Categories

## 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 questionâ€¦

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

We will recommend you to try the problem yourself.

Done?

Letâ€™s see the solution in the video below:

Categories

## 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 questionâ€¦

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

We will recommend you to try the problem yourself.

Done?

Letâ€™s see the proof in the video below:

Categories

## 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

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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## 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

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 , $1<x< \frac{3}{2}$.

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## 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

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, ${\sin^2 \theta} \geq \frac{1}{2}$ $\Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4}$.

Categories

## 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

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.

Categories

## 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 $n \rightarrow \infty$ .

## 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

## 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

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.

Categories

## 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

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

$\Rightarrow 29 values.$

Categories

## 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

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