:
Formula: 1Mark
Steps: 1Mark
Answer: 1Mark
Given,
The number 'n' is first increased by 25$\%$.
$25$
So, the new value of n
$=n+n\times \frac{25}{100}=n\times \frac{5}{4}$
Now, the value of is again decreased by $20$.
So, $20$of the new value
$=n\times \frac{5}{4}\times \frac{20}{100}=n\times \frac{1}{4}$
So, the new value
$=n\times \frac{5}{4}$ - $n\times \frac{1}{4}=n$
Net percent change
$=\frac{\text{amount changed}}{\text{original value}}\times 100$
Net change = n - n = 0.
So on substituting the values we get:
Percent change = $\frac{0}{n}\times 100$ = 0
So, there is no percentage change.

:
Each answer: 1 Mark
Given that,
Number circles = 26
Number of hearts = 1
Number of stars=1
$\therefore$ Number of circles: number of hearts
$\Rightarrow$ 26 : 1
$\therefore$ Number of hearts: number of stars
$\Rightarrow$ 1 : 1

:
4 drivers: 1hour
$\Rightarrow$ 4 : 1
20 drivers: 5hours
$\Rightarrow$ 20: 5$\Rightarrow$ 4 : 1
$\therefore$ Given ratios are equivalent.

:
Application: 1Mark
Steps: 2Marks
Answer: 1 Mark
Let the costprice of item sold at loss be₹ x
$\Rightarrow x-10\%ofx=45$
$\Rightarrow x-\frac{10}{100}\times x=45$
$\Rightarrow x-\frac{1}{10}\times x=45$
$\Rightarrow \frac{10x-x}{10}=45$
$\Rightarrow x=50\Rightarrow \text{Cost price is}$₹50
$\Rightarrow Loss$ = ₹(50 - 45) = ₹5
Similarly, letthe costprice of an item sold at profitbe ₹y.
$\Rightarrow y+25\%ofy=40$
$\Rightarrow y+\frac{25}{100}\times y=40$
$\Rightarrow y+\frac{1}{4}\times y=40$
$\Rightarrow \frac{4y+y}{4}=40$
$\Rightarrow y=32\Rightarrow \text{Cost price is}$₹32
$\Rightarrow \text{Profit}$ = ₹(40 - 32) = ₹8
Net gain = ₹(8 - 5) = ₹3
So, the shopkeeper makes a profit of ₹ 3.

:
Each Ratio: 1 Mark
Given ratio is,5:4
$\Rightarrow \frac{5}{4}$
$=\frac{5}{4}$ x$\frac{25}{25}$
=$\frac{125}{100}$
=125 $\%$
Now,
$9:25=\frac{9}{25}$ x$\frac{4}{4}$
$=\frac{36}{100}$
$=36\%$

:
Formula: 1 Mark
Application: 1Mark
Steps: 1 Mark
Answer: 1 Mark
Given that,
Population 2 years ago =62500.
Rate of decrease = 4 % p.a.
For population 1 year ago,
P = 62500 , R = 4 %, T = 1 year
$\therefore S.I=\frac{P\times R\times T}{100}$
$\Rightarrow I=\frac{62500\times 4\times 1}{100}$
$\Rightarrow I=2500$
$\Rightarrow$ After 1 year the population will decrease by 2500 people
$\therefore$ Population 1 year ago is = 62500 - 2500 = 60,000
For present population,
P = 60,000, R = 4 %, T = 1 year
$\therefore S.I=\frac{P\times R\times T}{100}$
$\Rightarrow I=\frac{60000\times 4\times 1}{100}$
$\Rightarrow I=2400$
$\Rightarrow$ Present population = 60,000 - 2400 = 57,600
[We are subtractingthe population because population is decreasing by 4% p.a.]
The present population is 57,600.

:
Formula: 1 Mark
Answer: 1 Mark
Given that,
Marked Price of the watch = ₹ 1250.
6% discount on MP
$\Rightarrow \frac{6}{100}\times 1250=₹75$
SP = MP - Discount.
SP = ₹ 1250 - ₹ 75.
SP = ₹ 1175.
The selling price of the watch is ₹ 1175.

:
Concept: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Let Nishu's pocket money be $n$
Given, she saves 10% of her pocket money.
$\Rightarrow$ 10% of n = ₹ 120
$\Rightarrow \frac{10}{100}\times n=120$
$\Rightarrow n=\frac{120\times 100}{10}$
$\Rightarrow n=1200$
So, Nishu's pocket money is ₹ 1200.

:
Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Given,
Principal = ₹ 56,000
Time = 2 years
Interest = ₹ 280
Let the rate of interest be R % p.a.
$S.I=\frac{P\times R\times T}{100}$
$\Rightarrow R=\frac{I\times 100}{P\times T}$
$\Rightarrow R=\frac{280\times 100}{56000\times 2}$
$\Rightarrow R=\frac{280}{560\times 2}$
$\Rightarrow R=\frac{1}{4}=0.25$
$\therefore$ The rate of interest is 0.25 % .
We have to find the interest if the rate of interest was 11 %,
Interest = $\frac{P\times R\times T}{100}$
On substituting the values we get,
Interest = $\frac{56000\times 11\times 2}{100}$ = ₹12320
Hence, the interest in two years if the rate was 11% is ₹ 12320.

:
Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Given, Selling price = ₹ 13,500
$Loss\%=20\%$
Let the cost price be $x.$
$\therefore Loss=20\%ofx$
Selling price = Cost price - Loss
$\Rightarrow 13500=x-\frac{20}{100}\times x$
$\Rightarrow 13500=x-\frac{1}{5}x$
$\Rightarrow 13500=\frac{4}{5}x$
$\Rightarrow x=16875$
Therefore, she bought it for ₹ 16875.
So, cost price of the article = ₹ 16875
20%of the cost price
$=\frac{20}{100}\times 16875=₹3375$
Net amount = 16875 + 3375 = ₹ 20250
So, in order to sell the washing machine at 20% profit, she needs to sell it at ₹ 20250.