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

 Let the monthly salary is $Rs.x$ Then pocket expenses is $\frac{1}{5}\text{x}=\frac{\text{x}}{5}$ And other expenses is $\frac{4}{5}$ of reminder $\frac{4\text{x}}{5}=\frac{16}{25}$ Then total expenses $=\frac{\text{x}}{5}+\frac{16\text{x}}{25}=\frac{21\text{x}}{25}$
Then savings $=\text{x} +\frac{21\text{x}}{25}=\frac{4\text{x}}{25}$ But saving given $\text{Rs }1200 \frac{ 4\text{x}}{25}=1200 \Rightarrow\text{x}=7500\text{ Rs}$
Then monthly salary $Rs.7500.$

$ 2n + 5 = 3(3n - 10)$
$⇒ 2n + 5 = 9n - 30$
$⇒ 9n - 2n = 5 + 30$
$⇒ 7n = 35$
$⇒ n = 5$

$ 3s = 0$
$⇒ s = \frac{0}{3}$
$⇒ s = 0 ($if we divide $0$ by any number except $0,$ we will always get $0)$

Given equation is $43m = 0.086$
On dividing the given equation by $43,$ we get
$\text{m}=\frac{0.086}{43}$
If we remove the decimal, we get $1000$ in denominator
$\text{m}=\frac{86}{43}\times\frac{1}{1000}=\frac{1}{1000}=0.002$

$x - 6 = 1$
$\Rightarrow x = 1 + 6 = 7.$

 Let the number is $x$
One-third of $\text{ x} = \frac{\text{x}}{3}$
Given, One-third of a number added to itself gives $10$
$\Rightarrow\frac{\text{x}}{3} + \text{x} = {10}$

 Average score of Maya, Madhura and Mohsina $= 19$
So $($Score of Maya $+$ Score of Madhura $+$ Score of Mohsina$) 3 = 19$
$\Rightarrow $ Score of Maya $+$ Score of Madhura $+$ Score of Mohsina $= 19 \times 3$
$\Rightarrow $ Score of Maya $+$ Score of Madhura $+$ Score of Mohsina $= 57$
$\Rightarrow 16 + 20 +$ Score of Madhura $= 57$
$\Rightarrow 36 +$ Score of Madhura $= 57$
$\Rightarrow $ Score of Madhura $= 57 - 36$
$\Rightarrow $ Score of Madhura $= 21$

 The given statement means $x$ is $7$ more than $3.$
So, the equation is $x - 7 = 3$
We can also write it as $x - 3 = 7.$

Let the number be $x.$
As, two-third of a number is greater than one-third of the number by $5.$
$\Rightarrow\frac{2}{3}\text{x}-\frac13\text{x}=5$
$\Rightarrow\frac{2\text{x}-\text{x}}{3}=5$
$\Rightarrow\frac{\text{x}}{3}=5$
$\Rightarrow\text{x}=5\times3$
$\therefore\text{x}=15$
So, the number is $15.$
Hence, the correct alternative is option $(c).$

 Let the two consecutive odd numbers be $x$ and $x + 2.$
As, the sum of the two consecutive odd numbers is $36.$
$\Rightarrow x + (x + 2) = 36$
$\Rightarrow 2x + 2 = 36$
$\Rightarrow 2x = 36 - 2$
$\Rightarrow 2x = 34$
$\Rightarrow\text{x}=\frac{34}2$
$\Rightarrow x = 17$
$\therefore x + 2 = 17 + 2 = 19$
So, the larger number is $19.$
Hence, the correct alternative is option $(c).$

Given, $\frac{\text{x}}{3}=4$ then the value of $2x + 5$ is
$\Rightarrow x = 4 \times 3$
$\Rightarrow x = 12$
Now, $2x + 5 = 2 \times 12 + 5 = 24 + 5 = 29$

$4x + 5 = 9$
$\Rightarrow 4x = 9 - 6 = 4$
$\Rightarrow\text{x} = \frac{4}{4} = 1$

 $\frac{2\text{x}-1}{3}=\frac{\text{x}-2}{3}+1$
$\Rightarrow\frac{2\text{x}-1=\text{x}-2 +3}{3}$
$\Rightarrow2\text{x}-\text{x}=-2+3+1$
$\Rightarrow\text{x}=2$
$\therefore\text{x}=2$

Given, $x$ exceed 4 by $9$
$\Rightarrow x = 9 + 4$
$\Rightarrow x - 4 = 9$

$5x = 10$
$\Rightarrow\text{x} = \frac{10}{5} = 2 $

Given, Add $9$ to $5$ times $n$ to get $3$
Now, $5$ times of $n = 5n$
Add $9,$ we get
$5n + 9$
This is equal to $3$
So, $5n + 9 = 3$

Let first whole number $= x$
Then second number $= x + 1$
And sum $= 53$
$x + x + 1 = 53$
$⇒ 2x = 53 - 1$
$⇒ 2x = 52$
$⇒ x = 26$
Smaller number $= 26$

$ 4p - 3 = 9$
$\Rightarrow 4p = 9 + 3 = 12$
$\Rightarrow\text{p} = \frac{12}{4} = 3.$

$\Rightarrow 10y + 20 = 50$
$\Rightarrow 10y = 50 - 20$
$\Rightarrow 10y = 30$
$ = \frac{30}{10}$
$\Rightarrow y = 3$

Given, $43x = 0.086$
${43}\text{x} = \frac{86}{1000}$
$\Rightarrow\text{x} = \frac {86}{(1000 \times43)}$
$\Rightarrow\text{ x} = \frac{2}{1000}$
$\text{x} = \frac{1}{500}$

Let the number be $x.$
As, the sum of a number and its two-fifth is $70.$
$\Rightarrow\text{x}+\frac25\text{x}=70$
$\Rightarrow\frac{\text{x}}1+\frac{\text{2x}}5=70$
$\Rightarrow\frac{\text{5x}}{5}+\frac{\text{2x}}{5}=70$
$\Rightarrow\frac{5\text{x}+2\text{x}}{5}=70$
$\Rightarrow\frac{7\text{x}}{5}=70$
$\Rightarrow7\text{x}=70\times5 ($By transposing $5$ to $R.H.S.)$
$\Rightarrow7\text{x}=350$
$\Rightarrow\text{x}=\frac{350}{7} ($By transposing $7$ to $R.H.S.)$
$\therefore\text{x}=50$
So, the number is $50.$
Hence, the correct alternative is option $(b).$

Let first odd number $= 2x + 1$
Second number $= 2x + 3$
$2x + 1 + 2x + 3 = 36$
$⇒ 4x + 4 = 36$
$⇒ 4x = 36 - 4 = 32$
$⇒ x = 8$
Smaller number $= 2x + 1 = 2 × 8 + 1 = 16 + 1 = 17$

$\frac{2}{3}\text{x} + 6 = 12$
$\Rightarrow\frac{2}{3}\text{x}=12-6$
$\Rightarrow\frac{2}{3}\text{x}=6$
$⇒ 2\text{x} = 6\text { x} 3$
$⇒\text{x}=\frac{6×3}{2}$
$⇒\text{x}=9$

Let $2$ number be $x$ and $y.$
So larger number be $x.$
Given difference of $2$ numbers is $21$
i.e larger number minus smaller number is $21 $
$​​​​​​​⇒ x - y = 21$
So the smaller number $y$ is given as $x - 21$

$3s + 12 = 0$
$\Rightarrow 3s = 0 - 12$
$\Rightarrow 3s = -12$
$\Rightarrow\text{s} = \frac{-12}{3}$
$\Rightarrow s = -4$

$\therefore\frac{2}{3}\text{x}=\text{x}-10$
$\Rightarrow\text{x}-\frac{2}{3}\text{x}=10$
$\Rightarrow\frac{1}{3}\text{x}=10$
$\Rightarrow\text{x}=30$

We have, $x = 0$
On multiplying both the sides by $3,$ we get
$3 \times x = 3 \times 0$
$\Rightarrow 3x = 0$
On adding $(-1)$ both the sides, we get
$3x + (-1) = 0 + (-1)$
$\Rightarrow 3x - 1 = -1$

two times $y = 2y$
Now the sum of two times $y$ and $10$ is $42$ is written as $2y + 10 = 42$

Subtraction $3$ from $2x = 2x - 3$

$2\text{z}+\frac{8}{3}=\frac{1}{4}\text{z}+5$
$\Rightarrow2\text{z}-\frac{1}{4}\text{z}=5-\frac{8}{3}$
$\Rightarrow\frac{8\text{z}-\text{z}}{4}=\frac{15-8}{3}$
$\Rightarrow\frac{7}{4}\text{z}=\frac{7}{3}$
$\Rightarrow\text{z}=\frac{7}{3}\times\frac{4}{7}$
$=\frac{4}{3}$

Given, $\frac{\text{x}}{3} = 4,$ then the value of $2x + 5$ is
$\Rightarrow x = 4 \times 3$
$\Rightarrow x = 12$
Now, $2x + 5 = 2 \times 12 + 5 = 24 + 5 = 29$

Let first angle $= x$
Then second $= 90^\circ - x$
$x - (90^\circ - x) = 10$
$\Rightarrow x - 90^\circ + x = 10^\circ $
$\Rightarrow 2x = 10^\circ + 90^\circ = 100^\circ $
$x = 50^\circ $
Second angle $= 90^\circ - 50^\circ = 40^\circ $
Larger angle $= 50^\circ $