Answer: Option A. -> 4 cm
:
A
Let us assume that her distance covered in cm in the map be x.

The scale and kilometer it representare directly proportional
$\frac{A}{B}$= Constant
$\frac{A_1}{B_1}$ =$\frac{A_2}{B_2}$
18 x = 72
x = $\frac{72}{18}=4cm$

Answer: Option A. -> 750
:
A
Let n be the number of boxes.
As there are 12 crayons in each box, 2400 crayons fits into
$\frac{2400}{12}=200boxes$
Now the number of boxes produced will be directly proportional to time taken.
$\frac{A}{B}$= Constant
$\frac{A_1}{B_1}$ =$\frac{A_2}{B_2}$
$\frac{No.ofcrayonsboxes}{time}=\frac{200}{4}=\frac{n}{15}$
So, n = 750

Answer: Option B. -> 3,500
:
B
Let the required number of toys produced by 7 machines in 5 minutes be n.
As the time increases, the number of toys increases. Hence they are in direct proportion.

n = 490 x 5= 2450
So, 7 machines can make 2,450 toys in 5 minutes.
As the number of machines increases, number of toys increases. They are in direct proportion.
Let k is thenumber of toys produced by 10 machines in 5 minutes.

Then, k = $\frac{2450\times 10}{7}$
$⟹$ k = 3500
So, the number of toys produced = 3,500

Answer: Option A. -> 15 days
:
A
Total number of work days required =10 x 20 = 200
Since 10 men worked for 10 days, 10 x 10 = 100 days work is already done.
So, 200 - 100 = 100 days of work is left.
Once, 10 more men joined, total number of workers=20
So, number of days required more $=\frac{100}{20}=5$
Thus, total number of days = 10 + 5 = 15.

Answer: Option D. -> 16 days
:
D
4 oxen = 12 cows
$\Rightarrow$ $1$ ox = $\frac{12}{4}$ cows
$\Rightarrow 7\text{oxen}\equiv \frac{12}{4}\times 7\text{cows}=21\text{cows}$
$\Rightarrow$ (7 oxen + 3 cows) $\equiv$ (21 cows + 3 cows) = 24 cows
Now, 12 cows can graze the field in 32 days.
So, 1 cow can graze the field in $(32\times 12)$ days.
[less cows, more days]
24 cows can graze the field in $\frac{32\times 12}{24}=16days$
[more cows, less days]
Hence, 9 oxen and 3 cows can graze the field in 16 days.

Answer: Option A. -> True
:
A
$$\begin{array}{ccc}Numberofgifts& 3& y\\ Price\left(InRupees\right)& 550& 1100\end{array}$$
$\frac{3}{550}=\frac{y}{1100}\Rightarrow y=\frac{3\times 1100}{550}=6$gifts

:
Remaining number of men = 250-50 =200 men
Remaining number of days = (350-25) days = 325 days.
250 men had provisions for 325 days
1 man had provisions for (250 × 325) days
200 men had provisions for $\frac{(250\times 325)}{200}$ days

= 406.25days=406days
hence, the remaining food will last for 406 days.

:
$$\begin{array}{ccc}Numberofbottles& 238& y\\ Time\left(Inhours\right)& 2& 7\end{array}$$
The time taken and the number of bottles filled are directly proportional.
$\frac{A}{B}$= Constant
$\frac{A_1}{B_1}$ =$\frac{A_2}{B_2}$
$\frac{2}{238}=\frac{7}{y}$
$\Rightarrow y=\frac{7\times 238}{2}$
$\Rightarrow y=833$bottles

Answer: Option B. -> 1 × 105
:
B
$$\begin{array}{ccc}Weightofsugar& 12& 5\\ Numberofcrystals& 24\times {10}^4& y\end{array}$$
This situation follows direct proportionality.
$\frac{12}{24\times {10}^4}=\frac{5}{y}$
$y={10}^5$

Answer: Option B. -> 105 km
:
B
Let the required distance be y.
$\begin{array}{ccc}Distance\left(inkm\right)& 363& y\\ Amountofpetrol\left(inl\right)& 24.2& 7\end{array}$
The distance travelled and the amount of petrol consumed are directly proportional
$\frac{A}{B}$= Constant
$\frac{A_1}{B_1}$ =$\frac{A_2}{B_2}$

$\frac{24.2}{363}$ =$\frac{7}{y}$
$24.2\times y=363\times 7$
$y=\frac{7\times 363}{24.2}=105$ km