Probability in Game | AMC-10A, 2005 | Problem 18

Try this beautiful problem from AMC 10A, 2005 based on Probability in Game.

Probability in Game - AMC-10A, 2005- Problem 18

Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

$\frac{1}{4}$
$\frac{1}{6}$
$\frac{1}{5}$
$\frac{2}{3}$
$\frac{1}{3}$

Key Concepts

Probability

combinatorics

Check the Answer

Answer: $\frac{1}{5}$

AMC-10A (2005) Problem 18

Pre College Mathematics

Try with Hints

Given that The first team to win three games wins the series, team B wins the second game and team A wins the series. So the Total number of games played=$5$. Now we have to find out the possible order of wins.....

Can you now finish the problem ..........

Possible cases :

If team B won the first two games, team A would need to win the next three games. Therefore the possible order of wins is BBAAA.
If team A won the first game, and team B won the second game, the possible order of wins is $A B B A A, A B A B A,$ and $A B A A X,$ where $X$ denotes that the 5th game wasn't played.
since ABAAX is dependent on the outcome of 4 games instead of 5, it is twice as likely to occur and can be treated as two possibilities.

According to the question, there is One possibility where team $\mathrm{B}$ wins the first game and 5 total possibilities, Therefore the required probability is $\frac{ 1}{5}$

Discontinuity Problem | ISI B.Stat Objective | TOMATO 734

Try this beautiful problem based on Discontinuity, useful for ISI B.Stat Entrance.

Discontinuity Problem | ISI B.Stat TOMATO 734

$\quad$ Let $f(x)=|| x-1|-1|$ if $x<1$ and $f(x)=[x]$ if $x \geq 1,$ where, for any real number $x,[x]$ denotes the largest integer $\leq x$ and $|y|$ denotes absolute value of $y .$ Then, the set of discontinuity-points of the function f consists of

all integers $\geq 0$
all integers $\geq 1$
all integers $> 1$
the integer 1

Key Concepts

Limit

Calculas

Continuous

Check the Answer

Answer: $C$

TOMATO, Problem 734

Challenges and Thrills in Pre College Mathematics

Try with Hints

Let us first check at $x=1$
$\lim f(x)$ as $x\to1-=\lim || x-1|-1|$ as $x\to 1-=1$
$\lim f(x)$ as $x\to1+=\lim [x]$ as $x\to1+=1$
So, continuous at $x=1$
Let us now check at $x=0$
$\lim f(x)$ as $x\to 0-=\lim || x-1|-1|$ as $x\to 0-=0$
$\lim f(x)$ as $x\to 0+=\lim || x-1|-1|$ as $x\to 0+=0$
So continuous at $x=0$

Can you now finish the problem ..........

Let us now check at $x=2$
$\lim f(x)$ as $x\to 2-=\lim [x]$ as $x\to 2-=1$
$\lim f(x)$ as $x\to 2+=\lim [x]$ as $x \rightarrow 2+=2$
Discontinuous at $x=2$
Option (c) is correct.

Quadratic Equation Problem | AMC-10A, 2005 | Problem 10

Try this beautiful problem from Algebra based on Quadratic Equation....

Quadratic equation - AMC-10A, 2005- Problem 10

There are two values of $a$ for which the equation $4 x^{2}+a x+8 x+9=0$ has only one solution for $x$. What is the sum of those values of $a$ ?

$5$
$20$
$-16$
$25$
$36$

Key Concepts

algebra

Quadratic equation

Equal roots

Check the Answer

Answer: $-16$

AMC-10A (2005) Problem 10

Pre College Mathematics

Try with Hints

The given equation is $4 x^{2}+a x+8 x+9=0$

$\Rightarrow 4 x^{2}+x(a+8)+9=0$

comparing the above equation with $Ax^2-Bx+C=0$ we will get $A=4$,$B=(a+8)$,$C=9$

Now for equal roots of a quadratic equation $B^2-4Ac=0$

Can you now finish the problem ..........

Now $B^2-4Ac=0$ becomes

$(a+8)^2-4\times 9 \times 4=0$

$\Rightarrow (a+8)^2=144$

$\Rightarrow (a+8)=\pm 12$

$\Rightarrow a=+4 $ & $-20$

Therefore The sum of the values of $a=-20+4=-16$

Ratio of the areas | PRMO-2019 | Problem 19

Try this beautiful problem from PRMO, 2019 based on Ratio of the areas.

Ratio of the areas | PRMO | Problem-19

Let $\mathrm{AB}$ be a diameter of a circle and let $\mathrm{C}$ be a point on the segment $\mathrm{AB}$ such that $\mathrm{AC}: \mathrm{CB}=6: 7 .$ Let $\mathrm{D}$ be a point on the circle such that $\mathrm{DC}$ is perpendicular to $\mathrm{AB}$. Let DE be the diameter through $\mathrm{D}$. If $[\mathrm{XYZ}]$ denotes the area of the triangle XYZ. Find [ABD] / $[\mathrm{CDE}]$ to the nearest integer.

$20$
$91$
$13$
$23$

Key Concepts

Geometry

Triangle

Area

Check the Answer

Answer:$13$

PRMO-2019, Problem 19

Pre College Mathematics

Try with Hints

$\angle \mathrm{AOC} \quad=\frac{6 \pi}{13}, \angle \mathrm{BOC}=\frac{7 \pi}{13}$

$\mathrm{Ar} \Delta \mathrm{ABD}=\mathrm{Ar} \Delta \mathrm{ABC}=\frac{1}{2} \mathrm{AB} \times \mathrm{OC} \sin \frac{6 \pi}{13}$

$\mathrm{Ar} \Delta \mathrm{CDE}=\frac{1}{2} \mathrm{DE} \times \mathrm{OC} \sin \left(\frac{7 \pi}{13}-\frac{6 \pi}{13}\right)$

$\frac{[\mathrm{ABD}]}{[\mathrm{CDE}]}=\frac{\sin \frac{6 \pi}{13}}{\sin \frac{\pi}{13}}=\frac{1}{2 \sin \frac{\pi}{26}}=\mathrm{p}$

because $\sin \theta \cong \theta$ if $\theta$ is small
$\Rightarrow \sin \frac{\pi}{26} \cong \frac{\pi}{26}$

$\mathrm{p}=\frac{13}{\pi} \Rightarrow$ Nearest integer to $\mathrm{p}$ is 4

Area of the Inner Square | AMC-10A, 2005 | Problem 8

Try this beautiful problem from Geometry: Area of the inner square

Area of Inner Square - AMC-10A, 2005- Problem 8

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$. What is the area of the inner square $EFGH$?

Area of the Inner Square - Problem Figure

$25$
$32$
$36$
$42$
$40$

Key Concepts

Geometry

Square

similarity

Check the Answer

Answer: $36$

AMC-10A (2005) Problem 8

Pre College Mathematics

Try with Hints

Area of the Inner Square - Shaded Figure

We have to find out the area of the region $EFGH$ Which is a square shape .so if we can find out one of it's side length then we can easily find out the area of $EFGH$. Now given that $BE=1$ i.e $BE=CF=DG=AH=1$ and side length of the square $ABCD=\sqrt {50}$.Therefore $(AB)^2=(\sqrt {50})^2=50$.so using this information can you find out the length of $EH$?

Can you find out the required area.....?

Since $EFGH$ is a square,therefore $ABH$ is a Right -angle Triangle.

Therefore,$(AH)^2+(BH)^2=(AB)^2$

$\Rightarrow (AH)^2+(HE+EB)^2=(AB)^2$

$\Rightarrow (1)^2+(HE+1)^2=50$

$\Rightarrow (HE+1)^2=49$

$\Rightarrow (HE+1)=7$

$\Rightarrow HE=6$

Therefore area of the inner square (red shaded region) =${6}^2=36$

Triangle and Quadrilateral | AMC-10A, 2005 | Problem 25

Try this beautiful problem from Geometry: Area of Triangle and Quadrilateral

Ratios of the areas of Triangle and Quadrilateral - AMC-10A, 2005- Problem 25

In $ABC$ we have $AB = 25$, $BC = 39$, and $AC=42$. Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$?

$\frac{19}{56}$
$\frac{19}{66}$
$\frac{17}{56}$
$\frac{11}{56}$
$\frac{19}{37}$

Key Concepts

Geometry

Triangle

quadrilateral

Check the Answer

Answer: $\frac{19}{56}$

AMC-10A (2005) Problem 25

Pre College Mathematics

Try with Hints

Given that $AB = 25$, $BC = 39$, and $AC=42$.we have to find out Ratios of the areas of Triangle$\triangle ADE$ and the quadrilateral $CBED$.So if we can find out the area the $\triangle ADE$ and area of the $\triangle ABC$ ,and subtract $\triangle ADE$ from $\triangle ABC$ then we will get area of the region $CBDE$.Can you find out the area of $CBDE$?

Can you find out the required area.....?

Now $\frac{\triangle ADE}{\triangle ABC}=\frac{AD}{AB}.\frac{AE}{AC}=\frac{19}{25}.\frac{14}{42}=\frac{19}{75}$

Therefore area of $BCED$=area of $\triangle ABC$-area of $\triangle ADE$.Now can you find out Ratios of the areas of Triangle and the quadrilateral?

can you finish the problem........

Now $\frac{\triangle ADE}{quad.BCED}$=$\frac{\triangle ADE}{{\triangle ABC}-{\triangle ADE}}$=$\frac{1}{\frac{\triangle ABC}{\triangle ADE}-1}=\frac{1}{\frac{75}{19}-1}=\frac{19}{56}$

Pentagon & Square Pattern | AMC-10A, 2001 | Problem 18

Try this beautiful problem from Geometry based on Pentagon and square Pattern.

Pentagon & Square Pattern - AMC-10A, 2001- Problem 18

The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to

$50$
$58$
$60$
$56$
$64$

Key Concepts

Geometry

Pentagon

square

Check the Answer

Answer: $56$

AMC-10A (2001) Problem 18

Pre College Mathematics

Try with Hints

The given square is the above square.we have to find out The percent of the plane that is enclosed by the pentagons.Notice that there are $9$ tiles in the square box.so if we can find out the area of pentagon and small square in single tile,then we can find out the total area of the pentagon in the total big square......

can you finish the problem........

Now consider a single tile from the big square,Let us take the side of the small square is $a$.There are four squares which is in the area $4a^2$ and there are five pentagons which are in areas $5a^2$.Then the area of the single tile is $9a^2$

Therefore we can say that exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane also

can you finish the problem........

Therefore for the whole square, expressed as a percentage,it becomes $55.\overline{5}\%$, and the closest integer to this value is $56$

Problem on Limit | ISI B.Stat Objective | TOMATO 728

Try this beautiful problem based on Limit, useful for ISI B.Stat Entrance

Problem on Limit | ISI B.Stat TOMATO 728

The limit lim $\int\frac {h}{(h^2 + x^2)}$dx (integration running from $x =-1$to $x = 1$) as$ h \to 0$

equals 0
equals $\pi$
equals $-\pi$
deoes not exist

Key Concepts

Limit

Calculas

trigonometry

Check the Answer

Answer: does not exist

TOMATO, Problem 728

Challenges and Thrills in Pre College Mathematics

Try with Hints

Now, $\int{h}{(h^2 + x^2)}$dx (integration running from $x = -1$ to $x = 1$)
Let, $x$ = h tany

$\Rightarrow dx = h sec^2y dy$
$\Rightarrow $ $x = -1$, $y = -tan^{-1}(1/h)$ and $x = 1$, $y = tan^{-1}(1/h)$

$\Rightarrow \int \frac{h}{(h^2 + x^2)}$dx =$\int \frac{h(hsec^2ydy)}{h^2sec^2y}$ (integration running from$ y = -tan^{-1}(1/h) $ to $y = tan^-1(1/h))$

= y (upper limit =$ tan-1(1/h)$) and lower limit = $-tan^-1(1/h)$

= $2tan^-1(1/h)$

Can you now finish the problem ..........

Now, lim $2tan^-(1/h)$ as$ h \to 0$ doesn‟t exist

Real valued function | ISI B.Stat Objective | TOMATO 690

Try this beautiful problem based on Real valued function, useful for ISI B.Stat Entrance

Real valued functions | ISI B.Stat TOMATO 690

Let $f(x)$ be a real-valued function defined for all real numbers x such that $|f(x) – f(y)|≤(1/2)|x – y|$ for all x, y. Then the number of points of intersection of the graph of $y = f(x)$ and the line $y = x$ is

0
1
2
none of these

Key Concepts

Limit

Calculas

Real valued function

Check the Answer

Answer: $1$

TOMATO, Problem 690

Challenges and Thrills in Pre College Mathematics

Try with Hints

Now,

$|f(x) – f(y)| ≤ (1/2)|x – y|$

$\Rightarrow lim |{f(x) – f(y)}/(x – y)|$( as x -> y ≤ lim (1/2)) as x - > y

$\Rightarrow |f‟(y)| ≤ ½$

$\Rightarrow -1/2 ≤ f‟(y) ≤1/2$

$\Rightarrow -y/2 ≤ f(y) ≤ y/2$ (integrating)

$\Rightarrow -x/2 ≤ f(x) ≤ x/2$

Can you now finish the problem ..........

Therefore from the picture we can say that intersection point is $1$

Sum of the numbers | AMC-10A, 2001 | Problem 16

Try this beautiful problem from Algebra based on Sum of the numbers....

Sum of the numbers - AMC-10A, 2001- Problem 16

The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5$. What is their sum?

$5$
$20$
$30$
$25$
$36$

Key Concepts

algebra

Mean

Median

Check the Answer

Answer: $30$

AMC-10A (2001) Problem 16

Pre College Mathematics

Try with Hints

Mean of three numbers means average of three numbers......

Let $x$ be the mean of three numbers then we can say that the least of the numbers is $m-10$ and the greatest is $m + 15$

Can you now finish the problem ..........

Given that The median of the three numbers is $5$. Now "median" means the middle of the three numbers

so we can write $\frac{1}{3}[(m-10) + 5 + (m + 15)] = m$,

$\Rightarrow m=10$

Therefore The sum of three numbers are $3(10)=30$