Linear Seating Arrangement facing North South Questions
Seating Arrangement facing North South Questions for IBPS PO Prelims. Linear Arrangement Puzzles for Bank Exam.Linear Arrangement Puzzles for IBPS PO 2018 .Welcome to the www.letsstudytogether.co online Reasoning section. If you are preparing for RRB PO/Clerk, IBPS PO/Clerk you will come across a section on Reasoning Ability Section. Here we are providing you with Reasoning quiz on “Linear Seating Arrangement facing North South Questions for IBPS PO Prelims” for your daily practice.
This “Linear Seating Arrangement facing North South Questions “ is also important for other banking exams such as SBI PO, IBPS Clerk, SBI Clerk, IBPS RRB Officer, IBPS RRB Office Assistant, IBPS SO, SBI SO and other competitive exams.
Linear Seating Arrangement facing North South Questions  Set 143
Quizsummary
0 of 10 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
Information
Help: Share this quiz to your Friends & FB Groups.
All the Best !!!
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Average score 

Your score 

Categories
 Not categorized 0%
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 Answered
 Review

Question 1 of 10
1. Question
P and Q can complete a work together in 18.75 days. If the product of the number of days taken by them to complete the work individually is 1500 days and the ratio of the number of days taken by P and Q was 5:3, then what was the number of days taken by P?
Correct
A. 50 days
Let the number of days taken by P and Q be ‘p’ and ‘q’ respectively. Then,
1/p + 1/q = 1/18.75
=>(p + q)/pq = 1/18.75
=>(p + q)/1500 = 1/18.75
=>p + q = 80
So, number of days taken by P to complete the work = (5/8) × 80 = 50 days
Incorrect
A. 50 days
Let the number of days taken by P and Q be ‘p’ and ‘q’ respectively. Then,
1/p + 1/q = 1/18.75
=>(p + q)/pq = 1/18.75
=>(p + q)/1500 = 1/18.75
=>p + q = 80
So, number of days taken by P to complete the work = (5/8) × 80 = 50 days

Question 2 of 10
2. Question
Superman and Spiderman won Rs. 600000 in a lottery and distributed it amongst them such that Superman’s share was Rs. ‘x’. Superman deposited his share at 6% simple interest for 3 years and Spiderman deposited his share at 4% simple interest for 2 years. If the sum of their accumulated amounts was Rs. 683000, then what was Superman’s share of lottery money?
Correct
C. Rs. 350000
According to the question,
[x × (1 + 0.06 × 3)] + [(600000 – x) × (1 + 0.04 × 2)] = 683000
[x × 1.18] + [(600000 – x) × 1.08] = 683000
1.18x + 648000 –1.08x = 683000
0.1x = 35000
x = Rs. 350000
Incorrect
C. Rs. 350000
According to the question,
[x × (1 + 0.06 × 3)] + [(600000 – x) × (1 + 0.04 × 2)] = 683000
[x × 1.18] + [(600000 – x) × 1.08] = 683000
1.18x + 648000 –1.08x = 683000
0.1x = 35000
x = Rs. 350000

Question 3 of 10
3. Question
Raj and Rohit started a business by investing Rs. ‘x + 15000’ and Rs. ‘x + 50000’ respectively. Raj withdrew Rs. 20000 at the end of the 3^{rd} month. Raj’s share of 1^{st} year’s profit of Rs. 16000 was Rs. 6000. Find ‘x’.
Correct
C. Rs. 75000
Profit sharing ratio between Raj and Rohit at the end of 1^{st} year
= [(x + 15000) × 3 + (x – 5000) × 9] : [(x + 50000) × 12]
= [12x] : [12x + 600000]
= (x) : (x + 50000)
According to the question,
(x/(x + x + 50000)) * 16000 = 6000
=>16000x = 12000x + 50000 * 6000
On solving x = Rs. 75000
Incorrect
C. Rs. 75000
Profit sharing ratio between Raj and Rohit at the end of 1^{st} year
= [(x + 15000) × 3 + (x – 5000) × 9] : [(x + 50000) × 12]
= [12x] : [12x + 600000]
= (x) : (x + 50000)
According to the question,
(x/(x + x + 50000)) * 16000 = 6000
=>16000x = 12000x + 50000 * 6000
On solving x = Rs. 75000

Question 4 of 10
4. Question
The total number of black cars and red cars in a showroom is 18. The number of red cars is 6 less than twice the number of black cars. If 3 cars are selected at random, what is the probability that at least one of them is Red?
Correct
B. 95/102
Let the number of black and red cars in the showroom be ‘x’ and ‘y’ respectively. So,
x + y = 18—(1)
y = 2x – 6—(2)
On solving (1) and (2), we get
x = 8 and y = 10
Required probability = P(atleast one of them is Red) = 1 – P(all of them are black)
= 1 – ((8/18) * (7/17) * (6/16))
= 1 – 7/102
= 95/102
Incorrect
B. 95/102
Let the number of black and red cars in the showroom be ‘x’ and ‘y’ respectively. So,
x + y = 18—(1)
y = 2x – 6—(2)
On solving (1) and (2), we get
x = 8 and y = 10
Required probability = P(atleast one of them is Red) = 1 – P(all of them are black)
= 1 – ((8/18) * (7/17) * (6/16))
= 1 – 7/102
= 95/102

Question 5 of 10
5. Question
3 cylinders of height ‘x’ cm and radius 42 cm are melted to form 4 cuboids of length, breadth, and height equal to 154 cm, 96 cm, and 9 cm respectively. What was the height of the cylinder?
Correct
D. 32 cm
According to the question,
The volume of 3 cylinders = Volume of 4 cuboids
3 × (22/7) × r^{2 }× h = 4 × l × b × h
3 × (22/7) × 42^{2 }× h = 4 × 154 × 96 × 9
16632 × h = 532224
So, h = 32 cm
Incorrect
D. 32 cm
According to the question,
The volume of 3 cylinders = Volume of 4 cuboids
3 × (22/7) × r^{2 }× h = 4 × l × b × h
3 × (22/7) × 42^{2 }× h = 4 × 154 × 96 × 9
16632 × h = 532224
So, h = 32 cm

Question 6 of 10
6. Question
A shopkeeper sold watch A costing Rs. 25000 at ‘x’% profit and watch B costing Rs. 24000 at ‘2x’% profit. If it is known that the sum of their selling prices was Rs. 63600, then what would have been the sum of their selling prices had each watch been sold at half of the original profit percentages?
Correct
B. Rs. 56300
According to the question,
25000 × (1 *+ x/100) + 24000 × (1 + 2x/100) = 63600
25000 + 250x + 24000 + 480x = 63600
730x = 14600
x = 20
Required sum = 25000 × (1 + 0.1) + 24000 × (1 + 0.2)
= 27500 + 28800 = Rs. 56300
Incorrect
B. Rs. 56300
According to the question,
25000 × (1 *+ x/100) + 24000 × (1 + 2x/100) = 63600
25000 + 250x + 24000 + 480x = 63600
730x = 14600
x = 20
Required sum = 25000 × (1 + 0.1) + 24000 × (1 + 0.2)
= 27500 + 28800 = Rs. 56300

Question 7 of 10
7. Question
Drum X and drum Y contain 300 litres and 500 litres of a mixture of milk and water in the ratio 3:7 and 11:9 respectively. ‘x%’ of liquid of drum X and drum Y is taken out and put into a bucket. If the total amount of water in the bucket is174 litres then find ‘x’.
Correct
A. 40
According to the question,
(x/100) × 300 × (7/10) + (x/100) × 500 × (9/20) = 174
2.1x + 2.25x = 174
4.35x = 174
x = 40
Incorrect
A. 40
According to the question,
(x/100) × 300 × (7/10) + (x/100) × 500 × (9/20) = 174
2.1x + 2.25x = 174
4.35x = 174
x = 40

Question 8 of 10
8. Question
A car is chasing a truck which is 60 km ahead of it. The speed of the car and the truck is ‘11x’ km/hr and ‘8x’ km/hr respectively. If the sum of the speed of the car and the truck is95 km/hr, then how long would it take the car to meet the truck?
Correct
B. 4 hours
According to the question,
11x + 8x = 95
19x = 95
x = 5
So, speed of the car and the truck is 55 km/hr and 40 km/hr respectively.
The time was taken by the car to meet the truck
= 60/(55 – 40)
= 60/15
= 4 hours
Incorrect
B. 4 hours
According to the question,
11x + 8x = 95
19x = 95
x = 5
So, speed of the car and the truck is 55 km/hr and 40 km/hr respectively.
The time was taken by the car to meet the truck
= 60/(55 – 40)
= 60/15
= 4 hours

Question 9 of 10
9. Question
M and N can complete a task in ‘8x’ days and ‘5x’ days respectively. They work together for 5 days and complete(13/40) part of the work. What is the difference between the number of days taken, when working alone, by M and N to complete the work?
Correct
D. 15 days
According to the question,
5 * (1/8x + 1/5x) = (13/40)
5 * (13x/40x^{2}) = 13/40
On solving, x = 5
So, the number of days taken by M and N are 40 and 25 respectively.
Required difference = 40 – 25 = 15 days
Incorrect
D. 15 days
According to the question,
5 * (1/8x + 1/5x) = (13/40)
5 * (13x/40x^{2}) = 13/40
On solving, x = 5
So, the number of days taken by M and N are 40 and 25 respectively.
Required difference = 40 – 25 = 15 days

Question 10 of 10
10. Question
The weights of A, B, C and D are 44 kgs, 40 kgs, 32 kgs and ‘x – 6’ kgs respectively. The average of weights of A and B was 2 kgs more than the average of weights of C and D. What is the average of weights of B, C and D?
Correct
B. 40 kgs
According to the question,
=>(44 + 40)/2 – (32 + (x – 6))/2 = 2
=>84/2 – (26 + x)/2 = 2
=>(84 – 26 – x) = 4
x = 54
So, D’s weight = 54 – 6 = 48 kgs
Required average = (40 + 32 + 48)/3 = 120/3 = 40 kgs
Incorrect
B. 40 kgs
According to the question,
=>(44 + 40)/2 – (32 + (x – 6))/2 = 2
=>84/2 – (26 + x)/2 = 2
=>(84 – 26 – x) = 4
x = 54
So, D’s weight = 54 – 6 = 48 kgs
Required average = (40 + 32 + 48)/3 = 120/3 = 40 kgs
Note:
 Click view Questions button to view Explanation
 Ask your doubt in comment section we’ ll clear your doubts in caring way.
 If you find any mistake, please tell us in the comment section and also mail us support@letsstudytogether.co
Download New Pattern Puzzle and Seating Arrangement Books PDF
300 High Level Data Interpretation (DI) Questions & Answers PDF – Download Here
SBI PO/Clerk  RRB PO/Clerk  Railway Exams  SSC CGL 2018
S. No.  Exams  Direct Links 
1.  SBI PO/Clerk 2018  Click Here 
2.  IBPS RRB PO/Clerk 2018  Click Here 
3.  Railway RRB ALP & Group D 2018  Click Here 
4.  SSC CGL 2018  Click Here 