QuAnT-mAsTerZ

Question of the day - 27th August, 2009

2009-08-27T05:06:00.002+05:30

Question 1:

How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?
a)1875
b)2000
c)2375
d)2500
e)3875

Question 2:

6 persons seat themselves at round table. What is the probability that 2 given persons are adjacent?
(A) 1/5
(B) 2/5
(C) 1/10
(D) 1/7
(E) 2/15

OA will be posted soon.

Thanks,

Quant-Master

Question of the day - 26th August, 2009

2009-08-26T05:24:00.003+05:30

Question 1:

If x is an integer, is y an integer?
(1) The average (arithmetic mean) of x, y and y – 2 is x.
(2) The average (arithmetic mean) of x and y is not an integer.

Question 2:

Three types of pencils, J,K, and L, cost $0.05, $0.10, and $0.25 each, respectively. If a box of 32 of these pencils costs a total of $3.40 and if there are twice as many K pencils as L pencils in the box, how many J pencils are in the box?
(A) 6
(B) 12
(C) 14
(D) 18
(E) 20

Question 3:

Forty percent of the rats included in an experiment were male rats. If some of the rats died during the experiment and 30 percent of the rats that died were male rats, what was the ratio of the death rate among the male rats to the death rate among the female rats?
(A) 9/14
(B) 3/4
(C) 9/11
(D) 6/7
(E) 7/8

OA will be posted soon.

Thanks,

Quant-Master

Question of the day - 25th August, 2009

2009-08-25T04:45:00.002+05:30

Question 1:

What is the probability for a family with three children to have a boy and two girls, assuming the probability of having a boy or a girl is equal?

a)1/8
b)¼
c)½
d)3/8
e)5/8

Question 2:

X,Y > 0 ,If X^3 = Y, is Y a fraction?

(1) X^2 is a fraction
(2) X >Y

OAs will be posted soon

Thanks,

Quant-Master

Finding last two digit of a number

2009-08-24T23:14:00.003+05:30

sorry folks! it has been a long time since I posted fundas (busy in my new coaching job). anyways here is a funda to cover up my long pending quota

I am dividing this method into four parts and we will discuss each part one by one:

a. Last two digits of numbers which end in one
b. Last two digits of numbers which end in 3, 7 and 9
c. Last two digits of numbers which end in 2
d. Last two digits of numbers which end in 4, 6 and 8

Before we start, let me mention Binomial theorem in brief as we need it for calculation

(x+a)^n = Nc0*a^n+Nc1*a^n-1*x+Nc2*a^n-2*x^2+….. where NcR =n!/(r!(n-r)!)

a. Last two digits of numbers which end in one

This is best explained with an example.

What are the last two digits of 31^786?

Solution: 31^786 = (30+1)^786 = 786C0*1^786+786C1*1^785*30+786C2*1^784*30^2…..

Note that in the above expression, after the first two terms, all the terms will have two or more 0’s at the right hand side (since the expression is multiplied by 30^x). Hence to find the last two digits, only first two terms needs to be evaluated. First term equals to 1 and second term equals to 786*30 = 23580 hence last two digits will be 81 (which is last two digits of 23580 and +1 hence 80+1)

Wiyee!! That’s really long isn’t it? Well those who need a shortcut of this read on!!

Multiply the tenth digit ( 3 in the above example) with the unit digit of the exponent (6 in the above example) to get the tenth digit. Unit digit is always one when a number ends in 1.

Now it’s a two second job isn’t it??

Few exercises for you:

41^2789

71^56747

51^456*61^567

Post your answers using comment option

Now I have covered 25% of this exercise i.e. part A alone. Based on your response, if you find this useful, I will go ahead and post the approach for below kind of numbers

b. Last two digits of numbers which end in 3, 7 and 9
c. Last two digits of numbers which end in 2
d. Last two digits of numbers which end in 4, 6 and 8

Thanks,

Quant-Master

Question of the day - 22nd August, 2009

2009-08-21T23:18:00.000+05:30

In a certain game played with red chips and blue chips, each red chip has a point value of X and each blue chip has a point value of Y, where X>Y and X and Y are positive integers. If a player has 5 red chips and 3 blue chips, what is the average (arithmetic mean ) point value of the 8 chips that the player has?

(1) The average point value of one red chip and one blue chip is 5.
(2) The average point value of the 8 chips that the player has is an integer.

OA will be posted soon.

Thanks,

Quant-Master

Question of the day - 21st August, 2009

2009-08-20T23:51:00.002+05:30

Question.1:

If x and y are positive, is x^3 > y

(1)square root of x >y
(2)x > y

Question.2:

Material A costs $3 per kilogram, and material B costs $5 per kilogram. If 10 kilograms of material K consists of x kilograms of material A and ykilograms of material B, is x > y?

(1) y > 4
(2) The cost of the 10 kilograms of material K is less than $40.

Answers will be posted in the blog after 12 hrs.

Thanks,

Quants-Master

Question of the day - 18th August, 2009

2009-08-18T00:59:00.001+05:30

In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

a. 19

b. 10

c. 9

d. 8

e. 7

OA will be posted very soon

Thanks,

Quant-Master

Permutation and Combination - Step by Step approach - 3

2009-08-18T00:23:00.004+05:30

Hi,

In continuation to approach in P&C here are some more questions to continue with

1. In how many ways can you post 10 letters in 4 letterboxes?

2.Four friends go to a city in which there are 10 hotels

a.In how many ways can they stay?

b.In how many ways can they stay if no two friends stay together?

c.In how many ways can they stay if atleast two friends stay together?

Post your answers to the above questions. OA will be posted soon.

Thanks,

Quant-Master

Permutation & Combination - Principle of counting - part 2

2009-08-15T12:21:00.004+05:30

Hi,

Always remember this basic concept of counting

If one operation can be performed in m ways and a second operation can be performed in n ways, the number of ways of performing both the operations will be m*n and the number of ways of performing either one of the operations will be m+n. So all you need to remember is
1. In case of OR we add
2. In case of AND we multiply

Examples based on above concept:
The routes between cities A, B, C, D, E and F is shown below

1. In how many ways can you travel from City A to City F via city B?

2. In how many ways can you travel from City A to City F via city C?

3. Total number of ways to travel from city A to city F?

Post your answers using comment option.

Note: Not giving options to above questions is intentional, as I don't want you to work from solution to the question

Thanks,

Quant-Master

Permutation and Combination - Step by Step approach - 1

2009-08-13T23:49:00.004+05:30

hi,

My posts on Permutation and combination begins from today. After serious thoughts, I concluded that the best way to go about P & C is to proceed question by question approach. I will post questions and P&C daily and while explaining the OA I will try to explain the basic concepts and fundas. The best way to start is to start simple

Let's get started than

Mr. X has come to a garment store to buy dress. Store has 20 trousers and 30 shirts. In how many ways he can buy

1) 1 shirt and 1 trouser

2) only one garment out of these

Post your solution and also let me know if you have any problem in approaching this problem

Thanks,

Quant-Master

Question of the day - 12th August

2009-08-12T23:59:00.003+05:30

Question:

What is the unit digit of 1007*1034*1221*15343*100009*10034*1091*1191*1233?

a. 6

b. 0

c. 4

d. 2

e. 8

Finding answer to this question should not take more than 15-20 secs. If you spend more than half a min than you need to check out the idea behind this question. Solution with the shortcut will be posted in about 12 hrs time in this blog while in other forums it will be posted in 24 hrs time.

Thanks,

Quant-Master

Finding unit digit - it's all this funda - PaRt-2

2009-08-11T22:49:00.002+05:30

Okay!! So you now know how to find the unit digit of a number of the form x^y that ends in 1, 2, 5, 6,.

Before proceeding just see if you can get answers to the below problems

1) 128975^478z

2) 1928321^67846y

3) 8726352^1234

4) 876192736^xyz

Wiyee!! You got all correct than let’s proceed. If you are not sure about any of the above answer than have a look at my post on finding unit digit of a number – Part 1

Now let’s have a look at the number 4 and 9.

Why I selected these two particular numbers?

Just like 1, 5 and 6 these two numbers also have something in common.

Both 4 and 9 have cycles of 2.

4^1 = 4

4^2 =16

4^3 =64

4^4 = 256

Ahh! There you see. Unit digit repeats itself for every two powers. So 4^2k+1 will have 4 as unit digit and 4^2k will have unit digit as 6. To put it in more simple words, 4 raised to even power will have 6 as unit digit and 4 raised to odd power will have 4 as unit digit.

Now if you calculate for 9 it will also yield similar pattern

9^1 = 9

9^2 = 81

9^3 =729

9^4 = 6561

So like you can see 9^x will have unit digit 9 or 1. 9 raised to odd power will have 9 as unit digit and 9 raised to even power will have 1 as unit digit.

So we are done with numbers 1,2,4,5,6 and 9.

Let’s have a look at 3,7 and 8.

What do these numbers have in common?

Just like 2, they have a cycle of 4.

Let’s try 3

3^1 = 3

3^2 = 9

3^3 =27

3^4 = 81

3^5 = 243

Okay! If you can see the unit digit repeating after 3^4, you got the trick.

3^4k+1 will end in 3

3^4k+2 will end in 9

3^4k+3 will end in 7

3^4k will end in 1

Similarly for 7 and 8

7^4k+1 will end in 7

7^4k+2 will end in 9

7^4k+3 will end in 3

7^4k will end in 1

8^4k+1 will end in 8

8^4k+2 will end in 4

8^4k+3 will end in 2

8^4k will end in 6.

There ends my responsibility. Now it’s your turn.

Find out the unit digit for following numbers

1625251^xyz

124352^5416

12735373^267255

1636284^2a

16273835^n

1763826^2n

172637^27635

1738273268^172282

17283639^2b+1

Let me know if you have any questions. My next post will be on permutation and combination.

Happy blogging,

Quant-Master

Quants- Weekend test - 1

2009-08-10T00:58:00.005+05:30

Hi Guys,

Here is a test to bring you back to preparation mode from weekend mood.

1)If x and y are integers and x*y^2 is a positive odd integer, which of the following must be true?

Ⅰ. xy is positive.
Ⅱ. xy is odd.
Ⅲ. x + y is even.

(A) Ⅰ only
(B) Ⅱ only
(C) Ⅲ only
(D) Ⅰ and Ⅱ
(E) Ⅱ and Ⅲ

2)If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20

B. 40

C. 50

D. 80

E. 120

3)Last Sunday a certain store sold copies of Newspaper A for $1.00 each and copies of Newspaper B for $1.25 each, and the store sold no other newspapers that day. If rpercent of the store’s revenues from newspaper sales was from Newspaper A and if ppercent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?

A. 100p / (125 – p)
B. 150p / (250 – p)
C. 300p / (375 – p)
D. 400p / (500 – p)
E. 500p / (625 – p)

4)Alice's take home pay last year was the same each month and she saved the same fraction of her take home pay each month.The total amount of money that she had saved at the end of the year was 3 times the amount of that portion of her monthly take home pay that she did not save. If all the money that she saved last year was from her take home pay, what fraction of her take home pay did save each month.

A 1/2
B 1/3
C 1/4
D 1/5
E 1/6

5)A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?

A. 42
B. 70
C. 140
D. 165
E. 315

6)On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation, the probability of rain on each day is 0.2, what is the probability that Malachi will return home at the end of the day on the following Monday?

(A) 0.008

(B) 0.128

(D) 0.512

(E) 0.640

7)The perimeter of a certain isosceles right triangle is 16+16 sqr root 2.
What is the lenght of the hypotenuse of the triangle.

A 8

B 16

C 4 sqr root 2
D 8 sqr root 2
E 16 sqr root 2

8)The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

A. 1,104
B. 972
C. 672
D. 294
E. 192

9)If n is a multiple of 5 and n=p^2q, where p and q are prime numbers, which of the following must be a multiple of 25?
(a) p^2
(b) q^2
(c) pq
(d) p^2q^2
(e) p^3q

10)The number 75 can be written as the sum of the squares of 3 different positive intergers.What is the sum of these 3 integers.

a 17
b 16
c 15
d 14
e 13

Official answers will be posted by friday. Post/Discuss the answers using comment option. Here is a comparison chart to gauge your performance

Correct	Remarks
9+	Good
7-8	Above Average
4-6	Average
<4	Need to work harder

Thanks,

Quant-Master

Question of the day - 9th August, 2009

2009-08-09T00:42:00.002+05:30

Q. How many integers are divisible by 3 between 10! and 10!+20 both inclusive?

A. 6
B. 7
C. 8
D. 9
E. 10

I haven't posted this in any forum. Discuss the answers using comment option. OA will be posted as soon as I see good amount of replies. Also do look at the funda posted below this post. That might be very tricky when it comes to number questions

Thanks,

Quant-Master

Finding unit digit - it's all this funda - PaRt-1

2009-08-08T22:12:00.001+05:30

I was heading to post “How to find the last two digit of a number that appears in weird form say 37^234566” then I realized I am yet to post “How to find unit digit of a number”!! So stupid of me:D

Okay! Let’s get on with the business:

First let me ask you few questions. If I ask you to find unit digit of the below mentioned numbers, how comfortable you are?

31^167535

6756425^243538

200019182^266435483

Very comfortable

Comfortable

Not that comfortable

Not comfortable at all

If you selected any of the last two options then read on!!

Well, for all the three questions I posted above I just typed the numbers as my finger went. I never bothered what number I am typing. Know why? Because no matter what the number is we approach it in very much the same way.

Let’s get on with the work now!!

All the numbers have its cycles. You should know that to proceed further

If we take the number 1: 1 raised to any power will have unit digit 1. This also holds true for any number that ends in 1. Example 41^x, 5011^x, 611^x etc will have unit digit as 1.

Now you have the answer for the first question I posted. Unit digit of 31^167535 ??

It’s 1 of course.

Same property applies for 0, 5 and 6. Any number that ends in 0, 5 or 6 will have same unit digit.

What if I get any other number which is not mentioned above? How do I find unit digit?

Here it is how:

Let’s take the cycles of each number and try to analyze them

Let’s take 2 :

2^1 is 2

2^2 is 4

2^3 is 8

2^4 is 16

2^5 is 32

2^6 is 64

Now if you notice the unit digit is repeating after the power 4. So the cycle of 2 is 2, 4, 8, 6.

Hence cycle can be expressed in 4k form. Why should I take it in 4k form? Here is why

Find the unit digit of 2^1676.

1676 is a multiple of 4. Hence 1676 can be expressed as 4k. Therefore 2^4k will have unit digit as 6.

Let’s take another example 132^1233.

Forget how big the number is. Just consider 2^1233. 1233 when divided by 4 will give a remainder of 1. Hence it is of the form 4k+1. When the number is of the form 4k+1 it means a cycle of 4 is complete and hence the first number of the cycle will be its unit digit which is 2.

Now you should have answers to all my questions posted above

Try these numbers and post the answer using comment option

Find the unit digit of

1230^56909374

122255^3279495763

7865436^129084645

1232^1201

18972^1223

9874562^1782

Once you are clear with this, based on your response, I will post the approach to find unit digit of numbers ending in 3,4,7,8 and 9.

Thanks,

Quant-Master

Question of the day - 8th August, 2009

2009-08-08T12:02:00.004+05:30

Question 1:

For a less than b, b less than c and c less than 0, what is the value of a+b / b+c ?

a) greater than 1

b) less than 1

c) greater than 0 but less than 1

d) greater than a/c

e) greater than -1

Question 2: 1.If the operation * is one of the four Arithmatic operation addition,Subtraction,Multiplication and division.Is (6*2)*4=6*(2*4)

A)3*2>3
B)3*1=3

Hats off to people who gets 2nd one correct. Answers will be posted in 24 hrs time. Discuss the solution using comment option.

Thanks,

Quant-Master

Question of the day - 6th August

2009-08-06T23:50:00.002+05:30

Question no.1:
3.2@#6 : If @ and # each represent single digits in the given decimal, what digit does @ represent?
(1) When the decimal is rounded to the nearest tenth, 3.2 is the result.
(2) When the decimal is rounded to the nearest hundredth, 3.24 is the result.

Question no.2:
If p is a positive integer, what is the remainder when p is divided by 4?
(1) When p is divided by 8, the remainder is 5.
(2) p is the sum of the squares of two positive integers.

Answer will be posted in 24 hrs time
Thanks,
Quant-Master

Question of the day - 5th August, 2009

2009-08-06T00:32:00.002+05:30

Question 1:

Are at least 10% of the people in Country X who are 65 years old or older employed?

1) In Country X, 11.3% of the population is 65 years old or older.

2) In Country X, of the population 65 years old or older, 20% of the men and 10% of the women are employed.

Question 2:

If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?

A. 5 to 4
B. 3 to 2
C. 2 to 1
D. 5 to 1
E. 6 to 1

Detailed explanation will be posted in 24 hrs time. Discuss the answers using comment option

Thanks,

Quant-Master

Small but worth knowing funda

2009-08-06T00:15:00.001+05:30

Why is sum of the factors of an perfect square is always odd?

If you are still guessing then read on

If Perfect Square N is odd: An odd number will not have even factors and since square number will have odd number of factors. Sum of all the odd factors will be odd

If Perfect Square N is even: This is bit tricky! We have to find number of even and odd factors. Lets say 100. 100=2²*5².

To find number of odd factors in 100 take 2² out of 100 (since odd factors will not have even multiple) we are left with 5² which will have 3 factors (2+1). Hence we have 3 odd factors in 100. Number of factors in 100 is (2+1)*(2+1) = 9 (I am adding one to the exponent of 2 and 5 and this is the standard method of finding number of factors)

We found that total number of factors is 9 and there are 3 odd factors hence we will have 6 even factors (9-3). Now 6 even factors added together will be an even number and 3 odd factors added together will be an odd number Hence Even+Odd = Odd

Hope this is clear

Thanks,
Quant-Master

Question of the day - 4th August, 2009 - Bonus

2009-08-05T02:19:00.001+05:30

The Question:

Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?

1.5

1.7

2.3

2.5

3.0

Discuss your answers using comment option. OA will be posted in 24 hrs time

Thanks,

Quant-Master

Question of the day - 4th August, 2009

2009-08-05T02:03:00.001+05:30

Question 1

If xyz ≠ 0, is x (y + z) ≥ 0?

(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|

Question 2

How many different prime factors does N have?

(1) 2N has 4 different prime factors.
(2) N ^2 has 4 different prime factors.

Answers will be posted in 24 hrs. Check back for detailed explanation

Thanks,

Quant-Master

In order to check prime number status...

2009-08-04T00:23:00.004+05:30

You must have been following this to check whether a number is prime

1. Pick a number n.
2. Start with the least prime number, 2. See if 2 is a factor
of your number. If it is, your number is not prime.
3. If 2 is not a factor, check to see if the next prime, 3, is a factor. If it
is, your number is not prime.
4. Keep trying the next prime number until you reach one that is a
factor (in which case n is not prime), or you reach a prime
number that is equal to or greater than the square root of n.
5. If you have not found a number less than or equal to the square
root of n, you can be sure that your number is prime.

Very few people would know the funda behind it, well for those who don't here it is

To understand this you need to understand the concepts of factor.
Every number is made up of factors and for each factor there is factor-pair. For example lets take the number 18

18 is made of one two and two 3s.

factors of 18 can be expressed as below

1-----18
2------9
3------6

In the above representation (1,18),(2,9) and (3,6) are factor-pairs. When you multiply the factor pairs it will lead to the original number itself, in this case it's 18.

Generally for a number N, N² will have k factors less than N and k factors more than N. Now lets come to the root of your question. Let N² be a square of a number say N. Than there will be k factors less than N and k factors more than N.

For example let's take the square of a a number 6 which is 36. Now its factors are 1,2,3,4,6,9,12,18,36. Now if you notice there are 4 factors less than 6 and 4 factors more than 6 and for every factor less than 6 there is a factor pair which is more than 6.

Now if you take a prime number P there must be k factors less than square root of P and k factors more than square root of P. Since prime number will have only 1 and P as its factors there is no factor apart from 1 which is less than square root of P, this means that there can't be any factor more than square root of P as this will violate factor-pair concept. Hence you need not check for the prime numbers which is more than square root of P

Hope I am clear.

Thanks,
Quant-Master

A number says it all

2009-08-03T00:39:00.000+05:30

Here is some more fundas on numbers

factorial based questions asking no. of zeroes and max power of sum integer.

Find the no. of zeroes at the right end of 300!

for every zero, we require 10..n every 10 is made up of 5x2.
in the expression 1x2x3...300, multiples of 2 wud obviously be more than the multiples of 5...so v need to find the maximum power of 5 in 300!

300/5 = 60 (because every fifth no. is a multiple of 5)

300/25 = 12(because every mutiple of 25 has two 5s in it) or, 60/5=12

300/125 = 3 (because multiples of 125 have three 5s in it) or,
12/5 = 2

now 2 cannot be further divided by 5 so add all the quotients...60 + 12 + 2 = 74.

we might also get the same type of questions in a different form,

500! is divisible by 1000^n...what is the max. integral value of n?

now every 1000 is made up of 3 5s and 3 2s....2s are redundant...we need to count no. of 5s....so find total no. of 5s and divide by 3

500/5 = 100
100/5 = 20
20/5 = 4

100 + 20 + 4 =124

124/3 = 41.33

max integral value is 41.

500! is divisible by 99^n...what is the max. integral value of n?

now every 99 is made of two 3s and one 11. obviously 11 will be the deciding factor. so count no. of 11s for the answer

500/11 = 45
45/11 = 4

ans will be 49.

so in such questions, just check which prime no. will be the deciding factor and count the no. of times it occurs. but please understand that highest prime no. is not necessarily always the deciding factor. see this example:

100! is divisible by 160^n...what is the max. integral value of n?

now 160 = 2^5 * 5^1. now although 5 is the biggest prime no. that 160 is made of, the deciding factor wud be 2. because five 2s occur less often than one 5 does. so we'll count the no. of 2s and divide by 5.

100/2 = 50
50/2 = 25
25/2 =12
12/2 = 6
6 /2 = 3
3/2 = 1

add 'em all...97.

97/5 = 19.

so the answer wud be 19

had v taken 5 as the deciding factor, the answer wud have been 100/5 + 100/25 = 24 which is more than 19...hence a wrong answer...

when in dilemma as to which prime no. wud be the deciding factor (e.g. a divisor like 144...its not possible to decide whether 3 or 2 will give the right answer) ....take out answer using both the prime nos...the one thats less is the right answer.

50! is divisible by 144^n...what is the max. integral value of n?

144 = 2^4 * 3^2...difficult to decide whether 3 or 2 will be the deciding factor...

count 2s

50/2=25
25/2=12
12/2=6
6/2=3
3/2=1

sum=47

answer = 47/4 = 11.

count 3s

50/3=17
17/3=5
5/3=1

sum = 23

23/2 = 11

a tie...else the smaller value wud have been the answer.

300! is divisible by (24!)^n. what is the max. possible integral value of n?

such questions are tricky...when u expand 24!...u get 1x2x3...24.

in this range the highest prime no. is 23...so maximum power of 23 in 300! will decide the max value of x...

when v expand 300!...v get a 23 in 23, 46,69,92....

total no of multiples of 23 in 300! will be 300/23 = 13,

forget the fractional part. so the maximum possible answer is 13. hope am clear...else, feel free to revert.

256! is expanded and expressed in base 576 . how many zeroes will this expression have on its right end?

such questions are same as finding maximum power of 576 in 256!

576 = 2^6 x 3^2
to get six 2s i have to travel eight places...1x2x3x4x5x6x7x8 has seven 2s. but to two 3s i have to travel only six places...1x2x3...6 has two 3s...hence 2 will be the constrain.

total 2s in 256! = 255

hence, no. of zeroes = 256/6 = 42.

just to check...3s = 126, 126/2 = 63>42

ans-42

Questions based on this concept

400! is divisible by x^n. what is the max. possible integral value of n if the value of x is:

Q1. 300
Q2. 99
Q3. 500
Q4. 320
Q5. 770
Q6. 5200
Q7. 270
Q8. 686
Q9. 338
Q10. 13000

Answers... 49, 39, 33, 66, 39, 32, 65, 22, 16, 32 (the answer is not 33, this one is actually tricky! )

200! is divisible by (x!)^n...whats the max. possible value of n when x =

Q11. 25
12. 35
13. 50
14. 100
15. 70
16. 300
17.15

answers... 8, 6, 4, 2, 3, 0, 16

300! is expanded and expressed in base x. find the no. of zeroes at the right end of this expression when x=

18. 25
19. 15
20. 35
21. 39
22. 98

answers...37, 74, 48, 23, 24

Let me know if you have any queries.

Thanks,

Quant-Master

Question of the day - 2nd August, 2009

2009-08-03T00:18:00.004+05:30

Question 1:
Train A leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Train B, which is making the trip from Boston to New York at a constant speed. If Train B left Boston at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in New York?
(1) Train B arrived in New York before Train A arrived in Boston.

(2) The distance between New York and Boston is greater than 140 miles.

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

Question 2:

baseball' world series matches 2 teams against each other in a best of 7 games. The first team to win four games wins the series and no subsequent games are played. If you have no special info about either of teams what is the probability that the world series will consist of fewer than 7 games?

a. 12.5
b. 25
c. 31.25
d. 68.75
e. 75

Discuss the answer using comment option.Answers will be posted in 24 hrs time

Thanks,

Quant-Master

Question of the day - 1st August, 2009

2009-08-01T23:26:00.002+05:30

How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?

A. 3
B. 16
C. 75
D. 24
E. 26

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box , if at least one black ball is to be included in the draw?

A. 32
B. 48
C. 64
D. 96
E. 55

Discuss the answer using comment option. Official answer with explanation will be posted after 24 hrs

Numbers - PlaY WiTh TheM

2009-08-01T22:03:00.001+05:30

Q) A prime number greater than 3 is divided by 6. Which of the following can be the remainder of the mentioned division?

a) 1

b)2

c)4

d)5

e)both a and d

soln : To solve this problem all you need to know is a prime number greater than 3 can be expressed as 6k+1 or 6k-1. 6k+1 will give remainder 1 and 6k-1 will give 5 as a remainder. Hence E

Factors:

I asked my friend today to list all factors of 100 and this is what he did

1,2,4,5,10,20,25,50,100. This is how an untrained mind calculates factor. One important thing to note in factors is for every single factor you have one more corresponding factor known as factor pair. A trained mind should list factors in the form of factor pair. A factor pair will appear something like this

1	100
2	50
4	25
5	20
10	10

If you multiply a factor by one more factor you should get the number itself. In the above example if you multiply 1 with 100, 2 with 50,…10 with 10 you get 100 in all cases. This is how a trained mind should list factors and should not waste time in finding each and every factor. Just remember that for every factor there is one more factor which if multiplied with will lead to number itself.

Easy way of finding factors:

You guys must be aware of the formula to calculate number of factors of a number. Just to confirm we are learning same maths :D here is the formula. A number N of the form a^p*b^q*C^r will have (p+1)(q+1)(r+1). Hence a number 140 = 2^2*5^1*7^1 will have (2+1)(1+1)(1+1) = 12 factors.

How does this formula work? What if I forget the formula? Did I actually understood the concept behind this?

Well if you are not sure to any of the above questions than read on…

Lets take the same number 140 = 2^2*5^1*7^1. I can represent the factors of 140 in the below presented way

	2^0	2^1	2^2
5^0	1	2	4
5^1	5	10	20
7	7	14	28
7	35	70	140

From the above table we can easily see all the 12 factors. How did I prepared this table?

If the number consists of 3 prime numbers than take the least two prime numbers and represent in the above format. For example: I have represented 2^2 as 2^0 2^1 and 2^2 and take the second least prime number and represent it as the way I did for 5. Now 2^0*5^0=1, 2^1*5^0=2 ….it goes on. So you updated the data for first two prime numbers and you have got two rows of data. Now for the 3^rd prime number in this case being 7 just multiply 7 with the values you have obtained in first and 2^nd row,

Try Doing the same for the number 150.

If N^2 has got k factors less than N than it will have k factors more than N. Hence the total factors will be 2k+1. We add one coz N will also be a factor of N^2

Example: N^2 = 36. Hence N =6. Number of factors of 36 less than 6 is 1,2,3,4. K=4 hence total number of factors is 2*4+1=9

Question:

If N^2 has got 3 factors less than N than N^4 will have how many factors?

a)7

b)8

c)12

d)13

e)14

Soln: N^2 has got 7 factors hence the number is of the form a^6.

N^4 will be square of a^6 hence a^12. a^12 will have (12+1) 13 factors

Let me know if you have any questions

Thanks,

Quant-Master

Uniqueness is the key

2009-07-31T23:05:00.005+05:30

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 donâ€™t?

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

Â· 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.

Â· 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.

Â· The number DDD = D Ã— 111 = D Ã— 3 Ã— 37 = 3D Ã— 37.

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

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

Â· 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 letâ€™s check our solution- Multilying 37 and 27, we get 37 Ã— 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 Ã— 4 = DCBA is different from ABCD Ã— 4 = EFGH or ABCD Ã— 7 = DCBA.

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

Â· 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?)

Â· 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

Â· What can be the value of D? looking at the first and last digits of the multiplicand, we can see that 4 Ã— D gives the unit digit of 2 and 4 Ã— 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:

Â· 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:

Â· 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 Ã— 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

2009-07-31T00:58:00.006+05:30

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

2009-07-31T00:15:00.001+05:30

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