WORD PROBLEMS ON PERCENTAGE INCREASE AND DECREASE

Always the original price should be considered as 100%.

Let x% be the percentage should be increased, then the new quantity will be (100+x)% of the old quantity.

Let x% be the percentage should be decreased, then the new quantity will be (100-x)% of the old quantity.

Problem 1 :

Last year, there was 20 students in a class. This year, there are 30% more students. How many students are in the class this year?

Solution :

Total number of students = 20

Since 20 is the original quantity, we consider it as 100%.

Increase of 30%.

Number of new students = (100 + 30)% of 20

= 1.30(20)

= 26

So, the new number of students is 26.

Problem 2 :

A TV normally costs $520. In a sale, all prices are reduced by 10% Calculate the sale price of the TV

Solution :

Original price of the TV = $520

The price of the TV is reduced by 10%.

Price of TV after reduction = (100 - 10)% of 520

= 90% of 520

= 0.90(520)

= $468

So, price of the TV after reduction is $468.

Problem 3 :

Over the past 10 years, the population of a town has increased by 25% The population of the town 10 years ago was 18000 What is the population of the town now?

Solution :

The population of the town = 18000

Percentage increase = 25%

New population = (100 + 25)% of 18000

= 125% of 18000

= 1.25(18000)

Now the population is 22500.

Problem 4 :

A standard bag of flour contains 600g of flour. A special edition bag contains 35% more flour. How much flour is in the special edition bag?

Solution :

Quantity of flour in the bag = 600 g = 100%

The special edition bag is containing 35% more flour.

= 135% of 600

= 1.35(600)

= 810 g

So, the special edition bag contains 810 g of flour.

Problem 5 :

Richard owns a coffee shop. In February, 4500 hot chocolates were sold. The number of hot chocolates sold in March was 3% less. How many hot chocolates are sold in March?

Solution :

Total number of chocolates = 4500

Reduction = 3%

Number of chocolates in the March = (100 - 3)% 4500

= 97% of 4500

= 0.97(4500)

= 4365

Problem 6 :

Gabriel’s salary is $24500. Next year he is due to get a 9% increase. What will his new salary be?

Solution :

Gabriel’s salary = $24500

Increase of 9%.

New salary = (100 + 9)% of 24500

= 109% of 24500

= 1.09(24500)

= 26705

Problem 7 :

Iris spends $40 a month on water. By changing company, Iris can save 16%. How much would Iris pay each month?

Solution :

Amount spent for water = $40

He can save 16%.

His pay each month = (100 - 16)% of 40

= 84% of 40

= 0.84(40)

= $33.6

Problem 8 :

Louis sees this special offer in a shop. Louis buys both items. How much does he pay?

Solution :

Original price of Ipad = £489

After 3% discount. 

New price = (100-3)% of 489

= 97% of 489

= 0.97 (489)

= $474.33

Original price of Case = $55

After 3% discount. 

New price = (100 - 3)% of 55

= 97% of 55

= 0.97 (55)

£53.35

Problem 9 :

An adult ticket for the cinema costs $13.40 A child ticket is half the price of an adult ticket. Mr and Mrs Henderson and their six children go to see a movie. Mrs Henderson has a voucher for 18% off. Work out how much Mrs Henderson pays for the tickets.

Solution :

Cost of tickets of Mr and Mrs Henderson's tickets = 2(13.40)

= $26.8

Cost of children tickets = 6(13.40/2)

= 6(6.7)

= $40.7

Total charge = 26.8 + 40.7

= $67.5

Offer for 18%.

So, 92% of charge = 0.92(67.5)

= $62.1

Problem 10 :

Zara wants to buy 72 candles. Each candle costs £4.80 There is a special offer Work out the cost of buying 72 candles using the special offer.  

Solution :

Cost of one candle = £4.80

Cost of 72 candles = 72(£4.80)

= 345.6

= (100 - 15)% of total cost 

= 85% of 345.6

= 0.85(345.6)

£293.76

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