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

A two digit number in such that the product of its digits is $8$. When $63$ is subtracted from the number, the digits interchange their places. Find the number.

  1. 18

  2. 72

  3. 27

  4. 81
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the digit at unit's place $= x$
$\displaystyle \therefore $ Digit at ten's place $\displaystyle \frac { 8 }{ x } $ and the number is $\displaystyle \left( \frac { 80 }{ x } +x \right) $
New number on interchanging the places of digits $\displaystyle =10x+\frac { 8 }{ x } $
$\displaystyle \therefore $ According to given condition
$\displaystyle \frac { 80 }{ x } +x-63=10x+\frac { 8 }{ x } $
$\displaystyle 80+{ x }^{ 2 }-63x=10{ x }^{ 2 }+8$
$\displaystyle { 9x }^{ 2 }+63x-72=0$
$\displaystyle { x }^{ 2 }-7x-8=0$
$\displaystyle { x }^{ 2 }+8x-x-8=0$
$\displaystyle x\left( x+8 \right) -1\left( x+8 \right) =0$
$\displaystyle \left( x+8 \right) \left( x-1 \right) =0$
$\displaystyle i.e.\quad x=-8$  and $ x=1$
Rejecting $\displaystyle x=-8$ and putting $\displaystyle x=1$ the required no. is $\displaystyle \left( \frac { 80 }{ 1 } +1 \right) =81$.

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

For the equation $|x|^{2}+|x|-6=0$, the roots are

  1. one and only one real number.

  2. real with sum one.

  3. real with sum zero.

  4. real with product zero.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For x>0 equation is $x^2+x-6=0$
$(x-2)(x+3)=0$
$x=2$
$x$ cant be equal to -3 as for this equation $x>0$
Now when $x <0$ equation becomes $x^2-x-6$
$(x+2)(x-3)=0$
Hence $x=-2$
So the roots are $2 and -2$
Thus sum of roots is zero and roots are real
So Option C

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

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

  1. real and distinct

  2. real and equal

  3. real

  4. imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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 

  1. $5$ min
  2. $10$ min
  3. $15$ min
  4. $15.66$ min
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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?

  1. $6$ seconds
  2. $7$ seconds
  3. $8$ seconds
  4. $9$ seconds
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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?

  1. $47\ yr$
  2. $49\ yr$
  3. $36\ yr$
  4. $48\ yr$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

  1. $\infty$
  2. $\dfrac{1}{4}$
  3. $\dfrac{-1}{6}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $

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

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

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\forall x\epsilon R$