Tag: applications of quadratic equations

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Questions Related to applications of quadratic equations

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

What is the value of $x$ in the equation $\displaystyle \sqrt{1+\sqrt{1-\frac{2176}{2401}}}=1+\frac{x}{7}$?

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\displaystyle \sqrt{1+\sqrt{1-\cfrac{2176}{2401}}}=1+\cfrac{x}{7}$
$\Rightarrow \sqrt{1+\sqrt{\cfrac{2401-2176}{2401}}}=1+\cfrac{x}{7}$
$\Rightarrow \sqrt{1+\sqrt{\cfrac{225}{2401}}}=1+\cfrac{x}{7}$
$\displaystyle \Rightarrow \sqrt{1+\cfrac{\sqrt{225}}{\sqrt{2401}}}=1+\cfrac{x}{7}$                                                                   
$\displaystyle \Rightarrow \sqrt{1+\cfrac{15}{49}}=1+\cfrac{x}{7}$
$\Rightarrow \sqrt{\cfrac{64}{49}}=1+\cfrac{x}{7}$
$\Rightarrow \cfrac{8}{7}=1+\cfrac{x}{7}$
$\Rightarrow \cfrac{x}{7}=\cfrac{8}{7}-1$
$\Rightarrow \cfrac{x}{7}=\cfrac{1}{7}$
$\Rightarrow x=\cfrac{1}{7}\times 7$
$\Rightarrow x=1$                                     
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

Two years ago Sam's age was $\displaystyle 4 \frac{1}{2}$ times the age of his son. Six years ago, his age was twice the square of the age of his son. What is the present age of Sam's son ?

  1. $20$
  2. $10$
  3. $15$
  4. $13$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $m$ be the age of Sam and $s$ be the age of his son
Then, 2 years ago,
$(m -2) = \dfrac{9}{2} (s-2)$
$2m - 4 = 9s - 18$
$2m = 9s - 14 ...(i)$
$6$ years ago,
$(m - 6) = 2(s- 6)^2$
$\left(\dfrac{9s-14}{2} - 6\right) = 2(s-6)^2$
$9s - 14 - 12 = 4(s^2 - 12s + 36)$
$9s - 26 = 4s^2 - 48s + 144$
$4s^2 - 57s + 170 = 0$
$s = \dfrac{57 \pm \sqrt{529}}{8}$ = $\dfrac{57 \pm 23}{8} = 10, 4.25$
Neglect the 4.25 which is a fraction. 
Hence, age of his son is $10$ years.

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

For the same amount of work , A takes 6 hours less than B. If together they complete the work in 13 hours 20 minutes; find how much time will B alone take to complete the work.

  1. $20$ hrs
  2. $30$ hrs
  3. $10$ hrs

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the time taken to complete the work  by B be  x.
A can take x - 6 hours.

A and B together can complete the work = 13 hours 20 minutes.
$\frac{1}{x-6}+\frac{1}{x}=13\frac{20}{60}$
$\frac{2x-6}{x^2-6x}=\frac{3}{40}$
$40(2x-6)=3x^2-18x$
$80x-240=3x^2-18x$
$3x^2-98x+240$
Using quadratic formula,
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\frac{98\pm\sqrt{(-98)^2-4\times 3\times 240}}{2\times 3}$
$\frac{98\pm\sqrt{6724}}{6}$
$\frac{98\pm 82}{6}$
So, x = 30, x = 2.6666..
x value cannot be negative.
B take 30 hours to complete the work alone.

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

 A can do a piece of work in $'x'$ days and $B$ can do the same work in $'x+16'$ days.If both working together can do it in $15$ days. Calculate $x$.

  1. $24$
  2. $25$
  3. $27$
  4. $None\ of\ the\ above$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given A do a piece of work in x days and B do work in x+16 days

Then A one day work = $\frac{1}{x}$ days
AND B one day work =$\frac{1}{x+16}$days
Then Both one day work =$\frac{1}{x}+\frac{1}{x+16}=\frac{x+16+x}{x(x+16)}=\frac{2x+16}{x^{2}+16x}$ 
So both do work in $\frac{x^{2}+16x}{2x+16} \ days$ 
But both do work in 15 days
$\therefore \frac{x^{2}+16x}{2x+16}=15$
$\Rightarrow x^{2}+16x=30x+240$
$\Rightarrow x^{2}-14x-240=0$
$\Rightarrow x^{2}-24x+10x-240=0$
$\Rightarrow x(x-24)+10(x-24)=0$
$\Rightarrow (x-24)(x+10)=0$
Then $x-24=0 , x=24$
And $x+10=0 , x=-10$
But work done is not negative
Then work done =$24$ days

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

One year ago, the father was $8$ times as old as his son. Now his age is square of the son's age. Find their present ages.

  1. Present age of father is $36$ years and that of his son is $6$ years.
  2. Present age of father is $49$ years and that of his son is $7$ years.
  3. Present age of father is $64$ years and that of his son is $8$ years.
  4. Present age of father is $25$ years and that of his son is $5$ years.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the present age of the son be $x$.
Let the father's present age be $y$.
One year ago, $y-1=8(x-1)$
$\Rightarrow  y-1=8x-8$
$\Rightarrow  y=8x-7$
Now applying the condition, we get
$(8x-7)=x^{2}$
$\Rightarrow x^{2}-8x+7=0$
$\Rightarrow (x-1)(x-7)=0$
$\Rightarrow x=1$ or $x=7$

Hence, the present age of son is either $1$ year or $7$ years.

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

A shopkeeper buys a number of books for Rs. $80$. If he had bought $4$ more for the same amount, each book would have cost Re. $1$ less. How many books did he buy?

  1. $8$
  2. $16$
  3. $24$
  4. $28$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the number of books he bought for Rs. $80$ be $x$

Now, the new number of books $=$ $(x+4)$

According to question, we have

$\dfrac{80}{x}-\dfrac{80}{x+4}=1$

$\Rightarrow \dfrac{80(x+4)-80x}{x(x+4)}=\dfrac{1}{1}$
$\Rightarrow \dfrac{80x+320-80x}{x^2+4x}=\dfrac{1}{1}$

$\Rightarrow x^2+4x=320$

$\Rightarrow x^2+4x-320=0$

$\Rightarrow x^2+20x-16x-320=0$

$\Rightarrow x(x+20)-16(x+20)=0$

$\Rightarrow (x+20)(x-16)=0$

$\Rightarrow x=-20$ or $x=16$

Since the number cannot be in negative, therefore $x=16$.

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

In a school hall, $460$ students were sitting in rows and columns in such a way that the number of students sitting in each column was three more than the number of students sitting in each row. The number of students in each column was: 

  1. $20$
  2. $23$
  3. $24$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the number of students in each row be $x$.
Hence, $x(x+3)=460$
$x^{2}+3x=460$
$x^{2}+3x-460=0$
$(x+23)(x-20)=0$
$x=-23$ and $x=20$
Hence, the number students in  each column is $20+3=23$.

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

The equation $x - \displaystyle{\frac{2}{x - 1}} = 1 - \displaystyle{\frac{2}{x - 1}}$ has

  1. no root

  2. one root

  3. two equal roots

  4. infinite roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

consider, $x-\cfrac{2}{x-1}=1-\cfrac{2}{x-1}$

$\Rightarrow x-1=\cfrac{2}{x-1}-\cfrac{2}{x-1}$

$\Rightarrow x-1=0$

$\Rightarrow x=1$.

However, as $x\rightarrow 1$,     

$(\cfrac{2}{x-1})\rightarrow \infty$

Hence, the above equation has no root.

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

If $\alpha, \beta$ are the roots of the equation $x^2 - 3x + 1=0$, then the equation with roots $\displaystyle \frac{1}{\alpha - 2}, \frac{1}{\beta - 2}$ will be-

  1. $x^2 - x - 1 = 0$
  2. $x^2 + x - 1 = 0$
  3. $x^2 + x + 2 = 0$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let y = 1/(x-2). Then x-2 = 1/y, so x = 2 + 1/y = (2y+1)/y. Substitute into x^2 - 3x + 1 = 0: ((2y+1)/y)^2 - 3((2y+1)/y) + 1 = 0. Simplifying leads to y^2 - y - 1 = 0.

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

If $\displaystyle \alpha ,\beta $ are the roots of $\displaystyle x^{2}+x+1=0$ and $\displaystyle \gamma ,\delta $ are the roots $\displaystyle x^{2}+3x+1=0,$ then $\displaystyle (\alpha -\gamma)(\beta +\delta )(\alpha +\delta )(\beta -\gamma )=$

  1. $2$`
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Using $\alpha \beta =1$     and       $\gamma \delta =1$

$(\alpha \beta + \alpha \delta - \beta \gamma - \gamma \delta) (\alpha \beta - \alpha \gamma +\beta \delta - \gamma \delta) $

$\Rightarrow (1+\alpha \delta - \beta \gamma - 1)(1-\alpha \gamma +\beta \delta - 1)$

$\Rightarrow (\alpha \delta - \beta \gamma) (\beta \delta - \alpha \gamma) $

$\Rightarrow \delta ^{2}-\alpha ^{2}-\beta ^{2}+\gamma ^{2}$

$\Rightarrow - (\alpha ^{2}+\beta ^{2})+(\gamma ^{2}+\delta ^{2})$

$\Rightarrow - (-1)+7=8$

Final answer is 8