Tag: solving linear equations

Questions Related to solving linear equations

Meera bought packs of trading cards that contain $10$ cards each. She gave away $7$ cards.
$x=$ Number of packs of trading cards
Which expression shows the number of cards left with Meera?

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

  1. $10x-7$

  2. $7x-8$

  3. $5-10x$

  4. $8-5x$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  We have given, $x$ is number of packs of trading cards.

$\Rightarrow$  Pack of trading cards contain $10$ cards each.
$\therefore$    Total number of cards = $10x$
$\Rightarrow$  From total cards she gave away $7$ cards. 
$\therefore$    Number of cards left with Meera = $10x-7$

Solve for x : $\dfrac{(x + 2)(2x - 3) - 2x^2 + 6}{x - 5} = 2.$

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

  1. $5$

  2. $10$

  3. $15$

  4. $\frac{20}{3}$

Show answer
Correct Option: B
Explanation:

$\dfrac{(x+2)(2x-3)-2x^2+6}{x-5}=2$


$\Rightarrow\dfrac{(2x^2+x-6)-2x^2+6}{x-5}=2$

$\Rightarrow\dfrac{x}{x-5}=2$

$\Rightarrow x=2x-10$

$\Rightarrow x=10$.  $[B]$

The present ages of a father and his son are in the ratio $7 : 3$ and the ratio of their ages will be $2 : 1$ after $10 $ years. Then, the present age of father (in years) is -

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

  1. $42$

  2. $56$

  3. $70$

  4. $77$

Show answer
Correct Option: C
Explanation:

Let the present ages of father and son be $7x$ and $3x$.

After $10$ years, their ages will be $7x+10$ and $3x+10$.
According to the question, we have
$\dfrac {7x+10}{3x+10}=\dfrac {2}{1}$
$1(7x+10)=2(3x+10)$
$7x+10=6x+20$
$7x-6x=20-10$
$x=10$
Then present age of father is $7x$, i.e. $7\times 10=70$ years.

A number when added to its half gives $36$. Find the number.

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

  1. $24$

  2. $28$

  3. $20$

  4. $18$

Show answer
Correct Option: A
Explanation:

Let the number be $x$

$\Rightarrow x+\cfrac { x }{ 2 } =36\ \Rightarrow \cfrac { 3x }{ 2 } =36\ \Rightarrow x=24$

$A$ has certain amount in his account. He gives half of this to his eldest son and one third of the remaining to his youngest son. The amount left with him now is

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

  1. $\cfrac { 1 }{ 3 } $ of the original

  2. $\cfrac { 2 }{ 5 } $ of the original

  3. $\cfrac { 3 }{ 4 } $ of the original

  4. $\cfrac { 1 }{ 6 } $ of the original

Show answer
Correct Option: A
Explanation:

$x-\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 6 } x=\cfrac { 1 }{ 3 } x$

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.