Class 8 Maths Exercise 8.2 Easy Solutions NCERT covers important concepts related to percentages, increase and decrease in quantities, and their real-life applications. This exercise helps students understand how to calculate percentage change, discounts, and profit or loss in easy steps. The solutions provided here are based on the NCERT textbook and explained in a simple, step-by-step manner. These easy solutions will help students grasp the concepts quickly and solve the problems with confidence.

Class 8 Maths Exercise 8.2 Easy Solutions NCERT

Class 8 Maths Exercise 8.2 Easy Solutions NCERT Book Full Chapter

Ex 8.2 Class 8 Maths Question 1.
A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.

Solution:
Let the original salary be ₹x.
The new salary after a 10% increase is:

(110/100) × x = ₹1,54,000

Now,
(11/10) × x = 1,54,000

Multiply both sides by 10:
11x = 15,40,000

Divide both sides by 11:
x = 15,40,000 ÷ 11 = ₹1,40,000

Ex 8.2 Class 8 Maths Question 2.
On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the Zoo on Monday?

Solution:
Number of people who visited the zoo on Sunday = 845
Number of people who visited the zoo on Monday = 169

Decrease in number of people = 845 − 169 = 676

Now,
Percentage decrease = (Decrease ÷ Original number) × 100
= (676 ÷ 845) × 100
= 0.799 × 100
= 79.9%

Ex 8.2 Class 8 Maths Question 3.
A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.

Solution:
Cost price of 80 articles = ₹2400
So, cost price of 1 article = 2400 ÷ 80 = ₹30

Profit = 16% of ₹30 = (16/100) × 30 = ₹4.80

Selling price of 1 article = Cost price + Profit
= ₹30 + ₹4.80 = ₹34.80

Ex 8.2 Class 8 Maths Question 4.
The cost of an article was ₹ 15,500. ₹ 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.

Solution:
Given:
Cost of article = ₹15,500
Repair cost = ₹450
Total cost price = 15,500 + 450 = ₹15,950

Profit = 15% of ₹15,950
= (15/100) × 15,950
= ₹2,392.50

Selling Price = Cost Price + Profit
= 15,950 + 2,392.50
= ₹18,342.50

Ex 8.2 Class 8 Maths Question 5.
A VCR and TV were bought for ₹ 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss per cent on the whole transaction.

Solution:
Given:
Cost Price of VCR = ₹8,000
Cost Price of TV = ₹8,000
Total Cost Price = 8,000 + 8,000 = ₹16,000

Loss on VCR = 4% of 8,000 = (4/100) × 8000 = ₹320
Profit on TV = 8% of 8,000 = (8/100) × 8000 = ₹640

Total Loss and Profit = ₹640 (gain) − ₹320 (loss) = ₹320 net gain

Now,
Overall Gain % = (Net Gain ÷ Total Cost Price) × 100
= (320 ÷ 16,000) × 100
= 0.02 × 100 = 2%

Ex 8.2 Class 8 Maths Question 6.
During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of Jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?

Solution: Marked price of Jeans = ₹1450
Marked price of one shirt = ₹850
Price of two shirts = 2 × 850 = ₹1700

Total marked price = 1450 + 1700 = ₹3150

Discount = 10% of ₹3150
= (10/100) × 3150 = ₹315

Amount to be paid = 3150 − 315 = ₹2835

Ex 8.2 Class 8 Maths Question 7.
A milkman sold two of his buffaloes for ₹ 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. [Hint: Find CP of each]

Solution:
Selling price of each buffalo = ₹20,000

Buffalo 1: Sold at 5% gain
Let the cost price be x.
Then,
105% of x = 20,000
⇒ (105/100) × x = 20,000
⇒ x = (20,000 × 100) / 105 = ₹19,047.62

Buffalo 2: Sold at 10% loss
Let the cost price be y.
Then,
90% of y = 20,000
⇒ (90/100) × y = 20,000
⇒ y = (20,000 × 100) / 90 = ₹22,222.22

Total cost price = 19,047.62 + 22,222.22 = ₹41,269.84
Total selling price = ₹20,000 + ₹20,000 = ₹40,000

Loss = 41,269.84 − 40,000 = ₹1,269.84

Loss % = (1269.84 / 41269.84) × 100 = 40/3%

Ex 8.2 Class 8 Maths Question 8.
The price of a TV is ₹ 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

Solution:
Given:
Price of the TV = ₹13,000
Sales tax = 12% of ₹13,000

Step 1: Calculate Sales Tax
Sales tax = (12/100) × 13,000 = ₹1,560

Step 2: Total amount to be paid
Total amount = Price + Sales tax = 13,000 + 1,560 = ₹14,560

Ex 8.2 Class 8 Maths Question 9.
Arun bought a pair of skates at a sale where the discount is given was 20%. If the amount he pays is ₹ 1,600, find the marked price.

Solution:
Let the MP of the skates be ₹ 100
Discount = ₹ 20% of 100 = ₹ 20
Sale price = ₹ 100 – ₹ 20 = ₹ 80
If SP is ₹ 80 then MP = ₹ 100
If SP is ₹ 1 then MP = ₹ 100/80
If SP is ₹ 1,600 then MP = ₹ 100/80 × 1600 = ₹ 2,000
Thus the MP = ₹ 2000.

Ex 8.2 Class 8 Maths Question 10.
I purchased a hair-dryer for ₹ 5,400 including 8% VAT. Find the price before VAT was added.

Solution:
Let the original price be ₹ 100
VAT = 8% of 100 = ₹ 8
Sale price = ₹ 100 + ₹ 8 = ₹ 108
If SP is ₹ 108 then original price = ₹ 100
If SP is ₹ 1 then the original price = ₹ 100/108
If SP is ₹ 5,400 then the original price = ₹ 100/108× 5,400 = ₹ 5,000
Thus, the price of hair-dryer before the addition of VAT = ₹ 5000.

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For the official Class 8 Mathematics Solutions, you can visit:

  1. NCERT Textbooks (for Class 8):
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Understanding concepts like profit, loss, discount, and tax becomes much simpler with Class 8 Maths Exercise 8.2 Easy Solutions NCERT. These solutions help students build a strong foundation in practical arithmetic used in daily life. By following the step-by-step methods in Class 8 Maths Exercise 8.2 Easy Solutions NCERT, learners can solve problems with confidence and accuracy. Whether you’re preparing for exams or revising important concepts, Class 8 Maths Exercise 8.2 Easy Solutions NCERT provides just the clarity you need. Keep practicing with Class 8 Maths Exercise 8.2 Easy Solutions NCERT to boost your marks and gain confidence in mathematics.