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Algebra Math Olympiad PRMO USA Math Olympiad

Pen & Note Books Problem| PRMO-2017 | Question 8

Try this beautiful Pen & Note Books Problem from Algebra, from PRMO 2017.

Pen & Note Books – PRMO 2017, Question 8


A pen costs $Rs 11$ and a notebook costs $Rs. 13 .$ Find the number of ways in which a person can spend exactly Rs.1000 to buy pens and notebooks.

  • $9$
  • $7$
  • $11$

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:$7$

PRMO-2017, Problem 8

Pre College Mathematics

Try with Hints


Given A pen costs Rs.\(11\) and a note book costs Rs.\(13\)

$11 x+13 y=1000$………………….(1)

Can you now finish the problem ……….

Now $11 x+13 y=1000$
$\Rightarrow 11 x=1000-13 y=(1001-11 y)-(2 y+1)$

=$11(91-y)-(2 y+1)$
$\Rightarrow 11 | 2 y+1$
Let $2 y+1=11(2 k-1), k \in I^{+}$
$\Rightarrow y=11 \mathrm{k}-6$
therefore $11 x=11(97-11 k)-11(2 k-1)$
$\Rightarrow x=98-13 k$
But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$

Can you finish the problem……..

But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$
therefore for each $\mathrm{k} \in{1,2, \ldots ., 7},$ we get a unique pair $(\mathrm{x}, \mathrm{y})=(98-13 \mathrm{k}, 11 \mathrm{k}-6)$ satisfying equation
Hence 7 ways are possible.

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Algebra Math Olympiad PRMO USA Math Olympiad

Rectangle Problem | Geometry | PRMO-2017 | Question 13

Try this beautiful Rectangle Problem from Geometry, from PRMO 2017.

Rectangle Problem – Geometry – PRMO 2017, Question 13


In a rectangle $A B C D, E$ is the midpoint of $A B ; F$ is a point on $A C$ such that $B F$ is perpendicular to $A C$; and FE perpendicular to BD. Suppose $\mathrm{BC}=8 \sqrt{3}$. Find AB.

  • $9$
  • $24$
  • $11$

Key Concepts


Geometry

Triangle

Trigonometry

Check the Answer


Answer:$24$

PRMO-2017, Problem 13

Pre College Mathematics

Try with Hints


Rectangle Problem

We have to find out the value of \(AB\). Join \(BD\). \(BF\) is perpendicular on \(AC\).

Let $\angle \mathrm{BAC}=\theta$
since $\mathrm{E}$ is mid point of hypotenous $\mathrm{AB}$ of right $\Delta \mathrm{AFB}$, therefore $A E=F E=B E$

Can you now finish the problem ……….

Rectangle Problem

Therefore

Therefore$ \angle E F A=\angle F A E=\theta$
and $\angle \mathrm{FEB}=\angle \mathrm{EAF}+\angle \mathrm{EFA}=2 \theta$
$\Rightarrow \angle E B D=90^{\circ}-\angle B E F=90^{\circ}-2 \theta$
But $\angle \mathrm{FAE}=\angle \mathrm{CAB}=\angle \mathrm{DBA}$
Therefore $\theta=90^{\circ}-2 \theta \Rightarrow \theta=30^{\circ}$
Therefore in $\Delta \mathrm{ABC}, \tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}$

Can you finish the problem……..

Therefore $A B=B C \cot \theta=8 \sqrt{3} \cot 30^{\circ}=24$

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Algebra Math Olympiad PRMO USA Math Olympiad

Chords in a Circle | PRMO-2017 | Question 26

Try this beautiful Problem based on Chords in a Circle, Geometry, from PRMO 2017.

Chords in a Circle – PRMO 2017, Question 26


Let $A B$ and $C D$ be two parallel chords in a circle with radius 5 such that the centre $O$ lies between these chords. Suppose $A B=6, C D=8 .$ Suppose further that the area of the part of the circle lying between the chords $A B$ and $C D$ is $(m \pi+n) / k,$ where $m, n, k$ are positive integers with gcd$(m, n, k)=1$ . What is the value of $m+n+k ?$

  • $9$
  • $75$
  • $11$

Key Concepts


Geometry

Triangle

Circle

Check the Answer


Answer:$75$

PRMO-2017, Problem 26

Pre College Mathematics

Try with Hints


Chords in a Circle

Draw OE $\perp A B$ and $O F \perp C D$

Clearly $\mathrm{EB}=\frac{\mathrm{AB}}{2}=3, \mathrm{FD}=\frac{\mathrm{CU}}{2}=4$

$\mathrm{OE}=\sqrt{5^{2}-3^{2}}=4$ and $\mathrm{OF}=\sqrt{5^{2}-4^{2}}=3$

Therefore $\Delta O E B \sim \Delta D F O$

Can you now finish the problem ……….

Chords in a Circle

Let $\angle \mathrm{EOB}=\angle \mathrm{ODF}=\theta,$ then

$\angle B O D=\angle A O C=180^{\circ}-\left(\theta+90^{\circ}-\theta\right)=90^{\circ}$

Now area of portion between the chords

= \(2 \times\) (area of minor sector BOD)+2 \times ar\((\triangle AOB)\)
$=2 \times \frac{\pi \times 5^{2}}{4}+2 \times \frac{1}{2} \times 6 \times 4=\frac{25 \pi}{2}+24=\frac{25 \pi+48}{2}$

Therefore $m=25, n=48$ and $k=2$

Can you finish the problem……..

Therefore $m+n+k=75$

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Algebra Math Olympiad PRMO USA Math Olympiad

Circle | Geometry Problem | PRMO-2017 | Question 27

Try this beautiful Problem from Geometry based on Circle from PRMO 2017.

Circle – PRMO 2017, Problem 27


Let $\Omega_{1}$ be a circle with centre 0 and let $A B$ be a diameter of $\Omega_{1} .$ Le $P$ be a point on the segment $O B$ different from 0. Suppose another circle $\Omega_{2}$ with centre P lies in the interior of $\Omega_{1}$. Tangents are drawn from A and B to the circle $\Omega_{2}$ intersecting $\Omega_{1}$ again at $A_{1}$ and $B_{1}$ respectively such that $A_{1}$, and $B_{1}$ are on the opposite sides of $A B$. Given that $A_{1} B=5, A B_{1}=15$ and $O P=10,$ find the radius of $\Omega_{1}$

  • $9$
  • $40$
  • $34$
  • $20$

Key Concepts


Geometry

Circle

Check the Answer


Answer:$20$

PRMO-2017, Problem 27

Pre College Mathematics

Try with Hints


Circle Problem

Let radius of $\Omega_{1}$ be $R$ and that of $\Omega_{2}$ be $r$
From figure, $\Delta \mathrm{ADP} \sim \Delta \mathrm{AA}_{1} \mathrm{B}$
[
\begin{array}{l}
\Rightarrow \frac{D P}{A, B}=\frac{A P}{A B} \
\Rightarrow \frac{r}{5}=\frac{R+10}{2 R}
\end{array}
]

Can you now finish the problem ……….

Circle Problem figure

Again, $\Delta B P E \sim \Delta B A B_{1}$
Therefore $\frac{P E}{A B_{1}}=\frac{B P}{B A}$
$\Rightarrow \frac{r}{15}=\frac{R-10}{2 R}$

Can you finish the problem……..

Dividing (1) by (2)

$3=\frac{R+10}{R-10} \Rightarrow R=20$

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Side of Square | AMC 10A, 2013 | Problem 3

Try this beautiful problem from Geometry: Side of Square.

Sides of Square – AMC-10A, 2013- Problem 3


Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$?

,

 i

Side of Square - Problem
  • $4$
  • $5$
  • $6$
  • $7$
  • \(8\)

Key Concepts


Geometry

Square

Triangle

Check the Answer


Answer: $8$

AMC-10A (2013) Problem 3

Pre College Mathematics

Try with Hints


Side of Square

Given that Square $ABCD$ has side length $10$ and area of $\triangle ABE$ is $40$.we have to find out length of \(BE\) where \(E\) is the point on \(BC\). we know area of the \(\triangle ABE=\frac{1}{2} AB.BE=40\)

Can you find out the side length of \(BE\)?

Can you now finish the problem ……….

Side of Square

\(\triangle ABE=\frac{1}{2} AB.BE=40\)

\(\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40\)

\(\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40\)

\(\Rightarrow BE=8\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Counting Days | AMC 10A, 2013 | Problem 17

Try this beautiful problem from Algebra: Counting Days

Counting Days – AMC-10A, 2013- Problem 17


Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365 -day period will exactly two friends visit her?

,

  • $48$
  • $54$
  • $60$
  • $65$
  • \(72\)

Key Concepts


Algebra

LCM

Check the Answer


Answer: $54$

AMC-10A (2013) Problem 17

Pre College Mathematics

Try with Hints


Given that Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day.

According to the questation , Let us assume that $A=3 x B=4 x C=5 x$
Now, we want the days in which exactly two of these people meet up
The three pairs are $(A, B),(B, C),(A, C)$
Can yoiu find out the LCM of each pair……….

Can you now finish the problem ……….

\(LCM(A, B)=12 x, LCM(B, C)=20 x, LCM(A, C)=15 x\)
Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.
Hence, $LCM(A, B, C)=60 x$

can you finish the problem……..

Now, we add all of the days up(including overcount).
We get $30+18+24=72 .$ Now, because $60(6)=360,$ we have to subtract 6 days from every pair. Hence, our answer is $72-18=54$

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Chosing Program | AMC 10A, 2013 | Problem 7

Try this beautiful problem from Combinatorics: Chosing Program

Chosing Program – AMC-10A, 2013- Problem 7


A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?

,

  • $6$
  • $8$
  • $9$
  • $12$
  • \(16\)

Key Concepts


Combinatorics

Check the Answer


Answer: $9$

AMC-10A (2013) Problem 7

Pre College Mathematics

Try with Hints


There are six programms: English, Algebra, Geometry, History, Art, and Latin. Since the student must choose a program of four course with the condition that there must contain English and at least one mathematics course. Therefore one course( i.e English) are already fixed and we have to find out the other subjects combinations…….

Can you now finish the problem ……….

There are Two cases :
Case 1: The student chooses both algebra and geometry.
This means that 3 courses have already been chosen. We have 3 more options for the last course, so there are 3 possibilities here.
case 2: The student chooses one or the other.
Here, we simply count how many ways we can do one, multiply by 2 , and then add to the previous.

Let us choose the mathematics course is algebra. so we can choose 2 of History, Art, and Latin, which is simply $3 \choose 2$=$3$. If it is geometry, we have another 3 options, so we have a total of 6 options if only one mathematics course is chosen.

can you finish the problem……..

Therefore the require ways are \(6+3=9\)

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AMC 10 Math Olympiad USA Math Olympiad

Order Pair | AMC-10B, 2012 | Problem 10

Try this beautiful problem from Algebra: Order Pair

Order Pair – AMC-10B, 2012- Problem 10


How many ordered pairs of positive integers (M,N) satisfy the equation $\frac{M}{6}=\frac{6}{N}$

  • \(31\)
  • \(78\)
  • \(43\)

Key Concepts


Algebra

Order Pair

Multiplication

Check the Answer


Answer: \(78\)

AMC-10A (2010) Problem 21

Pre College Mathematics

Try with Hints


Given that $\frac{M}{6}=\frac{6}{N}$ \(\Rightarrow MN=36\).Next we have to find out the the Possibilities to getting \(a \times b=36\)

can you finish the problem……..

Now the possibilities are ….

$1 \times 36=36$
$2 \times 18=36$
$3 \times 12=36$
$4 \times 9=36$
$6 \times 6=36$

We can reverse the order of the factors for all of the ones listed above, because they are ordered pairs except for 6*6 since it is the same back if you reverse the order.

can you finish the problem……..

Therefore the total Possible order pairs that satisfy the equation $\frac{M}{6}=\frac{6}{N}$=\(9\)

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AMC 10 Math Olympiad USA Math Olympiad

Triangle Area Problem | AMC-10A, 2009 | Problem 10

Try this beautiful problem from Geometry based on Area of Triangle.

Area of Triangle – AMC-10A, 2009- Problem 10


Triangle $A B C$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B, A D=3$, and $D C=4 .$ What is the area of $\triangle A B C$ ?

area of triangle - problem figure
Area of Triangle Problem
  • $4 \sqrt{3}$
  • $7 \sqrt{3}$
  • $14 \sqrt{3}$
  • \(21\)
  • \(42\)

Key Concepts


Triangle

Similarity

Geometry

Check the Answer


Answer: $7 \sqrt{3}$

AMC-10A (2009) Problem 10

Pre College Mathematics

Try with Hints


area of triangle - problem

We have to find out the area of the Triangle ABC where \(\angle B=90^{\circ}\) and \(BD \perp AC\)

Area of a Triangle = \(\frac{1}{2}\times \) Base \(\times\) Height.But we don know the value of \(AB\) & \(BC\). But we know \(AC=7\). So if we can find out the value of \(BD\) then we can find out the are of \(\triangle ABC\) by \(\frac{1}{2}\times AC \times BD\)

Can you now finish the problem ……….

area of triangle - problem

Let \(\angle C=\theta\), then \(\angle A=(90-\theta)\) (as \(\angle B=90^{\circ}\), Sum of the angles in a triangle is \(180^{\circ}\))

In \(\triangle ABD\), \(\angle ABD=\theta\) \(\Rightarrow \angle A=(90-\theta\))

Again In \(\triangle DBC\), \(\angle DBC\)=(\(90-\theta\)) \(\Rightarrow \angle C=\theta\)

From the above condition we say that , \(\triangle ABD \sim \triangle BDC\)

Therefore , \(\frac{BD}{CD}=\frac{AD}{BD}\) \(\Rightarrow {BD}^2=AD.CD=4\times 3\)

\(\Rightarrow BD=\sqrt {12}\)

can you finish the problem……..

area of triangle

Therefore area of the \(\triangle ABC=\frac{1}{2}\times AC \times BD=\frac{1}{2}\times 7 \times \sqrt{12}=7 \sqrt{3}\)

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AMC 10 Math Olympiad USA Math Olympiad

Side Length of Rectangle | AMC-10A, 2009 | Problem 17

Try this beautiful problem from Geometry based on Side Length of Rectangle.

Side Length of Rectangle – AMC-10A, 2009- Problem 17


Rectangle $A B C D$ has $A B=4$ and $B C=3 .$ Segment $E F$ is constructed through $B$ so that $E F$ isperpendicular to $D B$, and $A$ and $C$ lie on $D E$ and $D F$, respectively. What is $E F$ ?

  • $9$
  • $10$
  • $\frac{125}{12}$
  • \(\frac{103}{9}\)
  • \(12\)

Key Concepts


Triangle

Rectangle

Geometry

Check the Answer


Answer: $\frac{125}{12}$

AMC-10A (2009) Problem 10

Pre College Mathematics

Try with Hints


Finding Side Length of Rectangle

We have to find out the length of \(EF\)

Now $BD$ is the altitude from $B$ to $EF$, we can use the equation $BD^2 = EB\cdot BF$. ( as \(\triangle BDE \sim \triangle BDF\)).so we have to find out \(BE\) and \(BF\)

Can you now finish the problem ……….

Problem figure

Now Clearly, $\triangle BDE \sim \triangle DCB$. Because of this, $\frac{A B}{C B}=\frac{E B}{D B}$. From the given information and the Pythagorean theorem, $A B=4, C B=3$, and $D B=5 .$ Solving gives $E B=20 / 3$
We can use the above formula to solve for $B F . B D^{2}=20 / 3 \cdot B F$. Solve to obtain $B F=15 / 4$

can you finish the problem……..

Problem figure

Therefore $E F=E B+B F=\frac{20}{3}+\frac{15}{4}=\frac{80+45}{12}$

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