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

If both $x$ arid $y$ are in directly proportion, then and are in inverse proportion.

$\frac{60}{1}=\frac{150}{?}\Rightarrow?=2.5$

$\frac{15}{20}=\frac{36}{?}\Rightarrow?=\frac{20\times36}{15}=48$

$12 \times 50 = x \times 40$
$\text{x}=\frac{(12\times50)}{40}=15$

$3$ men = $5$ women
$1$ men $\frac{5}{3}$ women $5$
$6$ men $=\frac{5}{3}\times6=10$ women,
$\therefore$ Total women in second case,
$= 10 + 5 = 15$ women
Now,
$5$ women $: 15$ women $: 12$ days $: x$
$\therefore$ By inverse proportion,
$5 : 15 : x : 12$
$\text{x}=\frac{5\times12}{15}=4\ \text{days}$

$x = 20$ and $y = 40$
Clearly, $40 = 2 \times 20$
$y = 2x$
$\text{y}\propto\text{x},$ where $2$ is the proportionality constant.

Let each student get $x$ chocolates.
When number of students is less, each student will get more chocolates. It is a case of inverse proportion which follows $\frac{\text{x}_1}{\text{x}_2}=\frac{\text{y}_2}{\text{y}_1}$
$\frac{35}{20}=\frac{\text{x}}{3}$
$35\times3=20\text{x}$
$\text{x}=\frac{35\times3}{20}$
$x = 5$ Chocolates

For $6$ pipes, it takes $1$ hour $20$ minutes
$1$ hour $20$ minutes $= 60 + 20 = 80$ minutes
For $5$ pipes, let the time taken be $x.$
This is inverse proportion case:
$80 \times 6 = x \times 5$
$\text{x}=\frac{480}{5}=96$

Let $x$ km be the required distance. Now, $1h = 60$min

$100$ persons had food provision for $24$ days $1$ person had food provision for $24 \times 100$ i.e., $2400$ days.
If $20$ persons left the place, then remaining persons $= (100 − 20) = 80$
$80$ persons had food provision for$\frac{2400}{80}$ i.e,$30$ days.

For $12$ sheets, weight of paper is $40$ grams
Let number of sheets for $2500$ is $x.$
Using direct proportion concept:
$=\frac{12}{40}=\frac{\text{x}}{2500}$
$=\text{x}=\frac{(12\times2500)}{4}$
$=\text{x}=750$

The main idea in inverse variation is that as one variable increases the other variable decreases. That means that if $x$ is increasing $y$ is decreasing, and if $x$ is decreasing $y$ is increasing. The number $k$ is a constant so it's always the same number throughout the inverse variation problem.

$6 \times 60 = 5 \times ? \Rightarrow ? = 72$ minutes.

Since,$x \& y$ vary inversely,i.e., $\text{x}\times\text{y} = \text{k}$ (constant)
If $x = 10$ and $y= 6$
$xy=10\times 6=60$
In option $(a),12 \times 5 = 60$
In option $(b),15 \times 4 = 60$
In option $(c),25 \times 2.4 = 60$
But in option $(d),45 \times 1.3 = 58.3$

$\frac{25}{40}=\frac{40}{?}\Rightarrow?=\frac{40\times40}{25}=64$

$A$ and B's $1$ day's work $=\frac{1}{12}$
$B$ and C's $1$ day's work $=\frac{1}{20}$
$C$ and A's $1$ day's work $=\frac{1}{15}$
Adding we get,
$2(A, B$ and $C$)'s $1$ day's work,
$=\frac{1}{12}+\frac{1}{20}+\frac{1}{15}$
$=\frac{5+3+4}{60}=\frac{12}{60}=\frac{1}{5}$
$\therefore A, B$ and $C'$s $1$ days work,
$=\frac{1}{5}\times\frac{1}{2}=\frac{1}{10}$
$\therefore$ That will finish the work in $= 10$ days.

$\text{x}∝\text{y}$
$\text{x}=\text{ky}$
$\text{k}=\frac{\text{x}}{\text{y}},$ where $k$ is a constant.

$12$ men can paint walls in $3$ days. To paint the same walls, in $1$ day number of men required would be more.
It is a case of inverse proportion which follows $\frac{\text{x}_1}{\text{x}_2}=\frac{\text{y}_2}{\text{y}_1}$
$\frac{12}{\text{x}_2}=\frac{1}{3}$
$12\times3=1\times\text{x}_2$
$\text{x}_2=\frac{12\times3}{1}$
$\text{x}_2=36$

Let $y$ hours be required to finish the work by $3$ machines.
As number of machines is less so time required would be more. It is a case of inverse proportion which Follows $\frac{\text{x}_1}{\text{x}_2}=\frac{\text{y}_2}{\text{y}_2}$
$\frac{4}{3}=\frac{\text{y}}{90}$
$4\times90=3\text{y}$
$\text{y}=\frac{4\times90}{3}$
$y = 120$ minutes $= 2$ hours

$x = 10$ and $y = 20$
$y = 2 \times 10 = 2x$
$Y$ is directly proportional to $x$, where $2$ is the proportionality constant.

$\frac{20}{5}=\frac{32}{?}\Rightarrow?=8$

The distance travelled by a rickshaw in $1h = 10\ km$ The distance travelled by a rickshaw in $=\frac{10}{60}\text{km}=\frac{10\times1000}{60}\text{m}$ $=\frac{1000}{6}=\frac{500}{3}$

$40 \times 16 = ? \times 10 \Rightarrow ? = 64$

$\frac{7}{875}=\frac{16}{?}\Rightarrow?=\frac{875\times16}{7}=2000$

$\frac{15}{6}=\frac{6}{?}\Rightarrow?=2.4$

Let $x$ be number of days taken by $4$ persons to build the wall.

Rates at which taps $A$ and $B$ work $= 2 : 3$
The ratio of time taken by taps $A$ and $B$ to fill the cistern,
$=\frac{1}{\text{Rate at which taps A and B work}}$
$=\frac{1}{\frac{2}{3}}=\frac{3}{2}=3:2$

$\frac{300}{6000}=\frac{120}{?}\Rightarrow?=2400$