Tag: maths

Questions Related to 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.

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

  1. 18

  2. 72

  3. 27

  4. 81

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Correct Option: D
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 $\displaystyle \left( \frac { 80 }{ 1 } +1 \right) =81$.

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

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

  1. one and only one real number.

  2. real with sum one.

  3. real with sum zero.

  4. real with product zero.

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Correct Option: C
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

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

Show answer
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

Show answer
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

Show answer
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. $48\ yr$

Show answer
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. $11$

Show answer
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

Show answer
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. $3$

Show answer
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 $\forall x\epsilon R$