WORD PROBLEMS ON PERCENTAGE INCREASE AND DECREASE

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

Problem 11 :

The positive number a is 480% of the number b and a is 70% of the number c. If c is p% of b, which of the following is closest to the value of p?

a) 267      b) 336     c) 550     d) 685

Solution :

a = 480% of b -------(1)

a = 70% of c -------(2)

c = p% of b -------(3)

(1) = (2)

480% of b = 70% of c

4.8b = 0.7c

c = 4.8b/0.7

Applying the value of b, we get

4.8b/0.7 = p% of b

4.8b/0.7 = pb/100

Cancelling b on both sides.

4.8/0.7 = p/100

p = (4.8/0.7)(100)

p = 685.7

So, option d is correct.

Problem 12 :

684 is p% greater than 9. What is the value of p ?

Solution :

684 = (100+p)% of 9

684 = 9[(100 + p)/100]

684(100) = 9(100 + p)

68400/9 = 100 + p

7600 - 100 = p

p = 7500

So, the value of p is 7500.

Problem 13 :

How many liters of a 30% chlorine solution must be added to 12 liters of a 10%  chlorine solution to obtain a 15% chlorine solution?

Solution :

Let x be the number of liters of chlorine.

30% of x + 10% of 12 = 15% of (12 + x)

0.3x + 0.10(12) = 0.15(12 + x)

0.3x + 1.2 = 1.8 + 0.15x

1.2 - 1.8 = 0.15x - 0.3x

-0.6 = -0.15x

x = 0.6/0.15

x = 4

So, the required number of liters of chloring is 4 liters.

Problem 14 :

If 15% of 40 is greater than 25% of a number by 2, then the number is 

a)  12    b)  16     c)  24      d)  32

Solution :

Let x be the required number

15% of 40 = 25% of x + 2

0.15(40) = 0.25x + 2

6 = 0.25x + 2

0.25x = 6 - 2

0.25x = 4

x = 4/0.25

x = 16

So, the required number is 16.

Problem 15 :

If x% of y is the same as 4/5 of 80, then the value of xy is :

a)  320    b)  400     c)  6400      d)  none

Solution :

x% of y = 4/5 of 80

(x/100) y = (4/5)80

xy/100 = 320/5

xy = 64(100)

xy = 6400

So, option c is correct.