FRACTIONS Fraction could be written as P/Q, where P is numerator and Q is denominator. Created by Inna Shapiro ©2007

Problem 1 John put into his wallet 4/5 of his money, and then the remaining $12. How much money does he have?

Answer 1/5 of his money is $12. The whole amount is 12*5=60 John has $60 in his wallet.

Problem 2 Paul and Bill took equal loans. Paul paid 7/8 of his loan, and Bill paid 8/9. Who paid more?

Answer Paul has to pay 1/8, and Bill has to pay 1/9. 1/8>1/9, that means Bill already paid more money then Paul.

Problem 3 Compare two fractions: 48/49 and 49/50

Answer 1-48/49=1/ /50=1/50 1/49>1/50=> 48/49<49/50

Problem 4 Can you write two fractions, so that each is more than 3/7 and less than 4/7?

Answer 3/7=9/21 4/7=12/21 Now we can easily find 9/21<10/21<12/21 9/21<11/21<12/21 One of the answers is: 10/21, 11/21

Problem 5 How many irreducible fractions can you write with denominator 53?

Answer 53 is a prime number. This means that all fractions with denominator 53 are irreducible. 1/53, 2/53,…52/53 makes total of 52 different fractions.

Problem 6 How many irreducible fractions can you write with denominator 55?

Answer 55=5*11 This means that 5/55, 10/55, 11/55, 15/55, 20/55, 22/55, 25/55, 30/55, 33/55, 35/55, 40/55, 44/55, and 50/55 could be reduced We can write 54 fractions 1/55, 2/55, … 54/55 and 13 could be reduced. So we can write 41 irreducible fractions.

Problem 7 Water volume increases by 1/9 when it is frozen. How does ice volume decrease when ice is melted?.

Answer Let us suppose that the volume of water is equal to 1. When it is frozen we get the volume of ice 1+1/9=10/9 From the ice with volume 10/9 we receive water with volume 1. This means that ice is decreased by 1/9:10/9=1/10 of its volume.

Problem 8 Ann has 2/3 meters of a rope. How she can cut ½ meters if she does not have a ruler?

Answer Ann can bent a rope twice and measure 1/6m. Than she can cut 1/6m from 2/3m and get ½m (2/3-1/6=1/2)

Problem 9 Find the rule and exclude one number 1/31/823/7 4/111/ /134/11

Answer 0.125=1/8 Each fraction appears twice except 5/13 We can exclude 5/13

Problem 10 Rearrange fractions in the table so that al vertical, horizontal and diagonal sums would be equal to 1 1/52/53/5 1/152/154/15 7/151/38/15

Answer We can rewrite a table 3/154/159/15 1/152/154/15 7/151/38/15 =>

Answer (continue ) And than rearrange 1, 2, 3, …9 in the magic square 2/157/156/15 9/155/151/15 4/153/158/15 2/157/152/5 3/51/31/15 4/151/58/15 =>