Uniqueness is the key

Why is it that you know all the formulas and have gone through a lot of problems associated with a particular chapter, and yet you are unable to solve a new problem when it comes across to you? What is it that those math geniuses possess that you don’t?

There are two differences between a mathematical mind and an ordinary mind:

· A mathematical mind keeps its armory of formulas and theorems very handy and in an assorted manner, ready to be recalled at an instant notice.

· A mathematical mind identifies the uniqueness to a problem and subconsciously searches through its repository of formulas to find one that fits to that uniqueness.

Which brings me to the most essential part of problem solving-

A problem is solvable or worth solving because there is a uniqueness to it

Identifying that uniqueness can often help us solving the problem. Here is a specimen:

In the Eqn given below, each letter uniquely represents a different digit:

(AB).(CB)=DDD

What is the value of A+B+C+D?

Where will you start?

What is unique about this problem?

Would this problem make sense if it was something like ?

Yes. As the unique part of this problem is the number DDD, we start with this number itself.

· The number DDD = D × 111 = D × 3 × 37 = 3D × 37.

· 37 is a two digit number and since it cannot be reduced further it can be one of the numbers. Let it be AB.

· Then CB is a number ending in 7 because both AB and CB have the same unit digit.

· CB = 3D, i.e. CB is a multiple of 3.

Which is the first two digit multiple of 3 you can think of that ends in 7? You got it right its 27

Now let’s check our solution- Multilying 37 and 27, we get 37 × 27 = 999. Therefore, our logic is correct and A = 3, B = 7, C = 2, and D = 9. And A + B + C + D = 21.

Broadly speaking, here are the steps that you should try in solving a problem, regardless of the topic that problem came from:

Here is another simple problem:

Find the four digit number ABCD, such that ABCDx4 =DCBA

Where will you start?

What is unique about this problem?

There are two unique points about this problem- first that the digits of the number are getting reversed and second that the number is being multiplied by 4. Notice that ABCD × 4 = DCBA is different from ABCD × 4 = EFGH or ABCD × 7 = DCBA.

Right now you can handle that the number is multiplied by 4 because you know something about properties associated with number 4. Let’s start with that.

· Any number multiplied by 4 will give us an even number. Hence, the digit D when multiplied by 4 will give us an even number. Since A is the unit digit of the product it is even. Hence, A = 2, 4, 6 or 8 (It cannot be 0. Why?)

· A is also the first digit of the multiplicand and if A = 4, 6 or 8 the product will become a 5 digit number. Hence A = 2. Writing the value of A we get

· What can be the value of D? looking at the first and last digits of the multiplicand, we can see that 4 × D gives the unit digit of 2 and 4 × 2 gives the first digit of D. Yes, you got it right. D = 8. Writing the multiplication again with the value of D we get:

· What can be the value of B? From your repository of formulas associated with 4 recall the one about divisibility of 4. A number is divisible by 4 if the number formed by the last two digits is divisible by 4. Since the number 8CB2 is a multiple of 4, the number B2 should be divisible by 4. Or, the number B2 = 12, 32, 52, 72 or 92. Hence the original number ABCD is 21C8, 23C8, 25C8, 27C8 or 29C8. But the last 4 numbers when multiplied by 4 will not give you the first digit of 8 in the product! Therefore B = 1 and the original number is 21C8. We write the multiplication again:

· Can you identify C now? Notice that when you multiply 8, the unit digit of 21C8, by 4 you write 2 in the unit digit of the product and carry 3. The tenths digit of the product is 1. Therefore, 4 × C + 3 (carryover) gives a unit digit of 1. Hence, C is 2 or 7. You can easily check by the hundreds digit in the product (which is C again) that C = 7.

Therefore, our answer is

Notice how we grabbed a uniqueness of the problem, started unraveling it, and slowly discovered the complete solution to the problem.

The beauty of the above two problems is that they require elementary knowledge of numbers but they will confuse many of you because you do not know where to start looking. Now you have your answer:

Thanks,

Quant- Master

Courtesy TG

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.

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.

1^st Minute => A is opened => fills 9 L

2^nd Minute => B is opened =>fills another 6 L

3^rd 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 46^th minute A is opened and it fills 9 litres. In the 47^th 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

1^st Minute => inlet pipe opened => 5l filled

2^nd 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 51^st 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 4^th 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 3^rd 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 20^th hour.

Thanks

Quant-Master

courtesy : Total Gadha

Vedic Maths - Helpful speedy calculation techniques

Hi,

I have compiled some speedy calculation techniques, you ppl might find it useful

Multiplications:

*) To multiply a number with 5 multiply with 10 and divide it by 2. For example 550*5 multiply by 10 = 5500 now half of it (divided by 2) is 2750.

*) To multiply a number by 25 multiply by 100 and divide by 4 for example 555*25 multiply by 100 = 55500 and now divide by 4 which is 13875

*)To multiply a number by 11. This is best explained by an example. Lets say 7469*11. You need not perform all the traditional multiplication just assume 0s at both the ends of 7469 hence now the number will appear as 074690. Now starting from right start adding last two digits. Hence the product should be 82159

*) To multiply a number by 12. Lets take the same number 7469. Now as in the above case assume 0s hence number will appear as 074690. Now starting from right double each digit and add to the right digit so the number will appear as 89628.

*) To multiply by 13 rule is same but instead of doubling you will triple the digits.

*) Squares of numbers which is one more than or less than the known square can be written this way. For example take 31²=30²+31st odd number =900+61= 961

Note: to find xth odd number multiply x with 2 and subtract 1 from it. For example 30th odd number is 30*2-1 =59

Finding unit digit: Lets put it in example again

What is the unit digit of 12*2*3*17*13*16. It will take 5-7 seconds to say that unit digit is 2

How? here it is. when you are finding unit digit just consider unit digit of each multiplication, In the above case u multiply 2 of the number 12 with the next number 2 we get 4 now u multiply this with next number 3 we get 12 just take unit digit which is 2 we multiply this with 7 of the number 17 we get 14 take unit digit 4 and multiply by 3 of the number 13 we get 12 take unit digit 2 and multiply with 6 of the number 16 we get 12 hence unit digit is 2.

Let me know if you find these useful. I am in the process of compiling more shortcuts and will also be compiling some quick trickies in other chapters which will be useful for GMAT.

Thanks

Quant-Master