All questions of Infosys for Interview Preparation Exam

Let bullet one has = x
remaining bullets = 3x-12
atq:
3x-12=x
=> x=6
total bullets=3x
=3×6
=18(answer)

A/6+a/12+a/7 +5....

Given information:
- The ship starts its voyage and travels a distance of 180 miles.
- After the ship has traveled 180 miles, a plane starts with a speed that is 10 times the speed of the ship.

Approach to solve the problem:
To find the distance when the ship and the plane meet from the starting point, we need to determine the time it takes for the plane to catch up with the ship. Once we have the time, we can calculate the distance traveled by the plane.

Calculating the time taken:
Let's assume the speed of the ship is 'x' mph. Therefore, the speed of the plane is 10x mph.

To find the time taken, we can use the formula:
Distance = Speed * Time

For the ship:
Distance traveled by the ship = 180 miles
Speed of the ship = x mph
Time taken by the ship = Distance/Speed = 180/x hours

For the plane:
Distance traveled by the plane = Distance traveled by the ship (as they meet at the same point)
Speed of the plane = 10x mph
Time taken by the plane = Distance/Speed = 180/x hours

Calculating the distance when they meet:
Since the plane starts after the ship has traveled 180 miles, the distance traveled by the plane when they meet is the same as the distance traveled by the ship.

Distance when they meet = Distance traveled by the ship = 180 miles

Final answer:
Therefore, the distance when they meet from the starting point is 180 miles.

Explanation:

To solve this problem, we need to analyze the given conditions and eliminate the possibilities one by one.

Condition 1: If it were not green or if it were not of length 35, it is 55.

This means that if the snake is not green or not of length 35, then its length must be 55. So, we can eliminate the possibility of the snake being green or of length 35.

Condition 2: If it were not black or if it were not of length 45, it is 55.

This condition states that if the snake is not black or not of length 45, then its length must be 55. Therefore, we can eliminate the possibility of the snake being black or of length 45.

Condition 3: If it were not black or if it were not of length 35, it is 55.

Similarly, this condition implies that if the snake is not black or not of length 35, then its length must be 55. Hence, we can eliminate the possibility of the snake being black or of length 35.

By analyzing all the given conditions, we can conclude that the snake cannot be green, black, or of length 35 or 45. Therefore, the only possibility left is that the snake must be of length 55.

Conclusion:

Based on the given conditions, we can determine that the length of the snake is 55.

There are 2 series
1) 3, 13, 33, _
2) 6, 26, 66, _

16-4=12

Understanding the Problem
To solve the problem, we need to determine the total distance from ANYTOWN to ANYCITY based on the events that transpired during Mr. ANYMAN's journey.
Key Events
- Mr. ANYMAN drove for 2 hours before a puncture occurred.
- The wheel change took 10 minutes, totaling 2 hours and 10 minutes of travel time before resuming.
- After the wheel change, he traveled at a speed of 30 mph, arriving 30 minutes late.
Calculating the Distance
1. Initial Travel Time:
- In 2 hours, at an unknown speed, the distance covered is: Distance = Speed × Time.
2. Time Lost Due to Puncture:
- Total time before resuming: 2 hours + 10 minutes = 2 hours + 1/6 hour = 2.17 hours.
3. Remaining Time:
- If he was 30 minutes late, this means he should have arrived 30 minutes earlier than he did.
- Assuming he had a total time (T) for the journey, he effectively used (T + 0.5) hours.
4. Distance Calculation:
- If the puncture had occurred 30 miles later, he would have arrived only 15 minutes late, indicating a difference in time of 15 minutes (0.25 hours) due to reducing the distance he traveled at 30 mph.
- The equation can be set up as:
- (d - 30) / 30 = (T - 0.25) - (2.17 hours)
- (d / 30) = T - (2.17 hours) + 0.5.
Final Calculation
Through these equations, we can find that the total distance (d) can be calculated, leading to the conclusion that the total distance from ANYTOWN to ANYCITY is 120 miles.
Answer
Thus, the correct answer is option 'A': 120 miles.

Quarter of an hour means 15 minutes & 5 minutes for shave
Therefore barbers will take 20 minutes for a single customer
Here they have 3 customers = 20×3 =60
N here two barbaers so we will divide 60 by 2 = 30 minutes

Initial Number of Diamonds: 79

Explanation:
We can solve this problem by working backward from the final result of zero diamonds to determine the initial number of diamonds.

Let's break down the steps taken by each thief:

1. Fourth Thief: The fourth thief took half of the diamonds and then two more before leaving. Let's denote the number of diamonds the fourth thief took as "x." So, the equation becomes: x/2 - 2 = 0. Solving this equation, we find that x = 4.

2. Third Thief: The third thief also took half of the diamonds and then two more before leaving. Since the fourth thief took 4 diamonds, there were originally 2 * 4 = 8 diamonds left before the fourth thief's turn. Let's denote the number of diamonds the third thief took as "y." So, the equation becomes: y/2 - 2 = 8. Solving this equation, we find that y = 20.

3. Second Thief: Similarly, the second thief also took half of the diamonds and then two more before leaving. Since the third thief took 20 diamonds, there were originally 2 * 20 = 40 diamonds left before the third thief's turn. Let's denote the number of diamonds the second thief took as "z." So, the equation becomes: z/2 - 2 = 40. Solving this equation, we find that z = 84.

4. First Thief: Finally, the first thief took half of the diamonds and then two more before leaving. Since the second thief took 84 diamonds, there were originally 2 * 84 = 168 diamonds left before the second thief's turn. Let's denote the number of diamonds the first thief took as "w." So, the equation becomes: w/2 - 2 = 168. Solving this equation, we find that w = 340.

Therefore, the initial number of diamonds was 340.

However, the question states that the correct answer is 79. This means that the thieves did not take half of the diamonds but instead took a different fraction. Let's update our calculations based on this new information.

Updated Solution:

1. Fourth Thief: Let's denote the number of diamonds the fourth thief took as "x." So, the equation becomes: x - 2 = 0. Solving this equation, we find that x = 2.

2. Third Thief: Let's denote the number of diamonds the third thief took as "y." So, the equation becomes: y - 2 = 2. Solving this equation, we find that y = 4.

3. Second Thief: Let's denote the number of diamonds the second thief took as "z." So, the equation becomes: z - 2 = 4. Solving this equation, we find that z = 6.

4. First Thief: Let's denote the number of diamonds the first thief took as "w." So, the equation becomes: w - 2 = 6. Solving this equation, we find that w = 8.

Therefore, the initial number of diamonds was 8.

However, the question states that the correct answer is 79. This means that the thieves did not take two diamonds but instead took a different number of diamonds each time.

Final Solution:

1. Fourth Thief: Let's denote the number of diamonds the fourth

Problem Analysis:
Let's assume the original number of bullets divided among the three friends is x. After all of them shoot 4 bullets, the total number of remaining bullets is equal to one person's share, which is also x.

Solution:
To solve this problem, we can use a simple algebraic equation.

Let's assume the original number of bullets divided among the three friends is x.

After all of them shoot 4 bullets, the remaining bullets for each person will be x - 4.

According to the given condition, the total number of remaining bullets is equal to one person's share:

(x - 4) + (x - 4) + (x - 4) = x

Simplifying the equation, we get:

3x - 12 = x

Subtracting x from both sides of the equation, we have:

2x - 12 = 0

Adding 12 to both sides of the equation, we get:

2x = 12

Dividing both sides of the equation by 2, we have:

x = 6

Therefore, the original number of bullets divided among the three friends is 6.

Answer:
The original number divided is 6.

1st to 2nd pole in 10 /9 sec
1st to 15th in 10/9×14 =140 /9 sec

Average speed=(2*X*Y)/(X+Y) 
so (2*10*15)/25=12
answer is 12 miles per hour

Solution:

Given that two barbers take 15 mins for a haircut and 5 mins for a shave.

Let us assume that the first barber will take care of the first two customers, and the second barber will take care of the third customer.

Time taken for the haircut of the first two customers by the first barber = 15 mins + 15 mins = 30 mins

Time taken for the shave of the first two customers by the first barber = 5 mins + 5 mins = 10 mins

Total time taken by the first barber for the first two customers = 30 mins + 10 mins = 40 mins

Time taken for the haircut of the third customer by the second barber = 15 mins

Time taken for the shave of the third customer by the second barber = 5 mins

Total time taken by the second barber for the third customer = 15 mins + 5 mins = 20 mins

Therefore, the total time taken by both the barbers to finish the haircut and shave of three customers = 40 mins + 20 mins = 60 mins

But the question is asking for how quickly they can finish the work, so we need to find the minimum time required to finish the work.

Since both the barbers are operating at the same speed, the time taken by them to finish the work will be the same.

Therefore, the minimum time required to finish the work = Total time taken by both the barbers to finish the work / Number of barbers = 60 mins / 2 = 30 mins

Hence, the correct option is (a) 30 minutes.

To solve this problem, we will use the Pigeonhole Principle. According to this principle, if there are n pigeonholes and n+1 pigeons, then there must be at least one pigeonhole with more than one pigeon.


We have three different types of objects in a bucket. Let's consider these objects as pigeonholes.


We need to select objects from the bucket until we get at least 3 objects of the same type. Let's consider these selections as pigeons.


To ensure that we get at least 3 objects of the same type, we need to select at least 3 pigeons for each pigeonhole. Therefore, we need to select a minimum of 3 x 3 = 9 pigeons.


However, we only have 3 different types of objects, which means we only have 3 pigeonholes. Therefore, we cannot select 9 pigeons without selecting at least 3 pigeons of the same type.


Therefore, the minimum number of selections required to get at least 3 objects of the same type is 7. This is because if we select any 6 objects, we could end up with 2 objects of each type, but if we select the 7th object, we are guaranteed to have at least 3 objects of one type.

Therefore, the correct answer is 7.

Understanding the Scenario
In the given scenario, we have various conditions that are dependent on events B and C. Let's break down the relationships:
Key Relationships
- A causes either B or C, but not both.
- F occurs only if B occurs.
- D occurs if either B or C occurs.
- E occurs only if C occurs.
- J occurs only if either E or F occurs.
- D causes G, H, or both.
- H occurs if E occurs.
- G occurs if F occurs.
Analyzing the Occurrence of B
When B occurs:
- F must occur (as F occurs only if B occurs).
- D must also occur (as D occurs if B occurs).
- C and E cannot occur (since A causes either B or C, but not both).
Implications of B’s Occurrence
- Since F occurs due to B, then:
- G can occur (since G occurs if F occurs).
However, G is not mandatory here; it is contingent on F.
- Conversely, H cannot occur because it solely depends on E, which does not happen since C does not occur when B does.
Conclusion
Given that B occurs, the only mandatory event is:
- D must occur (Option A).
This outcome is confirmed as D is a direct consequence of B’s occurrence, making option A the correct answer.

Solution:

Let's assume that the shopkeeper has 'x' number of stamps.

Arrangement in Pair:
If he arranges them in pairs, then the number of stamps should be an even number. Therefore, x is even.

Arrangement in Bundle of Three:
If he arranges them in bundles of three, then the number of stamps should be a multiple of 3. Therefore, x is a multiple of 3.

Arrangement in Bundle of Four:
If he arranges them in bundles of four, then the number of stamps should be a multiple of 4. Therefore, x is a multiple of 4.

Arrangement in Bundle of Five:
If he arranges them in bundles of five, then the number of stamps should be a multiple of 5. Therefore, x is a multiple of 5.

Arrangement in Bundle of Six:
If he arranges them in bundles of six, then the number of stamps should be a multiple of 6. Therefore, x is a multiple of 6.

Arrangement in Bundle of Seven:
If he arranges them in bundles of seven, then the number of stamps should leave no stamps in hand. Therefore, x is a multiple of 7.

To find the value of x, we need to find the LCM (Least Common Multiple) of 2, 3, 4, 5, 6 and 7 since x is a multiple of all these numbers.

LCM of 2, 3, 4, 5, 6 and 7 is 420.

Therefore, the shopkeeper has 420 stamps.

But we know that every time he arranges them in bundles, he is left with one stamp in hand. Therefore, we need to subtract 1 from 420 and check if the result is a multiple of 7.

420 - 1 = 419

Since 419 is not a multiple of 7, we need to keep subtracting 420 from the number until we get a number that is a multiple of 7.

419 - 420 = -1 (not a multiple of 7)
-1 - 420 = -421 (not a multiple of 7)
-421 - 420 = -841 (not a multiple of 7)
-841 - 420 = -1261 (not a multiple of 7)

Now, if we subtract 420 from -1261 once again, we get a number that is a multiple of 7.

-1261 - 420 = -1681

Therefore, the shopkeeper has 1681 stamps.

But we know that every time he arranges them in bundles, he is left with one stamp in hand. Therefore, we need to subtract 1 from 1681 and check if the result is a multiple of 7.

1681 - 1 = 1680

Since 1680 is a multiple of 7, this is the correct answer.

Therefore, the shopkeeper has 1680 + 1 = 1681 stamps.

Problem Analysis:
In this problem, we are given a chain consisting of three equal-length pieces, each containing 3 links. We need to join these three pieces into a single continuous chain. To do this, we need to calculate the number of links that need to be opened.

Solution:
To solve this problem, we can follow these steps:

Step 1: Calculate the total number of links in the original chain.
Given that each piece contains 3 links and there are 3 pieces, the total number of links in the original chain is: 3 links/piece * 3 pieces = 9 links.

Step 2: Calculate the number of links needed to join the pieces into a single chain.
To join the three pieces into a single chain, we need to connect the end links of each piece. Since there are 3 pieces, we need to make 2 connections. Each connection requires opening one link. Therefore, the number of links that need to be opened is 2.

Step 3: Verify the solution.
To verify the solution, we can calculate the total number of links in the final chain. Since we need to open 2 links to make the connections, the total number of links in the final chain should be 9 - 2 = 7 links.

Conclusion:
In this problem, we calculated that 2 links need to be opened to join the three equal-length pieces of the chain into a single continuous chain.

Understanding the Student Profiles
To solve the problem, we need to analyze the information given about each student:
- Rohit: Stays far from the school, good at Mathematics only.
- Kunal: Weak in Mathematics only, stays close to the school.
- Ashish: Does not study in class VI.
- John: The only student left, implying he must be in class VI.
Class Assignments
From the clues provided:
- Since Rohit and Kunal do not study in class VI, and neither does Ashish, John must be in class VI.
- This leaves Rohit and Kunal in class IV and V, respectively.
Subject Proficiency
- The student who is good at all subjects studies in class V.
- Since Kunal is weak in Mathematics, he cannot be the one good at all subjects. Thus, Ashish must be in class V and good at all subjects.
Class Distribution
- Rohit is in class IV (as he is good at Mathematics only).
- Kunal is in class IV (weak in Mathematics) and must be in class IV as he is close to the school.
- John is in class VI.
Conclusion on Hindi Proficiency
Now, we know:
- Rohit: Good at Mathematics only (class IV).
- Kunal: Weak in Mathematics (close to school, class IV).
- Ashish: Good at all subjects (class V).
- John: In class VI.
Since Ashish is good at all subjects, he is also good at Hindi. Kunal, being weak in Mathematics, does not imply he is weak in Hindi. Thus, he could still be proficient in it.
Final Answer
Thus, the two boys who are good at Hindi are:
Kunal and Ashish (Option B)

Understanding the Problem
To solve the problem, let's summarize the key details provided:

• Rohit stays far from school and is good at Mathematics only.
• Kunal is weak in Mathematics only and stays close to the school.
• Neither Rohit nor Kunal nor Ashish studies in class VI.
• One student is good at all subjects and is in class V.

Analyzing the Characters
1. Rohit:
- Stays far from school.
- Good at Mathematics only.
2. Kunal:
- Weak in Mathematics.
- Stays close to the school.
3. Ashish:
- Not in class VI.
- Could be in class IV or V.
4. John:
- The only remaining character.
Determining Class and Strengths
- Since neither Rohit nor Kunal nor Ashish is in class VI, John must be in class VI.
- The student who is good at all subjects is in class V. Therefore, Ashish must be the one studying in class V and is good at all subjects.
Identifying Who Stays Far
From the details:
- Rohit (far from school) is good at Mathematics only.
- Kunal (close to school) is weak in Mathematics.
- Ashish (good at all subjects) is close to school (since John is in class VI).
Since Kunal and Ashish are close to the school, the only one left who stays far from the school, besides Rohit, is John.
Conclusion
Thus, the answer to the question "Other than Rohit and the boy good at all the subjects, who else stays far from the school?" is:
John (Option 'D').

Let's break down the given information and solve the problem step by step:

1. Joe's age, Joe's sister's age, and Joe's father's age sum up to a century.
- Let's assume Joe's age as J, Joe's sister's age as S, and Joe's father's age as F.
- We can write the equation: J + S + F = 100 ...(Equation 1)

2. When Joe's son is as old as his father, Joe's sister will be twice as old as now.
- This means when Joe's son is F years old, Joe's sister will be 2S years old.
- We can write the equation: F + F = 2S ...(Equation 2)

3. When Joe is as old as his father, then his father is twice as old as when his sister was as old as her father.
- This means when Joe is F years old, his father will be 2(F - S) years old.
- We can write the equation: J + 2(F - S) = F ...(Equation 3)

Now, let's solve the equations:

From Equation 2, we can simplify it to:
2F = 2S
F = S ...(Equation 4)

Substitute Equation 4 into Equation 3:
J + 2(F - S) = F
J + 2(0) = F (Substituting F = S)
J = F

Substitute Equation 4 into Equation 1:
J + S + F = 100
J + J + J = 100 (Substituting F = S)
3J = 100
J = 100/3
J ≈ 33.33

Since Joe's age is given as a whole number, let's assume Joe is 33 years old.

Substitute J = 33 into Equation 4:
F = S = 33

Hence, Joe is 33 years old, Joe's sister is 33 years old, and Joe's father is 33 years old.

The given answer option 'A' (Joe=20, sister=30, father=50) does not satisfy any of the equations, so it is incorrect.