Maths M.C.Q. [1 Marks Each]

Let the multiples of $2$ are $x$ and $x + 2$
Given, sum $= 18$
$\Rightarrow x + x + 2 = 18$
$\Rightarrow 2x + 2 = 18$
$\Rightarrow 2x = 18 - 2$
$\Rightarrow 2x = 16$
$\Rightarrow\text{ x} = \frac{16}{2}$
$\Rightarrow x = 8$
So, the numbers are $8, 8 + 2 = 8, 10$

$10y - 20 = 50$
$\Rightarrow 10y = 50 + 20$
$\Rightarrow 10y = 70$
$\Rightarrow\text{y} = \frac{70}{10}$
$\Rightarrow y = 7$

Given, $x = 2,$
Now $\frac{(1 - 3\text{x})}{3} = \frac{({1-3\times2)}}{3}$
$= \frac{(1−6)}{3}$
$= \frac{(−5)}{3}$

Let the larger angle be $x.$
Then, the smaller angle $= (x - 40^\circ )$
As, the sum of the two supplementary angles is always $180^\circ .$
$\Rightarrow x + (x - 40^\circ ) = 180^\circ $
$\Rightarrow 2x - 40^\circ = 180^\circ $
$\Rightarrow 2x = 180^\circ + 40^\circ $
$\Rightarrow 2x = 220^\circ $
$\Rightarrow\text{x}=\frac{220^\circ}{2} ($By transposing $2$ to $R.H.S.)$
$\therefore\text{x}=1106^\circ$
So, the measure of the larger angle is $110^\circ .$
Hence, the correct alternative is option $(c).$

Given, $5x - 8 = x + 4$
$\Rightarrow 5x - 8 - x = 4$
$\Rightarrow 4x - 8 = 4$
$\Rightarrow 4x = 4 + 8$
$\Rightarrow 4x = 12$
$\Rightarrow\text{x} = \frac{12}{4}$
$\Rightarrow x = 3$

$1400 \times\text{x} =1050$
$\Rightarrow{\text{x}}=\frac{1050}{1400}=\frac{3}{4}$

$ 7n + 5 = 12$
$\Rightarrow 7n = 12 - 5$
$\Rightarrow 7n = 7$
$\Rightarrow \text{n}= \frac{7}{7}$
$= 1$

Given equation is $ax = b$
On dividing the equation by $a,$ we get
$\text{x}=\frac{\text{b}}{\text{a}}$
Now, if $a$ and $b$ are positive integers, then the solution of the equation is also positive number as division of two positive integers is also a positive number.

Any even number is of the form $x = 2n$ where $n$ is an integer So then consecutive even number will be $2n + 2$ i.e $x + 2$

Let the number be $x.$
As, $23$ of the number is less than the original number by $20.$
$\Rightarrow\text{x}-\frac23\text{x}=20$
$\Rightarrow\frac{\text{x}}{1}-\frac{\text{2x}}{3}=20$
$\Rightarrow\frac{\text{3x}}{\text{3}}-\frac{\text{2x}}{3}=20$
$\Rightarrow\frac{\text{3x}-\text{2x}}{3}=20$
$\Rightarrow\frac{\text{x}}{3}=20$
$\Rightarrow\text{x}=20\times3 ($By transposing $3$ to $R.H.S.)$
 $\therefore\text{x}=60$
So, the number is $60.$
Hence, the correct alternative is option $(d).$

$2\text{x}+\frac{1}{3\text{x}}=15$ Multiply both side by $3x$ Then So,
$\frac{5\text{x}}{6\text{x}^2+20\text{x}+1}=\frac{5\text{x}}{\text{6x}^2-15\text{x}+1+\text{35x}}=\frac{5\text{x}}{0+\text{35x}}\Rightarrow\frac{5\text{x}}{35\text{x}}=\frac{1}{7}$

Let number $= x$ then
$3x + 6 = 24$
$\Rightarrow 3x = 24 - 6 = 18$
$\Rightarrow x = 6$
Number $= 6$

 Given equation is $7x + 4 = 25$
$⇒ 7x = 25 - 4 [$transposing $4$ to $RHS]$
$⇒ 7x = 21$
On dividing the above equation by $7,$ we get
$x = 3$
Hence, the solution of the given equation is $3.$

 Let the age of Sohan $= x$
Now, age of Mohan $= x + 3$
Given, sum of their ages $=$
$\Rightarrow x + x + 3 = 43$
$\Rightarrow 2x + 3 = 43$
$\Rightarrow 2x = 43 - 3$
$\Rightarrow 2x = 40$
$\Rightarrow\text{x} = \frac{40}{2}$
$\Rightarrow x = 20$
So, age of Sohan is $20$ years

Given, $a$ and $b$ are positive integers
Again, $ax = b$
$ \Rightarrow \text{x} = \frac{\text{b}}{\text{a}}$
Since $a$ and $b$ are positive integers, So $ \frac{\text{b}}{\text{a}}$ is also a positive number.

Given, $\frac{\text{x}}{2}=3$
On muliplying both sides by 2, we get $\frac{\text{x}}{2}\times2=3\times2$
$\Rightarrow\text{x}=3\times2=6$
Put $x = 6$ in the equation $3x + 2,$ we get
$3(6) + 2 = 18 + 2 = 20$

$m - 1 = 2$
$\Rightarrow m = 2 + 1 = 3.$

Let the two complementary angles be $x^\circ $ and $(90 - x)^\circ .$
According to the equation, we have:
$x - (90 - x) = 14$
$\Rightarrow 2x = 104$
$\Rightarrow x = 52$
$(90^\circ - x)^\circ = 90^\circ - 52^\circ = 38^\circ $
The larger angle is $52^\circ .$

As, $\frac{\text{x}+2}{\text{x}-2}=\frac{2}{3}$
$\Rightarrow3(\text{x}+2)=2(\text{x}-2)$ (By cross multiplication)
$\Rightarrow\text{3x}+6=2\text{x}-4$
$\Rightarrow\text{3x}-\text{2x}=-6+4 ($By transposing $2x$ to $L.H.S.$ and $6$ to $R.H.S.)$
$\therefore\text{x}=-10$
Hence, the correct alternative is option $(a).$

$10y - 20 = 30$
$\Rightarrow 10y = 30 + 20 = 50$
$\Rightarrow\text{y} = \frac{50}{10} = 5$

$\frac{\text{x}-1}{\text{x}+1}=\frac{7}{9}$
$\Rightarrow9\text{x}-9=7\text{x}+7$
$\Rightarrow9\text{x}-7\text{x}=7+9$
$\Rightarrow2\text{x}=16$
$\Rightarrow\text{x}=\frac{16}{2}=8$

Given equation is $ax + b = 0$
$\Rightarrow\text{ax}=-\text{b}[$ transposing b to $RHS]$
$\Rightarrow\text{x}=-\frac{\text{b}}{\text{a}} [$on dividing both sides by $a]$

$ x + 3 = 0$
$\Rightarrow x = -3.$

 As, $\frac{\text{x}}{6}+\frac{\text{x}}{4}=\frac{\text{x}}{2}+\frac34$
$\Rightarrow\frac{\text{x}}{6}+\frac{\text{x}}{4}-\frac{\text{x}}{2}=\frac{3}{4} ($By transposing $\frac{\text{x}}{2}$ to $L.H.S.)$
$\Rightarrow\frac{2\text{x}}{12}+\frac{3\text{x}}{12}-\frac{6\text{x}}{12}=\frac{3}{4}$
$\Rightarrow\frac{2\text{x}+3\text{x}-6\text{x}}{12}=\frac34$
$\Rightarrow\frac{-\text{x}}{12}=\frac34$
$\Rightarrow-\text{x}\times4=3\times12$ (By cross multiplication)
$\Rightarrow-4\text{x}=36$
$\Rightarrow\text{x}=\frac{36}{-4}$
$\therefore\text{x}=-9$
Hence, the correct alternative is option $(c).$

Consider the given intersects $2$ and $3$ on $x-$axis $x−$axis and $y-$axis $y−$axis respectively.
$\text{a}=\pm2,\text{b}=\pm3$
We know that the equation of the line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
So, $\frac{\text{x}}{\pm2}+\frac{\text{y}}{\pm3}-1$
$\pm3{\text{x}}\pm2{\text{y}}=6$
Hence, this is the answer.

$p + 4 = 4$
$\Rightarrow P = 4 - 4 = 0.$