Percentage Problems for Class VIII

Percentage = $\dfrac {2.7}{18}$ $\times 100$

$2, \displaystyle \frac{1}{2}, %$ of x = 0.2

Let the number is x
then $x\times 37\div 100 = 990.80$
$x = 2678 approx$
$ 2678 \times 19 \div 100 = 508 = 500 approx$

$\displaystyle\frac{p _1}{100}\times N _1=\frac{p _2}{100}\times N _2$

$\displaystyle\frac{N _1}{N _2}=\frac{p _2}{p _1}$, $\displaystyle N=\frac{p _2}{p _1}\times N _2$

Percent of $\displaystyle N _1=\left(\frac{p _2}{p _1}\times100\right)%$ of $N _2$

Percent of $\displaystyle N _2=\left(\frac{p _2}{p _1}\times100\right)%$ of $N _1$

Over all discount$=100%-(80%\times 90%\times 95%)=100%-68.4%=31.6%$

$88% \ of \ 370+24% \ of \ 210-x=118$

First hour: 62.5% of 192m = 120m. Remaining height: 192 - 120 = 72m. Second hour: 12.5% of 72m = (1/8) * 72 = 9m.

$Salary\quad of\quad Tulsiram$

$ =Rs\quad \dfrac { 720 }{ 4 } \times 100$

$=Rs\quad 18,000$

$Salary\quad of Kashyap\quad $

$=18000\times \dfrac { 100 }{ 120 } $

$ =Rs.15000$

Total students = 80 + 60 = 140. Passed in year 1: 60% of 80 = 48. Passed in year 2: 80% of 60 = 48. Total passed = 48 + 48 = 96. Average rate = (96 / 140) * 100 = 960/14 = 480/7 = 68 4/7%.

$50%$ of $(x-y)=30% (x+y)$

Percentage = $\dfrac {45}{54}$ $\times 100$

Good oranges are $ 800 - 50 = 750 $
So, percentage of good oranges $ = \cfrac { 750}{800} \times 100 $ %  $ = \cfrac {375}{4} $ % $ = 93 \cfrac {3}{4} $ %

Let B's income be $ 100 $
Then A's income is $ 125 $
So,  B's income is $ \cfrac {125-100}{125} \times 100 = 20 %$ less than of A.

Let the price of the sugar today be $ 100 $
Then its price tomorrow will be  $ 125 $

So,  to bring back the price to normal it should be decreased $ \cfrac {125-100}{125} \times 100 = 20 %$

Let Neetu's age  be $ x $ years
Then Mona's age is $ 0.8x$  years

So,  Neetu is $ \cfrac {x-0.8x}{0.8} \times 100 = 25 %$ older than Mona.

Let one number be $x$. 

$p = 6q$
$p - q = 6q - q = 5q$
$\therefore$ q is less than p by $\displaystyle \frac{p, -, q}{p}, \times, 100, %$
$=, \displaystyle \frac{5q}{6q}, \times, 100, %$ $=, 83, \displaystyle \frac{1}{3}, %$

$y, %, =, \displaystyle \frac{100, \times, x}{100, +, x} %$

$\displaystyle \frac{x}{100}, \times, 360, =, 60^{\circ}$

Ratio of circumference = 3 : 1
Ratio of radii = 3 : 1
$\therefore$ ratio of areas $=, 3^2, : 1^2, 9, :, 1$
% decrease in area $=, \displaystyle \frac{8}{9}, \times, 100$
$=, 88, \displaystyle \frac{8}{9}$ %

Given that, Rajan earns $33\dfrac{1}{3}$ percent  less than Ram.

Let, the income of Ram is 100 r.s.

Then the income of Rajan is$=100-33\dfrac{1}{3}=100-\dfrac{100}{3}=\dfrac{200}{3}$

Difference in income is $=100-\dfrac{200}{3}=\dfrac{100}{3}$

Now, required income in percent $=\dfrac{100\times \dfrac{100}{3}}{\dfrac{200}{3}}=50\,$ percent

Hence, this is the answer. 

Let the blank space be $x$ and we solve the given equality $12\dfrac { 1 }{ 2 }$% of $x=35$% of $700$ as follows:

$\cfrac{16}{12}=\cfrac{Percent}{100}$

Given that $30%$ of $140$ $=$ $x%$ of $840$

We need to find $25%$ of $25%$

$p=\cfrac{95q}{100}$