:
Concept: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Given that:
Marked price of the toy = ₹ 200
Discount = 20%.
Discount = Discount % of Marked Price
= $\frac{20}{100}\times 200$
= ₹ 40
Selling price = Marked Price - Discount
=₹( 200- 40 )
=₹ 160
It is given that even after the discount the shopkeeper was making a profit of 60%.
So, the cost price of the toy
$=\frac{100}{160}\times 160=₹100$
Hence, the actual price of the toy is ₹ 100.

:
Formula: 1Mark
Steps: 1Mark
Answer: 1 Mark
Here,
It is given that Soham bought a sofa for Rs6500.
He paid Rs 975 as interest in three years.
Let, Principal amount be 'P' = Rs 6500
Interest be 'I' = Rs 975
Time be 'T' = 3 years
The rate of interest =?
Let the rate of interest be 'R'.
We know that,
$I=\frac{P\times T\times R}{100}$
$\Rightarrow R=\frac{I\times 100}{P\times T}$
$\Rightarrow R=\frac{975\times 100}{6500\times 3}$
$\Rightarrow R=5$
Hence the rate of interest is 5%.

:
Steps: 2 Mark
Price before discount:1 Mark
Selling price: 1Mark
Given that,
Radhika bought a new mobile phone and saved ₹1000 when a discount of 10$\%$ was given.
We can infer from the above statement that 10$\%$ of theoriginal price= ₹ 1000
10$\%$ of the original price = ₹ 1000
Let the price be P.
$\Rightarrow 10\%\text{of P = 1000}$
$\Rightarrow \frac{10}{100}\times \text{P = 1000}$
$\Rightarrow \text{P}=\frac{1000\times 100}{10}$
$\Rightarrow P$ = ₹ 10000
Hence the original price of the mobile phone is ₹ 10000.
Now, the original price of the mobile phone is ₹ 10000.
Discount given = ₹ 1000
$\therefore$ The Selling price or the price at which the phone was sold = Original Price - Discount
On substituting the values we get:
Selling Price = 10000 - 1000 = ₹ 9000
Hence, the mobile phone was sold at ₹ 9000.

:
Steps: 1 Mark
Answer: 1 Mark
Total number of students = 860
Number of students who studySanskrit = 35%
Therefore, number of students who do not study Sanskrit =100% - 35% = 65%
$\therefore$ Number of students not studying Sanskrit,
$=\frac{65}{100}\times 860=559$
So, 559 students do not study Sanskrit.

:
Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Given,
Principal (P) = ₹ 1200
Rate (R) = $12\%$ p.a.
Time (T) = 3 years
$S.I.=\frac{P\times R\times T}{100}$
$S.I.=\frac{1200\times 12\times 3}{100}$
$S.I.=₹432$
Total amount to be paid at the end of three years = P + S.I.
Amount = ₹ 1200 + ₹ 432
Amount = ₹ 1632
₹ 1632 has to be paid at the end of three years.

:
Formula: 1Mark
Steps: 1 Mark
Answer: 1 Mark
Given that,
Original price = ₹ 3,50,000.
New price = ₹ 3,70,000.
Increase in the price = ₹3,70,000 -₹ 3,50,000.
$\therefore$Increase in the price = ₹ 20,000
$Percentageincrease=\frac{Increaseintheprice}{originalprice}\times 100$
$Percentageincrease=\frac{20000}{350000}\times 100$
$Percentageincrease=5.714\%$
The price of the car went up by 5.714$\%$

Answer: Option B. -> False
:
B
Expressing height of Raman to height of his younger brother as ratio, we have
$\frac{\text{Height of Raman}}{\text{Height of his younger brother}}=\frac{150}{125}=\frac{6}{5}$

Answer: Option A. -> True
:
A
Expressing $\frac{12}{16}$ as apercentage , we have
$\frac{12}{16}\times 100$=75%
Hence hisanswer is true.

Answer: Option C. -> 60 km
:
C
With 1 litre of petrol, distance travelled by car = $\frac{100}{50}$ = 2 km
Therefore, with 30 litres of petrol, it can travel 30× 2 = 60 km.

:
Simple Interest =$\frac{PRT}{100}$
Principal, P = 5000
Rate of Interest, R = 15
Time, T = 1
Simple Interest = $\frac{5000\times 15\times 1}{100}$ = Rs 750
Amount to be paid = 5000 + 750 = Rs. 5750