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I.S.I. and C.M.I. Entrance

What is the Area of Quadrilateral? | AMC 12 2018 | Problem 13

Here is a video solution for a Problem based on finding the area of a quadrilateral. This question is from American Mathematics Competition, AMC 12, 2018. Watch and Learn!

Here goes the question…

Connect the centroids of the four triangles in a square. Can you find the area of the quadrilateral?

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Done?

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I.S.I. and C.M.I. Entrance

Solving Weird Equations using Inequality | TOMATO Problem 78

Here is a video solution for ISI Entrance Number Theory Problems based on solving weird equations using Inequality. Watch and Learn!

Here goes the question…

Solve: 2 \cos ^{2}\left(x^{3}+x\right)=2^{x}+2^{-x}

We will recommend you to try the problem yourself.

Done?

Let’s see the proof in the video below:

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Greatest Positive Integer | AIME I, 1996 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Greatest Positive Integer.

Positive Integer – AIME I, 1996


For each real number x, Let [x] denote the greatest integer that does not exceed x,find number of positive integers n is it true that \(n \lt 1000\) and that \([log_{2}n]\) is a positive even integer.

  • is 107
  • is 340
  • is 840
  • cannot be determined from the given information

Key Concepts


Inequality

Greatest integer

Integers

Check the Answer


Answer: is 340.

AIME I, 1996, Question 2

Elementary Number Theory by Sierpinsky

Try with Hints


First hint

here Let \([log_{2}n]\)=2k for k is an integer

\(\Rightarrow 2k \leq log_{2}n \lt 2k+1\)

\(\Rightarrow 2^{2k} \leq n \lt 2^{2k+1}\) and \(n \lt 1000\)

Second Hint

\(\Rightarrow 4 \leq n \lt 8\)

\(16 \leq n \lt 32\)

\(64 \leq n \lt 128\)

\(256 \leq n \lt 512\)

Final Step

\(\Rightarrow 4+16+64+256\)=340.

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Algebra I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB

ISI MStat 2015 PSA Problem 17 | Basic Inequality

This is a problem from ISI MStat 2015 PSA Problem 17. First, try the problem yourself, then go through the sequential hints we provide.

Basic Inequality – ISI MStat Year 2015 PSA Question 17


Let \( X=\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}\). Then,

  • x<1
  • \( x>\frac{3}{2} \)
  • \( 1<x< \frac{3}{2} \)
  • None of these

Key Concepts


Basic Inequality

Check the Answer


Answer: is \( 1<x< \frac{3}{2} \)

ISI MStat 2015 PSA Problem 17

Precollege Mathematics

Try with Hints


Take it easy. Group things up. Use \(\frac{1}{n+k} < \frac{1}{n}\) for all natural (k)

\( \frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{1999} < 1000 \times \frac{1}{1000} \)
\( \frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\dots+\frac{1}{3001} < 1000 \times \frac{1}{2000}\)

\( \Rightarrow x < 1+ \frac{1}{2}= \frac{3}{2} \)

Again see that we can write , x=\( (\frac{1}{1001}+\frac{1}{3001}) + (\frac{1}{1002}+\frac{1}{3000})+ \dots + (\frac{1}{2000}+\frac{1}{2002}) + \frac{1}{2001} \) > \( \frac{2}{2001} +\frac{2}{2001} + \dots + \frac{2}{2001} +\frac{1}{2001} > \frac{2001}{2001}=1 \)

Hence , \( 1<x< \frac{3}{2} \).

ISI MStat 2015 PSA Problem 17
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Coordinate Geometry I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB

ISI MStat 2016 PSA Problem 9 | Equation of a circle

This is a problem from ISI MStat 2016 PSA Problem 9 based on equation of a circle. First, try the problem yourself, then go through the sequential hints we provide.

Equation of a circle- ISI MStat Year 2016 PSA Question 9


Given \( \theta \) in the range \( 0 \leq \theta<\pi,\) the equation \( 2 x^{2}+2 y^{2}+4 x \cos \theta+8 y \sin \theta+5=0\) represents a circle for all \( \theta\) in the interval

  • \( 0 < \theta <\frac{\pi}{3} \)
  • \( \frac{\pi}{4} < \theta <\frac{3\pi}{4} \)
  • \( 0 < \theta <\frac{\pi}{2} \)
  • \( 0 \le \theta <\frac{\pi}{2} \)

Key Concepts


Equation of a circle

Trigonometry

Basic Inequality

Check the Answer


Answer: is \( \frac{\pi}{4} < \theta <\frac{3\pi}{4} \)

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

Try with Hints


Complete the Square.

We get ,

\(2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3)) \)
\(6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2} \)

We are given that \( 0 \leq \theta<\pi,\) . So, \( {\sin^2 \theta} \geq \frac{1}{2} \) \( \Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4} \).

ISI MStat 2016 PSA Problem 9
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Consecutive positive Integers | AIME I, 1990| Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on consecutive positive integers.

Consecutive positive integer – AIME I, 1990


Someone observed that 6!=(8)(9)(10). Find the largest positive integer n for which n! can be expressed as the product of n-3 consecutive positive integers.

  • is 107
  • is 23
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Inequality

Algebra

Check the Answer


Answer: is 23.

AIME I, 1990, Question 11

Elementary Number Theory by David Burton

Try with Hints


First hint

The product of (n-3) consecutive integers=\(\frac{(n-3+a)!}{a!}\) for a is an integer

Second Hint

\(n!=\frac{(n-3+a)!}{a!}\) for \(a \geq 3\) \((n-3+a)! \geq n!\)

or, \(n!=\frac{n!(n+1)(n+2)….(n-3+a)}{a!}\)

Final Step

for a=4, n+1=4! or, n=23 which is greatest here

n=23.

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IIT JAM Statistics ISI M.Stat PSB Probability

ISI MStat 2016 Problem 5 | Order Statistics | PSB Sample

This is a beautiful problem from ISI MStat 2016 Problem 5 (sample) PSB based on order statistics. We provide a detailed solution with the prerequisites mentioned explicitly.

Problem– ISI MStat 2016 Problem 5

Let \( n \geq 2,\) and \( X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed Poisson \( (\lambda) \) random variables for some \( \lambda>0 .\) Let \( X_{(1)} \leq\) \( X_{(2)} \leq \cdots \leq X_{(n)}\) denote the corresponding order statistics.
(a) Show that \( \mathrm{P}\left(X_{(2)}=0\right) \geq 1-n\left(1-e^{-\lambda}\right)^{n-1}\)
(b) Evaluate the limit of \( \mathrm{P}\left(X_{(2)}>0\right)\) as the sample size \( n \rightarrow \infty \) .

Prerequisites

Solution

(a) Given , \( n \geq 2,\) and \( X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed Poisson \( (\lambda) \) random variables for some \( \lambda>0 .\) Let \( X_{(1)} \leq\) \( X_{(2)} \leq \cdots \leq X_{(n)}\) denote the corresponding order statistics.

Let , F(j) be the CDF of \( X_{1}, X_{2}, \ldots, X_{n}\) i.e CDF of Poisson \( (\lambda) \)

Then , Pmf of k-th Order Statistic i.e \( x_{(k)} \)

\( P(x_{(k)} = j)= F_{k} (j)-F_{k} (j-0) \) , where \( F_{k} (j) =P(x_{(k)} \le j) \) i.e the CDF of k-th Order Statistic

\( F_{k} (j) = \sum_{i=k}^{n} \) \({n \choose i} \) \( {(F(j))}^{i} {(1-F(j))}^{n-i} \)

So, \( P(x_{(k)} = j) = \sum_{i=k}^{n} {n \choose i} [ {(F(j))}^{i} {(1-F(j))}^{n-i}-{(F(j-0))}^{i} {(1-F(j-0))}^{n-i}] \)

Here we have to find , \( P(x_{(2)} = 0)= \sum_{i=2}^{n} {n \choose i} [{(F(0))}^{i} {(1-F(0))}^{n-i} – 0] \)

since , Poisson random variable takes values 0 ,1,2,…. i.e it takes all values < 0 with probabiliy 0 , that’s why \( {(F(j-0))}^{i} {(1-F(j-0))}^{n-i} =0\) here for j=0 .

And , \( F(0)=P(x \le 0) = P(X=0)={e}^{- \lambda} \frac{{\lambda}^{0}}{0!} ={e}^{- \lambda} \) , as X follows Poisson \( (\lambda) \) .

So, \( {(F(0))}^{i} {(1-F(0))}^{n-i}= {({e}^{- \lambda})}^{i} {(1-{e}^{- \lambda})}^{n-i} \)

Therefore , \( P(x_{(2)} = 0)= \sum_{i=2}^{n} {n \choose i} [{({e}^{- \lambda})}^{i} {(1-{e}^{- \lambda})}^{n-i} ] \)

\( = {({e}^{- \lambda}+1-{e}^{- \lambda})}^{n} – {n \choose 0}[{({e}^{- \lambda})}^{0} {(1-{e}^{- \lambda})}^{n-0} ]- {n \choose 1}[{({e}^{- \lambda})}^{1} {(1-{e}^{- \lambda})}^{n-1}] =1-{(1-{e}^{- \lambda})}^{n}- n {e}^{- \lambda} {(1-{e}^{- \lambda})}^{n-1} \)

\( =1-{(1-{e}^{- \lambda})}^{n-1}[1-{e}^{- \lambda} +n{e}^{- \lambda}] \)

\( =1-{(1-{e}^{- \lambda})}^{n-1}[1+(n-1){e}^{- \lambda}] \ge 1- n{(1-{e}^{- \lambda})}^{n-1} \) .

Since , \( 1+(n-1){e}^{- \lambda} \le n \Longleftrightarrow {e}^{ \lambda} \ge 1 \) for \( n \ge 2\) and \( \lambda >0 \) which is true hence our inequality hold’s true (proved)

Hence , \( \mathrm{P}\left(X_{(2)}=0\right) \geq 1-n\left(1-e^{-\lambda}\right)^{n-1}\) (proved )

(b) \( 0 \le P(x_{(2)} >0) =1-P(x_{(2)}= 0) \) \( \le 1-1+n\left(1-e^{-\lambda}\right)^{n-1}\) ( Using inequality in (a) )

So, \( 0 \le P(x_{(2)} >0) =1-P(x_{(2)}= 0) \) \( \le n\left(1-e^{-\lambda}\right)^{n-1}\) —–(1)

As \( 0< 1-{e}^{- \lambda} <1\) for \( \lambda >0 \) i.e it’s a fraction so it can be written as \( \frac{1}{a} \) for some \( a>1\) , Hence \( \lim_{n\to\infty} n\left(1-e^{-\lambda}\right)^{n-1} = \lim_{n\to\infty} \frac{n}{a^n} =0 \) (Proof -Use l’hospital rule or think intutively that as n tends to infinity the exponential functions grows more rapidly than any polynomial function ).

Now taking limit \( n \to \infty \) in (1) , we get by squeeze (or sandwichtheorem

\( \lim_{n\to\infty} P(x_{(2)} >0) =0 \)

Categories
Algebra Arithmetic Functions I.S.I. and C.M.I. Entrance ISI Entrance Videos

Periodic Function | TOMATO B.Stat Objective 710

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Periodic Function.

Periodic Function (B.Stat Objective Question )


If f(x) = \(a_0+a_1cosx+a_2cos2x+….+a_ncosnx\) where \(a_0,a_1,….,a_n\) are non zero real numbers and \(a_n > |a_0|+|a_1|+….+|a_{n-1}|\), then number of roots of f(x)=0 in \( 0 \leq x \leq 2\pi\), is

  • 0
  • at least 2n
  • 53361
  • 5082

Key Concepts


Periodic

Real Numbers

Inequality

Check the Answer


Answer:at least 2n

B.Stat Objective Problem 710

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

f is periodic with period \(2\pi\)

here \(0 <|\displaystyle\sum_{k=0}^{n-1}a_kcos(kx)| \leq \displaystyle\sum_{k=0}^{n-1}|a_k|<a_n, x\in\)set of reals

Second Hint

for points \(x_k=\frac{k\pi}{n}\) \(1 \leq k \leq 2n\)

\(f(x_k)=a_n[cos(k\pi)+\theta], |\theta|<1 for 1 \leq k \leq 2n\)

[ since cos \(\theta\) is periodic and f(x) is expressed for every point x=\(x_k\)]

Final Step

f has at least 2n points in such a period interval where f has alternating sign.

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Algebra Arithmetic Math Olympiad PRMO

Integers and Inequality | PRMO 2017 | Question 7

Try this beautiful problem from the Pre-RMO, 2017 based on Integers and Inequality.

Integers and Inequality – PRMO 2017


Find the number of positive integers n such that \(\sqrt{n}+\sqrt{n+1} \lt 11\)

  • is 107
  • is 29
  • is 840
  • cannot be determined from the given information

Key Concepts


inequality

Integers

Algebra

Check the Answer


Answer: is 29.

PRMO, 2017, Question 7

Elementary Algebra by Hall and Knight

Try with Hints


First hint

here \(\sqrt{n}+\sqrt{n+1} \lt 11\) for n=1,2,3,4,5,6,7,8,….,16,…..25

taking \(\sqrt{n}+\sqrt{n+1}=11\) is first equation

Second Hint

\(\Rightarrow \frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{1}{11}\)

\(\Rightarrow \sqrt{n+1}-\sqrt{n}=\frac{1}{11}\) is second equation

adding both equations \(2\sqrt{n+1}\)=\(\frac{122}{11}\)

Final Step

\(\Rightarrow n+1 = \frac{3721}{121}\)

\(\Rightarrow n=\frac{3600}{121}\)

=29.75

\(\Rightarrow 29 values.\)

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Algebra Arithmetic Calculus I.S.I. and C.M.I. Entrance ISI Entrance Videos

Series and Integers | B.Stat Objective | TOMATO 81

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Series and Integers.

Series and Integers (B.Stat Objective)


The value of \(\sum ij\), where the summation is over all i and j such that \(1 \leq i \lt j \leq 10\)

  • 2640
  • 1320
  • 3025
  • none of these

Key Concepts


Series

Integers

Inequality

Check the Answer


Answer: 1320

B.Stat Objective Question 81

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

required sum=1(2+3+4+….+10)+2(3+4+….+10)+3(4+5+….+10)+4(5+6+….+10)+5(6+7+….+10)+6(7+8+..+10)+7(8+9+10)+8(9+10)+9(10)

Second Hint

=54+104+147+180+200+204+189+152+90

Final Step

=1320

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