5.3B Multiply 3 digit by 2 digit numbers

Generate all of these 25 questions Part A: Each correct answer is worth 5. 1. The regular pentagon shown has a side length of 2 cm. The perimeter of the pentagon is (A) 2 cm (B) 4 cm (C) 6 cm (D) 8 cm (E) 10 cm 2 cm 2. The faces of a cube are labelled with 1, 2, 3, 4, 5, and 6 dots. Three of the faces are shown. What is the total number of dots on the other three faces? (A) 6 (B) 8 (C) 10 (D) 12 (E) 15 3. The equation that best represents \a number increased by _ve equals 15" is (A) n 􀀀 5 = 15 (B) n _ 5 = 15 (C) n + 5 = 15 (D) n + 15 = 5 (E) n _ 5 = 15 4. The line graph shows the number of bobbleheads sold at a store each year. The sale of bobbleheads increased the most between (A) 2016 and 2017 (B) 2017 and 2018 (C) 2018 and 2019 (D) 2019 and 2020 (E) 2020 and 2021 Number of 2016 2017 2018 2019 2020 Year Sale of Bobbleheads 2021 Bobbleheads 20 40 60 80 5. Starting at 72, Aryana counts down by 11s: 72; 61; 50; : : : . What is the last number greater than 0 that Aryana will count? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 6. In the diagram, \ABC = 90_. The value of x is (A) 68 (B) 23 (C) 56 (D) 28 (E) 26 Day of the Week 44° x° A B C x° 7. Which of the following values is closest to zero? (A) 􀀀1 (B) 5 4 (C) 12 (D) 􀀀4 5 (E) 0:9 Grade 8 8. A jar contains 267 quarters. One quarter is worth $0.25. How many quarters must be added to the jar so that the total value of the quarters is $100.00? (A) 33 (B) 53 (C) 103 (D) 133 (E) 153 9. A package of 8 greeting cards comes with 10 envelopes. Kirra has 7 cards but no envelopes. What is the smallest number of packages that Kirra needs to buy to have more envelopes than cards? (A) 3 (B) 4 (C) 5 (D) 6 (E) 7 10. For the points in the diagram, which statement is true? (A) e > c (B) b < d (C) f > b (D) a < e (E) a > c y x (e, f ) (a, b) (c, d ) Part B: Each correct answer is worth 6. 11. The 26 letters of the English alphabet are listed in an in_nite, repeating loop: ABCDEFGHIJKLMNOPQRSTUVWXY ZABC : : : What is the 258th letter in this sequence? (A) V (B) W (C) X (D) Y (E) Z 12. A public holiday is always celebrated on the third Wednesday of a certain month. In that month, the holiday cannot occur on which of the following days? (A) 16th (B) 22nd (C) 18th (D) 19th (E) 21st 13. A circular spinner is divided into three sections. An arrow is attached to the centre of the spinner. The arrow is spun once. The probability that the arrow stops on the largest section is 50%. The probability it stops on the next largest section is 1 in 3. The probability it stops on the smallest section is (A) 1 4 (B) 2 5 (C) 1 6 (D) 2 7 (E) 3 10 14. A positive number is divisible by both 3 and 4. The tens digit is greater than the ones digit. How many positive two-digit numbers have this property? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 15. A rectangular pool measures 20 m by 8 m. There is a 1 m wide walkway around the outside of the pool, as shown by the shaded region. The area of the walkway is (A) 56 m2 (B) 60 m2 (C) 29 m2 (D) 52 m2 (E) 50 m2 20 m 8 m 1 m Grade 8 16. The results of asking 50 students if they participate in music or sports are shown in the Venn diagram. What percentage of the 50 students do not participate in music and do not participate in sports? (A) 0% (B) 80% (C) 20% (D) 70% (E) 40% Music Sports 15 5 20 17. There are 2 3 as many golf balls in Bin F as in Bin G. If there are a total of 150 golf balls, how many fewer golf balls are in Bin F than in Bin G? (A) 15 (B) 30 (C) 50 (D) 60 (E) 90 18. In the sequence shown, Figure 1 is formed using 7 squares. Each _gure after Figure 1 has 5 more squares than the previous _gure. What _gure has 2022 squares? (A) Figure 400 (B) Figure 402 (C) Figure 404 (D) Figure 406 (E) Figure 408 Figure 1 Figure 2 Figure 3 19. Mateo's 300 km trip from Edmonton to Calgary passed through Red Deer. Mateo started in Edmonton at 7 a.m. and drove until stopping for a 40 minute break in Red Deer. Mateo arrived in Calgary at 11 a.m. Not including the break, what was his average speed for the trip? (A) 83 km/h (B) 94 km/h (C) 90 km/h (D) 95 km/h (E) 64 km/h 20. Equilateral triangle ABC has sides of length 4. The midpoint of BC is D, and the midpoint of AD is E. The value of EC2 is (A) 7 (B) 6 (C) 6:25 (D) 8 (E) 10 Part C: Each correct answer is worth 8. 21. The positive factors of 6 are 1, 2, 3, and 6. There are two perfect squares less than 100 that have exactly _ve positive factors. What is the sum of these two perfect squares? (A) 177 (B) 80 (C) 145 (D) 52 (E) 97 22. In the list p; q; r; s; t; u; v, each letter represents a positive integer. The sum of the values of each group of three consecutive letters in the list is 35. If q + u = 15, then p + q + r + s + t + u + v is (A) 85 (B) 70 (C) 80 (D) 90 (E) 75 Grade 8 23. The net shown is folded to form a cube. An ant walks from face to face on the cube, visiting each face exactly once. For example, ABCFED and ABCEFD are two possible orders of faces the ant visits. If the ant starts at A, how many possible orders are there? (A) 24 (B) 48 (C) 32 (D) 30 (E) 40 A D B C E F 24. The number 385 is an example of a three-digit number for which one of the digits is the sum of the other two digits. How many numbers between 100 and 999 have this property? (A) 144 (B) 126 (C) 108 (D) 234 (E) 64 25. Student A, Student B, and Student C have been hired to help scientists develop a new avour of juice. There are 4200 samples to test. Each sample either contains blueberry or does not. Each student is asked to taste each sample and report whether or not they think it contains blueberry. Student A reports correctly on exactly 90% of the samples containing blueberry and reports correctly on exactly 88% of the samples that do not contain blueberry. The results for all three students are shown below. Student A Student B Student C Percentage correct on samples 90% 98% (2m)% containing blueberry Percentage correct on samples 88% 86% (4m)% not containing blueberry Student B reports 315 more samples as containing blueberry than Student A. For some positive integers m, the total number of samples that the three students report as containing blueberry is equal to a multiple of 5 between 8000 and 9000. The sum of all such values of m is (A) 45 (B) 36 (C) 24 (D) 27 (E) 29

Quale dei seguenti numeri è un numero naturale? a) -3 b) 0 c) -1/2 d) √16 Qual è l'opposto di -10? a) 10 b) -20 c) 0 d) -5 Quale dei seguenti numeri è un numero intero relativo maggiore di -8? a) -9 b) -7 c) -6 d) 0 Quale dei seguenti numeri è un multiplo di 5? a) -15 b) 12 c) 7 d) 25 Qual è il risultato di 7 - (-3)? a) 4 b) 10 c) -4 d) -10 Quale delle seguenti affermazioni è vera? a) La somma di due numeri pari è sempre pari. b) Il prodotto di due numeri dispari è sempre pari. c) La somma di un numero pari e un numero dispari è sempre pari. d) Il prodotto di due numeri pari è sempre dispari. Se moltiplichiamo un numero intero per zero, qual è il risultato? a) 1 b) 0 c) Il risultato è indefinito. d) Dipende dal numero intero. Se sommiamo un numero intero a se stesso, qual è il risultato? a) Il risultato è sempre zero. b) Il risultato è sempre positivo. c) Il risultato dipende dal numero intero. d) Il risultato è sempre negativo. Quale dei seguenti numeri è un numero primo? a) 6 b) 11 c) 15 d) 20 Qual è il minimo comune multiplo (MCM) di 8 e 12? a) 6 b) 24 c) 48 d) 10

5.3B: Multiplying Whole Numbers

SPANISH STUDENTS 10/20/25 STANDARDS LC.RL.5.2c Determine the theme of a story, drama, or poem including how characters in a story or drama respond to challenges or how the speaker in a poem reflects upon a topic LC.RL.5.3a Compare characters, settings, events within a story; provide or identify specific details in the text to support the comparison. LC.RL.5.3b Compare and contrast two or more characters, settings, or events in a story or drama, drawing on specific details in the text (e.g., how characters interact). LC.RL.5.4 Determine the meaning of words and phrases as they are used in a text including figurative language such as metaphors and similes. LC.RL.5.5a Use signal words (e.g., meanwhile, unlike, next) to identify common types of text structure (e.g., sequence, compare/contrast, cause/effect, description) within a text. LC.W.5.1a Produce an opinion piece which has an introduction that states an opinion and has an organizational structure in which ideas are logically grouped to support the writer's opinion

A solution is composed of a solute dissolved in a solvent. In the sugar water described in Figure 5-1, the solute was sugar and the solvent was water, and the solute molecules diffused through the solvent. It is also possible for solvent molecules to diffuse. In the case of cells, the solutes are organic and inorganic compounds, and the solvent is water. The process by which water molecules diffuse across a cell membrane from an area of higher concentration to an area of lower concentration is called osmosis (ahs-MOH-sis). Because water is moving from a higher to lower concentration, osmosis does not require cells to expend energy. Therefore, osmosis is the passive transport of water. Direction of Osmosis The net direction of osmosis depends on the relative concentra- tion of solutes on the two sides of the membrane. Examine Table 5-1. When the concentration of solute molecules outside the cell is lower than the concentration in the cytosol, the solution outside is hypotonic to the cytosol. In this situation, water diffuses into the cell until equilibrium is established. When the concentration of solute molecules outside the cell is higher than the concentration in the cytosol, the solution outside is hypertonic to the cytosol. In this situation, water diffuses out of the cell until equilibrium is established. Observing Diffusion Materials 600 mL beaker, 25 cm dialysis tubing, funnel, 15 mL starch solution (10 percent), 20 drops Lugol’s solution, 300 mL water, 100 mL graduated cylinder, 20 cm piece of string (2) Procedure 1. Put on your disposable gloves, lab apron, and safety goggles. 2. Pour 300 mL of water in the 600 mL beaker. 3. Add 20 drops of Lugol’s solution to the water. CAUTION: Lugol’s solution is a poison and eye and skin irritant. 4. Open the dialysis tubing, and tie one end tightly with a piece of string. 5. Using the funnel, pour 15 mL of 10 percent starch solution into the dialysis tubing. 6. Tie the other end of the dialysis tubing tightly with the second piece of string, forming a sealed bag around the starch solution. 7. Place the bag into the solution in the beaker, and observe the setup for a color change. Analysis What happened to the color in the bag? What happened to the color of the water around the bag? Explain your observations. Quick Lab www.scilinks.org Topic: Osmosis Keyword: HM61090 mb06se_homs01.qxd 11/27/07 8:52 AM Page 98 HOMEOSTASIS AND CELL TRANSPORT 99 When the concentrations of solutes outside and inside the cell are equal, the outside solution is said to be isotonic to the cytosol. Under these conditions, water diffuses into and out of the cell at equal rates, so there is no net movement of water. Notice that the prefixes hypo-, hyper-, and iso- refer to the relative solute concentrations of two solutions. Thus, if the solution outside the cell is hypotonic to the cytosol, then the cytosol must be hyper- tonic to that solution. Conversely, if the solution outside is hypertonic to the cytosol, then the cytosol must be hypotonic to the solution. Water tends to diffuse from hypo- tonic solutions to hypertonic solutions. How Cells Deal with Osmosis Cells that are exposed to an isotonic external environment usually have no difficulty keeping the movement of water across the cell membrane in balance. This is the case with the cells of ver- tebrate animals on land and of most other organ- isms living in the sea. In contrast, many cells function in a hypotonic environment. Such is the case for unicellular freshwater organisms. Water constantly diffuses into these organisms. Because they require a relatively lower concentration of water in the cytosol to function normally, unicel- lular organisms must rid themselves of the excess water that enters by osmosis. Some of them, such as the paramecia shown in Figure 5-2, do this with contractile vacuoles (kon-TRAK-til VAK-y ̄ ̄o ̄ ̄o-OL), which are organelles that remove water. Contractile vacuoles collect the excess water and then contract, pumping the water out of the cell. Unlike diffusion and osmosis, this pumping action is not a form of passive trans- port because it requires the cell to expend energy. Copyright © by Holt, Rinehart and Winston. All rights reserved. (a) (b) Vacuole filling with water Vacuole contracting TABLE 5-1 Direction of Osmosis Condition External solution is hypotonic to cytosol External solution is hypertonic to cytosol External solution is isotonic to cytosol Net movement of water into the cell out of the cell none H2O H2O H2O H2O H2O H2O The paramecia shown below live in fresh water, which is hypotonic to their cytosol. (a) Contractile vacuoles collect excess water that moves by osmosis into the cytosol. (b) The vacuoles then contract, returning the water to the outside of the cell. (LM 315) FIGURE 5-2 100 CHAPTER 5 (a) HYPOTONIC Cell walls (b) HYPERTONIC (a) ISOTONIC (b) HYPOTONIC (c) HYPERTONIC Other cells, including many of those in multicellular organisms, respond to hypotonic environments by pumping solutes out of the cytosol. This lowers the solute concentration in the cytosol, bring- ing it closer to the solute concentration in the environment. As a result, water molecules are less likely to diffuse into the cell. Most plant cells, like animal cells, live in a hypotonic environ- ment. In fact, the cells that make up plant roots may be surrounded by water. This water moves into plant cells by osmosis. These cells swell as they fill with water until the cell membrane is pressed against the inside of the cell wall, as Figure 5-3a shows. The cell wall is strong enough to resist the pressure exerted by the water inside the expanding cell. The pressure that water molecules exert against the cell wall is called turgor pressure (TER-GOR PRESH-er). In a hypertonic environment, water leaves the cells through osmosis. As shown in Figure 5-3b, the cells shrink away from the cell walls, and turgor pressure is lost. This condition is called plasmolysis (plaz-MAHL-uh-sis), and is the reason that plants wilt if they don’t receive enough water. Some cells cannot compensate for changes in the solute con-

5.NBT.3b

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