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

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

  1. $270$ cm
  2. $272$ cm
  3. $274$ cm
  4. $280$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. Rs. $2,000$
  2. Rs. $24,000$
  3. Rs.$ (8x + 400)$
  4. Rs. $(12x + 400)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.$400 + 8x$

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $a\in (-\infty ,\cfrac { 1 }{ 2 } )$

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

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

  1. $1$
  2. $2$
  3. $3$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $x=2$.

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

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?

  1. $20$
  2. $25$
  3. $30$
  4. $40$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $-3.464$
  2. $-1.978$
  3. $-0.918$
  4. $1.978$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$
$\Rightarrow  m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { 3 }  } =\dfrac { -12.928 }{ 6.536 } =-1.98$

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

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?

  1. $32$ years
  2. $36$ years
  3. $44$ years
  4. $40$ years
Reveal answer Fill a bubble to check yourself
D Correct answer
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.   

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

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:

  1. $45m$
  2. $60m$
  3. $75m$
  4. $120m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Parallel sides $= 5x,\, 3x$
area $=\dfrac{24}{2}(5x+3x)=1440 $
$12(8x)=1440$
$x=\dfrac{120}{8}=15$ 
$5x=15\times 5$ 
     $=75m$
Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

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

  1. $\dfrac { 2 } { 3 }$
  2. $\dfrac { 3 } { 4 }$
  3. $\dfrac { 2 } { 5 }$
  4. $\dfrac { 1 } { 4 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
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]