Tag: maths

Questions Related to maths

H.C.F. of $6, 72$ and $120$ is:

maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

  1. $6$

  2. $2$

  3. $3$

  4. $1$

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Correct Option: A
Explanation:

Factors of 6 are $1,2,3,6$
Factors of 72 are $1,2,3,4,6,8,9,12,18,24,36,72$
Factors of 120 are $1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120$
The highest factor in all the three is 6 .

Sum of all two digit numbers divisible by $7$ leaves remainder $2$ or $5$ is 

maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

  1. 1300

  2. 1345

  3. 1465

  4. 1356

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Correct Option: D

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

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: B
Explanation:
$\displaystyle \sqrt{1+\sqrt{1-\cfrac{2176}{2401}}}=1+\cfrac{x}{7}$
$\Rightarrow \sqrt{1+\sqrt{\cfrac{2401-2176}{2401}}}=1+\cfrac{x}{7}$
$\Rightarrow \sqrt{1+\sqrt{\cfrac{225}{2401}}}=1+\cfrac{x}{7}$
$\displaystyle \Rightarrow \sqrt{1+\cfrac{\sqrt{225}}{\sqrt{2401}}}=1+\cfrac{x}{7}$                                                                   
$\displaystyle \Rightarrow \sqrt{1+\cfrac{15}{49}}=1+\cfrac{x}{7}$
$\Rightarrow \sqrt{\cfrac{64}{49}}=1+\cfrac{x}{7}$
$\Rightarrow \cfrac{8}{7}=1+\cfrac{x}{7}$
$\Rightarrow \cfrac{x}{7}=\cfrac{8}{7}-1$
$\Rightarrow \cfrac{x}{7}=\cfrac{1}{7}$
$\Rightarrow x=\cfrac{1}{7}\times 7$
$\Rightarrow x=1$                                     

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 ?

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

  1. $20$

  2. $10$

  3. $15$

  4. $13$

Show answer
Correct Option: B
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.

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.

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

  1. $20$ hrs

  2. $30$ hrs

  3. $10$ hrs

  4. None of the above

Show answer
Correct Option: B
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.

 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 $x$.

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

  1. $24$

  2. $25$

  3. $27$

  4. $None\ of\ the\ above$

Show answer
Correct Option: A
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

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.

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

  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.

Show answer
Correct Option: B
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.

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?

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

  1. $8$

  2. $16$

  3. $24$

  4. $28$

Show answer
Correct Option: B
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 $x=16$.

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: 

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

  1. $20$

  2. $23$

  3. $24$

  4. None of these

Show answer
Correct Option: B
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 $20+3=23$.

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

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

  1. no root

  2. one root

  3. two equal roots

  4. infinite roots

Show answer
Correct Option: A
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.