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MATH 5: HOMEWORK1 SEPTEMBER 16, 2012 Please try to do as many problems as you can and bring completed solutions with you to the next class. Some of these problems are similar to those we discussed in class and some are new. It is OK if you can not solve some problem, but do not give up before you make an effort - perhaps you can try it again another day. Start the homework early. Write the homework on a separate sheet of paper. Solutions should include explanations, I want to see how you arrived at this answer, show your work. This homework will help you review some topics that you studied last year. These topics are: divisibility, computing the LCM and GCF (also called GCD), operations with fractions, negative numbers and some word problems involving distances and speeds. Summary Of Divisibility, LCM and GCF • Any whole number can be written as a product of prime numbers, and in only one way (except that the same primes can be written in different orders). • All factors of a number are obtained by writing products of some of the factors in the number’s prime factorization. For example, factors of 45 = 3 × 3 × 5 are 1, 3, 5, 3 × 3 = 9, 3 × 5 = 15 and 3 × 3 × 5 = 45. • To find the greatest common factor (GCF) of two numbers, write the prime factorization for each of them, then box the factors they have in common and take the product of boxed factors in one of the numbers. For example, for 90 = 2 × 3 × 3 × 5 12 = 2 × 2 × 3 The GCF is 2 × 3 = 6 • To find the least common multiple (LCM) of two numbers, write the prime factorization for each of them. For each prime factor, see how many times it occurs in each of the numbers, and take the larger multiplicity - this is how many times the prime factors occur in the LCM. For example, LCM of 90 and 12 is 2 × 2 × 3 × 3 × 5 = 180. All common multiples will be given by LCM, 2× LCM, 3× LCM,.... For example, for 90 and 12, the common multiples will be 180, 2 × 180 = 360, ... • The product of LCM and GCF is the product of the two original numbers. For example, for 90 and 12 we get GCF × LCM= 6 × 180 = 1080 = 12 × 90. Homework problems on the back Homework 1. Compute: 2 5 (a) + = 3 7 2 5 (b)1 − = 11 3 22 (c) − 4= 7 5 4 (d) × = 7 3 5 (e) 1 ÷ 4= 11 22 13 (f) × (1 − ) = 7 11 2. Two shelves together contain 25 books. One has 3 books more than the other. How many books are there on each shelf? 3. Find the LCM and GCF of the following numbers • • • • 12 and 32 16 and 10 15 and 20 365 and 30 4. A messenger is sent from one city to another far away. He can travel 40km a day. Next day, another messenger is sent, who can travel 45km a day. When will he overtake the first messenger? 5. A boy is buying candy. If he buys 11 pieces of candy, he will have 15 cents left. If he tries to buy 15 pieces of candy, he will be 21 cents short. How much does one piece of candy cost? 6. Two bells ring together at 10 : 45am. One bell rings every 9 minutes and the other every 12 minutes. When will they next ring together? 7. Find the following sums (without using a calculator). Do it in the most efficient way: 1 + 2 + 3... + 29= 20 − 19 + 18 − 17 + ... − 3 + 2 − 1= 21 + 20 + 21 + 24 + 19 + 26= 7 × 19 + 7 × 11 = 8. An apple costs 9 cents, and an orange costs 15 cents. Elena bought some apples and oranges, 20 fruit in all, and paid $2.64. How many apples and how many oranges did she buy?