Quadratic equation Problem | AMC-10A, 2002 | Problem 12

Try this beautiful problem from Algebra based Quadratic equation.

Quadratic equation Problem - AMC-10A, 2002- Problem 12

Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is

$0$
$1$
$2$
$4$
more than $4$

Key Concepts

Algebra

Quadratic equation

prime numbers

Check the Answer

Answer: $1$

AMC-10A (2002) Problem 12

Pre College Mathematics

Try with Hints

The given equation is $x^2 - 63x + k = 0$. Say that the roots are primes...

Comparing the equation with $ax^2 +bx+c=0$ we get $a=1 , b=-63 , c=k$.. Let $m_1$ & $m_2 $ be the roots of the given equation...

using vieta's Formula we may sat that...$m_1 + m_2 =-(- 63)=63$ and $m_1 m_2 = k$

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

Now the roots are prime. Sum of the two roots are $63$ and product is $k$

Therefore one root must be $2$ ,otherwise the sum would be even number

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

So other root will be $63-2$=$61$. Therefore product must be $m_1m_2=122$

Hence the answer is $1$

Problem on Real numbers | Algebra | PRMO-2017 | Problem 18

Try this beautiful problem from Algebra PRMO 2017 based on real numbers.

Problem on Real numbers | PRMO | Problem 18

If the real numbers $x$, $y$, $z$ are such that $x^2 + 4y^2 + 16z^2 = 48$ and $xy + 4yz + 2zx = 24$. what is the
value of $x^2 + y^2 + z^2$ ?

$24$
$21$
$34$

Key Concepts

Algebra

Equation

Check the Answer

Answer:$21$

PRMO-2017, Problem 18

Pre College Mathematics

Try with Hints

The given equation are

$x^2 + 4y^2 + 16z^2 = 48$
$\Rightarrow (x)2 + (2y)2 + (4z)2 = 48$
$2xy + 8yz + 4zx = 48$
adding tis equations we have to solve the problem....

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

Now we can say that
$(x)^2 + (2y)^2 + (4z)^2 – (2xy) – (8yz) – (4zx) = 0$
$\Rightarrow [(x – 2y)2 + (2y – 4z)2 + (x – 4y)2)] = 0$
$x = 2y = 4z $

$\Rightarrow \frac{x}{4}=\frac{y}{2}=z$

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

Therefore we may say that,

$(x, y, z) = (4m, 2m, m)$
$x^2 + 4y^2 + 16z^2 = 48$

$16m^2 + 16m^2 + 16m^2 = 48$
so $m^2 = 1$
$x^2 + y^2 + z^2 = 21m^2 = 21$

Problem on Cylinder | AMC-10A, 2004 | Problem 11

Try this beautiful problem from AMC 10A, 2004 based on Mensuration: Cylinder

Problem on Cylinder - AMC-10A, 2004- Problem 11

A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume, by what percent must the height be decreased?

$16$
$18$
$20$
$36$
$25$

Key Concepts

Mensuration

Cylinder

Percentage

Check the Answer

Answer: $36$

AMC-10A (2004) Problem 11

Pre College Mathematics

Try with Hints

Let the radius of the jar be $x$ and height be $h$.then the volume (V) of the jar be$V$= $\pi (x)^2 h$. Diameter of the jar increase $25 $% Therefore new radius will be $x +\frac{x}{4}=\frac{5x}{4}$ .Now the given condition is "after increase the volume remain unchange".Let new height will be $h_1$.Can you find out the new height....?

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

Let new height will be $H$.Therefore the volume will be $\pi (\frac{5x}{4})^2 H$.Since Volume remain unchange......

$\pi (x)^2 h$=$\pi (\frac{5x}{4})^2 H$ $\Rightarrow H=\frac{16h}{25}$.

height decrease =$h-\frac{16h}{25}=\frac{9h}{25}$.can you find out the decrease percentage?

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

Decrease Percentage=$ \frac {\frac {9h}{25}}{h} \times 100=36$%

Length of a Tangent | AMC-10A, 2004 | Problem 22

Try this beautiful problem from American Mathematics Competitions, AMC 10A, 2004 based on Geometry: Length of a Tangent.

Length of a Tangent - AMC-10A, 2004- Problem 22

Square $ABCD$ has a side length $2$. A Semicircle with diameter $AB$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $AD$at $E$. What is the length of $CE$?

$\frac{4}{3}$
$\frac{3}{2}$
$\sqrt 3$
$\frac{5}{2}$
$1+\sqrt 3$

Key Concepts

Square

Semi-circle

Geometry

Check the Answer

Answer: $\frac{5}{2}$

AMC-10A (2004) Problem 22

Pre College Mathematics

Try with Hints

We have to find out length of $CE$.Now $CE$ is a tangent of inscribed the semi circle .Given that length of the side is $2$.Let $AE=x$.Therefore $DE=2-x$. Now $CE$ is the tangent of the semi-circle.Can you find out the length of $CE$?

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

Since $EC$ is tangent,$\triangle COF$ $\cong$ $\triangle BOC$ and $\triangle EOF$ $\cong$ $\triangle AOE$ (By R-H-S law).Therefore $FC=2$ & $ EC=x$.Can you find out the length of $EC$?

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

Shaded Triangle to find the length of the tangent

Now the $\triangle EDC$ is a Right-angle triangle........

Therefore $ED^2+ DC^2=EC^2$ $\Rightarrow (2-x)^2 + 2^2=(2+x)^2$ $\Rightarrow x=\frac{1}{2}$

Hence $EC=EF+FC=2+\frac{1}{2}=\frac{5}{2}$

Largest possible value | AMC-10A, 2004 | Problem 15

Try this beautiful problem from Number system: largest possible value

Largest Possible Value - AMC-10A, 2004- Problem 15

Given that $ -4 \leq x \leq -2$ and $2 \leq y \leq 4$, what is the largest possible value of $\frac{x+y}{2}$

$\frac {-1}{2}$
$\frac{1}{6}$
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{9}$

Key Concepts

Number system

Inequality

divisibility

Check the Answer

Answer: $\frac{1}{2}$

AMC-10A (2003) Problem 15

Pre College Mathematics

Try with Hints

The given expression is $\frac{x+y}{x}=1+\frac{y}{x}$

Now $-4 \leq x \leq -2$ and $2 \leq y \leq 4$ so we can say that $\frac{y}{x} \leq 0$

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

Therefore, the expression $1+\frac{y}x$ will be maximized when $\frac{y}{x}$ is minimized, which occurs when $|x|$ is the largest and $|y|$ is the smallest.

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

Therefore in the region $(-4,2)$ , $\frac{x+y}{x}=1-\frac{1}{2}=\frac{1}{2}$

Points of Equilateral triangle | AIME I, 1994 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Points of Equilateral triangle.

Points of Equilateral triangles - AIME I, 1994

The points (0,0), (a,11), and (b,37) are the vertices of equilateral triangle, find the value of ab.

is 107
is 315
is 840
cannot be determined from the given information

Key Concepts

Integers

Complex Number

Equilateral Triangle

Check the Answer

Answer: is 315.

AIME I, 1994, Question 8

Complex Numbers from A to Z by Titu Andreescue

Try with Hints

Let points be on complex plane as b+37i, a+11i and origin.

then $(a+11i)cis60=(a+11i)(\frac{1}{2}+\frac{\sqrt{3}i}{2})$=b+37i

equating real parts b=$\frac{a}{2}-\frac{11\sqrt{3}}{2}$ is first equation

equating imaginary parts 37=$\frac{11}{2}+\frac{a\sqrt{3}i}{2}$ is second equation

solving both equations a=$21\sqrt{3}$, b=$5\sqrt{3}$

ab=315.

Problem on Ratio | PRMO 2017 | Question 12

Try this beautiful problem from the Pre-RMO, 2017 based on ratio.

Problem on Ratio - PRMO 2017

In a sports team, the total number of boys and girls are in the ratio 4:3. On one day it was found that 8 boys and 14 girls were absent from the team and that the number of boys was the square of the number of girls, find the total number of members in the sports team.

is 107
is 42
is 840
cannot be determined from the given information

Key Concepts

Ratio

Integers

Algebra

Check the Answer

Answer: is 42.

PRMO, 2017, Question 12

Elementary Algebra by Hall and Knight

Try with Hints

here ratio = 4:3

$\Rightarrow boys =4x, girls=3x$

given $(4x-8)=(3x-14)^{2}$

$\Rightarrow 4x-8=9x^{2}-84x+196$

$\Rightarrow 9x^{2}+196-84x=4x-8$

$\Rightarrow 9x^{2}-88x+204=0$

$x=\frac{88\pm\sqrt{88^{2}-4(9)(204)}}{18}$

$\Rightarrow x=6 or \frac{34}{9}$

$\Rightarrow 7x=42$ or non-integer

$\Rightarrow 42 members.$

Complex roots and equations | AIME I, 1994 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Complex roots and equations.

Complex roots and equations - AIME I, 1994

$x^{10}+(13x-1)^{10}=0$ has 10 complex roots $r_1$, $\overline{r_1}$, $r_2$,$\overline{r_2}$.$r_3$,$\overline{r_3}$,$r_4$,$\overline{r_4}$,$r_5$,$\overline{r_5}$ where complex conjugates are taken, find the values of $\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}$

is 107
is 850
is 840
cannot be determined from the given information

Key Concepts

Integers

Complex Roots

Equation

Check the Answer

Answer: is 850.

AIME I, 1994, Question 13

Complex Numbers from A to Z by Titu Andreescue

Try with Hints

here equation gives ${13-\frac{1}{x}}^{10}=(-1)$

$\Rightarrow \omega^{10}=(-1)$ for $\omega=13-\frac{1}{x}$

where $\omega=e^{i(2n\pi+\pi)(\frac{1}{10})}$ for n integer

$\Rightarrow \frac{1}{x}=13- {\omega}$

$\Rightarrow \frac{1}{(x)(\overline{x})}=(13-\omega)(13-\overline{\omega})$

=$170-13(\omega+\overline{\omega})$

adding over all terms $\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}$

=5(170)

=850.

Length and Inequalities | AIME I, 1994 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Length and Inequalities.

Length and Inequalities - AIME I, 1994

A fenced, rectangular field measures 24 meters by 52 meters.An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field, find the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence.

is 107
is 702
is 840
cannot be determined from the given information

Key Concepts

Integers

Inequalities

Length

Check the Answer

Answer: is 702.

AIME I, 1994, Question 12

Inequalities (Little Mathematical Library) by Korovkin

Try with Hints

Number of squares in a row=$\frac{52n}{24}$=$\frac{13n}{6}$ squares in every row

each vertical fence lengths 24 for $\frac{13n}{6}-1$ vertical fences

each horizontal fence lengths 52 for n-1 such fences

total length of internal fencing 24 ($\frac{13n}{6}-1$)+52(n-1)=104n-76 $ \leq 1994$

$\Rightarrow n \leq \frac{1035}{52}$

$\Rightarrow n \leq 19$

the largest multiple of 6 that is $ \leq 19 $

$\Rightarrow n=18$

required number =$\frac{13n^{2}}{6}$=702.

Trigonometry & natural numbers | PRMO 2017 | Question 11

Try this beautiful problem from the Pre-RMO, 2017 based on Trigonometry & natural numbers.

Trigonometry & natural numbers - PRMO 2017

Let f(x) =$sin\frac{x}{3}+cos\frac{3x}{10}$ for all real x, find the least natural number x such that $f(n\pi+x)=f(x)$ for all real x.

is 107
is 60
is 840
cannot be determined from the given information

Key Concepts

Trigonometry

Least natural number

Functions

Check the Answer

Answer: is 60.

PRMO, 2017, Question 11

Plane Trigonometry by Loney

Try with Hints

here f(x) =$sin\frac{x}{3}+cos\frac{3x}{10}$

period of$sin\frac{x}{3}$ is $6\pi$

period of $cos\frac{3x}{10}$ is $\frac{20\pi}{3}$

Lcm=$\frac{60\pi}{3}$ $\Rightarrow n=60$.