Tag: solving (simple) problems

Questions Related to solving (simple) problems

Solve the following equations:
$x^{2} + 2xy + 3xz = 50$,
$2y^{2} + 3yz + yx = 10$,
$3z^{2} + zx + 2zy = 10$.

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

  1. $x=\pm 4; y=\pm 2; z=\pm 2$

  2. $x=\pm -4; y=\pm -2; z=\pm 2$

  3. $x = \pm 5; y = \pm 1; z = \pm 1$

  4. None of these

Show answer
Correct Option: C
Explanation:

Given equations are ${ x }^{ 2 }+2xy+3xz=50$

$\Rightarrow  x(x+2y+3z)=50$    ........(i),
$ 2{ y }^{ 2 }+3yz+yx=10$
$\Rightarrow  y(2y+3z+x)=10$    ........(ii)
and $ 3{ z }^{ 2 }+zx+2zy=10$
$\Rightarrow  z(3z+x+2y)=10$    ..........(iii)
Dividing (i) by (ii), we get

$\dfrac { x }{ y } =\dfrac { 50 }{ 10 } =5\\ \Rightarrow x=5y$    ......(a)
Dividing (ii) by (iii), we get
$\dfrac { y }{ z } =\dfrac { 10 }{ 10 } =1\\ \Rightarrow z=y$    ..........(b)
Substituting (a) and (b) in (ii)
$y(2y+3y+5y)=10\\ \Rightarrow 10{ y }^{ 2 }=10\\ \Rightarrow y=\pm 1$
From (a), we have
$x=\pm 5$
From (b), we have
$z=\pm 1$

Solve the following equations:
$x + 2y - z = 11$,
$x^{2} - 4y^{2} + z^{2} = 37$,
$xz = 24$.

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

  1. $x=2, -5; y=2; z=2,-4$

  2. $x=8,-3; y=3; z=-3,-8$

  3. $x=-3, 5; y=4; z=2,5$

  4. $x=2,4; y=3; z=3, -5$

Show answer
Correct Option: B
Explanation:

${ x }^{ 2 }-4{ y }^{ 2 }+{ z }^{ 2 }=37$    ......(i)

$xz=24$    .......(ii)
$ x+2y-z=11$    .....(iii)
$\Rightarrow  x-z=11-2y$
On squaring both sides, we have

${ x }^{ 2 }+{ z }^{ 2 }-2xz=121+4{ y }^{ 2 }-44y\\ \Rightarrow { x }^{ 2 }+{ z }^{ 2 }-4{ y }^{ 2 }-2xz=121-44y\\ \Rightarrow 37-2(24)=121-44y\\ \Rightarrow -44y=-132\\ \Rightarrow y=3$
Substituting $y$ in (iii), 
$x+6-z=11\\ \Rightarrow x-z=5$
From (ii), $z=\dfrac { 24 }{ x } $
Thus $x-\dfrac { 24 }{ x } =5$
$ \Rightarrow { x }^{ 2 }-24=5x\\ \Rightarrow { x }^{ 2 }-5x-24=0\\ \Rightarrow { x }^{ 2 }-3x+8x-24=0\\ \Rightarrow x(x-3)+8(x-3)=0\\ \Rightarrow (x+8)(x-3)=0\\ \Rightarrow x=-8,3$
Putting in $ z=\dfrac { 24 }{ x } $
Thus $ z=-3,8$
So, the values of $x$ are $-8,3$, values of $z$ are $-3,8$ and value of $y$ is $3$.

If the zeroes of the rational expression $ (ax+b)(3x+2)$ are $-\dfrac{2}{3}$ and $ \dfrac{1}{2}$, then $ a+b=$

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

  1. $4$

  2. $0$

  3. $-b$

  4. None of these

Show answer
Correct Option: B
Explanation:

$(ax+b)(3x+2)=0$

$\Rightarrow$  $3ax^2+2ax+3bx+2b$
$\Rightarrow$  $3ax^2+(2a+3b)x+2b=0$
It is given that $\dfrac{-2}{3}$ and $\dfrac{1}{2}$ are zeros of the equation $3ax^2+(2a+3b)x+2b=0$
Here, $A=3a,\,B=2a+3b,\,C=2b$
Let $\alpha=\dfrac{-2}{3}$ and $\beta=\dfrac{1}{2}$
We know,
$\Rightarrow$  $\alpha+\beta=\dfrac{-B}{A}$
$\Rightarrow$  $\dfrac{-2}{3}+\dfrac{1}{2}=\dfrac{-(2a+3b)}{3a}$

$\Rightarrow$  $\dfrac{-4+3}{6}\times 3a=-2a-3b$

$\Rightarrow$  $\dfrac{-a}{2}=-2a-3b$
$\Rightarrow$  $-a=-4a-6b$
$\Rightarrow$  $3a+6b=0$              ----- ( 1 )
Now,
$\Rightarrow$  $\alpha.\beta=\dfrac{CA}{A}$

$\Rightarrow$  $\dfrac{-2}{3}\times\dfrac{1}{2}=\dfrac{2b}{3a}$

$\Rightarrow$  $\dfrac{-1}{3}\times 3a=2b$

$\Rightarrow$  $-a+2b=0$          ----- ( 2 )
Adding equation ( 1 ) and ( 2 ) we get,
$b=0$
Put $b=0$ in ( 2 ) we get,
$a=0$
$\therefore$  $a+b=0+0=0$

In a bangle shop, if the shopkeeper displays the bangles in the form of a square then he is left with 38 bangles with him. If he wanted to increase the size of square by one unit each side of the square he found that 25 bangles fall short of In completing the square. The actual number of bangles which he had with him in the shop was ________.

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

  1. 1690

  2. 999

  3. 538

  4. can't be determined

Show answer
Correct Option: B
Explanation:

Let the length of the side of the square initially be $l$.

Number of bangles it can store is $l^{2}$
The total number of bangles will be $l^{2}+38$
In the second case,
The length of the side of the square is $l+1$
The number of bangles it can now store is $(l+1)^{2}$
Now the total number of bangles equals to $(l+1)^{2}-25$
$\therefore$ $(l+1)^{2}-25$=$l^{2}+38$
$\Rightarrow$ $l^{2}+2l+1-25=l^{2}+38$
$\Rightarrow$ $2l=62$
$\Rightarrow$ $l=31$
$\therefore$ The total number of bangles=$(l+1)^{2}-25$=$999$
$\therefore$ Option B is correct

A man walks a distance of 48 km in a given time. If he walks 2 km/hr faster, he will perform the journey 4 his before. His normal rate of walking is _______.

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

  1. 3 km/hr

  2. 4 km/hr

    • 6 km/hr or 4 km/hr

  4. 5 km/hr

Show answer
Correct Option: B
Explanation:

Let us assume that the man walks at a speed of $x$ $kmph$

Then the time taken by the man to complete the journey is $t=\frac{48}{x}$  ($\because time=\frac {distance}{speed}$)
Now in the second case, $t'=t-3$ and $x'=x+2$.
Then, 
$t-4=\frac{48}{x+2}$
$\Rightarrow$ $t-\frac{48}{x+2}=4$
$\Rightarrow$ $\frac{48}{x}-\frac{48}{x+2}=4$
$\Rightarrow$ $\frac{96}{x(x+2)}=4$
$\Rightarrow$ $x^{2}+2x=24$
$\Rightarrow$ $x^{2}+2x-24=0$
$\Rightarrow$ $x^{2}-4x+6x-24=0$
$\Rightarrow$ $(x-4)(x+6)=0$
Speed cannot be a negative value and hence $x=4kmph$
$\therefore$ Option B is correct.

 Choose the correct answer from the alternatives given.
If $\alpha \, and \, \beta$ are the roots of the equation $x^2$ - 7x + 12 = 0, then $\alpha^2 \, + \, \beta^2$ equals.

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

  1. 19

  2. 25

  3. 14

  4. 24

Show answer
Correct Option: B
Explanation:
: Let $\alpha \, and \, \beta$ are the roots of the equation
$ax^2 + bx + c = 0$
We know that,
$\displaystyle \alpha \, + \, \beta \, = \, \frac{-b}{a} \, = \, \frac{- (-7)}{1} \, = \, 7$
$\displaystyle \alpha^2 \, + \, \beta^2 \, = \, (\alpha \, + \, \beta)^2 \, - \, 2\alpha \beta$
$\displaystyle \alpha^2 \, + \, \beta^2 \, = \, (7)^2 \, - \, 2 \, \times \, 12 \, = \, 49 \, - \, 24 \, = \, 25$

A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Their present ages are 6 years and 12 years.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  Let present age of girl be $2x$.

$\Rightarrow$  Then, age of her sister be $x.$
According to the question,
$\Rightarrow$  $(2x+4)(x+4)=160$
$\Rightarrow$  $2x^2+8x+4x+16=160$
$\Rightarrow$  $2x^2+12x+16-160=0$
$\Rightarrow$  $2x^2+12x-144=0$
$\Rightarrow$  $2(x^2+6x-42)=0$
$\Rightarrow$  $x^2+6x-72=0$
$\Rightarrow$  $x^2+12x-6x-72=0$
$\Rightarrow$  $x(x+12)-6(x+12)=0$
$\Rightarrow$  $(x+12)(x-6)=0$
$\Rightarrow$  $x+12=0$ and $x-6=0$
$\therefore$  $x=-12$ and $x=6$
$\Rightarrow$  Age cannot be negative.
$\therefore$  $x=6$
$\Rightarrow$  $2x=2\times 6=12$
$\therefore$  The present ages of girls are $6\,years$ and $12\,years.$

If  a,b,c are distinct and the roots of $\left( b-c \right) { x }^{ 2 }+\left( c-a \right) x+(a-b)=0$ are equal, then a,b,c are in

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

  1. Arithmetic progression

  2. Geometric progression

  3. Harmonic progression

  4. Arithmetico-Geometric progression

Show answer
Correct Option: A
Explanation:

$Clearly\quad x=1\quad is\quad a\quad solution$
$\therefore \quad Product\quad of\quad the\quad roots=\frac { a-b }{ b-c } $$\therefore \quad \left( 1 \right) \left( 1 \right) =\frac { a-b }{ b-c } $
$\Rightarrow b-c=a-b$
$\Rightarrow 2b=a+c\Rightarrow a,b,c\quad are\quad in\quad A.P.$






If the harmonic mean of the roots of$\sqrt { 2 } { x }^{ 2 }-bx+\left( 8-2\sqrt { 5 }  \right) =0$ is 4, the the value of b=

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

  1. 2

  2. 3

  3. $4-\sqrt { 5 } $

  4. $4+\sqrt { 5 } $

Show answer
Correct Option: C
Explanation:

$Let\quad \alpha ,\beta \quad be\quad the\quad roots$ 
$\Rightarrow \frac { 2\alpha \beta  }{ \alpha +\beta  } =4\quad \quad \Rightarrow \frac { \frac { 2\left( 8-2\sqrt { 5 }  \right)  }{ \sqrt { 2 }  }  }{ \frac { b }{ \sqrt { 2 }  }  } =4$
$\Rightarrow \frac { 2\left( 8-2\sqrt { 5 }  \right)  }{ 4 } =b$
$\therefore b=4-\sqrt { 5 } $