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

 Let the original number be $x.$
According to the question we can write as $\Big(\frac{2}{3}\Big)\text{x}+\text{20x}$
On rearranging $\text{x}-\Big(\frac{2}{3}\Big)\text{x}=20$
Now taking the $L.C.M$ od $1$ and $3$ is $3$
$\frac{(\text{3x - 2x})}{ 3=20}$
$\frac{\text{x}}{3=20}$
Again by transposing $x = 60$
So the original number is $60$

 If $x$ is one number, then $x + 1$ would be the next consecutive number. Since the sum of the two consecutive numbers is $71,$ we can say,
$x + (x + 1) = 71$
$2x + 1 = 71$
$2x = 70$
$x = 35,$ so $x + 1 = 36$
$35 + 36 = 71$

 Perimeter of a rectangle $= 40\ cm$
Width $= 10\ cm$
Let the length be $x.$
Perimeter of rectangle $= 2($length $+$ width$)$
$40 = 2(x + 10)$
$\frac{40}{2} = \text{x} + 10$
$20 = x + 10$
$x = 20 - 10 = 10$
Thus, the length is also $10\ cm.$
Hence, we can say, that the given rectangle is basically a square, with all its sides equal.

 $\frac{\text{5x}}{3}= 30$
$\Rightarrow \text{5x}= 3 \times 30 = 90$
$ \Rightarrow \text{x}= \frac{90}{5}=18$

 Let the number be $x.$
$x + x = 24$
$2x = 24$
$\text{x} = \frac{24}{2}$
$x = 12$

 $3\text{x} + 8 = 14$
$\Rightarrow 3\text{x} = 14 - 8 = 6$
$\Rightarrow\text{x}= \frac{6}{3}= 2$

Let $CP = x$
$\text{Gain}=\frac{16}{100}\text{x}=\frac{\text{4x}}{25}$
$\text{CP}=\frac{\text{4x}}{25}+\text{x}=\frac{\text{29x}}{25}$
$\frac{\text{29x}}{25}=1885$
$\text{x}=\text{Rs. 1625}$

 Taking, $\text{m}-\frac{\text{m}-1}{3}=1-\frac{\text{m}-2}{2}$
$\Rightarrow\frac{\text{m}-\text{m}+1}{3}=\frac{2-\text{m}+2}{2}$
$\Rightarrow\frac{\text{2m}+1}{3}=\frac{4-\text{m}}{2}$
$\Rightarrow2{\text({2\text{m}}+1})=3(4-\text{m})$
$\Rightarrow{\text{4m}+2}=12-\text{3m}$
$\Rightarrow{\text{4m}+2}\text{3m}=12-2$
 $\Rightarrow\text{7m}=10$
$\Rightarrow\text{m}=\frac{10}{7}$

 $2\text{y} = 5(\text{7} - \text{y})$
$\Rightarrow2\text{y} = 35 - 5\text{y} $
$\Rightarrow2\text{y} + 5\text{y}= 35$
$\Rightarrow\text{7y = 35 }\Rightarrow\text{y}= - \frac{35}{7}= 5$

 Let $'a' a + 1,$ and $a + 2$ are the three consecutive integers.
According to the question,
$a + a + 1 + a + 2 = 180$
$\Rightarrow 3a + 3 = 180$
$\Rightarrow a + 1 = 160$
$\Rightarrow a = 60 - 1$
$\Rightarrow a = 59$
Therefore, the required integers are $59, 60$ and $61.$

 $3\text{y} + 4 = 5\text{y} - 4$
$\Rightarrow5\text{y} - 3\text{y} = 4 + 4$
$\Rightarrow\text{2y}=8 \Rightarrow\frac{8}{2}=4$

Let the number be $x.$

Before $10$ Years $x$
Today after $10$ Years $- x + 10$
After $10$ Years $x + 10 + 10 = x + 20$

 $3\text{x} + 4 = 13$
$\Rightarrow3\text{x}= 13 – 4 = 9$
$ \Rightarrow \text{x}= \frac{9}{3}=3$

Given, a number increased by $8\%$ of itself gives $135.$
Thus, $x + 8\%$ of $x = 135$
$\Rightarrow\frac{\text{108x}}{100}=135$
$\Rightarrow\text{x}=135\times\frac{100}{108}=125$
Thus, the number is $125.$

Let the number be $x.$

The number is $10x + y$
When digits are reversed, the number becomes $10y + x.$
We have,
$x + y = 8 (i)$
$10x + y + 18 = 10y + x$
$x - y + 2 = 0 (ii)$
Solving $(i)$ and $(ii),$ we get
$x = 3, y = 5$
$\therefore$ The number is $35.$

Adding $4$ times $x$ to $16$ is $45$ is $4x\ 16 = 45.$

$5\text{t}-3=3\text{t}-5$
$\Rightarrow5\text{t}-3\text{t}=3-5$
$\Rightarrow2\text{t}=-2$
$\Rightarrow\text{t}=\frac{-2}{2}=-1$

 Let the number be $x.$
Let $x$ be the common multiple of the ages of A$$ and $B$.Then. the ages of $A$ and $B$ would be $5x$ and $7x,$ respectively.  
$\therefore\frac{5\text{x}+4}{7\text{x}-4}=\frac{3}{4}$
$\Rightarrow4( 5\text{x} +4 ) = 3 ( 7\text{x} + 4 )$
$\Rightarrow20\text{x} + 16 = 21\text{x} + 12$
$\Rightarrow16 - 12 = 21\text{x} - 20\text{x}$
$\Rightarrow4 = \text{x}$
$\Rightarrow\text{x} = 4 $
$\therefore$ Age of $\text{B} = 7(\text{x}) = 7\times4 $
$= 28 \text{ years}$

$ x -$ smaller number $= 21$
$\Rightarrow $ smaller number $= x - 21$

 $-\frac{1}{3}= \text{2x}\Rightarrow \text{x}= - \frac{1}{3}$

 Let the number be $x.$

 Perimeter of rectangle $= 2($Length $+$ Breadth$)$
$20 = 2(6 + x)$
$6 + x =\frac{20}{2}$
$6 + x = 10$
$x = 10 - 6$
$x = 4\ cm$