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

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

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

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

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

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## 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 $a^2 + a^{-2}=14$ $\Rightarrow (a^2 + a^{-2})^2=(14)^2$ $\Rightarrow a^4 + a^{-4} +2=196$ $\Rightarrow a^4 + a^{-4}=194$

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

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## 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=$19+29+39+49+59+69+79+89+99=531$

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## Quadratic Equation Problem | AMC-10A, 2005 | Problem 10

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## Quadratic equation – AMC-10A, 2005- Problem 10

There are two values of $a$ for which the equation $4 x^{2}+a x+8 x+9=0$ has only one solution for $x$. What is the sum of those values of $a$ ?

• $5$
• $20$
• $-16$
• $25$
• $36$

### Key Concepts

algebra

Equal roots

Answer: $-16$

AMC-10A (2005) Problem 10

Pre College Mathematics

## Try with Hints

The given equation is $4 x^{2}+a x+8 x+9=0$

$\Rightarrow 4 x^{2}+x(a+8)+9=0$

comparing the above equation with $Ax^2-Bx+C=0$ we will get $A=4$,$B=(a+8)$,$C=9$

Now for equal roots of a quadratic equation $B^2-4Ac=0$

Can you now finish the problem ……….

Now $B^2-4Ac=0$ becomes

$(a+8)^2-4\times 9 \times 4=0$

$\Rightarrow (a+8)^2=144$

$\Rightarrow (a+8)=\pm 12$

$\Rightarrow a=+4$ & $-20$

Therefore The sum of the values of $a=-20+4=-16$

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## Quadratic equation | ISI-B.stat | Objective Problem 198

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## Quadratic equation | ISI B.Stat Entrance | Problem 198

Consider the quadratic equation of the form $x^2 + bx + c = 0$. The number of such equations that have real roots and coefficients b and c from the set $\{1, 2, 3, 4, 5\}$ (b and c may be equal) is

• (a) $18$
• (b) $15)$
• (c) $12$
• (d) None of these

### Key Concepts

Algebra

Roots of the nature

Answer: (c) $12$

TOMATO, Problem 198

Challenges and Thrills in Pre College Mathematics

## Try with Hints

The given equation is $x^2 + bx + c = 0$.

we know that in the equation $ax^2+bx+c=0$ ,the condition for real root is $b^2-4ac \geq 0$

Can you now finish the problem ……….

Now, $b^2 > 4c$
b cannot be equal to 1.
If b = 2, c = 1
If b = 3, c = 1, 2
If b = 4, c = 1, 2, 3, 4
If b = 5, c = 1, 2, 3, 4, 5
Total number of equations = $1 + 2 + 4 + 5 = 12$

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## Roots of Equations | PRMO-2016 | Problem 8

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## Roots of Equations | PRMO | Problem 8

Suppose that $a$ and $b$ are real numbers such that $ab \neq 1$ and the equations $120 a^2 -120a+1=0$ and $b^2-120b+120=0$ hold. Find the value of $\frac{1+b+ab}{a}$

• $200$
• $240$
• $300$

### Key Concepts

Algebra

Roots

Answer:$240$

PRMO-2016, Problem 8

Pre College Mathematics

## Try with Hints

The given equations are $120 a^2 -120a+1=0$ and $b^2-120b+120=0$.we have to find out the values of $a$ and $b$….

Let $x,y$ be the roots of the equation $120 a^2 -120a+1=0$then $\frac{1}{x},\frac{1}{y}$ be the roots of the equations of $b^2-120b+120=0$.can you find out the value of $a$ & $b$

Can you now finish the problem ……….

From two equations after sim[lificatiopn we get…$a=x$ and $b=\frac{1}{y}$ (as $ab \neq 1)$

Can you finish the problem……..

$\frac{1+b+ab}{a}$=$\frac { 1+\frac{1}{y} +\frac{x}{y}}{x}$=$\frac{(x+y)+1}{xy}=240$

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## Problem based on Integer | PRMO-2018 | Problem 6

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## Algebra based on Integer PRMO Problem 6

Integers a, b, c satisfy $a + b – c = 1$ and $a^2 + b^2 – c^2 = –1$. What is the sum of all possible values of
$a^2 + b^2 + c^2$ ?

• $24$
• $18$
• $34$

### Key Concepts

Algebra

Factorization

Answer:$18$

PRMO-2018, Problem 6

Pre College Mathematics

## Try with Hints

Use $(a + b – 1 )= c$ this relation

Can you now finish the problem ……….

$(a + b – 1 )^2= c^2$ (squaring both sides…….)

Can you finish the problem……..

Given that a,b,c are integer satisfy $a + b – c = 1$ and $a^2 + b^2 – c^2 = –1$.

Now

$(a + b – c )= 1$

$\Rightarrow (a + b – 1 )= c$

$\Rightarrow (a + b – 1 )^2= c^2$ (squarring both sides…….)

$\Rightarrow (a^2 + b^2 +1^2+2ab-2a-2b)= c^2$

$\Rightarrow (a^2 + b^2 –c^2+1+2ab)= 2(a+b)$

$\Rightarrow (-1)+1+2ab= 2(a+b)$ ( as $a^2 + b^2 – c^2 = –1$.

$\Rightarrow ab=a+b$

$\Rightarrow (a-1)(b-1)=1$

Therefore the possibile cases are $(a-1)=\pm 1$ and $(b-1)=\pm 1$

Therefore $a=1$,$b=1$ or $a=0$,$b=0$

From the equation $(a + b – c )= 1$ ,C =3 (as a=b=2) and C=-1(as a=b=0)

Therefore $(a^2 +b^2 +c^2)=4+4+9=17$ and $(a^2 +b^2 +c^2)=0+0+1=1$

sum of all possible values of $a^2 + b^2 + c^2=17+1=18$

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## Quadratic Equation Problem | PRMO-2018 | Problem 9

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## Algebra based on Quadratic equation PRMO Problem 9

Suppose a,b are integers and a + b is a root of $x^2 +ax+b=0$.What is the maximum possible
values of $b^2$?

• $49$
• $81$
• $64$

### Key Concepts

Algebra

Factorization

Answer:$81$

PRMO-2018, Problem 9

Pre College Mathematics

## Try with Hints

(‘a+b”) is the root of the equation therefore (“a+b”) must satisfy the given equation

Can you now finish the problem ……….

Discriminant is a perfect square

can you finish the problem……..

Given that $‘a+b”$ is the root of the equation therefore $“a+b”$ must satisfy the given equation

Therefore the given equation becomes ……

$(a+b)^2 +a(a+b)+b=0$

$\Rightarrow a^2 +2ab+b^2+a^2+ab+b=0$

$\Rightarrow 2a^2 +3ab+b^2+b=0$

Now since “a” is an integer,Discriminant is a perfect square

$\Rightarrow 9b^2 -8(b^2+b)=m^2$ (for some $m \in \mathbb Z)$

$\Rightarrow (b-4)^2 -16=m^2$

$\Rightarrow (b-4+m)(b-4-m)=16$

Therefore the possible cases are  $b-4+m=\pm 8$, $b-4-m=\pm 2$,$b-4+m=b-4-m=\pm 4$

i.e b-4=5,-5,4,-4

$\Rightarrow b =9,-1,8,0$

Therefore $(b^2)_{max} = 81$

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## Quadratic equation Problem | AMC 8, 2009 | Problem 23

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## Algebra based on Quadratic equation – AMC-8, 2009 – Problem 23

On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought  400 jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?

• $34$
• $28$
• $25$

### Key Concepts

Algebra

Factorization

Answer:$28$

AMC-8 (2009) Problem 23

Pre College Mathematics

## Try with Hints

Let the number of girls be x

so the number of boys be x+2

Can you now finish the problem ……….

She gave away  400-6=394  jelly beans

$x^2 + (x+2)^2=394$

can you finish the problem……..

Let the number of girls be x

so the number of boys be x+2

she gave each girl x jellybeans and each boy x+2 jellybeans,

Therefore she gave total number of  jelly beans to girls be $x^2$

Therefore she gave total number of  jelly beans to boys be $(x+2)^2$

She gave away  400-6=394  jelly beans

$x^2 + (x+2)^2=394$

$\Rightarrow x^2 + x^2 +4x+4=394$

$\Rightarrow 2x^2 +4x-390=0$

$\Rightarrow x^2 +2x -195=0$

$(x+15)(x-13)=0$

i.e x=-15 , 13 (we neglect negetive as number of students can not be negetive )

Therefore x=13 i.e number os girls be 13 and number of boys be 13+2=15

total number of students ne 13+15=28