Free Comparing Quantities 03 Practice Test - 8th Grade 

For a sum whose interest is calculated by compounding, the following things need to be defined:

-Principal

-Rate of interest

-Conversion periods (annual, half-yearly)

The time taken to return the loan is not fixed. Based on the time taken and the other fixed details above, the final interest is calculated.

Principal = ₹$12800,$ Rate $=10\%$

Interest for one year $=\frac{P\times r\times t}{100}=\frac{(12800\times 10\times 1)}{100}=₹1280$

So, sum due after one year $=12800+1280=₹14080$

Marked price is same as the list price.

20% discount means that on Rs 100 (MP), the discount is Rs 20.

By unitary method, on Re 1 the discount will be Rs $\frac{20}{100}$

On Rs 220, discount = Rs $\frac{20}{100}$ x 220 = Rs 44

The sale price = (Rs 220 - Rs 44) or Rs 176

OR

A discount of 20% means for MP of Rs 100, the discount is Rs 20.

Hence, the sale price is Rs 80.

Using unitary method, when MP is Rs 100, sale price is Rs 80;

When MP is Re 1, sale price is Rs  $\frac{80}{100}$ 

Hence, when MP is Rs 220, sale price = Rs  $\frac{80}{100}$ x 220 = Rs 176. 

Let the number to be added be $x,$ then $(2+x):(3+x)=4:5$

$\Rightarrow \frac{2+x}{3+x}=\frac{4}{5}\Rightarrow 5(2+x)=4(3+x)\Rightarrow 10+5x=12+4x\Rightarrow 5x-4x=12-10\Rightarrow x=2$

Let the strength of the class be $x$.

Given that $60\%$ are girls, so $40\%$ are boys.
Now, $40\%\times x=20$   
(Since number of boys in the class is 20)
$\Rightarrow \frac{40}{100}\times x=20$
$\Rightarrow x=\frac{20\times 100}{40}$
$\Rightarrow x=50$

Therefore, total number of students in the class is 50.

Here, amount spent for its transportation is the overhead charge (amount spent for repairs/transportation/labour charges) which should always be added to the cost price of the item while calculating profit made on the sale.

Therefore, effective cost price is the sum of cost price of the painting and amount spent for its transportation.

Given that, principal, $P=₹15000$ and time period, $n=2$ years . 
Let the rate of interest per annum be $r$. 
Simple interest in 2 years 
$S.I=\frac{15000\times r\times 2}{100}$
$\Rightarrow S.I=300r$

Compound interest $C.I=A-P=15000(1+\frac{r}{100}{)}^2-15000$, here, $A$ is the amount.
Given that $C.I-S.I=96$   
(Since C.I > S.I)
$\Rightarrow 15000(1+\frac{r}{100}{)}^2-15000-300r=96$
$\Rightarrow 15000(1+\frac{r}{100}{)}^2-300r=15096$
$\Rightarrow 15000(1+\frac{r^2}{10000}+\frac{2r}{100})-300r=15096$
$\Rightarrow 15000+\frac{15r^2}{10}+300r-300r=15096$
$\Rightarrow \frac{15r^2}{10}=96$
$\Rightarrow 15r^2=960$
$\Rightarrow r^2=64$
$\Rightarrow r=\pm 8$
$\because$ Rate of interest cannot be negative, so $r=8\%$    

Let the printed price of the article be $x$.

$\text{GST}=7\%\text{of}x=\left(\frac{7}{100}\right)\times x=\frac{7x}{100}$

$\text{Selling price}=x+\frac{7x}{100}=\frac{107x}{100}$

As per question, $\frac{107x}{100}=₹856$

$\Rightarrow x=\frac{856\times 100}{107}=₹800$

Hence, $\text{printed price}=₹800$

Discount = 30% of ₹ 800 $=\left(\frac{30}{100}\right)\times 800=₹240$

$\text{Cost price of the article}=800-240=₹560$

$\text{Overhead}=\text{cost of transportation}=₹40$

$\text{Actual cost price}=560+40=₹600$

GST is charged by the government on the sale of an item. It is collected by the shopkeeper from the customer and given to the government. Hence, while calculating profit we will not include GST amount.

$\text{Profit}=\text{Printed price}-\text{Actual cost price}=800-600=₹200.$

$\text{Profit \%}=\left(\frac{\text{profit}}{\text{cost price}}\right)\times 100=\left(\frac{200}{600}\right)\times 100$

$=\frac{100}{3}\%=33\frac{1}{3}\%$

Let the present value of the car be $V$ which is ₹ 35600,
Time, $n=2$ years, and rate of interest, $r=20\%$

Let the purchase price be $V_0$

$\text{Using,}V=V_0{(1-\frac{r}{100})}^n$
$\Rightarrow 35600=V_0{(1-\frac{20}{100})}^2\Rightarrow 35600=V_0{\left(\frac{80}{100}\right)}^2=V_0\left(\frac{16}{25}\right)\Rightarrow V_0=35600\times \frac{25}{16}=₹55625$