Tag: maths

Questions Related to maths

When an iron rod is cut into equal pieces of $30$ cm each, a piece of $4$ cm is left out. When cut into equal pieces of $29$ cm, a piece of $13$ cm is left out. The minimum length of rod is

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $270$ cm

  2. $272$ cm

  3. $274$ cm

  4. $280$ cm

Show answer
Correct Option: C
Explanation:

Let the length of rod be x.

If rod cuts in 30 cm each, 4cm from left
$\implies x=(n _1)(30)+4-------(1)$
If cuts in 29 cm each 13 cm left
$x=(n _2)(29)+13------------(2)$
$\implies n _1(30)+4=(n _2)(29)+13$
$(n _1)(30)=n _2(29)+9---------(3)$
From (1) , for different values of n,
$x=34,64,94,124,154,184,214,244,274,304,.......$
Similarly from (2)
$x=42,71,100,129,158,187,216,245,274,303,.......$
From above it is clear that $274$ is minimum common value.So, minimum length must be $274$ cm.

It costs Rs. $10$ a kilometer to fly and Rs. $2$ a kilometer to drive. If one travels $200$ km covering $x$ km of the distance by flying and the rest by driving, then the cost of the trip is

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. Rs. $2,000$

  2. Rs. $24,000$

  3. Rs.$ (8x + 400)$

  4. Rs. $(12x + 400)$

Show answer
Correct Option: C
Explanation:

Let distance travelled by flying is $x$ km.
Hence, the distance  travelled  by  driving $= (200-x)$km.
The cost of flying $ = $ Rs. $ 10 $ per km
Hence, the cost of flying $x$ km $ = 10 \times x =10x$
The cost of driving $ = $ Rs. $ 2 $ per km
The cost of driving $(200-x)$ km $ = 2 \times (200-x) = 400 -2x$
Hence, the total cost $ =10x + 400-2x = $Rs.$400 + 8x$

The values of a so that the equation $\Vert x - 2\vert - 1\vert = a \vert x \vert$ does not contain any solution lying in the interval {2, 3} are

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $a \ \epsilon(-\infty \dfrac{1}{2})$

  2. $a \ \epsilon (1, \infty)$

  3. $a \ \epsilon (-\infty, 0) \cup (\dfrac{1}{2}, \infty )$

  4. None of these

Show answer
Correct Option: A
Explanation:

The values of a so that the equation $\left| \left| x-2 \right| -1 \right| =a\left| x \right| $ does not contains any solution lying in the interval $\left{ 2,3 \right} $ are

$a\in (-\infty ,\cfrac { 1 }{ 2 } )\ \left| \left| x-2 \right| -1 \right| =a\left| x \right| \ =>\left| \left| 2-2 \right| -1 \right| =a\left| 2 \right| \ =>\left| -1 \right| =a\left| 2 \right| \ =>1=a\left| 2 \right| \ =>a=\cfrac { 1 }{ 2 } $
Obviously, it does not contains any solution lying in the interval $\left{ 2,3 \right} $ are $a\in (-\infty ,\cfrac { 1 }{ 2 } )$

The minimum value of $\displaystyle f(x)=|x-1|+|x-2|+|x-3|$ is equal to 

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $1$

  2. $2$

  3. $3$

  4. $0$

Show answer
Correct Option: B
Explanation:

Solution:- (B) 2

The function $f$ is linear on each of the intervals $\left( - \infty, 1 \right], \left[ 1, 2\right], \left[2, 3\right] \text{ and } \left[ 3, \infty \right)$. Since a linear function on an interval always attains its minimum at one of the endpoints of the interval, and $f \left( x \right) = +\infty \text{ as } x = \pm \infty$, the function $f$ must attain its minimum at one of $x = 1, 2, 3$. Since $f(1)=3,  f \left( 2 \right) = 2 \text{ and } f \left( 3 \right) = 3$, the function $f$ attains a minimum of 2 at $x=2$.

A Gym sells two types of memberships. One packages costs $ $325$ for one year of membership with an unlimited number of visits. The second package has a $ $125$ enrolment fee, includes five free visits, and costs an additional $ $8$ per visit after first five. How many visits would a person need to use for each type of membership to cost the same amount over a one-year period?

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $20$

  2. $25$

  3. $30$

  4. $40$

Show answer
Correct Option: C
Explanation:

Let the total number of visits be x. Hence equating the costs give us 

$ $325=$ 125+(x-5) $ 8$ $\rightarrow 200=8x-40$
$240=8x$ or $x=30$. Hence the person needs a total of 30 visits.

If $\cfrac{7}{m-\sqrt{3}} = \cfrac{\sqrt{3}}{m} + \cfrac{4}{2m}$, calculate the value of $m$.

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $-3.464$

  2. $-1.978$

  3. $-0.918$

  4. $1.978$

Show answer
Correct Option: B
Explanation:

Given $\dfrac { 7 }{ m-\sqrt { 3 }  } =\dfrac { \sqrt { 3 }  }{ m } +\dfrac { 4 }{ 2m } =\dfrac { 4+2\sqrt { 3 }  }{ 2m } $
$\Rightarrow  14m=(4+2\sqrt { 3 } )m-\sqrt { 3 } (4+2\sqrt { 3 } )$
$ \Rightarrow (10-2\sqrt { 3 } )m=-4\sqrt { 3 } -6$
$\Rightarrow  m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { 3 }  } =\dfrac { -12.928 }{ 6.536 } =-1.98$

Four years ago, the father's age was three times the age of his son. The total of the age of the father and the son after four years will be $64$ years. What is the father's age at present?

maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

  1. $32$ years

  2. $36$ years

  3. $44$ years

  4. $40$ years

Show answer
Correct Option: D
Explanation:

fathers present age $= x$ years.

sons present age $= y$ years.  

four years ago, fathers age $=3 \times $ sons age.
$(x-4)=(y-4)\times 3$
$x=3y-8$.....(i) 

after 4 years, fathers age$ +$ sons age $= 64 $
$x+4+y+4=64 $
$x+y=56$.....(ii) 

solve for $x$
$x=3\times 56-3x-8$ 
$4x=160$ 
$x=40$ years  

D is correct.   

The area of a field in the shape of a trapezium measures $1440{m}^{2}$. The perpendicular distance between its parallel sides is $24m$. If the ratio of the parallel sides is $5:3$, the length of the longer parallel side is:

maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

  1. $45m$

  2. $60m$

  3. $75m$

  4. $120m$

Show answer
Correct Option: C
Explanation:
Parallel sides $= 5x,\, 3x$
area $=\dfrac{24}{2}(5x+3x)=1440 $
$12(8x)=1440$
$x=\dfrac{120}{8}=15$ 
$5x=15\times 5$ 
     $=75m$

If $\left(\dfrac { 3 } { 4 }\right)^{th}$ of $x$ of $\left(\dfrac { 1 } { 4 }\right)^{th}$ of $35600 = 1668.75 ,$ find $x$

maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

  1. $\dfrac { 2 } { 3 }$

  2. $\dfrac { 3 } { 4 }$

  3. $\dfrac { 2 } { 5 }$

  4. $\dfrac { 1 } { 4 }$

Show answer
Correct Option: D
Explanation:

$\dfrac { 3 }{ 4 } \times x\times \dfrac { 1 }{ 4 } \times 35600=1668.75$

$\Rightarrow x=\dfrac { 1668.75\times 16 }{ 3\times 35600 } =\dfrac { 1 }{ 4 } $      [D]

If a function $f$ is linear with $f(0)=5$ and $f(2)=9$ then find $f(9)$

maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

  1. 23

  2. 25

  3. 26

  4. none of these

Show answer
Correct Option: A
Explanation:

$f$ is a linear function 

So Let $f=ax+b$
$f(0)=a(0)+b=5\\implies b=5\cdots(1)\f(2)=9\a(2)+5=9\\implies a=2\f(x)=2x+5\f(9)=2(9)+5\18+5=23$