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## Area of the Trapezium | PRMO-2017 | Question 30

Try this beautiful Problem from Geometry based on Area of the Trapezium from PRMO 2017.

## Area of the Trapezium – PRMO 2017, Problem 30

Consider the areas of the four triangles obtained by drawing the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium ABCD. The product of these areas, taken two at time, are computed. If among the six products so obtained, two products are 1296 and 576 , determine the square root of the maximum possible area of the trapezium to the nearest integer.

• $9$
• $40$
• $13$
• $20$

### Key Concepts

Geometry

Triangle

Trapezium

Answer:$13$

PRMO-2017, Problem 30

Pre College Mathematics

## Try with Hints

Let $x, y, z, w$ be areas of the four triangles as shown in figure.
then area of $\triangle ADB$= Area of $\triangle ACB$
$\Rightarrow x+y=x+w \Rightarrow y=w$

Also $\frac{AE}{EC}$=$\frac{area of \triangle ADE}{area of \triangle DEF}$=$\frac{area of \triangle AEB}{area of \triangle BEC}$
$\Rightarrow \frac{y}{z}=\frac{x}{w}=\frac{x}{y} \Rightarrow y^{2}=z x$
$\Rightarrow z, y, x$ are in G.P.

Can you now finish the problem ……….

Let $y=z r$ and $x=z r^{2},$ where $r \geq 1$ To make area of trapezium ABCD maximum, we take $z y=z^{2} r=576$
and $y w=z^{2} r^{2}=1296$
As $( z \leq y \leq x)$
Therefore $\frac{z^{2} r^{2}}{z^{2} r}=\frac{1296}{576} \Rightarrow r=\frac{9}{4} \Rightarrow z=16$

Can you finish the problem……..

Therefore area of trapezium $\mathrm{ABCD}$
$=x+y+z+w=z r^{2}+2 z r+z$
$=z(1+r)^{2}=16\left(1+\frac{9}{4}\right)^{2}=13^{2}$
Therefore Answer is $13 .$

Categories

## Positive Integer | PRMO-2017 | Question 1

Try this beautiful Positive Integer Problem from Algebra from PRMO 2017, Question 1.

## Positive Integer – PRMO 2017, Question 1

How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by $3 ?$

• $9$
• $7$
• $28$

### Key Concepts

Algebra

Equation

multiplication

Answer:$28$

PRMO-2017, Problem 1

Pre College Mathematics

## Try with Hints

Let $n$ be the positive integer less than 1000 and $s$ be the sum of its digits, then $3 \mid n$ and $7 \mid s$
$3|n \Rightarrow 3| s$
therefore$21| s$

Can you now finish the problem ……….

Also $n<1000 \Rightarrow s \leq 27$
therefore $\mathrm{s}=21$
Clearly, n must be a 3 digit number Let $x_{1}, x_{2}, x_{3}$ be the digits, then $x_{1}+x_{2}+x_{3}=21$
where $1 \leq x_{1} \leq 9,0 \leq x_{2}, x_{3} \leq 9$
$\Rightarrow x_{2}+x_{3}=21-x_{1} \leq 18$
$\Rightarrow x_{1} \geq 3$

Can you finish the problem……..

For $x_{1}=3,4, \ldots ., 9,$ the equation (1) has $1,2,3, \ldots ., 7$ solutions
therefore total possible solution of equation (1)

=$1+2+\ldots+7=\frac{7 \times 8}{2}=28$

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

## 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

## Sum of digits Problem | PRMO 2016 | Question 6

Try this beautiful problem from Number system based on sum of digits.

## Sum of digits | PRMO | Problem 6

Find the sum of digits in decimal form of the number $(9999….9)^3$ (There are 12 nines)

• $200$
• $216$
• $230$

### Key Concepts

Number system

Digits

counting

Answer:$216$

PRMO-2016, Problem 6

Pre College Mathematics

## Try with Hints

we don’t know what will be the expression of $(9999….9)^3$. so we observe….

$9^3$=$729$

$(99)^3$=$970299$

$(999)^3$=$997002999$

……………..

……………

we observe that,There is a pattern such that…

In $(99)^3$=$970299$ there are 1-nine,1-seven,1-zero,1-two,2-nines & $(999)^3$=$997002999$ there are 2- nines,1-seven,2-zeros,1-two,3-nines….so in this way…..$(999….9)^3$ will be 11-nines,1-seven,11-zeros,1-two,12-nines……….

Therefore $(999….9)^3$=$(99999999999) 7 (00000000000) 2(999999999999)$

can you finish the problem?

Therefore $(999….9)^3$=$(99999999999) 7 (00000000000) 2(999999999999)$…….

total numbers of Nines are (11+12) and (7+2)=9(another one) …..so total (11+12+1)=24 nines and the sum be $(24\times 9)$=$216$

Categories

## Finding side of Triangle | PRMO-2014 | Problem 15

Try this beautiful problem from PRMO, 2014 based on Finding side of Triangle.

## Finding side of Triangle | PRMO | Problem 15

Let XOY be a triangle with angle XOY=90 degrees. Let M and N be the midpoints of the legs OX and OY, respectively. Suppose that XN=19 and YM=22. what is XY?

• $28$
• $26$
• $30$

### Key Concepts

Geometry

Triangle

Pythagoras

Answer:$26$

PRMO-2014, Problem 15

Pre College Mathematics

## Try with Hints

Given that $\angle XOY=90^{\circ}$ .Let M and N be the midpoints of the legs OX and OY, respectively, and that XN=19 and YM=22. Now $\triangle XON$ & $\triangle MOY$ are Right angle Triangle. Use Pythagoras theorem …….

Can you now finish the problem ……….

Let $XM=MO=p$ and $ON=NY=q$.Now using Pythagoras theorm on $\triangle XON$ & $\triangle MOY$ we have…

$OX^2 +ON^2=XN^2$ $\Rightarrow 4p^2 +q^2=19^2$ $\Rightarrow 4p^2 +q^2=361$………..(1) and $OM^2 +OY^2=MY^2$ $\Rightarrow p^2 +4q^2=22^2$ $\Rightarrow p^2 +4q^2=484$……(2)

Now Adding (1)+(2)=$(4p^2 +q^2=361)$+$(p^2 +4q^2=484$ $\Rightarrow 5(p^2+q^2)=845$ $\Rightarrow (p^2+q^2)=169$ $\Rightarrow 4(p^2+q^2)=676$ $\Rightarrow (OX)^2+(OY)^2=(26)^2$ $\Rightarrow (XY)^2=(26)^2$ $\Rightarrow XY=26$.

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