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## 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

## 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

## 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

## Sum of reciprocals Problem | AMC-10A, 2003 | Problem 18

Try this beautiful problem from Algebra based on Sum of reciprocals in the quadratic equation.

## Sum of reciprocals in equation – AMC-10A, 2003- Problem 18

What is the sum of the reciprocals of the roots of the equation $\frac{2003}{2004} x^2 +1+\frac{1}{x}=0$?

• $\frac{1}{2}$
• $-1$
• $-\frac{1}{2}$

### Key Concepts

Algebra

root of the equation

Answer: $-1$

AMC-10A (2003) Problem 18

Pre College Mathematics

## Try with Hints

The given equation is $\frac{2003}{2004} x^2 +1+\frac{1}{x}=0$.after simplification we will get,

$2003 x^2 +2004 x+2004=0$ which is a quadratic equation .we have to find out the roots of the equation.

suppose there is a quadratic equation $ax^2 +bx+c=0$ (where a,b,c are constant) and roots of the equation are $p_1$ & $p_2$ then we know that

$p_1 +p_2$=-$\frac{b}{a}$ & $p_1 p_2$=$\frac{c}{a}$.now can you find out sum of the reciprocals of the roots of the given equation..?

can you finish the problem……..

The given equation is $2003 x^2 +2004 x+2004=0$.Let $m_1$ &$m_2$ are the roots of the given equation

Then $m_1 +m_2$=-$\frac{2004}{2003}$ & $m_1 m_2=\frac{2004}{2003}$.Now can you find out sum of the reciprocals of the roots of the given equation?

can you finish the problem……..

Now $\frac{m_1 +m_2}{m_1m_2}$=$\frac{1}{m_1} +\frac{1}{m_2}$=$\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}$=$-1$

Categories

## Number system | AMC-10A, 2007 | Problem 22

Try this beautiful problem from Number system based on digit problem

## Number system – AMC-10A, 2007- Problem 22

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$?

• $31$
• $37$
• $43$

### Key Concepts

Number system

multiplication

Answer: $37$

AMC-10A (2007) Problem 22

Pre College Mathematics

## Try with Hints

The given condition is “A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term,And also another codition that the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term” so we may assume four integers that be $(xyz,yzm,zmp,qxy)$ i.e$(100x+10y+z,100y+10z+m,100z+10m+p,100q+10x+y)$

Now the sum of the digits be$(110x+111y+111z+11m+p+100q)$

can you finish the problem……..

But “the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term”……so we may say that in last integer $qxy$…$q=m$ & $p=x$.Therefore the sum becomes $(110x+111y+111z+11q+x+100q)$=$111(x+y+z+m)$ i.e $111 K$ (say)

can you finish the problem……..

N ow in $111K$= $3.37.K$………So in the given answers the largest prime number is 37

Categories

## Pen & Note Books Problem | PRMO-2019 | Question 16

Try this beautiful Problem from Algebra based on Pen & Note Books from PRMO 2019.

## Pen & Note Books – PRMO 2019, Problem 16

A pen costs Rs.13 and a notebook costs Rs.17. A school spends exactly Rs.10000 in the year 2017 – 18 to buy x pens and y notebooks such that $x$ and $y$ are as close as possible (i.e. $|x-y|$ is minimum). Next year, in $2018-19,$ the school spends a little more than Rs.10000 and buys y pens and x notebooks. How much more did the school pay?

• $9$
• $40$
• $34$

### Key Concepts

Algebra

Equation

multiplication

Answer:$40$

PRMO-2019, Problem 16

Pre College Mathematics

## Try with Hints

Given A pen costs Rs.$13$ and a note book costs Rs.$17$

Nunber of pens $x$ and numbers of notebooks $y$

According to question

$13 x+17 y=10,000$ ……………………..(1)
$17 x+13 y=10000+P$………………….(2)

Where $\mathrm{P}$ is little more amount spend by school in $2018-19$.we have to find out the value of $P$.Can you find out?

Can you now finish the problem ……….

$x+y=\frac{20,000+P}{30}$

Subtract (1) form (2)

$x-y=\frac{P}{4}$

Let $P=4 a$
\begin{aligned} \text { So, } x+y=& \frac{20,000+4 a}{30} \ &=666+\frac{10+2 a}{15} \end{aligned}

Therefore $a$=$10$

Can you finish the problem……..

Since $P=4 a$ $\Rightarrow P=40$

So school had to pay Rs. 40 extra in 2018 – 2019

Categories

## Problem on Equation | AMC-10A, 2007 | Problem 20

Try this beautiful problem from Algebra based on quadratic equation

## Problem on Equation – AMC-10A, 2007- Problem 20

Suppose that the number $a$ satisfies the equation $4 = a + a^{ – 1}$. What is the value of $a^{4} + a^{ – 4}$?

• $174$
• $194$
• $156$

### Key Concepts

Algebra

Linear equation

multiplication

Answer: $194$

AMC-10A (2007) Problem 20

Pre College Mathematics

## Try with Hints

Given that $4 = a + a^{ – 1}$. we have to find out the value $a^{4} + a^{ – 4}$

Squarring both sides of $a^{4} + a^{ – 4}$ …then opbtain…

can you finish the problem……..

$(a + a^{ – 1})^2=4^2$ $\Rightarrow (a^2 + a^{-2} +2)=16$ $\Rightarrow a^2 + a^{-2}=14$ and now squarring both side again………….

can you finish the problem……..

Squarring both sides of $a^2 + a^{-2}=14$ $\Rightarrow (a^2 + a^{-2})^2=(14)^2$ $\Rightarrow a^4 + a^{-4} +2=196$ $\Rightarrow a^4 + a^{-4}=194$

Categories

## Quadratic equation Problem | AMC-10A, 2003 | Problem 5

Try this beautiful problem from Algebra based on quadratic equation.

## Quadratic equation – AMC-10A, 2003- Problem 5

Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$. What is the value of $(d-1)(e-1)$?

• $2$
• $0$
• $\frac{7}{2}$

### Key Concepts

algebra

Roots

Answer: $0$

AMC-10A (2003) Problem 5

Pre College Mathematics

## Try with Hints

To find out the value of $(d-1)(e-1)$,at first we have to find out the value of $d$ and $e$.Given that $d$ and $e$ are the solutions of the equations $2x^{2}+3x-5=0$ that means $d$ and $e$ are the roots of the given equation.so if we find out the values of roots from the given equation then we will get $d$ and $e$.Can you find out the roots?

Can you now finish the problem ……….

To find out the roots :

The given equation is $2x^{2}+3x-5=0$ $\Rightarrow (2x+5)(x-1)=0$ $\Rightarrow x=1 or \frac{-5}{2}$

Therefore the values of $d$ and $e$ are $1$ and $\frac{-5}{2}$ respectively

can you finish the problem……..

Therefore $(d-1)(e-1)$=$(1-1)(\frac{-5}{2} -1)$=$0$

Categories

## Sum of two digit numbers | PRMO-2016 | Problem 7

Try this beautiful problem from Algebra based on Sum of two digit numbers from PRMO 2016.

## Sum of two digit numbers | PRMO | Problem 7

Let s(n) and p(n) denote the sum of all digits of n and the products of all the digits of n(when written in decimal form),respectively.Find the sum of all two digits natural numbers n such that $n=s(n)+p(n)$

• $560$
• $531$
• $654$

### Key Concepts

Algebra

number system

Answer:$531$

PRMO-2016, Problem 7

Pre College Mathematics

## Try with Hints

Let $n$ is a number of two digits ,ten’s place $x$ and unit place is $y$.so $n=10x +y$.given that $s(n)$= sum of all digits $\Rightarrow s(n)=x+y$ and $p(n)$=product of all digits=$xy$

now the given condition is $n=s(n)+p(n)$

Can you now finish the problem ……….

From $n=s(n)+p(n)$ condition we have,

$n=s(n)+p(n)$ $\Rightarrow 10x+y=x+y+xy \Rightarrow 9x=xy \Rightarrow y=9$ and the value of$x$ be any digit….

Can you finish the problem……..

Therefore all two digits numbers are $19,29,39,49,59,69,79,89,99$ and sum=$19+29+39+49+59+69+79+89+99=531$

Categories

## Integer based Problem | PRMO-2018 | Question 20

Try this beautiful Integer based Problem from Algebra, from PRMO 2018.

## Integer based Problem – PRMO 2018, Question 20

Determine the sum of all possible positive integers n, the product of whose digits equals $n^2 -15n – 27$

• $9$
• $17$
• $34$

### Key Concepts

Algebra

Integer

multiplication

Answer:$17$

PRMO-2018, Problem 17

Pre College Mathematics

## Try with Hints

Product of digits = $n^2 – 15n – 27 = n(n – 15) – 27$

so at first we observe when n=one digit ,two digit and 3 digit numbers…..

If n is a more than 2-digit number, say 3-digit number, then product has to be$\leq 9 × 9 × 9 = 729$ but $(n(n – 15) – 27)$ is more than 729 (in fact it a more than 3-digit numbers for any 3-digit n). Hence, n can be either one-digit or 2-digit.

If n is 1-digit then $n^2 – 15n – 27 = n$ $\Rightarrow n$= not an integer , so n is a two digit number

now we will observe for 2-digit numbers…..

Can you now finish the problem ……….

For Two-digit numbers:

As product is positive so n(n-15)-27>0$\Rightarrow n\geq 17$

Now two digit product is less than equal to 81

so $n(n-15)-27\leq 1$$\Rightarrow n(n-15)\leq 108$ $\Rightarrow n\leq 20$

Therefore n can be $17$,$18$,$19$ or $20$

Can you finish the problem……..

For $n$= $17$,$18$,$19$ or $20$

when n=17,then $n(n-15)-27=7=1 \times 7$

when n=18,then $n(n-15)-27=27\neq 1\times 8$

when n=19,then $n(n-15)-27=49=1 \neq 9$

when n=20,then $n(n-15)-27=73=1 \neq 0$

Therefore $n$=17