Tag: maths

Questions Related to maths

Find the value of $y$ in the equation : 
$\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$

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

  1. $18$

  2. $9$

  3. $-38$

  4. $-32$

Show answer
Correct Option: C
Explanation:

Given, $\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$
$\Rightarrow \displaystyle \frac{2+y}{y-7}=\frac{4}{5}$

$\Rightarrow 5(2+y)=4(y-7) $ ....( cross multiplying )
$\Rightarrow 10+5y=4y-28$
$\Rightarrow y=-38$
Hence,. the solution is $x=-38$

A combination of locks requires 3 numbers to open. The second number is $\displaystyle 2d + 5$ greater than the first number. The third number is $\displaystyle 3d - 20$ less than the second number. The sum of the three numbers is $\displaystyle 10d + 9$. The first number is 

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

  1. $\displaystyle 5d-11$

  2. $\displaystyle 3d-7$

  3. $\displaystyle 2d+19$

  4. $\displaystyle 3d-11$

Show answer
Correct Option: B
Explanation:

Let the first number be $x$
Then, second number = $x + 2d + 5$
Third number = $x + 2d + 5 -(3d -20)$ = $x -d + 25$
Sum of the three numbers = $ x + x+2d + 5 + x - d +25$ = $ 3x + d +30$

Thus, $ 3x + d +30$ = $10d + 9$
$3x = 10d + 9 - d - 30$ = $9d - 21$
$x = 3d - 7$

If $\displaystyle \sqrt{\left ( x-1 \right )\left ( y+2 \right )}=7$, $x$ and $y$ being positive whole numbers, then the values of $x$ and $y$ are, respectively

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

  1. $8,5$

  2. $15,12$

  3. $22,19$

  4. $6,8$

Show answer
Correct Option: A
Explanation:

$ \sqrt{(x-1)(y+2)}=7\Rightarrow (x-1)(y+2)=7^{2}$
$ \Rightarrow (x-1)=7:and:(y+2)=7$
$ x=8$ and $\displaystyle y=5$

If $\displaystyle 4=\sqrt{x+\sqrt{x+\sqrt{x+....,}}}$ then the value of x will be 

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

  1. 20

  2. 16

  3. 12

  4. 8

Show answer
Correct Option: C
Explanation:

$\displaystyle 4 = \sqrt{x+\sqrt{x+\sqrt{x+....}}}$
$\displaystyle \Rightarrow 4=\sqrt{x+4}$
$\displaystyle \Rightarrow 16={x+4}$
$\displaystyle \Rightarrow 16=x+4\Rightarrow x=12.$

$\displaystyle \sqrt{6+\sqrt{6+\sqrt{6+...}}}$ equals 

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

  1. $\displaystyle 6^{\frac{2}{3}}$

  2. 6

  3. $\displaystyle 6^{\frac{1}{3}}$

  4. 3

Show answer
Correct Option: D
Explanation:

Let  $ x =\sqrt{6+\sqrt{6+\sqrt{6+...}}}$


sqare it on both sides


$x^2=\sqrt{6+\sqrt{6+\sqrt{6+...}}} = 6+x$


$\Rightarrow x^2=6+x $


$\Rightarrow x^2-x-6=0 $


$\Rightarrow x^2-3x+2x-6=0 $


$\Rightarrow x(x-3)+2(x-3)=0 $


$\Rightarrow (x-3)(x+2)=0 $


$either  x-3=0---> x=3 $


$ or  x+2=0 ----> x=-2 $


since result of sqrt of anything will be positive only, therefore,


Answer $x=3$





If $\displaystyle \frac{x^2\, -\, (x\, +\, 1)(x\, +\, 2)}{5x\, +\, 1}\, =\, 6$, then $x$ is equal to

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

  1. $\displaystyle \frac{8}{33}$

  2. $\displaystyle \frac{8}{3}$

  3. $\displaystyle \frac{-8}{33}$

  4. $\displaystyle \frac{-6}{33}$

Show answer
Correct Option: C
Explanation:

Given, $\displaystyle \frac{x^2\, -\, (x\, +\, 1)(x\, +\, 2)}{5x\, +\, 1}\, =\, 6$
$\Rightarrow x^2\, -\, (x^2\, +\, 3x\, +\, 2)\, =\, 6(5x\, +\, 1)$, .....(on cross multiplying )
$\Rightarrow x^2\, -\, x^2\, -\, 3x\, -\, 2\, =\, 30x\, +\, 6$
$\Rightarrow -3x - 2 = 30x + 6$
$\Rightarrow -3x - 30x = 6 + 2 \Rightarrow -33x = 8$
$\Rightarrow x\, =\, \displaystyle \frac{-8}{33}$
Hence, the solution is, $x=-\cfrac{8}{33}$.

After receiving two successive raises Hrash's salary became $\dfrac {15}{8}$ times of his initial salary. By how much percent was the salary raised the first time if the second raise was twice as much as high (in percent) as the first ?

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

  1. $15 \%$

  2. $20 \%$

  3. $25 \%$

  4. $30 \%$

Show answer
Correct Option: C
Explanation:

Let initial salary was Rs. $100$
After two raise it become $\dfrac {15}{8}$ i.e. $\dfrac {(15 \times 100) }{ 8} =$ Rs. $187.5$
Raise $= 187.5 - 100 = 87.5$
Using formula,
[( first raise  + second raise) + (first raise * 2nd raise) / 100]  = 87.5
$x + 2x +\dfrac { (2x ^2)}{100} = 87.53$
$300x + 2x ^2 = 8750$
$x ^2 + 150x = 4375$
$x ^2 + 150x - 4375 = 0$
$x ^2 + 175x - 25x - 4375 = 0$
$x = -175, 25$ .....(Negative value is not possible)
So, $x= 25\%$
Second raise was $2x = 2 \times 25 = 50\%$.

Reduce the following linear equation: $6t - 1 = t - 11$

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

  1. $t=-1$

  2. $t=-2$

  3. $t=-3$

  4. $t=-4$

Show answer
Correct Option: B
Explanation:

$6t - 1 = t - 11$
On transposing $t$ to the L.H.S.,  we obtain
$6t - t = -11 + 1$
$5t = -10$
$t = -2$

Solve the equation: $\dfrac{a+4}{6-3a}=\dfrac{1}{3}$

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

  1. $a=1$

  2. $a=-1$

  3. $a=2$

  4. $a=-2$

Show answer
Correct Option: B
Explanation:

Given, $\dfrac{a+4}{6-3a}=\dfrac{1}{3}$


On multiplying $6 - 3a$ on both sides, we get
$a+4=\dfrac{6-3a}{3}$
$3a + 12 = 6 - 3a$
$6a = 6 - 12$
$6a = -6$
$a = -1$

Solve the linear equation: $5x - 12 = 2x + 18$

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

  1. $x=8$

  2. $x=9$

  3. $x=10$

  4. $x=11$

Show answer
Correct Option: C
Explanation:

$5x - 12 = 2x + 18$
On transposing $2x$ to the L.H.S and $12$ to RHS, we obtain
$5x - 2x = 18 + 12$
$3x = 30$
$x = 10$