotels = 3 × (4 star hotels)
Let, 3 star hotels M, 4 star hotels N and 5 star hotels P. 
N = 2M
P = 3N
M + 2M + 6M = 720
9M = 720
M = 80
N = 160
P = 480
Total 200 hotels listed on sites A. Out of which, 30% are 3 star.
60 hotels are there in 3 star category on Site A. 
Total 5 star hotels are 480. 

Ratio of the hotels on site A, B and C is 1:1:2. 
4x = 480
x = 120
Number of 5 star hotels on B website is 20% more than number of 4 star hotels on the same website.
Number of 5 Star Hotels on site B = 120 
So, number of 4 star hotels on site B would be 100. 
Total 4 star hotels are 160.
So 4 star hotels on site C will be 160 – 20 – 100 = 40
Number of 3 star hotels on website B and C are same. 

Total 80
So 3 star and 4 star hotels on site B and C should be 10. 

The number of students in DU in the year 2017 = 20 thousands 
 The number of students in JNU in the year 2014 = 30 thousands 
 Required difference = 30 – 20 = 10 thousands 
 Hence, option A is correct.

From the chart it is clear it is highest in 2012

The number of students in JNU in the year 2012 = 13 thousands 

The number of students in DU in the year 2016 = 24 thousands 

Required sum = 13 + 24 = 37 thousands 

Total number of students in JNU in the year 2017 = 30 thousand

Number of male students = (100 – 30) % of 30 thousand = 70% of 30 thousand = 21 thousand

Number of students in DU,
In 2012, 8 thousand
In 2013, 15 thousand
In 2014, 22 thousand
In 2015, 28 thousand
In 2016, 24 thousand
In 2017, 20 thousand
From here it is clear that in 2013 it was highest percentage increase over the previous year

Question bank of English contains (5 + 25 + 10) 40 questions 
Question bank of Maths contains (7 + 15 + 8) 30 questions 
Question bank of GA contains (5 + 6 + 14) 25 questions 
Question bank of Computer contains (7 + 6 + 12) 25 questions 
Question bank of Reasoning contains (9 + 12 + 9) 30 questions 
A paper of 60 questions are to be made that it has 5% questions from GA, 45% of questions from Maths and at least 28 questions from reasoning. 
∴ Questions of GA in the paper = 5% of 60 = 3 and Questions of Maths in the paper = 45% of 60 = 27 
The different ways that we can choose questions to form such paper are: 
i. 3GA × 27Maths × 28reasoning × 2(Computer +English) = ²⁵C₃ × ³⁰C₂₇ × ³⁰C₂₈  × ⁶⁵C₂ 
⇒ 3GA × 27Maths × 28reasoning × 2(Computer +English) = 2300 × 4060 × 435 × 2080 
ii. 3GA × 27Maths × 29reasoning × 1(Computer +English) = ²⁵C₃ × ³⁰C₂₇ × ³⁰C₂₉ × ⁶⁵C₁ 
⇒ 3GA × 27Maths × 28reasoning × 1(Computer +English) = 2300 × 4060 × 30 × 65 
iii. 3GA × 27Maths × 30reasoning = ²⁵C₃ × ³⁰C₂₇ × ³⁰C₃₀ 
⇒ 3GA × 27Maths × 30reasoning = 2300 × 4060 × 1 
∴ Total number of ways in which a paper of 60 questions can be made so that it has 3 GA questions, 27 Maths questions and at least 28 reasoning questions = (2300 × 4060 × 435 × 2080) + (2300 × 4060 × 30 × 65) + (2300 × 4060 × 1) 
⇒ Required ways = 2300 × 4060 (435 × 2080 + 30 × 65 + 1) 
⇒ Required ways = 2300 × 4060 (904800 + 1950 + 1) 
⇒ Required ways = 2300 × 4060 × 906751 
Hence, the total number of ways in which a paper of 60 questions can be made so that it has 3 GA questions, 27 Maths questions and at least 28 reasoning questions is 2300 × 4060 × 906751

Hence, option C is correct.

Question bank contains 5 easy level English questions, 7 easy level Maths questions, 9 easy level Reasoning questions, 5 easy level G.A questions and 7 easy level computer questions 
⇒ There are a total of 33 easy level questions. 
But all Maths and Reasoning questions are set next to each other 
⇒ 16 questions are grouped and can be considered as 1 question 
∴ We can assume a total of (33 – 16 + 1) 18 questions which need to be arranged. 
⇒ 18 questions can be arranged in 18! Ways 
Now, the number of ways in which Maths and reasoning question can be arranged = 16! 
∴ Required number of ways = 18! 16! 
Hence, in 18! 16! Ways questions can be arranged if Maths and reasoning questions are always set next to each other.

Hence, option E is correct.

Question bank of Maths contains 30 questions. 
Out of 30 questions, 7 are easy, 15 are moderate and 8 are hard. 
8 hard level questions can be set in 8! Ways 
In 8 places of hard questions, there are 9 places in which 7 easy questions are to be set, so that no two easy questions are set to each other 
_H1_H2_H3_H4_H5_H6_H7_H8_
⇒ Easy questions can be set in 9P7 ways 

∴ Number of ways in which the test can be made if no two easy questions are set next to each other = 8! × 9!/2! = 20160 × 9! 
Hence, 20160 × 9! Ways in which the test can be made if no two easy questions are set next to each other
Hence, option B is correct.

Question bank of English contains 40 questions. 
Out of 40 questions, 5 are of the easy level, 25 are of moderate level and 10 are of hard level questions. 
The tutor has to choose questions for an English test of 28 questions which contain 3 easy questions, 23 moderate questions and at most 4 hard questions 
The test can be prepared in following way: 
3 easy question, 23 moderate questions, 2 hard question = ⁵C₃ × ²⁵C₂₃ × ¹⁰C₂ 
⇒ Number of ways in which the test can be prepared = 10 × 300 × 45 
⇒ Number of ways in which the test can be prepared = 135000 
Hence, in 135000 ways a tutor can choose questions for an English test of 28 questions which contain 3 easy questions, 23
moderate questions and at most 4 hard questions.

Hence, option D is correct.

Question bank of reasoning contains 30 questions. 
Out of 30 questions, 9 are of easy level, 12 are of moderate level and 9 are of hard level questions. 
The student has to answer 18 questions. 
According to the given information: 
Numbers of possibilities are: 
i. 12 from moderate level, 5 from hard level and 1 from easy level = ¹²C₁₂ × ⁹C₅ × ⁹C₁ 
ii. 10 from moderate level, 7 from hard level and 1 from easy level = ¹²C₁₀ × ⁹C₇ × ⁹C₁ 
iii. 11 from moderate level, 6 from hard level and 1 from easy level = ¹²C₁₁ × ⁹C₆ × ⁹C₁ 
∴ Required number of ways = (¹²C₁₂ × ⁹C₅ × ⁹C₁) + (¹²C₁₀ × ⁹C₇ × ⁹C₁) + (¹²C₁₁ × ⁹C₆ × ⁹C₁) 
⇒ Required number of ways = (1 × 126 × 9) + (66 × 36 × 9) + (12 × 84 × 9) 
⇒ Required number of ways = (1134) + (21384) + (9072) 
⇒ Required number of ways = 31590 
Hence, in 31590 ways a student can answer 18 questions of reasoning from its question bank, choosing at least 10 from moderate level questions, at least 5 from hard level questions and 1 from easy level.

Hence, option A is correct.