Categories

## 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

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.

Categories

## 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

Answer:$24$

PRMO-2017, Problem 13

Pre College Mathematics

## Try with Hints

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 ……….

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

Categories

## 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

Answer:$75$

PRMO-2017, Problem 26

Pre College Mathematics

## Try with Hints

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 ……….

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 $m+n+k=75$

Categories

## 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

Answer:$20$

PRMO-2017, Problem 27

Pre College Mathematics

## Try with Hints

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 ……….

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)

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

Categories

## 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

• $4$
• $5$
• $6$
• $7$
• $8$

### Key Concepts

Geometry

Square

Triangle

Answer: $8$

AMC-10A (2013) Problem 3

Pre College Mathematics

## Try with Hints

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 ……….

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

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

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

$\Rightarrow BE=8$

Categories

## 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

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 $72-18=54$

Categories

## 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

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 $6+3=9$

Categories

## 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

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 $\frac{M}{6}=\frac{6}{N}$=$9$

Categories

## 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 $\triangle A B C$ ?

• $4 \sqrt{3}$
• $7 \sqrt{3}$
• $14 \sqrt{3}$
• $21$
• $42$

### Key Concepts

Triangle

Similarity

Geometry

Answer: $7 \sqrt{3}$

AMC-10A (2009) Problem 10

Pre College Mathematics

## Try with Hints

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 ……….

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……..

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

Categories

## 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

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

AMC-10A (2009) Problem 10

Pre College Mathematics

## Try with Hints

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 ……….

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……..

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