QA #9 Factors and Multiples

Questions and Answers Sheet 9 Factors and Multiples Question #1 Find the LCM of 48𝑥, 24𝑦𝑥 and 42𝑥 ? Answer: Step 1 Let us first factorize all monomials 48𝑥 = 2 × 2 × 2 × 2 × 3 × 𝑥 24𝑦𝑥 = 2 × 2 × 2 × 3 × 𝑦 × 𝑥 42𝑥 = 2 × 3 × 7 × 𝑥 Observe that this uses 2, 3, 7, x and y and in a single monomial, 2 occurs maximum 4 times and 3, 7, x and y occur not more than once Hence Lease Common Multiple is 2 × 2 × 2 × 2 × 3 × 7 × 𝑥 × 𝑦 = 336 𝑥y Question #2 How do you find the LCM of 20𝑥 3 , 16𝑥 4 ? Answer: Step 1 20𝑥 3 = 2 × 2 × 5 × 𝑥 × 𝑥 × 𝑥 16𝑥 4 = 2 × 2 × 2 × 2 × 𝑥 × 𝑥 × 𝑥 × 𝑥 So the smallest expression that both wil go into is 2×2×2×2×5×𝑥×𝑥×𝑥×𝑥 Answer: 80𝑥 4 Question #3 How do you find the LCM of 18𝑥𝑦 2 , 15𝑦 3 ? Answer: Step 1 18𝑥𝑦 2 = 2𝑡𝑖𝑚𝑒𝑠32 × 𝑥 × 𝑦 2 15𝑦 3 = 3𝑡𝑖𝑚𝑒𝑠5 × 𝑦 3 (For LCM, you should choose the greatest power of each factor. ) ∴ 𝐿𝐶𝑀 = 2 × 32 × 5 × 𝑥 × 𝑦 3 = 90𝑥𝑦 3 Question #4 How do you find the LCM of 24 36 ? Answer: Step 1 The LCM of 24 and 36 is the smallest positive integer that divides the numbers 24 and 36 without a remainder. If you just want to know what is the least common multiple of 24 and 36, it is 72. Usually, this is written as. 𝐿𝐶𝑀(24,36) = 72 Question #5 20 How do you find the LCM of 30 ? Answer: Step 1 To solve this, I used my head, but if you were to use extremely high numbers, what you could try is multiplying the two numbers together. For this case, that is 600. Then, you find the GCF of the two numbers. In this case, that would be 10. You take the product and divide it by the GCF. This will yield 600 10 = 60 and that's your answer. Question #6 How do you find the Least common multiple of 16𝑥 2 𝑦, 32𝑥 ? Answer: Step 1 The least common multiple in this case can be found by writing the factors of both the terms. 16𝑥 2 𝑦 = 2 × 2 × 2 × 2 × 𝑥 × 𝑥 × 𝑦 32𝑥 = 2 × 2 × 2 × 2 × 2 × 𝑥 We find the common terms which are 24 and x in this case. The terms left are 2, x and y. To get the LCM, we multiply the common terms with the terms that are left. So, 𝐿𝐶𝑀 = 24 × 𝑥 × 2 × 𝑥 × 𝑦 Therefore, 𝐿𝐶𝑀 = 32𝑥 2 𝑦 Question #7 How do you find the Least common multiple of 1? 23𝑢2 , 14𝑣 2 ? Answer: Step 1 Write the terms as a product of ' factors. The LCM is the product of common and non common factors each taken once with the highest exponent. 23𝑢2 is already factorized as 23 is '. 14𝑣 2 = 2 × 7 × 𝑣 2 There are no common factors so the LCM is just the product of all factors: 𝐿𝐶𝑀 = 322𝑢2 𝑣 2 Question #8 How do you find the Least common multiple of 25𝑥 2 , 25𝑦? Answer: Step 1 The only way you can make this work is as follows: Let 25𝑥 2 = 25𝑦 × 𝑘 Divide both ides by 25 𝑥 2 = 𝑘𝑥 Let the least common multiple be m giving 𝑚𝑥 2 = 𝑘𝑚𝑦 If this is not the case then there is no common multiple Question #9 How do you find the Least common multiple of 3𝑥 2 , 10? Answer: Step 1 The L.C.M of 3 and 10 is 30 and 𝑥 2 is alone. So L.C.M of 3𝑥 2 , 10 = 30𝑥 2 Question #10 How do you find the Least common multiple of 20𝑥 3 , 16𝑥 4 ? Answer: Step 1 Given: 20𝑥 3 , 16𝑥 4 4𝑥 3 [5, 4𝑥 ] 𝐿𝐶𝑀 = 4𝑥 3 × 5 × 4𝑥 = 80𝑥 4 Question #11 How do you find the Least common multiple of 18𝑥𝑦 2 , 15𝑦 3 ? Answer: Step 1 To find the LCM, first find the ' factorization (a composite number being divided into ' numbers being multiplied together). 18𝑥𝑦 2 = 3 × 3 × 2 × 𝑥 × 𝑦 × 𝑦 15𝑦 3 = 3 × 5 × 𝑦 × 𝑦 × 𝑦 Now, circle the ones that they have in common. They each have at least one 3 and 2 y's. Multiply these together to get your LCM. Question #12 I gave Maya and Yosief an equal amount of money. Maya spent $20 each day and Yosief spent $25 each day. When Maya had $157 left, Yosief had $82 left. How much money do I give Yosief? Answer: Step 1 Yoseif has $82 left and Maya has been left with $157 So Yoseif has $157 − $82 = $75 short. Though they started with same amount, Yoseif spends $25 − $20 = $5 extra each day. Hence it is 75 5 = 15 days since money was given to Yoseif. In 15 days Yoseif spent 15 × $25 = $375 and he has $82 left Hence, Questioner gave Yoseif $375 + 82 = $457ZK Question #13 The local school raises by selling tickets to play, during two days. In the equations 5𝑥 + 2𝑦 = 48 and 3𝑥 + 2𝑦 = 32𝑥 represents the cost for each adult ticket and y represents the cost for each student ticket, what is the cost for each adult ticket? Answer: Step 1 5𝑥 + 2𝑦 = 48 indicates that five adult tickets and two student tickets cost $48 Similarly 3𝑥 + 2𝑦 = 32 indicates that three adult tickets and two student tickets cost $32 As number of students is same, it is obvious that additional charge of 48 − 32 = $16 is due to two additional adult tickets. Hence, each adult ticket must cost 16 2 =8 Question #14 Grandfather is 15 time the age of his grandson today, in 10 years the grandson will be grandfather. How old are the grandfather and grandson? Answer: Step 1 Let the grandfather's age be x and the grandson's age be y. We are told 𝑥 = 15𝑦 and 3 17 the age of the 3 𝑦 + 10 = 17 (𝑥 + 10) → 17𝑦 + 170 = 3𝑥 + 30 = 45𝑦 + 170 = 3𝑥 + 30 = 45𝑦 + 30 (since 𝑥 = 15𝑦 ) → 140 = 28𝑦 →𝑦=5 𝑥 = 15𝑦 = 65 Question #15 What is the lowest common multiple of 12 & 21? Answer: Step 1 12 = 1 × 22 × 3 21 = 1 × 3 × 7 The LCM will be the product of multiplying the highest power of each ' number together: 1 × 22 × 3 × 7 𝐿𝐶𝑀(12, 21) = 84 Question #16 What is the Least common Denominator (LCD) of 7 8 and 1 6 ? Answer: The least common multiple of 6 and 8 is 24. Hence: 7 8 1 7⋅3 1⋅4 21 4 25 + 6 = 8⋅3 + 6⋅4 = 24 + 24 = 24. Question #17 How do you find the Least common multiple of 30𝑎𝑏3 , 20𝑎𝑏3 ? Answer: The least common multiple (or LCM) of two numbers is the product of the largest amounts of each (') factor that appear in either number. In other words, it is the smallest value that we can guarantee will have both numbers as factors. Step 1: Factor both numbers. 30𝑎𝑏3 has the factors  2 3 5 a b3  . 20𝑎𝑏3 has the factors  22 5 a b3  . Step 2: Compare the powers of each factor that appear in both numbers, and circle the one that's bigger. The factor 2 appears once in 30𝑎𝑏3 , and it appears twice in 20𝑎𝑏3 . Circle the 22 .  2 3 5 a b3   2  5 a b3  (2 ) The factor 3 appears once in 30𝑎𝑏3 , and not at all in 20𝑎𝑏3 . Circle the 3.  2 (3) 5 a b3   2  5 a b3  (2 ) The remaining three factors (5, 𝑎, 𝑏3 ) appear the same number of times in both numbers. Circle either appearance of these factors.  2 (3) 5 a b3   2  (5) (a) (b3 )  (2 ) Step 3: Multiply these circled values together. The circled values are (22 )(3)(5)(𝑎)(𝑏3 ) . The product of these values is 60𝑎𝑏3 . This is our least common multiple. Question #18 How do you find the Least common multiple of 18𝑚2 , 24𝑛𝑚 ? Answer: The lcm is the product of all the factors raised to the highest power they have in the given terms. We can rewrite 18𝑚2 as 2 ⋅ 32 ⋅ 𝑚2 , and we can rewrite 24𝑛𝑚as 23 ⋅ 3 ⋅ 𝑛 ⋅ 𝑚 . The highest power of 2 appears in the 2nd term: 23 . The highest power of 3 appears in the 1st term: 32 . The highest power of m appears in the 1st term: 𝑚2 . The highest power of n appears in the 2nd term: n. The 𝑙𝑐𝑚(18𝑚2 , 24𝑛𝑚) is just the product of all of these pieces: 𝑙𝑐𝑚(18𝑚2 , 24𝑛𝑚) = 23 ⋅ 32 ⋅ 𝑚2 ⋅ 𝑛 = 72𝑚2 𝑛 . Question #19 Yosie, Maya, Noah & Datan shared $168. Datan received Yosie, Noah & Maya. Yosie received received 2 3 1 7 of the total amount of money received by of the total amount of money received by Noah and Maya. Maya 4 as much as Noah. How much did Maya receive? 5 Answer: Suppose Maya received 2x. As she received 2 5 as much as Noah, Noah must have received 5x and they together received 2𝑥 + 5𝑥 = 7𝑥 . As Yosie received 3 4 × 7𝑥 = 21 4 3 4 of the total amount of money received by Noah and Maya, he must have received 𝑥, and all three Yosie, Noah and Maya together received 7𝑥 + 21 4 𝑥= 7×4 4 𝑥+ 21 4 𝑥= 28 4 𝑥+ 21 4 𝑥= 49 4 𝑥 1 As Datan received 7 4 of what Yosie, Noah and Maya together received. Hence Datan received 7 1 7 × 49 4 𝑥= 𝑥 and all of them together received 49 4 7 56 4 4 𝑥+ 𝑥= 𝑥 = 14𝑥 Now as they together received $168, 14𝑥 = 168, 𝑥 = 168 14 = 12 Hence Maya eceived 2 × 12 = $24 . Question #20 A tank of water is 5 9 3 full of water. If a 6 4 liters of water is poured it will be 2 3 full. What is the capacity of the tank? Answer: As the question uses fractions the answer should be in fractions. 3 The total volume of the tank is 60 4 litres This solution is using algebra but it is in disguise Explanation: The question is not complete in its description, in that it does not state if the water is removed or added. The change is 5 to 9 2 3 which is the same as As 6 1 9 to 9 6 9 is more than the starting point then it is added. 9 The difference is So 5 1 9 3 as a consequence of adding 6 4 litres. 3 × tank = 6 4 litres. Multiply both sides by 9. 1 3 9 × 9 × tank = 9 × 6 4 1 × tank = 9 × 27 4 3 = 60 4 litres