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

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

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

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

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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.

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

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

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

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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

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

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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.

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

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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.

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

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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.

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

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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)

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

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

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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

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

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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 =>

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