Free Direct and Inverse Proportions 02 Practice Test - 8th Grade 

When number of periods is 10, time for each period is 40 mins.
Let the time for each period be x mins when number of periods is changed to 8.

AB = constant
$A_1\times B_1$ = $A_2\times B_2$
 

$\Rightarrow x=\frac{10\times 40}{8}=50minutes$.

The principal should make each period of 50 mins.

$$\begin{array}{ccc}Weightofsugar& 12& 5\\ Numberofcrystals& 24\times {10}^4& y\end{array}$$
This situation follows direct proportionality.

$\frac{12}{24\times {10}^4}=\frac{5}{y}$

$y={10}^5$

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.

Let the number of rounds done by SUV be $x$, car be $y$, jeep be $z$ and auto be p.
Thus the number of passengers transported will be $7x+5y+8z+2p$ ....   (1)
Checking options respectively:
1$\to x=1,z=2,p=1$
       Substituting these in (1)
       7+8 $\times$ 2+2=25.
       So option A is correct.
2$\to x=2,y=2$
       Substituting these in (1)
       7$\times$2+85$\times$2=24
       So option B is incorrect.
3$\to x=1,y=2,z=1$
       Substituting these in (1)
       7+5$\times$2+8=25.
       So option C is correct.
4$\to$ Similarly substituting the values of option D. We will find number of people transported=26
 

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

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 =406 days

        hence, the remaining food will last for 406 days.
 

$$\begin{array}{ccc}Numberofgifts& 3& y\\ Price\left(InRupees\right)& 550& 1100\end{array}$$
$\frac{3}{550}=\frac{y}{1100}\Rightarrow y=\frac{3\times 1100}{550}=6$ gifts

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.

Let the number of pipes be y.

$\begin{array}{ccc}\text{Number of pipes}& 7& y\\ \text{Time (in minutes)}& 220& 81\end{array}$

The number of pipes and the time are inversely proportional.

AB = constant
$A_1\times B_1$ = $A_2\times B_2$

$\frac{7}{y}$ = $\frac{81}{220}$

$\Rightarrow y=\frac{7\times 220}{81}=19.01\approx 19$

Shane has to use 19 pipes to fill the tank in 1 hour 21 minutes.

$$\begin{array}{ccc}Numberofbottles& 238& y\\ Time\left(Inhours\right)& 2& 7\end{array}$$

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