Tag: solving (simple) problems

Questions Related to solving (simple) problems

If $a, b, c$ are in A.P., then the roots of the equation $ax^{2}+2bx+c=0$ are

applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

  1. real and distinct

  2. real and equal

  3. real

  4. imaginary

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Correct Option: C
Explanation:

If $a,b,c$ are in A.P, then

$2b=a+c$.
Hence the give equation transforms into
$ax^{2}+(a+c)x+c=0$
Hence
$D$
$=B^{2}-4AC$

$=(a+c)^{2}-4ac$

$=a^{2}+c^{2}+2ac-4ac$

$=a^{2}+c^{2}-2ac$

$=(a-c)^{2}$
Now 
$(a-c)^{2}\geq 0$
Hence 
$D\geq 0$.
Or 
$B^{2}-4AC\geq 0$.
Since the discriminant is greater than 0, hence the roots real.

$a^{nm}\ne a^{m^n}$ if

applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

  1. $a=1$

  2. $m=n=1$, $a=6$

  3. $m=n=2$, $a=3$

  4. $m=n=a=0$

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Correct Option: D

To fill a cistern, pipes $P, Q$ & $R$ take $20, 15$ and $12$ minutes respectively.  The time in minutes that the three pipes together will take to fill the cistern is 

applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

  1. $5$ min

  2. $10$ min

  3. $15$ min

  4. $15.66$ min

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Correct Option: A
Explanation:

Given: pipes $P, Q, R$ takes $20, 15, 12$ minutes respectively to fill a cistern

To find the time in minutes that the three pipes together will take to fill the cistern
Sol: By taking the LCM of $12, 15,20$ we will be able to find the actual capacity of the cistern, i.e., the actual capacity of cistern is 60 L(lets assume the unit to be litres)
IF pipe P can fill the cistern in $20$ minutes. Then in 1 min it can fill $\dfrac {60}{20}=3$ liters

Similarly in 1 min pipe Q can fill $\dfrac {60}{15}=4$ liters
And pipe R can fill $\dfrac {60}{12}=5$ liters
Therefore in 1 min the cistern will be filled by (P in 1 min)+(Q in 1 min)+(R in 1 min)= $(3+4+5) =12$ liters
Now, 
Time taken to fill $12 $ liters=$1$ minute
Therefore, 
Time taken to fill $60$ litres=$\dfrac {60}{12}$ (apply unitary method) =$5$ mins
Therefore, all the three pipes can fill the cistern together in 5 mins

A ball is thrown upwards from a rooftop, $28$m above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time $t$ is $h$, which is given by, $h =$ $(-t^{2}+2t +35) +28$

How long will it take before hitting the ground?

applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

  1. $6$ seconds

  2. $7$ seconds

  3. $8$ seconds

  4. $9$ seconds

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Correct Option: D
Explanation:

$h =$ $-t^{2}+2t +35 + 28=0$
$=>$ $t^{2}-2t - 63=0$
$=>(t - 9)(t + 7) = 0$
$=>t = 9$ or $-7$
The time cannot be negative, so the time is $9$ seconds.

The age of a man is the square of his son's age. A year ago, the man's age was $8$ times the age of his son. What is the present age of the man?

applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

  1. $47\ yr$

  2. $49\ yr$

  3. $36\ yr$

  4. $48\ yr$

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Correct Option: B
Explanation:

let present age of son be $x$

present age of father be ${ x }^{ 2 }$
$1$ year ago
age of son $=x-1$
age of father = ${ x }^{ 2 }$-1
so 
${ x }^{ 2 }-1=8(x-1)$
${ x }^{ 2 }-8x+7=0$
$(x-7)(x-1)=0$
$x=7$
son's age $= 7 years$
man's age = ${ x }^{ 2 }=49$ years

Solve for $x:2\sqrt { x+5 } =8$

applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

  1. $4$

  2. $6$

  3. $9$

  4. $11$

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Correct Option: D
Explanation:

Given, $2\sqrt{x+5}=8$

$\Rightarrow \sqrt{x+5}=4$
On squaring both sides, we get 
$x+5 = 16$
$\Rightarrow x=16-5$
$\Rightarrow x=11$

If $a$ and $b$ are the roots of the quadratic equation $x^2-4x+3=0$, then $(1+a+a^2+a^3...)(1+b+b^2+b^3+....)$ equal to

applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

  1. $\infty$

  2. $\dfrac{1}{4}$

  3. $\dfrac{-1}{6}$

  4. none of these

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Correct Option: A
Explanation:

${ x }^{ 2 }-4x+3=0\ (x-1)(x-3)=0\ x=1\quad or\quad 3\ \therefore a=3,\quad b=1\ a+b=4,\quad ab=3\ 1+a+{ a }^{ 2 }+{ a }^{ 3 }+........+\infty =\cfrac { 1 }{ 1-a } (sum\quad of\quad infinite\quad G.P.)\ \therefore (1+a+{ a }^{ 2 }+{ a }^{ 3 }+........+\infty )(1+b+{ b }^{ 2 }+{ b }^{ 3 }+.......+\infty )\ =(\cfrac { 1 }{ 1-a } )(\cfrac { 1 }{ 1-b } )\ =\cfrac { 1 }{ 1-(a+b)+ab } \ =\cfrac { 1 }{ 1-4+3 } =\cfrac { 1 }{ 0 } =\infty $

The number of real solution of the equation $(\dfrac{9}{10})^x=-3+x-x^2$ is-

applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

  1. $0$

  2. $1$

  3. $2$

  4. $3$

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Correct Option: A
Explanation:
Let $f\left( x \right) ={ \left( \cfrac { 9 }{ 10 }  \right)  }^{ x }$
$g\left( x \right) =-3+x-{ x }^{ 2 }$
For $x<0$
$f\left( x \right) >1$
$g\left( x \right) -1=-{ x }^{ 2 }+x-4$
$\triangle =1-4\left( -1 \right) \left( -4 \right) $
$=-15<0$
and co efficient of ${ x }^{ 2 }=-1<0$
$\therefore g\left( x \right) -1<0$
$\therefore g\left( x \right) <1$
$\therefore f\left( x \right) -g\left( x \right) $ has no solution if $x<0$
For $x>0$
$f\left( x \right) <1$
$-3+x-{ x }^{ 2 }=f\left( x \right) $
$-{ x }^{ 2 }+x-3-f\left( x \right) =0$
$\triangle =1-4\left( -1 \right) \left( -3-f\left( x \right)  \right) $
$=1-12+4f\left( x \right) $
$\triangle =-11+4f\left( x \right) $
$f\left( x \right)<0$
$4f\left( x \right)-11<-11$
$\therefore \triangle <0$
$f\left( x \right)=g\left( x \right)$ has no real solution for $x>0$
At $x=0$
$f\left( x \right)=1,g\left( x \right)=-3$
$\therefore f\left( x \right)=g\left( x \right)$ has no real solution $\forall x\epsilon R$

If $a,b,c,d$ are four consecutive terms of an increasing A.P., then the roots of the equation
$(x-a)(x-c)+2(x-b)(x-d)=0$ are

applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

  1. $\text{real and distinct}$

  2. $\text {non-real complex}$

  3. $\text {real and equal}$

  4. $\text {integers}$

Show answer
Correct Option: A
Explanation:
$a,b,c,d$ are $4$ consecutive terms of A

Let $a=m-3n,b=m-n.c=m+n,d=m+3n$

$'2n'$ is common difference

$(x-a)(x-c)+2(x-b)(x-d)=0$

$3{ x }^{ 2 }-\left( a+c+2b+2d \right) x+\left( ac+2bd \right) =0$

$=6m+2n$

$ac+2bd={ m }^{ 2 }-2mn-3{ n }^{ 2 }+2{ m }^{ 2 }+4mn-6{ n }^{ 2 }$

$=3{ m }^{ 2 }+2mn-9{ n }^{ 2 }$

$3{ x }^{ 2 }-\left( 6m+2n \right) x+\left( 3{ m }^{ 2 }+2mn-9{ n }^{ 2 } \right) =0$

$\triangle ={ \left( 6m+2n \right)  }^{ 2 }-4\left( 3 \right) \left( 3{ m }^{ 2 }+2mn-9{ n }^{ 2 } \right) $

$=4\left( 9{ m }^{ 2 }+6mn+{ n }^{ 2 }-9{ m }^{ 2 }-6mn+36{ n }^{ 2 } \right) $

$=4\left( 37 \right) { n }^{ 2 }$

$\triangle >0$

$\therefore $ Roots of $(x-a)(x-c)+2(x-b)(x-d)$ are real and distinct.

If roots of equation $ x^2 - (2n+ 18) x - n-1 = 0 ( n \epsilon Z ) $ are rational, then number of possible value of $n $ is :

applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

  1. $1$

  2. $2$

  3. $0$

  4. Infinite

Show answer
Correct Option: B