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

 $\frac{\text{5}}{12}\text{ x}= -27 \Rightarrow\text{x}=- \frac{\text{27}\times 2}{3}=-18$

 Let the number of boys in the class be $x.$
Then, the number of girls will be $(x - 8).$
The equation becomes:
$\Rightarrow\frac{\text{x}}{\text{x}-8}=\frac{7}{5}$
$\Rightarrow5\text{x} = 7\text{x} - 56$
$\Rightarrow5\text{x} -7\text{x} = -56$
$\Rightarrow -2\text{x} = -56$
$\Rightarrow \text{x} = \frac{-56} {-2} = 28$
Therefore, the number of boys is $28.$
Number of girls $= ( \text{x}- 8) = 28 -8 =20$
Total strength of the class $= 28 + 20 = 48$

$ x^\circ = (90^\circ - x^\circ ) + 10^\circ$
$\Rightarrow 2x^\circ = 100^\circ $
$\Rightarrow x^\circ = 50^\circ$
$\Rightarrow 90^\circ - x^\circ = 40^\circ .$

 Let the three consecutive multiplies of $7$ be $7x, (7x + 7), (7x + 14)$ where $x$ is a natural number.
According to question,
$7x + (7x + 7) + (7x + 14) = 357$
$21x + 21 = 357$
$21(x + 1) = 357$
$\frac{21(\text{x}+1)}{21}=\frac{357}{21}$
$x + 1 = 17$
$x = 17 - 1$
$x = 16$
Hence, the smallest multiple of $7$ is $7 × 16$ i.e., $112.$

Let the present ages of three persons be $x$ years, $y$ years & $z$ years, respectively.
According to the question, $x + y + z = 50$
After $5$ years,
The ages of the 3 persons will be is $(x + 5)$ years, $(y + 5)$ years and $(z + 5)$ years, respectively.
To find:$(x + 5) + (y + 5) + (z + 5) = x + y + z + 15$
$= 50 + 15$
$= 65$

 Consider $ax + b = O$ where $a$ and $b$ can take any value.
$x$ is the only variable with power one. Therefore, such equations are called linear equation in one variable.

Let the numbers be $x$ and $x + 15.$
$\therefore x + x + 15 = 95$
$\Rightarrow 2x + 15 = 95$
$\Rightarrow 2x = 95 - 15$
$\Rightarrow 2x = 80$
$\Rightarrow x = 40$
The smaller number is $40.$

$ X's$ one day's work $=\frac{1}{12}$
Let, $Y$ work for $x$ days
$\therefore$ $Y's$ one day's work $=\frac{1}{\text{x}}$​
One day work by $X$ and $Y$ together
$\frac{1}{12}+\frac{1}{\text{x}}=\frac{1}{8}$
$\frac{1}{\text{x}}=\frac{1}{8}-\frac{1}{12}=\frac{1}{24}$
$\therefore x = 24$ days
$\therefore Y$ can complete work alone in $24$ days.

 $\text{2x}-3=\text{x}+2$
$\Rightarrow2​​\text{x}-\text{x}=3+2$
$\Rightarrow​​\text{x}=5$

$\text{5x} -8 =7 \Rightarrow \text{5x = 8 + 7 = 15}$
$\Rightarrow \text{x}= \frac{15}{5}= 3.$

Let the greater number be $x$
Smaller number be $x - 1$
$\frac{\text{a}}{\text{q}}$
$x + x - 1 = 15$
$2x – 1 = 15$
$2x = 16$
$X = 8.$
Smaller number $= 8 - 1 = 7.$

Let the required number be $'x '.$
Then, $\frac{\text{a}}{9}=-2$
$⇒ a = -18$
The required number is $-18.$

$\frac{\text{y}}{5} = 10$
$\text{y} = 5\times10 = 50$

Let the number be $x$
$-\frac{7}{3}+\text{x} = \frac{3}{7}$
$\text{x}=\frac{3}{7}+\frac{7}{3} = \frac{(9+49)}{21} = \frac{58}{21}$

Given, $-\frac{4}{3}\text{y}=-\frac{3}{4}$
$\text{y}=-\frac{3}{4}\times-\frac{3}{4}$
$\text{y}=\Big(\frac{3}{4}\Big)^2$
Hence, the value of $y$ is $\Big(\frac{3}{4}\Big)^2.$

According to the question,
$\frac{\text{x}}{100}\times200=10$
$\therefore\frac{\text{x}}{100}\times200=10$
$\Rightarrow\text{2x}=10$
$\Rightarrow\text{x}=\frac{10}{2}$
$\therefore\text{x}=5$

$3x + x = 4$
when we transpose $-x$ to $LHS$ we get $+x$
$3x + x = 4$

 $13\text{x} - 14 = 9\text{x} + 10$
$\Rightarrow1\text{x}- 9\text{x} = 10 + 14$
$\Rightarrow\text{4x = 24 }\Rightarrow\text{x}= - \frac{24}{4}= 6$

$ 11x - 5 - x + 6 = 2x + 17$
$⇒ 8x = 16 $
$⇒ x = 2.$

Given that,
Two year ago, my age was $x$ years
Therefore,
Present age $=$ age two years ago $+\ 2$
Present age $= x + 2$
What was my age $5$ years ago$?$
Age $5$ years ago $=$ present age $- 5$
Age $5$ years ago $= x + 2 - 5$
Age $5$ years ago $= x - 3$
Therefore, my age $5$ years ago is $x - 3$

 Let $X$ be the number
$\frac{\text{X}}{8} = 6$
$\text{X} = 8 \times 6 = 48$

 Given that the equation is $7x + 15 = 50$
To find the root of the equation.
That is the root must satisfy the given equation.
Put $x = 5$ in given equation we get,
$7(5) + 15 = 50$
$35 + 15 = 50$
$50 = 50$
$\therefore 5 $ is a root.
$\therefore 5$ is the root of the given equation $7x + 15 = 50.$

 Let the first number be $x$ and the second number be $x + 15.$
According to the question,
$x + x + 15 = 95$
$2x = 80$
$x = 40$
Hence, the first number is $40$ and the second number is $55.$

$ 2x + 9 = 67 ⇒ x = 29.$

 Let the number be $x$
$-\frac{7}{3}+\text{x} =\frac{3}{7}$
$\text{x}=\frac{3}{7}+\frac{7}{3}= \frac{(9+49)}{21} = \frac{58}{21}$

 $\frac{\text{x}-1}{3}=\frac{\text{x}-2}{4}$
$\Rightarrow 4(x + 1) = 3(x - 1)$
$\Rightarrow 4x + 4 = 3x - 6$
$\Rightarrow 4x - 3x = -6 - 4$
$\Rightarrow x = -10$

 $\frac{2}{3}\text{y}=\frac{5}{12}$
$\Rightarrow\text{ y}=\frac{5}{12} \times\frac{3}{2}=\frac{5}{8}$

$ x^\circ + 2x^\circ = 180^\circ$
$\Rightarrow 3x^\circ = 180^\circ$
$\Rightarrow x^\circ = 60^\circ$
$\Rightarrow 2x^\circ = 120^\circ$