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Sunday, January 9, 2011
Irregular verbs (Difference between Raise vs Rise , Lay vs Lie)
Raise vs. Rise
Both words can mean "to move upwards", but they are not interchangeable. "Rise" is an intransitive verb and "raise" is a transitive verb.
Intransitive verbs, like rise, do not require an object. "Rise" does not require an object to do the motion. E.g., the sun rises every morning; she rose from her nap around 2 o'clock. The sun is rising on its own as did the napping girl.
Transitive verbs, like raise, require an object. "Raise" requires an object to cause the motion. E.g., she raised her hand to answer the question; She raised the girl up onto his shoulders. The girl moved her own arm to answer the question and physically lifted the child onto his shoulders.
- Raise is a regular verb: raise, raised, raised
- Rise is an irregular verb: rise,rose, risen
Common Uses of Raise
- To elevate: She raised the bar in the competition.
- To lift something: Please raise your hand.
- To set upright by building: They raised the statue in her honor.
- To bring to maturity: She raised him all by herself.
- To increase: He raised his bet by five dollars.
Raise/Rise and Lay/Lie
There are similarities between "raise/rise" and "lay/lie".
"Raise" and "lay" both require an outside object to do the action. You raise something else up and lay something else down.
"Rise" and "lie" are done to oneself. You rise yourself (or the subject of the sentence) up and lie yourself (or the subject of the sentence) down.
Both of these examples require an object to do the motion.
- She raised up her arms in disgust.
- She laid the pillow on top of the bed.
Both of these examples do not have an object causing the motion
- They will rise up against their oppressors.
- All she wanted was to lie down on the sofa for 15 minutes.
Tuesday, July 28, 2009
Reading Comprehension (RC) Strategy- 2: Attacking the Passage
Once in a while during his CAT preparations, every MBA aspirant goes through a phase I call the ‘RC Madness’. Propelled by the battle cries of ‘start reading’, ‘increase your RC speed’, ‘build your vocabulary’, ‘read newspapers’ etc. he tries one thing after another to gain that elusive accuracy in reading comprehension passages. Things become worse if he joins a coaching institute; his classmates at his coaching are completely convinced that ˜RC ke bina to kucch nahin ho sakta” and his English trainer, who never cracked CAT himself or went to an IIM, keeps feeding him with ideas to attempt the RC passages. The trainer also keeps pressurizing him to become a good reader. The trainer believes that he is right about the reading part. The better reader the student becomes, the better he will be at RC passages.
Are better readers better at RC passages?
My answer is yes. But to me, this is NOT the question that a student should be asking. The correct question to ask is “how do I achieve cut-offs in verbal section of CAT?” Is this question not same as “how do I score well in RC passages?” Not really.The worst thing about the verbal section is that a student can attempt most of the given questions. This is not same as Quant or DI sections where he can attempt only a limited number of questions. Why is it a bad thing? Because, a student wastes time reading and doing many extra questions in which his accuracy will be poor rather than selecting some relatively easier questions and spending more time on them.
Let’s do some mathematics- In CAT 2006, there were 25 questions of 4 marks each. To achieve 96 percentile in the verbal section (the cut offs are at 93 percentile) you needed to score 37 marks. Let’s make it 40 marks. To achieve 40 marks you needed to attempt 10 questions at 100 percent accuracy. If you got one question wrong out of these 10, you would have still achieved your cut off. In essence, if you attempted 12 questions, you could afford to get 2 of them wrong. Attempting 12 questions meant attempting 50% of the section only. There were 15 RC questions, 5 sentence completion questions, and 5 inference, judgment questions. Technically speaking, if you spent your time only on sentence completion and inference questions, you had ample time to find the right answers to these questions. Since these 10 questions would have barely taken 20 seconds to read, nearly all of the time could have been spent in solving them. Which means that you could have avoided the RC completely! But is it a sane advice? No. My real point is that if a student has CHOSEN one RC passage (completely ignoring the two others) from the three given, he would have 15 questions at his disposal and he would have had ample time to attempt these questions with a good accuracy. Therefore, in this article, I hereby present the first point of my RC technique-
CHOOSE YOUR RC PASSAGE: This is the most important point in the verbal section. Don’t attempt every RC passage. Verbal section is like any other section- you need to select questions that you are going to answer. Remember that it takes 10 minutes on average to answer an RC passage completely. Attempting a tough RC passage consumes more time than you take for an average RC passage, makes you unsure of your answers, leaves you unsettled and nervous about the paper. It even tires you out mentally and increases the panic situation. The best strategy is to give a brief glance to each RC passage, and then select the ones you are going to attempt. Make a firm resolve that you will NOT look at the passages that you have excluded. How do you select a passage? I have only one criterion- I select passages that are easier to read. Different people will find different passages easier to read depending on their backgrounds.
In CAT 2006, had you attempted only one RC passage and ignored the two others completely, you would have 15 questions at your disposal and you would have 50 minutes to solve them. You would have had more time to think in every question and it would have increased your accuracy remarkably. Even a 75% percent accuracy in these 15 questions would have fetched you the cut offs.
Which also does not mean that you have to leave some RCs necessarily. In CAT 2008, I found the RCs the most easy. I attempted all of them and achieved my cutoffs in the verbal section easily. So modify my advice a little- Choose your questions.·
START SLOOOWWWLYYY: Most students start with an RC passage with a timer in their heads. Therefore, they try to hit the ground running. They want to finish the first paragraph as soon as they can, then move on to the second paragraph and finish it quickly… and so on. In the process, they lose track of the main idea of the passage, tone of the author, the structure of the passage etc. By the time they reach halfway through the passage, they lose track of the argument and get confused. These students do not understand that the most important part of a passage is the first paragraph. The first paragraph introduces the topic and the main idea of the passage, and very often states the author’s position on it. Start with the first paragraph SLOWLY! Do not move on to the second paragraph until you have understood the first one completely.
FORM AN OPINION AND TAKE A STAND: Do you realize how passively you absorb every line of a passage that was written by some highly opinionated jerk? Do you ever question whether the author of the passage is talking sense or feeding you crap? You become a punching bag whom someone is hitting with his ideas. For a change, why not start questioning the content of the passage? Why not start forming opinions about every paragraph in the passage and start taking sides? Trying to judge every paragraph will have the following effect:
In order to form an opinion, you will be forced to read the contents of a passage very minutely.
You will become aware of the logical points the author is trying to put across to convince you.
You will be forced to paraphrase every paragraph and summarize what the author is trying to say.
You will become a highly active reader, become interested in the passage and be able to maintain your focus throughout the passage.
CONNECT THE DOTS: While moving from one paragraph to another, be VERY aware of the logical flow of the passage. Be conscious about how the passage is structured. At any point in the passage, you should be able to say, ‘the author started from X, then moved to Y, then discussed Z to prove his point, then discussed some points against Z’… and connect every part of the passage with the previous parts. Practice this technique again and again until it becomes a habit with you.
GO BACK TO THE PASSAGE TO ANSWER EVERY QUESTION: Many errors in RC passages happen because a student facing crunch of time marks answers relying on his memory of the passage. He marks the options which he thinks are “similar to something he read in the passage.” Don’t do it. The person who made the test is counting on you to do that and he has deliberately put similar sounding options in the questions. Go back to the passage and check out every option. Yes. EVERY option. Let’s say that you are somewhat shaky about option b and even after checking the passage feel that it is probably right. But if you keep checking out the options you might find that option d is better than option b. So check every thing out. Be a suspicious bastard.
In the end, realize that practicing these techniques will save you more time during your CAT exam than will any other technique. Ignore those ‘skim and scan’ cries coming from your English instructor. He never cleared the CAT exam. I did. In 2005, I cracked every test paper I took; held all India ranks in top 10 throughout my mocks, cracked CAT, IIFT, and GMAT. And my marks were always the highest in the verbal section.
It was because I never listened to my English instructor. The best advice I got was from a fellow test taker- “RC is similar to cheating. All the information is given in the passage. You just have to copy it from there.”
And I kept cheating. I kept reading the passage completely.
Are better readers better at RC passages?
My answer is yes. But to me, this is NOT the question that a student should be asking. The correct question to ask is “how do I achieve cut-offs in verbal section of CAT?” Is this question not same as “how do I score well in RC passages?” Not really.The worst thing about the verbal section is that a student can attempt most of the given questions. This is not same as Quant or DI sections where he can attempt only a limited number of questions. Why is it a bad thing? Because, a student wastes time reading and doing many extra questions in which his accuracy will be poor rather than selecting some relatively easier questions and spending more time on them.
Let’s do some mathematics- In CAT 2006, there were 25 questions of 4 marks each. To achieve 96 percentile in the verbal section (the cut offs are at 93 percentile) you needed to score 37 marks. Let’s make it 40 marks. To achieve 40 marks you needed to attempt 10 questions at 100 percent accuracy. If you got one question wrong out of these 10, you would have still achieved your cut off. In essence, if you attempted 12 questions, you could afford to get 2 of them wrong. Attempting 12 questions meant attempting 50% of the section only. There were 15 RC questions, 5 sentence completion questions, and 5 inference, judgment questions. Technically speaking, if you spent your time only on sentence completion and inference questions, you had ample time to find the right answers to these questions. Since these 10 questions would have barely taken 20 seconds to read, nearly all of the time could have been spent in solving them. Which means that you could have avoided the RC completely! But is it a sane advice? No. My real point is that if a student has CHOSEN one RC passage (completely ignoring the two others) from the three given, he would have 15 questions at his disposal and he would have had ample time to attempt these questions with a good accuracy. Therefore, in this article, I hereby present the first point of my RC technique-
CHOOSE YOUR RC PASSAGE: This is the most important point in the verbal section. Don’t attempt every RC passage. Verbal section is like any other section- you need to select questions that you are going to answer. Remember that it takes 10 minutes on average to answer an RC passage completely. Attempting a tough RC passage consumes more time than you take for an average RC passage, makes you unsure of your answers, leaves you unsettled and nervous about the paper. It even tires you out mentally and increases the panic situation. The best strategy is to give a brief glance to each RC passage, and then select the ones you are going to attempt. Make a firm resolve that you will NOT look at the passages that you have excluded. How do you select a passage? I have only one criterion- I select passages that are easier to read. Different people will find different passages easier to read depending on their backgrounds.
In CAT 2006, had you attempted only one RC passage and ignored the two others completely, you would have 15 questions at your disposal and you would have 50 minutes to solve them. You would have had more time to think in every question and it would have increased your accuracy remarkably. Even a 75% percent accuracy in these 15 questions would have fetched you the cut offs.
Which also does not mean that you have to leave some RCs necessarily. In CAT 2008, I found the RCs the most easy. I attempted all of them and achieved my cutoffs in the verbal section easily. So modify my advice a little- Choose your questions.·
START SLOOOWWWLYYY: Most students start with an RC passage with a timer in their heads. Therefore, they try to hit the ground running. They want to finish the first paragraph as soon as they can, then move on to the second paragraph and finish it quickly… and so on. In the process, they lose track of the main idea of the passage, tone of the author, the structure of the passage etc. By the time they reach halfway through the passage, they lose track of the argument and get confused. These students do not understand that the most important part of a passage is the first paragraph. The first paragraph introduces the topic and the main idea of the passage, and very often states the author’s position on it. Start with the first paragraph SLOWLY! Do not move on to the second paragraph until you have understood the first one completely.
FORM AN OPINION AND TAKE A STAND: Do you realize how passively you absorb every line of a passage that was written by some highly opinionated jerk? Do you ever question whether the author of the passage is talking sense or feeding you crap? You become a punching bag whom someone is hitting with his ideas. For a change, why not start questioning the content of the passage? Why not start forming opinions about every paragraph in the passage and start taking sides? Trying to judge every paragraph will have the following effect:
In order to form an opinion, you will be forced to read the contents of a passage very minutely.
You will become aware of the logical points the author is trying to put across to convince you.
You will be forced to paraphrase every paragraph and summarize what the author is trying to say.
You will become a highly active reader, become interested in the passage and be able to maintain your focus throughout the passage.
CONNECT THE DOTS: While moving from one paragraph to another, be VERY aware of the logical flow of the passage. Be conscious about how the passage is structured. At any point in the passage, you should be able to say, ‘the author started from X, then moved to Y, then discussed Z to prove his point, then discussed some points against Z’… and connect every part of the passage with the previous parts. Practice this technique again and again until it becomes a habit with you.
GO BACK TO THE PASSAGE TO ANSWER EVERY QUESTION: Many errors in RC passages happen because a student facing crunch of time marks answers relying on his memory of the passage. He marks the options which he thinks are “similar to something he read in the passage.” Don’t do it. The person who made the test is counting on you to do that and he has deliberately put similar sounding options in the questions. Go back to the passage and check out every option. Yes. EVERY option. Let’s say that you are somewhat shaky about option b and even after checking the passage feel that it is probably right. But if you keep checking out the options you might find that option d is better than option b. So check every thing out. Be a suspicious bastard.
In the end, realize that practicing these techniques will save you more time during your CAT exam than will any other technique. Ignore those ‘skim and scan’ cries coming from your English instructor. He never cleared the CAT exam. I did. In 2005, I cracked every test paper I took; held all India ranks in top 10 throughout my mocks, cracked CAT, IIFT, and GMAT. And my marks were always the highest in the verbal section.
It was because I never listened to my English instructor. The best advice I got was from a fellow test taker- “RC is similar to cheating. All the information is given in the passage. You just have to copy it from there.”
And I kept cheating. I kept reading the passage completely.
Sunday, July 12, 2009
Time and Work
Problems on Time and Work are a common feature in most of the standard MBA exams. If you are well versed with the basics and have practised these problems during your preparation, they give you an easy opportunity to score and also save time. Here, I will try and give you the basic fundas with the help of examples. Let us start with a very basic problem:
Problem 1: A takes 5 days to complete a piece of work and B takes 15 days to complete a piece of work. In how many days can A and B complete the work if they work together?
Standard Solution: Let us consider Work to be 1 unit. So if W = 1 Unit and A takes 5 days to complete the work then in 1 day A completes 1/5th of the work. Similarly B completes 1/15th of the work.
If they work together, in one day A and B can complete (1/5 + 1/15 = 4/15) of the work. So to complete 1 unit of work they will take 15/4 days.
New method: Let us assume W = 15 units, which is the LCM of 5 and 15.
Given that total time taken for A to complete 15 units of work = 5 days
--> A’s 1 day work = 15/5 = 3 units
Given that total time taken for B to complete 15 units of work = 15 days
--> B’s 1 day work = 15/15 = 1 unit
-->(A + B)’s 1 day work = 3 + 1 = 4 units
-->15 units of work can be done in 15/4 days.
Many solve Time and Work problems by assuming work as 1 unit (first method) but I feel it is faster to solve the problems by assuming work to be of multiple units (second method). This would be more evident when we solve problems which are little more complex than the above one.
Problem 2: X can do a work in 15 days. After working for 3 days he is joined by Y. If they complete the remaining work in 3 more days, in how many days can Y alone complete the work?
Solution: Assume W = 15 units.
(Note: You can assume work to be any number of units but it is better to take the LCM of all the numbers involved in the problem so that you can avoid fractions)
X can do 15 units of work in 15 days
-->X can do 1 unit of work in 1 day
(Note: If I had assumed work as 13 units for example then X’s 1 day work would be 13/15, which is a fraction and hence I avoided it by taking work as 15 units which is easily divisible by 15 and 3)
Since X worked for 6 days, total work done by X = 6 days × 1 unit/day = 6 units.
Units of work remaining = 15 – 6 = 9 units.
All the remaining units of work have been completed by Y in 3 days
-->Y’s 1 day work = 9/3 = 3 units.
If Y can complete 3 units of work per day then it would take 5 days to complete 15 units of work. So Y takes 5 days to complete the work.
Problem 3: A, B and C can do a piece of work in 15 days. After all the three worked for 2 days, A left. B and C worked for 10 more days and B left. C worked for another 40 days and completed the work. In how many days can A alone complete the work if C can complete it in 75 days?
Solution: Assume the total work to be 600 units. (LCM of all the numbers)
Then C’s 1 day work = 8 units.
-->(A + B + C)’s 1 day work = 40 units.
A, B, C work together in the first 2 days
-->Work done in the first 2 days = 40 × 2 = 80 units
C alone works during the last 40 days
-->Work done in the last 40 days = 40 × 8 = 320 units
Remaining work = 600 – (320 + 80) = 200 units
This work is done by B and C in 10 days.
-->(B + C)’s 1 day work = 20 units
-->A’s 1 day work = (A + B + C)’s 1 day work – (B + C)’s 1 day work = 40 units – 20 units = 20 units
-->A can do the work of 600 units in 30 days.
Problem 4: Gerrard can dig a well in 5 hours. He invites Lampard and Rooney who can dig 3/4th as fast as he can to join him. He also invites Walcott and Fabregas who can dig only 1/5th as fast as he can (Inefficient gunners you see ) to join him. If the five person team digs the same well and they start together, how long will it take for them to finish the job?
Solution: Let the work be 100 units.
Gerrard’s 1 hour work = 100/5 = 20 units
Lampard and Rooney’s 1 day work = 3/4 × 20 = 15 units.
Fabregas and Walcott’s 1 day work = 1/5 × 20 = 4 units.
Þ In one day all five of them can do = 20 + 15 + 15 + 4 + 4 = 58 units of work. Hence they can complete the work in 100/58 days.
I hope you got the knack of it. Let us now see how to solve the second kind of problems in Time and Work – the MANDAYS problems.
In these kinds of problems we need to remember that the number of men multiplied by the number of days that they take to complete the work will give the number of mandays required to complete the work. The number of mandays required to complete a piece of work will remain constant. We will try and understand this concept by applying it to the next three problems.
A Very simple problem to start with:
Problem 5: If 10 men take 15 days to complete a work. In how many days will 25 men complete the work?
Solution: Given that 10 men take 15 days to complete the work. So the number of mandays required to complete the work = 10 × 15 mandays. So assume W = 150 mandays.
Now the work has to be done by 25 men and since W = 150 mandays, the number of days to complete the work would be 150/25 = 6 days.
Problem 6: A piece of work can be done by 8 boys in 4 days working 6 hours a day. How many boys are needed to complete another work which is three times the first one in 24 days working 8 hours a day?
Solution: Assume the first piece of work to be 8 × 4 × 6 = 192 boy-day-hours.
The second piece of work = 3 (The first piece of work) = 3 × 192 = 576 boy-day-hours. So W = 576 boy-day-hours.
If this work has to be completed in 24 days by working 8 hours a day the number of boys required would be 576/(24 × 8) = 3 boys.
Problem 7: X can do a piece of work in 20 days working 7 hours a day. The work is started by X and on the second day one man whose capacity to do the work is twice that of X, joined. On the third day another man whose capacity is thrice that of X, joined and the process continues till the work is completed. In how many days will the work be completed, if everyone works for four hours a day?
Solution: Since X takes 20 days working 7 hours a day to complete the work, the number of day-hours required to complete this work would be 140 day-hours. Like in the two problems above, this is going to be constant throughout. So, W = 140 day-hours.
Amount of work done in the 1st day by X = 1day × 4 hours = 4 day-hours
2nd day, X does again 4 day-hours of work. The second person is twice as efficient as X so he will do 8 day-hours of work. Total work done on second day = 8 + 4 = 12 day-hours. Amount of work completed after two days = 12 + 4 = 16 day-hours.
3rd day, X does 4 day-hours of work. Second Person does 8 day-hours of work. Third person who is thrice as efficient as X does 12 day-hours of work. Total work done on 3rd day = 4 + 8 + 12 = 24 day-hours
Amount of work completed after 3 days = 16 + 24 = 40 day-hours
Similarly on 4th day the amount of work done would be 4 + 8 + 12 + 16 = 40 day-hours
Work done on the 5th day = 4 + 8 + 12 + 16 + 20 = 60 day-hours
Total work done after 5 days = 4 + 12 + 24 + 40 + 60 = 140 day-hours = W.
So it takes 5 days to complete the work.
Remember that whenever there is money involved in a problem, the money earned should be shared by people doing the work together in the ratio of total work done by each of them. Again I will explain this with the help of an example:
Problem 8: X can do a piece of work in 20 days and Y can do the same work in 30 days. They finished the work with the help of Z in 8 days. If they earned a total of Rs. 5550, then what is the share of Z?
Solution: Let work W = 120 units. (LCM of 20, 30 and 8)
X’s 1 day work = 6 units
Y’s 1 day work = 4 units
(X + Y + Z)’s 1 day work = 15 units.
So Z’s 1 day work = 15 – (6 + 4) = 5 units
In 8 days Z would have completed 5 units/day × 8 days = 40 units of work
Since Z does 40/120 = 1/3rd of the work, he will receive 1/3rd of the money, which is 1/3 x 5550 = Rs. 1850.
Pipes and Cisterns
Problem 9: There are three hoses, A, B and C, attached to a reservoir. A and B can fill the reservoir alone in 20 and 30 mins, respectively whereas C can empty the reservoir alone in 45 mins. The three hoses are kept opened alone for one minute each in the the order A, B and C. The same order is followed subsequently. In how many minutes will the reservoir be full?
Solution: These kinds of problems can be solved in the same way as we solve problems where one or more men are involved. A, B and C are equivalent to three people trying to complete a piece of work.
The amount of work to be done would be the capacity of the reservoir. Lets assume capacity of the reservoir = W = 180 (LCM of 20, 30, 45) litres.
A can fill the reservoir in 20 mins Þ In 1 min A can fill 180/20 = 9 L. B can fill 180/60 = 6 L in a minute.
In one minute C can empty 180/45 = 4 L from the reservoir.
1st Minute => A is opened => fills 9 L
2nd Minute => B is opened =>fills another 6 L
3rd Minute => C is opened => empties 4 L
Hence every 3 minutes => (9 + 6 – 4 =) 11 litres are filled into the reservoir.
So in 45 minutes (11 × 15 =) 165 litres are filled.
In the 46th minute A is opened and it fills 9 litres. In the 47th minute B is opened and it fills 6 litres.
Hence the reservoir will be full in 47 minutes.
Problem 10: There is an empty reservoir whose capacity is 30 litres. There is an inlet pipe which fills at 5 L/min and there is an outlet pipe which empties at 4 L/min. Both the pipes function alternately for 1 minute. Assuming that the inlet pipe is the first one to function, how much time will it take for the reservoir to be filled up to its capacity?
Solution: The work to be done = Capacity of reservoir = W = 30 litres
1st Minute => inlet pipe opened => 5l filled
2nd minute => inlet pipe closed; outlet pipe opened => 4l emptied
In 2 minutes (5 litres -4 litres =) 1l is filled into the reservoir.
It takes 2 minutes to fill 1l => it takes 50 minutes to fill 25 litres into the tank.
In the 51st minute inlet pipe is opened and the tank is filled.
Problem 11: Sohan can work for three hours non-stop but then needs to rest for half an hour. His wife can work for two hours but rests for 15 min after that, while his son can work for 1 hour before resting for half an hour. If a work takes 50 man-hours to get completed, then approximately how long will it take for the three to complete the same? Assume all of them all equally skilled in their work.
(a) 15 (b) 17 (c) 20 (d) 24
Solution: W = 50 man-hours
Since all of them are equally skilled; in 1 hour they can do 3 man-hours of work if no one is resting.
It will take them 50/3 = 16.6 hours to complete the work if they work continuously.
But, since they take breaks the actual amount of time would > 17 hours.
Option (a) and (b) are ruled out.
Now let us calculate the amount of work done in 20 hours.
Sohan does 3 man-hours in every 3.5 hours (because he takes rest for half an hour on the 4th hour)
In 20 hours (3.5 × 5 + 2.5) Sohan completes => 3 × 5 + 2.5 = 17.5 man-hours ---- (1)
His wife completes 2 man-hours every 2.25 hours (because she rests on the 3rd hour)
In 20 hours (2.25 × 8 + 2) she completes => 2 × 8 + 2 = 18 man-hours. ---- (2)
Child completes 1 man-hours every 1.5 hour.
In 20 hours (1.5 × 13 + 0.5) he completes 1 × 13 + 0.5 = 13.5 man-hours of work. ------ (3)
Adding 1, 2 & 3
In approximately 20 hours 49 man-hours will be completed; so the work can be completed in 20th hour.
Problem 1: A takes 5 days to complete a piece of work and B takes 15 days to complete a piece of work. In how many days can A and B complete the work if they work together?
Standard Solution: Let us consider Work to be 1 unit. So if W = 1 Unit and A takes 5 days to complete the work then in 1 day A completes 1/5th of the work. Similarly B completes 1/15th of the work.
If they work together, in one day A and B can complete (1/5 + 1/15 = 4/15) of the work. So to complete 1 unit of work they will take 15/4 days.
New method: Let us assume W = 15 units, which is the LCM of 5 and 15.
Given that total time taken for A to complete 15 units of work = 5 days
--> A’s 1 day work = 15/5 = 3 units
Given that total time taken for B to complete 15 units of work = 15 days
--> B’s 1 day work = 15/15 = 1 unit
-->(A + B)’s 1 day work = 3 + 1 = 4 units
-->15 units of work can be done in 15/4 days.
Many solve Time and Work problems by assuming work as 1 unit (first method) but I feel it is faster to solve the problems by assuming work to be of multiple units (second method). This would be more evident when we solve problems which are little more complex than the above one.
Problem 2: X can do a work in 15 days. After working for 3 days he is joined by Y. If they complete the remaining work in 3 more days, in how many days can Y alone complete the work?
Solution: Assume W = 15 units.
(Note: You can assume work to be any number of units but it is better to take the LCM of all the numbers involved in the problem so that you can avoid fractions)
X can do 15 units of work in 15 days
-->X can do 1 unit of work in 1 day
(Note: If I had assumed work as 13 units for example then X’s 1 day work would be 13/15, which is a fraction and hence I avoided it by taking work as 15 units which is easily divisible by 15 and 3)
Since X worked for 6 days, total work done by X = 6 days × 1 unit/day = 6 units.
Units of work remaining = 15 – 6 = 9 units.
All the remaining units of work have been completed by Y in 3 days
-->Y’s 1 day work = 9/3 = 3 units.
If Y can complete 3 units of work per day then it would take 5 days to complete 15 units of work. So Y takes 5 days to complete the work.
Problem 3: A, B and C can do a piece of work in 15 days. After all the three worked for 2 days, A left. B and C worked for 10 more days and B left. C worked for another 40 days and completed the work. In how many days can A alone complete the work if C can complete it in 75 days?
Solution: Assume the total work to be 600 units. (LCM of all the numbers)
Then C’s 1 day work = 8 units.
-->(A + B + C)’s 1 day work = 40 units.
A, B, C work together in the first 2 days
-->Work done in the first 2 days = 40 × 2 = 80 units
C alone works during the last 40 days
-->Work done in the last 40 days = 40 × 8 = 320 units
Remaining work = 600 – (320 + 80) = 200 units
This work is done by B and C in 10 days.
-->(B + C)’s 1 day work = 20 units
-->A’s 1 day work = (A + B + C)’s 1 day work – (B + C)’s 1 day work = 40 units – 20 units = 20 units
-->A can do the work of 600 units in 30 days.
Problem 4: Gerrard can dig a well in 5 hours. He invites Lampard and Rooney who can dig 3/4th as fast as he can to join him. He also invites Walcott and Fabregas who can dig only 1/5th as fast as he can (Inefficient gunners you see ) to join him. If the five person team digs the same well and they start together, how long will it take for them to finish the job?
Solution: Let the work be 100 units.
Gerrard’s 1 hour work = 100/5 = 20 units
Lampard and Rooney’s 1 day work = 3/4 × 20 = 15 units.
Fabregas and Walcott’s 1 day work = 1/5 × 20 = 4 units.
Þ In one day all five of them can do = 20 + 15 + 15 + 4 + 4 = 58 units of work. Hence they can complete the work in 100/58 days.
I hope you got the knack of it. Let us now see how to solve the second kind of problems in Time and Work – the MANDAYS problems.
In these kinds of problems we need to remember that the number of men multiplied by the number of days that they take to complete the work will give the number of mandays required to complete the work. The number of mandays required to complete a piece of work will remain constant. We will try and understand this concept by applying it to the next three problems.
A Very simple problem to start with:
Problem 5: If 10 men take 15 days to complete a work. In how many days will 25 men complete the work?
Solution: Given that 10 men take 15 days to complete the work. So the number of mandays required to complete the work = 10 × 15 mandays. So assume W = 150 mandays.
Now the work has to be done by 25 men and since W = 150 mandays, the number of days to complete the work would be 150/25 = 6 days.
Problem 6: A piece of work can be done by 8 boys in 4 days working 6 hours a day. How many boys are needed to complete another work which is three times the first one in 24 days working 8 hours a day?
Solution: Assume the first piece of work to be 8 × 4 × 6 = 192 boy-day-hours.
The second piece of work = 3 (The first piece of work) = 3 × 192 = 576 boy-day-hours. So W = 576 boy-day-hours.
If this work has to be completed in 24 days by working 8 hours a day the number of boys required would be 576/(24 × 8) = 3 boys.
Problem 7: X can do a piece of work in 20 days working 7 hours a day. The work is started by X and on the second day one man whose capacity to do the work is twice that of X, joined. On the third day another man whose capacity is thrice that of X, joined and the process continues till the work is completed. In how many days will the work be completed, if everyone works for four hours a day?
Solution: Since X takes 20 days working 7 hours a day to complete the work, the number of day-hours required to complete this work would be 140 day-hours. Like in the two problems above, this is going to be constant throughout. So, W = 140 day-hours.
Amount of work done in the 1st day by X = 1day × 4 hours = 4 day-hours
2nd day, X does again 4 day-hours of work. The second person is twice as efficient as X so he will do 8 day-hours of work. Total work done on second day = 8 + 4 = 12 day-hours. Amount of work completed after two days = 12 + 4 = 16 day-hours.
3rd day, X does 4 day-hours of work. Second Person does 8 day-hours of work. Third person who is thrice as efficient as X does 12 day-hours of work. Total work done on 3rd day = 4 + 8 + 12 = 24 day-hours
Amount of work completed after 3 days = 16 + 24 = 40 day-hours
Similarly on 4th day the amount of work done would be 4 + 8 + 12 + 16 = 40 day-hours
Work done on the 5th day = 4 + 8 + 12 + 16 + 20 = 60 day-hours
Total work done after 5 days = 4 + 12 + 24 + 40 + 60 = 140 day-hours = W.
So it takes 5 days to complete the work.
Remember that whenever there is money involved in a problem, the money earned should be shared by people doing the work together in the ratio of total work done by each of them. Again I will explain this with the help of an example:
Problem 8: X can do a piece of work in 20 days and Y can do the same work in 30 days. They finished the work with the help of Z in 8 days. If they earned a total of Rs. 5550, then what is the share of Z?
Solution: Let work W = 120 units. (LCM of 20, 30 and 8)
X’s 1 day work = 6 units
Y’s 1 day work = 4 units
(X + Y + Z)’s 1 day work = 15 units.
So Z’s 1 day work = 15 – (6 + 4) = 5 units
In 8 days Z would have completed 5 units/day × 8 days = 40 units of work
Since Z does 40/120 = 1/3rd of the work, he will receive 1/3rd of the money, which is 1/3 x 5550 = Rs. 1850.
Pipes and Cisterns
Problem 9: There are three hoses, A, B and C, attached to a reservoir. A and B can fill the reservoir alone in 20 and 30 mins, respectively whereas C can empty the reservoir alone in 45 mins. The three hoses are kept opened alone for one minute each in the the order A, B and C. The same order is followed subsequently. In how many minutes will the reservoir be full?
Solution: These kinds of problems can be solved in the same way as we solve problems where one or more men are involved. A, B and C are equivalent to three people trying to complete a piece of work.
The amount of work to be done would be the capacity of the reservoir. Lets assume capacity of the reservoir = W = 180 (LCM of 20, 30, 45) litres.
A can fill the reservoir in 20 mins Þ In 1 min A can fill 180/20 = 9 L. B can fill 180/60 = 6 L in a minute.
In one minute C can empty 180/45 = 4 L from the reservoir.
1st Minute => A is opened => fills 9 L
2nd Minute => B is opened =>fills another 6 L
3rd Minute => C is opened => empties 4 L
Hence every 3 minutes => (9 + 6 – 4 =) 11 litres are filled into the reservoir.
So in 45 minutes (11 × 15 =) 165 litres are filled.
In the 46th minute A is opened and it fills 9 litres. In the 47th minute B is opened and it fills 6 litres.
Hence the reservoir will be full in 47 minutes.
Problem 10: There is an empty reservoir whose capacity is 30 litres. There is an inlet pipe which fills at 5 L/min and there is an outlet pipe which empties at 4 L/min. Both the pipes function alternately for 1 minute. Assuming that the inlet pipe is the first one to function, how much time will it take for the reservoir to be filled up to its capacity?
Solution: The work to be done = Capacity of reservoir = W = 30 litres
1st Minute => inlet pipe opened => 5l filled
2nd minute => inlet pipe closed; outlet pipe opened => 4l emptied
In 2 minutes (5 litres -4 litres =) 1l is filled into the reservoir.
It takes 2 minutes to fill 1l => it takes 50 minutes to fill 25 litres into the tank.
In the 51st minute inlet pipe is opened and the tank is filled.
Problem 11: Sohan can work for three hours non-stop but then needs to rest for half an hour. His wife can work for two hours but rests for 15 min after that, while his son can work for 1 hour before resting for half an hour. If a work takes 50 man-hours to get completed, then approximately how long will it take for the three to complete the same? Assume all of them all equally skilled in their work.
(a) 15 (b) 17 (c) 20 (d) 24
Solution: W = 50 man-hours
Since all of them are equally skilled; in 1 hour they can do 3 man-hours of work if no one is resting.
It will take them 50/3 = 16.6 hours to complete the work if they work continuously.
But, since they take breaks the actual amount of time would > 17 hours.
Option (a) and (b) are ruled out.
Now let us calculate the amount of work done in 20 hours.
Sohan does 3 man-hours in every 3.5 hours (because he takes rest for half an hour on the 4th hour)
In 20 hours (3.5 × 5 + 2.5) Sohan completes => 3 × 5 + 2.5 = 17.5 man-hours ---- (1)
His wife completes 2 man-hours every 2.25 hours (because she rests on the 3rd hour)
In 20 hours (2.25 × 8 + 2) she completes => 2 × 8 + 2 = 18 man-hours. ---- (2)
Child completes 1 man-hours every 1.5 hour.
In 20 hours (1.5 × 13 + 0.5) he completes 1 × 13 + 0.5 = 13.5 man-hours of work. ------ (3)
Adding 1, 2 & 3
In approximately 20 hours 49 man-hours will be completed; so the work can be completed in 20th hour.
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