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Percent Change & Discounts – Comparing Quantities | Class 8 Maths
• Last Updated : 31 Oct, 2020

Percentage change refers to the concept of variation in the measurement basis of an item, that is by what quantity the net worth of an article increased or decreased. The change in percentage occurs when the value of a commodity change, that is, increased or decreased by some numerical value. Three cases may arise:

### Percent Increase

When the new value is greater than the old value, in this case, an increase in the value of the number is noticed.

Increase = New value – old value
Percentage Increase is given by,
Percentage Increase = Increase × 100
Old value

### Percent Decrease

When the old value is greater than the new value, in this case, a decrease, in the value of the number is noticed.

Decrease =  Original value – New value
Percentage Decease =  Decrease × 100
Original value

### No Change

When the old value is equal to the new value, in this case, there is neither an increase nor decrease, and the net percentage change = 0%.

### Sample Problems on Percent Change

The following examples illustrate the concept of Percentage change:

Problem 1: A pencil initially was brought by Ram at ₹10. Later, the price was made ₹5. Tell whether it was an increase or decrease and by what percentage.

Solution:

Original/Old value = ₹10
New value = ₹5
Since, old value > new value, there is a decrease in value and therefore, percentage decrease is given by,

Old value – New value  × 100
Old value
(10 – 5) × 100
10
1 × 100 = 50%
2

Therefore, there is a net percentage decrease of 50%.

Problem 2: The new value of a commodity is ₹20 and the percentage increase is -40%. Find the original value of the commodity.

Solution:

Now, new value = ₹20
Percentage increase = -40%, which means there is a net decrease

Let original value = x
Percentage decrease = x – 20
x
⇒ 40 = x – 20 × 100
x
⇒ 40x = (x – 20)×100
⇒ 2000 = 60x
⇒ x =  2000
60
⇒ 33.333
≈ 33.34
Therefore, original price was ₹33.34.

Problem 3: In 2009, the annual rainfall was 300 sq.cm. However, in 2010, the rainfall was heavy and was observed at 600 sq.cm. By what amount the rainfall increased?

Solution:

Here,
2009 Annual rainfall = 300 sq.cm
2010 Annual rainfall = 600 sq.cm
Here we can easily see there is increase in the annual rainfall
So,
Percentage increase =  2010 Annual rainfall – 2009 Annual rainfall  × 100
2009 Annual rainfall
600 sq.cm – 300 sq.cm  × 100
300 sq.cm
300 sq.cm  × 100
300 sq.cm
⇒ 100%
Therefore, the Annual rainfall is increased by 100%.

Problem 4: The side of the square initially was 4 cm. After making amendments, it was made to 3 cm. By what percentage, the area of the square changed?

Solution:

Here,
Initial area of square = Side × Side
= 4 cm × 4 cm
= 16 sq.cm
Amendment area of square = 3 cm × 3 cm
= 9 sq.cm
Here we can see the area has been decreased.
So,
Percentage Decreased =  Initial area – Amendment area  × 100
Initial area
⇒  16 – 9 × 100
16
⇒    × 100
16
⇒ 43.75%
Therefore, the area of square after amendment is decreased by 43.75%.

## How Discount Works?

Discounts are given on the marked price which reduces the selling price of the goods for the customer.

Discount(D) = Marked price(M.P.) – Selling price(S.P.)
Also,
Discount percentage =    Discount   × 100
Marked price

### Sample Problems on Discounts

Problem 1: Finding discount percentage and discount. A toy had a M.P of ₹1000 and was sold to Shyam after a discount at ₹800. Find how much less did the toy cost to Shyam.

Solution:

Now, M.P = ₹1000
S.P  = ₹800
Discount = M.P -S.P
= ₹200
Also, discount percentage = Discount×100
M.P
= 200 × 100
1000
= 20%

Problem 2: Sita purchased an article from Mr. Ramlal at a cost of ₹500, after Mr. Ramlal offered a discount of 15%. How much did the article cost to Mr. Ramlal originally?

Solution:

Now, Selling Price, S.P = ₹500
Discount percentage = 15%
Let Market Price = x
Now, Discount percentage =  x – 500 ×100
x
15 =  (x – 500) ×100
x
15x = (x – 500)×100
15x = 100x – 50000
85x = 50000
x = 588.23
Therefore, the article costed Mr Ramlal ₹588.23.

Problem 3: The marked price of a jacket was ₹2500. After bargaining, Mallika brought it for ₹1750. At what discount, Mallika got the jacket?

Solution:

Here,
Marked Price, M.P. = ₹2500
Selling Price, S.P. = ₹1750
Now, Discount Percentage =  ₹2500 – ₹1750  × 100
2500
⇒  ₹750  × 100
₹2500
⇒ 30%
Therefore, Mallika brought jacket at 30% discount.

Problem 4: A shopkeeper sold a Refrigerator for ₹15500. He gives a discount of 15% on its marked price and still gains 8%. Find the marked price of the Refrigerator?

Solution:

Cost Price of the refrigerator =  ₹15500,
Gain% = 8%.
Therefore, Selling Price = [{(100 + gain%)/100} × CP]
= ₹ [{(100 + 8)/100} × 15500]
= ₹ [(108/100) × 15500]
= ₹ 16740.
Let the Marked Price be ₹ ‘x’.
Then, the discount = 15% of ₹x
= ₹ {x × (15/100)}
= ₹ 3x/20
Therefore, SP = (Marked Price) – (Discount)
= ₹(x –  3x )
20
= ₹  17x
20
But, the SP = ₹ 16740.
Therefore,  17x  = ₹16740
20
⇒ x = (16740 ×  20 )
17
⇒ x = ₹19695.
Hence, the marked price of the refrigerator is ₹19695.

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