Free Comparing Quantities 01 Practice Test - 8th Grade 

If there are 1000 bacteria now, 1 hour ago there would have been 500 since the number of bacteria doubles every hour. 2 hours ago there would have been half of 500 bacteria, that is 250 bacteria. 3 hours ago there would have been 125 bacteria.

Since the tenure is 1 year,  the compound interest will be same as simple interest.
Amount = Principal + Interest
Interest = 5600 - 5000 = 600
Interest for one year $=\frac{PRT}{100}=600$

$R=\frac{100\times 600}{5000\times 1}=12\%$

$\text{Discount \%}=\frac{\text{Marked price}-\text{Selling price}}{\text{Marked price}}\times 100$

                     $=\frac{15000-14400}{15000}\times 100=4\%$

There are 6 apples and 6 oranges.

The ratio is $\frac{6}{6}$, that is 1:1.

Let the number of apples added be $x.$
Then, $\frac{6+x}{6}=\frac{3}{2}$
$\Rightarrow 6+x=\frac{3}{2}\times 6$
$\Rightarrow x=9-6$
$\Rightarrow x=3$
Hence, 3 apples should be added.
Thus, the given statement is true.

200% of 12 

$=\frac{200}{100}\times 12=2\times 12=24$

$\text{We know that,}C.I=P[{(1+\frac{r}{100})}^n-1]\text{where P is the principal, r is the rate}\text{of interest and t is the time period.}$

$\text{Hence,}246=P[{(1+\frac{5}{100})}^2-1]$

$246=P[\left(\frac{21}{20}\right)\times \left(\frac{21}{20}\right)-1]246=P\times \left(\frac{41}{400}\right)$

$\Rightarrow P=\frac{246\times 400}{41}=₹2400$

$\text{Now, P=}₹2400,r=6\%\text{per annum and t =}3\text{years}$

$S.I=\frac{P\times r\times t}{100}=\frac{2400\times 6\times 3}{100}=₹432$

Goods and Service Tax (GST) is levied on selling price.

Given that A = ₹ 3,15,600 , r = 20 % ; n = 2 years.
 
Let P be the value of the car two years ago. 

 $A=P(1-\frac{20}{100}{)}^2$
$315600=P(1-\frac{1}{5}{)}^2$
$315600=P\left(\frac{4\times 4}{5\times 5}\right)$

$P=\frac{315600\times 5\times 5}{4\times 4}$

$P=₹4,93,125$

$\therefore \text{Purchase value of the car}=₹4,93,125$