Thursday, October 16, 2008
Mini Concept: Multinomial Theorem
x+y+z=n the number of non negative solutions of this equation will be
(n+3-1)C(3-1)=(n+2)C2
if we extend this to r variables then the formula becomes (n+r-1)C(r-1)
if we remove 0, means we need only positive integral solutions to the equation then we get formula as (n-1)C(r-1)
Lets take up one example.
On the occasion of Diwali, PAPA CHIPS is offering one of five prizes with every packet( the prize is inside the packet). the prizes include a pen, pencil, a CD, a movie ticket and a small game. Banta Singh is a fan of PAPA chips and he keeps buying the chips, what is the probability that Banta Singh gets all the five prizes by buying 12 packets of PAPA chips
Solution:
Chuck the story, the question is there are 5 variables and we need the solutions to the equation
a+b+c+d+e=12 ( non negative)
and a+b+c+d+e=12 ( positive integral)
The first case comes as every packet has a prize, and those 5 are the only kinds of prizes.
Second comes from that we need each kind of prize.
so the answer is 11C4/16C4=11!12!/(7!16!)=11.10.9.8/13.14.15.16= 33/182
Lets take another example
If the sum of 101 distinct terms in arithmetic progression is zero , in how many ways can three of these terms be selected such that their sum is zero?
Solution
it is obvious that the middle term is zero
so a(51)=0
so the terms are
-50D, -49D,....,-D, 0, D, ....49D, 50D
now the sum of 3 numbers to be zero
Case 1) if we pick 0, then we have to pick one positive and one negative, which must be equal except for the sign . so 50 ways
case 2) we leave 0 and pick two positive and one negative
then xD+yD-zD=0
x+y=z
z can vary from 1 to 50
we need positive solutions to the equation
which comes 0C1+1C1+2C1...+49C1
add this it will come to 50C2
case 3 it will be same as case 2
we get 50C2
hence total is 2.50C2+50=2500
Wednesday, October 15, 2008
Problem Of The Week 53
Friday, October 10, 2008
Problem Of The Week 52
,
what is the value of ?
Problem Of The Week 51
Let P(x)=ax^4 +bx^3+cx^2+dx+e be a polynomial with all integer coefficients and a=1. If√2+√5 is one of the roots of P(x)=0 , which of the following can be the value of |(b+c+d+e)|?
1) 103
2) 89
3) 63
4) 23
5) 5
Thursday, October 9, 2008
Problem Of The Week 50
1) 0
2) 1
3) 2
4) 3
5) None of these
Problem Of The Week 49
1) 6√2
2) 2√6
3) 4√6
4) 3√2
5) None of these
Wednesday, October 8, 2008
Problem Of The Week 48
1) 360
2) 60
3) 1296
4) 256
5) None of these
Problem Of The Week 47
1) 1
2 11
3) 22
4) 33
5) None of these
Problem Of The Week 46
Source: ALtheman's problem column
Monday, October 6, 2008
MockaMania Oct 5 2008
Question 1) For what value of n>0 does the following pair of equations yield exactly three solutions for y?
|x|=|y|
x=y^2+n(n+3)-4
1) 0 2) 1 3) 4 5) no unique vale 6) none of these.
Question 2
The sum of 2k+1 consecutive natural numbers is 2n such that the sum of first k+1 natural numbers among these (2k+1) numbers is same as next k numbers. Which of the following cannot be the 7th least natural number among the 2k+1 natural numbers, if k>=3
1) 231 2) 175 3) 535 4) 325 5) none of these
Question 3
if x=1+1/(x+1/(1+1/(x+...))) ; then which of the following bet represents x
1) 1<x<2 2) 1.5<x<2 3) 1<x<1.5 4) 1<x<1.2 5) 0<x<1
Question 4
let a(n) be q sequence such that n is an inetger and n>=1
a(n)-a(n+1)=a(n+4)-a(n+5), find a(80)-a(84) if a(75)=103 and a(83)=205
1)- 51 2) -36 ) -17 4) 75 5) cannot be determined.
Question 5
Maya has six indentical pots, which she is planning to arrange in a straight line in her showcase. before that each of these pots is to be colored either red or yellow or green or blue, such that at least one pot is coloured with each of the four colours. In how many different ways can she arrange the pots in the showcase so that now two pots of the same colour are adjacent?
1) 120 2) 84 3) 840 4) 600 5) 936
Question 6
The second rightmost digit of (102)^33 is
1) 8 2) 2 3) 4 4) 6 5) none of these
7) A number is said to be crazy number is the product of the digits is equal to products of the distinct primes in its prime factorisation. How many crazy numbers less than 100 have less than 3 distinct prime factors in their prime factorisation?
1) 4 2_ 5 3)8 4) 6 5) none of these
More to follow
Sunday, October 5, 2008
Problem Of The Week 45
Problem Of The Week 44
Saturday, October 4, 2008
Problem Of The Week 43
Friday, October 3, 2008
Problem Of The Week 42
Problem Of The Week 41
Thursday, October 2, 2008
Problem Of The Week 40
A) 3 B) 6 C) 9 D) 17 E) 20
Source : AMC 12
Problem Of The Week 39
A) 1 B) 3 C) 4 D) 5) E) Cannot be determined
Problem Of The Week 38
Problem Of The Week 37
Wednesday, October 1, 2008
Concept 6 Data Sufficiency
"The ultimate goal of mathematics is to eliminate any need for intelligent thought."-A. N. Whitehead
Well, for starters try to think what's the meaning of the above quote, it has a nice and beautiful meaning. To make things so simple that an average mind is able to see it. So lets put this into practice, with another concept lesson.
Today, we will discuss DATA SUFFICIENCY, one of the very scoring problems in cat and other mba entrance exams. The good thing about DS is we get DS both in quant and DI, and it makes for 6-8 problems in almost every paper. As there are no theorems in DS, we will take things to note
Things To Note:
1) DS problems, we need to answer if it is sufficient information to answer, means, there should be one conclusive answer. We do not need to find the answer, just if it can be found or not?
2) Some questions ask , is this true? So if we can find that the information available is enough to prove that it is not, we are still able to answer the question, that it is not true. Hence we are able to answer the question.
3) Check for all possibilties, that is using one statement, using second, then only combine the two.
Lets take up an example.
Instructions For DS Questions
Each question is followed by two statements, X and Y. Answer each question using the following instructions:
Mark (A) If the question can be answered by using the statement X alone but not by using the statement Y alone.
Mark (B) If the question can be answered by using the statement Y alone but not by using the statement X alone.
Mark (C) If the question can be answered by using either of the statements alone.
Mark (D) If the question can be answered by using both the statements together but not by either of the statements alone.
Mark (E) If the question cannot be answered on the basis of the two statements.
Example 6.1
In a triangle ABC, D is a point on the side BC and M, N are length of perpendicular dropped on line AD from the vertices B and C respectively. Is M > N?
(X) AB > AC
(Y) BD < DC
The problem is simple, but again see this, this question asks is M>N? The answer may be yes or no, but what we are concerned with is our ability to answer it and not the actual answer.
AB>AC gives us no idea, just imagine a few figures and you will know this.
Look at the other statement it says BD<DC, if BP and CQ are the perpendiculars then, BDP and CDQ are similar
so BD/DC=BP/CQ so we know the ratio and thus are able to answer and see this teh answer came NO.
so, we dont really want the answer, but the ability to give a unique answer.
So, answer is B)
This question was taken from QQAD practice test 4, here is the official solution.
Let D' be the midpoint of BC and let X' and Y' be the feet of the perpendicular from B and C to AD' respectively => X' = Y'. As D' shifts to right or left, we can know which of M or N is bigger. Thus, (Y) answers
(X) doesn't tell us anything
=> choice (B) is the right answer
Practice Problem 6.1
Is x=y?
X: (x+y)(1/x+1/y)=4?
Y: (x-50)^2=(y-50)^2
Practice Problem 6.2
What is the value of m and n?
X: n is an even number, m is an odd number, m>n
Y: mn=30
Note: Both the practice problems are old cat problems, enjoy!
Example 6.2
The distance of point P=(x,y,z) from the origin is sqrt(62) units, then find the coordinates of point P.
X: x+y+z=12
Y: x,y, and z are positive integers.
From original questions x^+y^2+z^2=62
From statement X
(x+y+z)^2=144
2(xy+yz+zx)=62
can't say for sure
From statement Y:
positive integers
but we can easily find two pairs (1,5,6), (2,3,7). can't find a unique solution
Option E
This question is taken from SIMCAT 9
Practice Problem 6.3
Rahim plans to draw a square JKLM with a point O on the side JK, but is not successful. Why is Rahim unable to draw the square?
X: The length of OM is twice that of OL
Y: The length of OM is 4 cm
This problem is taken for CAT 07 paper.
(X) p(i) – p(j) is not a composite number
(Y) p(2i) + p(2j) is a composite number
One of my favourite problems !
p(i) – p(j) is not a composite number
=>i^2-j^2 is a prime as i ,j are positive integers and i >j, ( i^2-j^2) can't be 1
=>(i+j)(i-j)= prime
so i-j=1
let p be the prime so i=(p+1)/2
j=(p-1)/2
clearly p is not 2 hence all p is odd
p(i) + p(j)=80 +(p^2+1)/2
now p^2=6k+1 ( Refer my lession on prime numbers for this )
therefore
p(i) + p(j)=80 +(p^2+1)/2
becomes
80+(6k+2)/2=81+3k=3(27+k)
so not a prime => can be answered by using X
now, p(2i) + p(2j) is a composite number
4(i^2+j^2+20) is composite
now i and j can be anything
can't make any conclusions
=> choice (A) is the right answer
This is QQAD pratice test problem!
Thats all, do post your queries and suggestions !
Good Luck!