Math in Focus

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Review Test Answer Key

Concepts and Skills

State whether each event is a simple or compound event.

Question 1.
Drawing 2 yellow marbles ¡n a row from a bag of yellow and green marbles.
Answer:
Drawing 2 yellow marbles in a row from a bag of yellow and green marbles are compound events. Because the number of marbles is 3 so, the result is the number of outcomes.

Question 2.
Drawing 1 red pebble and 1 yellow pebble in a row from a bag of red and yellow pebbles.
Answer:
Drawing 1 red pebble and 1 yellow pebble in a row from a bag of red and yellow pebbles is compound. Because the number of pebbles is 2 so, the result is the number of outcomes

Question 3.
Tossing a coin once.
Answer:
Tossing a coin once is a simple event because it has a result of one outcome.

Draw the possibility diagram and state the number of possible outcomes for each compound event.

Question 4.
From three cards labeled A, B, and C, draw two cards, one at a time with replacement.
Answer:

Question 5.
From a pencil case with 1 red pen, 1 green pen, and 1 blue pen, select two pens, one at a time without replacement.
Answer:

Question 6.
Toss a fair four-sided number die, labeled 1 to 4, and a coin.
Answer:

Draw the tree diagram for each compound event.

Question 7.
Spinning a spinner divided into 4 equal areas labeled 1 to 4, and tossing a coin.
Answer:

Question 8.
Picking two green apples randomly from a basket of red and green apples.
Answer:
The probability of picking red and green apples is 1/2.

State whether each compound event consists of independent events or dependent events.

Question 9.
From a pencil case, two-color pencils are randomly drawn, one at a time without replacement.
Answer: Dependent event

Question 10.
From two classes of 30 students, one student ¡s selected randomly from each class for a survey.
Answer: Independent event

Problem Solving

Solve. Show your work.

Question 11.
There are two tables in a room. There are 2 history textbooks and 1 math textbook on the first table. There are 1 history workbook and 1 math workbook on the second table. Use a possibility diagram to find the probability of randomly selecting a history textbook from the first table and a math workbook from the second table.
Answer:
Given,
There are two tables in a room. There are 2 history textbooks and 1 math textbook on the first table.
1 + 2 = 3
The probability of randomly selecting a history textbook from the first table and a math workbook from the second table is 1/3.

Question 12.
A fair four-sided number die is marked 1, 2, 2, and 3. A spinner equally divided into 3 sectors is marked 3, 4, and 7. Jamie tosses the number die and spins the spinner.
a) Use a possibility diagram to find the probability that the sum of the two resulting numbers is greater than 5.
Answer:
Note that there are 4 possible outcomes in rolling a die which is 1, 2, 2, and 3, and 3 outcomes for the spinner which are 3, 4 and 7. Add these possible resuLts, then mark the answers that are bigger the 5.

Observe that there are 12 possible outcomes and 8 are greater than 5 then

Therefore, the probability of obtaining a number that is larger than 5 is \(\frac{2}{3}\).

b) Use a possibility diagram to find the probability that the product of the two resulting numbers is odd.
Answer:
Notice that there are 4 possible outcomes in rolling a die which is 1, 2, 2. and 3, and 3 outcomes for the spinner which are 3, 4 and 7. Multiply these possible results, then mark the answers that odd numbers.

Observe that there are 12 possibLe outcomes and 4 are odd numbers then

Therefore, the probability of obtaining an odd number is \(\frac{1}{3}\).

Question 13.
A juggler is giving a performance by juggling a red ball, a yellow ball, and a green ball. All 3 balls have equal chance of dropping. If one ball drops, the juggler will stop and pick up the ball and resume juggling. If another ball drops again, the juggler will stop the performance.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
Answer:

b) Find the probability of dropping the same colored ball twice.
Answer: 1/3

c) Find the probability of dropping one green and one yellow ball.
Answer:
P(G)P(R) = 1/3 × 1/3= 1/9

Question 14.
In a marathon, there is a half-marathon and a full-marathon. There are 60 students who participated in the half-marathon and 80 participated in the full-marathon. Half of the students in the half-marathon warm up before the run, while three-quarters of the students in the full-marathon warm up. Assume that warming up and not warming up are mutually exclusive and complementary.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
Answer:
Since, there are 2 types of marathon, then start the tree diagram with 2 branches. Now, at the end of these branches put the possible outcomes: H for half-marathon, and F for full marathon. Note that the probability of full marathon is \(\frac{80}{140}\) and the probability of half-marathon is \(\frac{60}{140}\). Next, there are a group who warms up and who does not, then make another 2 branches. At the end of these branches put the possible outcomes: W for warming up and NW for not warming up. Finally, conclude the possible results. Therefore, the tree diagram for the said event is as depicted below.

b) What is the probability of randomly picking a marathon participant who warms up before running a full-marathon?
Answer: full marathon: 80 runners

c) What is the probability of randomly picking a marathon participant who does not warm up before running?
Answer:
80 + 60 = 140
total: 140 runners; 90 warm-up.
The probability that a randomly selected runner warms up is 90/140.

Question 15.
The probability of Cindy waking up after 8 A.M. on a weekend day is p. Assume the events of Cindy waking up after 8 A.M. and by 8 A.M. are mutually exclusive and complementary.
a) If p = 0.3, find the probability that she will wake up after 8 A.M. on two consecutive weekend days.
Answer:
P(waking up after 8 A.M. two weekend days in a row) = p²
If p=0.3
p² = (0.3)²
p² = 0.09

b) If p = 0.56, find the probability that she will wake up by 8 A.M. on two consecutive weekend days.
Answer:
P(waking up before 8 A.M two weekend days in a row) = (1 – p)²
If p = 0.56, 1-p = 0.44 and
(1 – p)² = (0.44)² = 0.1936.

Question 16.
In a jar, there are 2 raisin cookies and 3 oat cookies. Steven takes two cookies one after another without replacement.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
Answer:
Since there are two different kinds of muffins, which are bran and pumpkin, then start the tree diagram with 2 branches. Now, at the end of these branches put the possible outcomes: P for pumpkin muffin, and B for bran muffins. Note that there are 2 pumpkin muffins out of the total of 5 muffins, and 3 bran muffins out of the 5 total muffins. So the probability for selecting a pumpkin is \(\frac{2}{5}\), and the probability of selecting a bran is \(\frac{3}{5}\).

Next, there are stilt 2 different kinds of muffins, then again, at the end of these branches put the possible outcomes: P for pumpkin muffin, and B for bran muffins. If a pumpkin muffin is obtained in the first pick, then on the second pick, there are now 1 pumpkin muffin and a totat of 4 muffins left since a muffin is already taken, then the probability of choosing a pumpkin is \(\frac{1}{4}\). Also, there are still 3 bran muffin out of 4 muffins left, thus, the probability of selecting a bran muffin is \(\frac{3}{4}\).

If a bran muffin is obtained in the first pick, then on the second pick, there are still a pumpkin muffin and a total of 4 muffins left since a muffin is already taken, then the probability of choosing a pumpkin is \(\frac{2}{4}\). Also, there are only 2 bran muffin left out of 4 muffins, thus, the probability of selecting a bran muffin is \(\frac{2}{4}\). Therefore, the tree diagram for the said event is as depicted below.

b) Find the probability of Steven randomly getting two of the same type of cookie.
Answer:
The probability of seLecting 2 muffins of a matching type is the probability of obtaining 2 pumpkin muffins, P(P, P), or the probability of choosing 2 bran muffins, P(B, B), then
P(P, P) or P(B, B) = P(P, P) + (B, B)
= P(P) · P(P) +P(B) · P(B)
= \(\frac{2}{5}\) · \(\frac{1}{4}\) + \(\frac{3}{5}\) · \(\frac{2}{4}\)
= \(\frac{2}{20}\) + \(\frac{6}{20}\)
= \(\frac{8}{20}\)
Therefore, the probability of picking 2 muffins of a matching type is \(\frac{8}{20}\).

c) Find the probability of Steven randomly getting at least one raisin cookie.
Answer:
For, the probability of selecting a minimum of one pumpkin muffin, note that the probability of choosing 2 bran muffins, P(B, B), and the probability of picking at least one pumpkin muffin is complementary.
P(B, B) + P(Minimum of One Pumpkin) = 1
P(Minimiun of One Pumpkin) = 1 – P(B, B)
P(Minimum of One Pumpkin) = 1 – P(B) · P(B)
P(Minimum of One Pumpkin) = 1 – \(\frac{3}{5}\) · \(\frac{2}{4}\)
P(minimum of One Pumpkin) = 1 – \(\frac{6}{20}\)
P(Minimum of One Pumpkin) = \(\frac{7}{10}\)
Therefore, the probabiUty of picking a minimum of one pumpkin is \(\frac{7}{10}\).

Question 17.
Out of 100 raffle tickets, 4 are marked with a prize. Matthew randomly selects two tickets from the box.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
Answer:

b) What is the probability that Matthew does not win any prizes?
Answer: 152/165

c) What is the probability that Matthew gets exactly one of the prizes?
Answer: 64/825

Question 18.
The tree diagram shows the probability of how Shane spends his day gaming or cycling, depending on the weather. The probability of rain is denoted by a. Assume that gaming and cycling are mutually exclusive.

a) If a = 0.4, find the probability that he will spend his day gaming.
Answer:
Given that a = 0.4
Here the G represents gaming.
The probability that he will spend his day on gaming =(1- 0.4) × 1/4 = 0.15.

b) If a = 0.75, find the probability that he will spend his day cycling.
Answer:
Given that a = 0.75
Here C represents cycling.
The probability that he will spend his day in cycling = (1-0.75) × 3/4 = 0.1875.

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Lesson 11.4 Dependent Events detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Lesson 11.4 Answer Key Dependent Events

Math in Focus Grade 8 Chapter 11 Lesson 11.4 Guided Practice Answer Key

Solve. Show your work.

Question 1.
A deck of four cards with the letters D, E, E, D are placed facing down on a table. Two cards are turned at random to show the letter. Draw a tree diagram to represent the possible outcomes in this compound event.

Let D represent letter D and E represent letter E.
1st draw
P(D) = \(\frac{2}{4}\)
P(E) = \(\frac{2}{4}\)

2nd draw
P(D after D)4 = \(\frac{?}{?}\) There is D left after 1 D is drawn.
P(E after D) = \(\frac{?}{?}\) There are E still after 1 D is drawn.
P(D after E) = \(\frac{?}{?}\) There are D still after 1 E is drawn.
P(E after E) = \(\frac{?}{?}\) There is E left after 1 E is drawn.

Answer:

P(D after D)4 = There is 1/3 D left after 1 D is drawn.
P(E after D) = There are 2/3 E still after 1 D is drawn.
P(D after E) = There are 2/3 D still after 1 E is drawn.
P(E after E) = There is 1/3 E left after 1 E is drawn

Question 2.
There are 16 different color pebbles in a jar. 11 of them are blue and the rest are orange. Two pebbles are randomly selected from the jar, one at a time without replacement.
a) Find the probability of taking an orange pebble followed by a blue pebble.

The probability of randomly taking an orange pebble followed by a blue pebble is
Answer:
P(B,O) = P(B) x P(O)
= 11/16 x 5/15
= 55/240
= 11/48
~0.23
Therefore, the probability of selecting a blue than a orange pebble is 11/48 or approx 0.23.

b) Find the probability of taking two orange pebbles.
P(O, O) = P(O) • P(O after O)
= \(\frac{?}{?} \cdot \frac{?}{?}\)
= \(\frac{?}{?}\)
The probability of randomly taking two orange pebbles is .
Answer:
P(O, O) = P(O) • P(O)
= 5/16 x 4/15
= 20/240
= 1/12
~0.08
Therefore, the probability of selecting two orange pebbles is 1/12 or approx 0.08.

c) Find the probability of taking two blue pebbles.
P(B, B) = P(B) • P(B after 8)
= \(\frac{?}{?} \cdot \frac{?}{?}\)
= \(\frac{?}{?}\)
The probability of randomly taking two blue pebbles is .
Answer:
P(B, B) = P(B) • P(B)
= 11/16 x 10/15
= 110/240
= 11/24
~0.46
The probability of randomly taking two blue pebbles is 11/24 or approx 0.46.

Question 3.
The tree diagram below shows how passing an examination depends on whether a student studies (S) or does not study (NS) for the exam. The probability that a student studies is denoted by p. Assume that S and NS are mutually exclusive events.

a) If p = 0.4, find the probability that a student passes the examination.
P(S) = \(\frac{?}{?}\) Write the fraction for 0.4.
P(NS) = 1 – P(S) Events S and NS are complementary.
= 1 – \(\frac{?}{?}\)
= \(\frac{?}{?}\)
P(P) = \(\frac{?}{?}\) • \(\frac{?}{?}\) + \(\frac{?}{?}\) • \(\frac{?}{?}\) Evaluate P(S, P) + P(NS, P).
= \(\frac{?}{?}\)
If the probability of studying is 0.4, then the probability of passing is .
Answer:
P(NS) + P(S) = 1
P(NS) = 1 – P(S)
P(NS) = 1 – 4/10
P(NS) = 6/10
Therefore, the probability of not studying is 6/10.
P(Pa) = P(S,P) + P(NS,P)
= P(S) x P(P) + P(NS) x P(P)
= 4/10 x 2/3 + 6/10 x 1/5
= 8/30 + 6/50
= 29/75
~0.39
Therefore, the probability of passing the exam is 29/75 or approx 0.39.

b) If p = 0.75, find the probability that a student fails the examination.
P(S) = \(\frac{?}{?}\) Write the fraction for 0.75.
P(NS) = 1 – P(S) Events S and NS are complementary.
= 1 – \(\frac{?}{?}\)
= \(\frac{?}{?}\)
P(F) = \(\frac{?}{?}\) • \(\frac{?}{?}\) + \(\frac{?}{?}\) • \(\frac{?}{?}\) Evaluate P(S, F) + P(NS, F).
= \(\frac{?}{?}\)
If the probability of studying is 0.75, then the probability of failing is .
Answer:
P(NS) + P(S) = 1
P(NS) = 1 – P(S)
P(NS) = 1 – 3/4
P(NS) = 1/4
Therefore, the probability of not studying is 1/4.
P(Fa) = P(S,F) + P(NS,F)
= P(S) x P(F) + P(NS) x P(F)
= 3/4 x 1/3 + 1/4 x 4/5
= 3/12 + 4/20
= 9/20
= 0.45
Therefore, the probability of failing the exam is 9/20 or approx is 0.45

Math in Focus Course 3B Practice 11.4 Answer Key

State whether each event is a dependent or independent event.

Question 1.
Drawing 2 red balls randomly, one at a time without replacement, from a bag of six balls.
Answer: Dependent event

Question 2.
Tossing a coin twice.
Answer: Independent event

Question 3.
Reaching school late or on time for two consecutive days.
Answer: Independent event

Question 4.
Flooding of roads during rainy or sunny days.
Answer: Dependent event

Draw the tree diagram for each compound event.

Question 5.
2 balls are drawn at random, one at a time without replacement, from a bag of 3 green balls and 18 red balls.
Answer:

Question 6.
The probability of rain on a particular day is 0.3. If it rains, then the probability that Renee goes shopping is 0.75. If it does not rain, then the probability that she goes jogging is 0.72. Assume that shopping and jogging are mutually exclusive and that rain and no rain are complementary.
Answer:
The probability of raining is 0.3, and the probability of raining P(R) and the probability of not raining P(NR) is equal to 1, then
P(NR) + P(R) = 1
P(NR) = 1 – P(R)
P(NR) = 1 – 0.3
P(NR) = 0.7
Therefore, the probability of not raining is 0.7
The probability of shopping when rain comes is 0.75, and the probability of shopping P(S), when it rains and the probability of jogging P(J) during rainy day is equal to 1, then
P(S) + P(J) = 1
P(J) = 1 – P(S)
P(J) = 1 – 0.75
P(J) = 0.25
Therefore, the probability of jogging when rains is 0.25
The probability of jogging when the rain did not come is 0.72, and the probability of shopping P(S) when it did not rain and the probability of jogging P(J) during a non-rainy day is equal to 1, then
P(S) + P(J) = 1
P(S) = 1 – P(J)
P(S) = 1 – 0.72
P(S) = 0.28
Therefore, the probability of shopping when it did not rain is 0.28.

Solve. Show your work.

Question 7.
Geraldine has a box of 13 colored pens: 3 blue, 4 red, and the rest black. What is the probability of drawing two blue pens randomly, one at a time without replacement?

Answer:
P(B, B) = P(B) x P(B)
= 3/13 x 2/12
= 1/26.
Therefore, the probability of picking blue pens twice is 1/26.

Question 8.
A box contains 8 dimes, 15 quarters, and 27 nickels. A student is to randomly pick two items, one at a time without replacement, from the bag. Find the probability that 2 quarters are picked.
Answer:
P(Q,Q) = P(Q) x P(Q)
= 15/50 x 14/49
= 210/2450
= 3/35
Therefore, the probability of picking a quarter twice is 3/35.

Question 9.
There are 9 green, 2 yellow, and 5 blue cards in a deck. Players A and B each randomly pick a card from the deck. Player A picks a card first before player B picks. Find the probability that both players pick the same color cards.
Answer:
P(B,B) or P(Y,Y) or P(G,G) = P(B,B) + P(Y,Y) + P(G,G)
= P(B) x P(B) + P(Y) x P(Y) + P(G) x P(G)
= 5/16 x 4/15 + 2/16 x 1/15 + 9/16 x 8/15
= 20/240 + 2/240 + 72/240
= 94/240
= 47/120
Therefore, the probability of obtaining a matching color of cards is 47/120.

Question 10.
The probability diagram below shows the probability of Xavier going to library or park depending if the weather is sunny or rainy. The probability of rain on a particular day is denoted by a. Assume that going to the library and going to the park are mutually exclusive and complementary.

a) If a = 0.3, find the probability that Xavier goes to the park on any day.
Answer:
To convert decimals into fractions,
= 0.3/1
Multiplying both numerator and denominator to 100
= 0.3/1 x 100/100
= 30/100
= 3/10
Therefore, 0.3 is 3/10 in fractional form.
P(S) + P(R) = 1
P(S) = 1 – P(R)
P(S) = 1- 3/10
P(S) = 7/10
Therefore, the probability of having a sunny day is 7/10.
P(L) + P(P) = 1
P(P) = 1 – P(L)
P(P) = 1 – 1/5
P(P) = 4/5
Therefore, the probability of going to the park during the sunny day is 4/5.
P(S,P) or P(R,P) = P(S,P) + P(R,P)
= P(S) x P(P) + P(R) x P(P)
= 7/10 x 4/5 + 3/10 x 1/3
= 28/50 + 3/30
= 33/30
= 0.66
Therefore, the probability of going to park at any day is 0.66.

b) If a = 0.75, find the probability that he goes to the library on any day.
Answer:
Converting the decimals into fractions,
= 0.75/1
Multiply both the numerator and denominator to 100,
= 0.75/1 x 100/100
= 75/100
= 3/4
Therefore, 3/4 is the fractional form of 0.75
P(S) + P(R) = 1
P(S) = 1 – P(R)
P(S) = 1 – 3/4
P(S) = 1/4
Therefore, the probability of having a sunny day is 1/4.
P(S,L) or P(R,L) = P(S,L) + P(R,L)
= P(S) x P(L) + P(R) x P(L)
= 1/4 x 1/5 + 3/4 x 2/3
= 1/20 + 6/12
= 11/20
= 0.55
Therefore, the probability of going to library at any day is 0.55.

Question 11.
There are 15 apples in a fruit basket. 6 of them are red apples and the rest green apples. Two apples are picked randomly, one at a time without replacement.

a) Draw a tree diagram to represent the possible outcomes.
Answer:

b) Find the probability of picking a green apple and then a red apple.
Answer:
P(G,R) = P(G) x P(R)
= 9/15 x 6/14
= 54/210
= 9/35
Therefore, the probability of picking a green apple then a red apple is 9/35.

c) Find the probability of picking two green apples.
Answer:
P(G,G) = P(G) x P(G)
= 9/15 x 8/14
= 72/210
= 12/35
Therefore, the probability of getting two green apples is 12/35.

d) Find the probability of picking two red apples.
Answer:
P(R,R) = P(R) x P(R)
= 6/15 x 5/14
= 30/210
= 1/70
Therefore, the probability of obtaining two red apples is 1/7.

Question 12.
There are 8 people in a room: 3 of them have red hair, 2 have blonde hair, and the rest have dark hair. Two people are randomly selected to leave the room, one after another, and they do not re-enter the room.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

b) What is the probability of a person with dark hair leaving the room first?
Answer:
P(D,D) or P(D,R) or P(D,B) = P(D,D) + P(D,R) + P(D,B)
= P(D) x P(D) + P(D) x P(R) + P(D) x P(B)
= 3/8 x 2/7 + 3/8 x 3/7 + 3/8 x 2/7
= 6/56 + 9/56 + 6/56
= 21/56
= 3/8
Therefore, the probability that a dark haired person is the first one to leave is 3/8

c) What is the probability of a person with red hair leaving the room, followed by a person with blonde hair?
Answer:
P(R,B) = P(R) x P(B)
= 3/8 x 2/7
= 6/56
= 3/28
Therefore, the probability that a red haired person is the first one to leave then a blonde is 3/28.

d) What is the probability of two people with the same hair color leaving the room?
Answer:
P(D,D) or P(R,R) or P(B,B) = P(D,D) + P(R,R) + P(B,B)
= P(D) x P(D) + P(R) x P(R) + P(B) x P(B)
= 3/8 x 2/7 + 3/8 x 2/7 + 2/8 x 1/7
= 6/56 + 9/56 + 2/56
= 17/56
Therefore, the probability that a dark haired person is the first to leave is 17/56.

Question 13.
Along a stretch of road there are 2 traffic light intersections. Having red or green light for the first intersection is equally likely. Having a red light at the second intersection is twice as likely as a green light, if the first intersection traffic light was red. What is the probability of having a red light on the first intersection and a green light on the second intersection? Draw a tree diagram to show the possible outcomes.
Answer:

P(R,G) = P(R) x P(G)
= 1/2 x 1/3
= 1/6
Therefore, the probability of encountering a red then a green light is 1/6.

Question 14.
To get to work, Mr. Killiney needs to take a train and then a bus. The probability that the train breaks down is 0.1. When the train breaks down, there is a 0.7 probability that the bus will be overcrowded. When the train is operating normally, there is a 0.2 probability that the bus will be overcrowded. What is the probability of getting a seat in the bus? Draw a tree diagram to show the possible outcomes.
Answer:
P(SW) + P(W) = 1
P(W) = 1 – P(SW)
P(W) = 1 – 0.1
P(W) = 0.9
Therefore, the probability of the train working normally is 0.9.
P(NF) + P(F) = 1
P(NF) = 1 – P(F)
P(NF) = 1 – 0.7
P(NF) = 0.3
Therefore, the probability that the bus is not full is 0.3 when the train is not working properly.
P(NF) + P(F) = 1
P(NF) = 1- P(F)
P(NF) = 1 – 0.2
P(NF) = 0.8
Therefore, the probability that the bus is not full is 0.8 when the train is not working properly.

P(SW,NF) or P(W,NF) = P(SW) x P(NF) + P(W) x P(NF)
= 0.1 x 0.3 + 0.9 x 0.8
= 0.03 + 0.72
= 0.75
Therefore, the probability that the bus is not full is 0.75.

Brain @ Work

Question 1.
If there are 12 green and 6 red apples, find the probability of randomly choosing three apples of the same color in a row, without replacement. Show your work.
Answer:
P(G,G,G) = P(G) x P(G) x P(G)
= 12/18 x 11/17 x 10/16
= 55/204
Therefore, the probability of obtaining green apples is 55/204.
P(R,R,R) = P(R) x P(R) x P(R)
= 6/18 x 5/17 x 4/16
= 5/204
Therefore, the probability of obtaining red apples is 5/204.
P(R,R,R) or P(G,G,G) = P(R) x P(R) x P(R) + P(G) x P(G) x P(G)
= 5/204 + 55/204
= 60/204
= 5/17
Therefore, the probability of obtaining of picking 3 apples of a matching color is 5/17.

Question 2.
William has five $1 bills, ten $10 bills, and three $20 bills in his wallet. He picks three bills randomly in a row, without replacement. What is the probability of him picking three of the same type of bills? $how your work.
Answer:
P($20,$20,$20) = P($20) x P($20) x P($20)
= 3/18 x 2/17 x 1/16
= 6/4896
= 1/816
Therefore, the probability of obtaining three $20 bills is $1/816.
P($10,$10,$10) = P($10) x P($10) x P($10)
= 10/18 x 9/17 x 8/16
= 720/4896
= 5/34
Therefore, the probability of obtaining three $10 bills is $5/34.
P($1,$1,$1) = P($1) x P($1) x P($1)
= 5/18 x 4/17 x 3/16
= 60/4896
= 5/408
Therefore, the probability of obtaining three $1 bills is 5/408.
P($20,$20,$20) or P($10,$10,$10) or P($1,$1,$1) = P($20,$20,$20) + P($10,$10,$10) + P($1,$1,$1)
= 1/816 + 5/34 + 5/408
= 131/816
Therefore, the probability of having three bills of a matching type is 131/816.

Question 3.
Daniel plans to visit Australia. Whether he goes alone or with a companion is equally likely. If he travels with a companion there is a 40% chance of joining a guided tour. If he travels alone, there is an 80% chance of joining a guided tour.

a) What is the probability of traveling with a companion and not joining a guided tour?
Answer:
P(NT) + P(T) = 1
P(NT) = 1 – P(T)
P(NT) = 1 – 0.4
P(NT) = 0.6
Therefore, the probability of not going to tour is 0.6 if he travels with others.
P(NT) + P(T) = 1
P(NT) = 1 – P(T)
P(NT) = 1 – 0.8
P(NT) = 0.2
Therefore, the probability of not going to tour is 0.2 if he travels without the other.
P(C,NT) = P(C) x P(NT)
= 0.5 x 0.6
= 0.3
Therefore, the probability of travelling with others and not going with a tour is 0.3.

b) What is the chance of joining a guided tour?
Answer:
P(C,T) or P(WC,T) = P(C) x P(T) + P(WC) x P(T)
= 0.5 x 0.4 + 0.5 x 0.8
= 0.2 + 0.4
= 0.6
Therefore, the probability of being in a tour is 0.6

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Lesson 11.3 Independent Events detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Lesson 11.3 Answer Key Independent Events

Math in Focus Grade 8 Chapter 11 Lesson 11.3 Guided Practice Answer Key

Solve. Show your work.

Question 1.
A game is played with a bag of 6 color tokens and a bag of 6 letter tiles. The 6 tokens consist of 2 green tokens, 1 yellow token, and 3 red tokens. The 6 letter tiles consist of 4 tiles of letter A and 2 tiles of letter B. To win the game, you need to randomly get a yellow token and a tile of letter B from a random selection in each bag.
a) Copy and complete the tree diagram.

Answer:

b) Use the multiplication rule of probability to find the probability of winning the game in one try.
P(winning the game) = P(Y, B)
= P(Y) • P(B)
= •
=
The probability of winning the game is .
Answer:
P(winning the game) = P(Y, B)
= P(Y) • P(B)
P(Y) = 1/6 and P(B) = 2/6
= 1/6 • 2/6
= 2/36
= 1/18
The probability of winning the game is 1/18.

Technology Activity

Materials:
spreadsheet software

SIMULATE RANDOMNESS

Work in pairs.

Background
Two fair six-sided number dice are thrown. Using a spreadsheet, data will be generated to investigate how frequently the outcome of doubles (1 and 1, 2 and 2, … , 6 and 6) occurs.
STEP 1: Label your spreadsheet as shown.

STEP 2: To generate a random integer between 1 and 6, in cell A2, enter the formula = INT(RAND()*6 + 1) to simulate rolling a die. A random number from 1 to 6 should appear in the cell.

STEP 3: To model 100 rolls, select cells A2 to A101 and choose Fill Down from the Edit menu.

STEP 4: Repeat STEP 2 and STEP 3 for cells B2 to B101.

STEP 5: In cell C2, enter the formula = A2 – B2. Select cells C2 to C101 and choose Fill Down from the Edit menu. This column serves as a check to see if the random numbers generated in columns A and B are the same. If the numbers are the same, their difference is 0. A zero difference indicates doubles outcome.

STEP 6: To see how many times the data shows double occurring, in cell E1, enter the formula = COUNTIF(C2:C101,0).

STEP 7: Find the experimental probability of the occurrence of two number dice showing the same number by dividing the number you get in cell E1 by the total, 100 rolls.

Math Journal Find the theoretical probability of rolling doubles with 2 fair number dice. Compare this theoretical probability with the experimental probability you obtained in the spreadsheet simulation. Are these two values the same?

Answer:

Question 2.
In a bag, there are 9 magenta balls and 1 orange ball. Two balls are randomly drawn, one at a time with replacement.
a) Find the probability of drawing two magenta balls.

P(M, M) = P(M) • P(M)
= •
=
The probability drawing two magenta balls is .
Answer:

P(M) = 9/10
P(M, M) = P(M) • P(M)
= 9/10 • 9/10
= 81/100
The probability drawing two magenta balls is 81/100.

b) Find the probability of drawing an orange ball followed by a magenta ball.
P(O, M) = P(O) • P(M)
= •
=
The probability drawing an orange ball followed by a magenta ball is .
Answer:
P(O) = 1/10
P(M) = 9/10
P(O, M) = P(O) • P(M)
= 1/10 • 9/10
= 9/100
The probability drawing an orange ball followed by a magenta ball is .

c) Find the probability of drawing both orange balls.
P(O, O) = P(O) • P(O)
= •
=
The probability drawing both orange balls is .
Answer:
P(O) = 1/10
P(O, O) = P(O) • P(O)
= 1/10 • 1/10
= 1/100
The probability drawing both orange balls is 1/100.

Question 3.
On weekends, Carli either jogs (J) or plays tennis (T) each day, but never both. The probability of her playing tennis is 0.75.
a) Find the probability that Carli jogs on both days.
Because J and T are complementary,
P(J) = 1 – P(T)
= 1 –
=

P(J, J) = P(J) • P(J)
P(J, J) = •
=
The probability that Carli jogs both days is .
Answer:
P(J) = 1 – P(T)
= 1 – 0.75
= 0.25

P(J, J) = P(J) • P(J)
P(J, J) = 0.25 • 0.25
= 0.0625
The probability that Carli jogs both days is 0.0625.

b) Find the probability that Carli jogs on exactly one of the days.
Using the addition rule of probability:
P(J, J) + P(T, J) = P(J) • P(T) + P(T) • P(J)
= • + •
=
The probability that Carli jogs on exactly one of the days is .
Answer:
Using the addition rule of probability:
P(J, J) + P(T, J) = P(J) • P(T) + P(T) • P(J)
= 0.25 • 0.75 + 0.75 • 0.25
= 0.125 + 0.125
= 0.25
The probability that Carli jogs on exactly one of the days is 0.25.

Math in Focus Course 3B Practice 11.3 Answer Key

Draw a tree diagram to represent each compound event.

Question 1.
Tossing a fair coin followed by drawing a marble from a bag of 3 marbles:
1 yellow, 1 green, and 1 blue.
Answer:

H represents heads
T represents  tails
Y represents yellow
G represents green
B represents blue

Question 2.
Drawing two balls randomly with replacement from a bag with 1 green ball and 1 purple ball.
Answer:

B represents ball
G represents green ball
P represents purple ball

Question 3.
Drawing a ball randomly from a bag containing 1 red ball and 1 blue ball, followed by tossing a fair six-sided number die.
Answer:

R represents red ball
B represents blue ball

Question 4.
Tossing a fair coin twice.
Answer:
If we toss a fair coin twice, we have the following possible outcomes, or events: {(H,H), (H,T),(T,H), (T,T). The total number of possible outcomes is therefore 4 and the number of outcomes where the result is two heads is 1. The probability of getting two heads in tossing a fair coin twice is therefore 1/4.

Question 5.
Reading or playing on each day of a weekend.
Answer:

R represents reading
P represents playing

Question 6.
On time or tardy for school for two consecutive days.
Answer:

Solve. Show your work.

Question 7.
Mindy is playing a game that uses the spinner shown below and a fair coin. An outcome of 3 on the spinner and heads on the coin wins the game.

a) Draw a tree diagram to represent the possible outcomes of this game.
Answer:

H represents Heads
T represents Tails

b) Find the probability of winning the game in one try.
Answer:
1/2 × 1/3 = 1/6
Thus the probability of winning the game in one try is 1/6.

c) Find the probability of losing the game in one try.
Answer:
1/2 + 1/3
= 3/6 + 2/6
= 5/6
Thus the probability of losing the game in one try is 5/6.

Question 8.
There are 2 blue balls and 4 yellow balls in a bag. A ball is randomly drawn from the bag, and it is replaced before a second ball is randomly drawn.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

b) Find the probability that a yellow ball is drawn first, followed by another yellow ball.
Answer:
The probability of obtaining 2 yellow balls is 4/9.
Explanation:
P(Y,Y) = P(Y) x P(Y)
= 4/6 x 4/6
= 16/36
= 4/9

c) Find the probability that a yellow ball is drawn after a blue ball is drawn first.
Answer:
The probability of picking a blue than a yellow ball is 2/9.
Explanation:
P(B,Y) = P(B) x P(Y)
= 2/6 x 4/6
= 8/36
= 2/9

Question 9.
Jasmine has 3 blue pens and 2 green pens in her pencil case. She randomly selects a pen from her pencil case, and replaces it before she randomly selects again.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

B represents blue
G represents green

b) Find the probability that she selects 2 blue pens.
Answer:
P(B) =3/5
P(B) × P(B) = 3/5 × 3/5 = 9/25

c) Find the probability that she selects 2 green pens.
Answer:
P(G) = 2/5
P(G) × P(G) = 2/5 × 2/5 = 4/25
Thus the probability that she selects 2 green pens is 4/25

d) Find the probability that she selects 2 pens of the same color.
Answer:
P(B, B) + P(G, G) = 9/25 + 4/25 = 13/25
Therefore the probability that she selects 2 pens of the same color is 13/25

Question 10.
Henry has 4 fiction books, 6 non-fiction books, and 1 Spanish book on his bookshelf. He randomly selects two books with replacement.

a) Draw a tree diagram to represent the possible outcomes.
Answer:

b) Find the probability that he selects a fiction book twice.
Answer:
The probability of obtaining 2 fictional books is 16/121 or around 0.13
Explanation:
P(F,F) = P(F) x P(F)
= 4/11 x 4/11
= 16/121
~ 0.13

c) Find the probability that he first selects a non-fiction book, and then a Spanish book.
Answer:
The probability of obtaining a non-fiction book and a spanish book is 6/121 or around 0.05
Explanation:
P(N,S) = P(N) x P(S)
= 6/11 x 1/11
= 6/121
~ 0.05

d) Find the probability that he first selects a fiction book, and then a non-fiction book.
Answer:
The probability of obtaining a fiction book and a non-fiction book is 24/121 or around 0.20
P(F,N) = P(F) x P(N)
= 4/11 x 6/11
= 24/121
~ 0.20

Question 11.
Andy tosses a fair six-sided number die twice. What is the probability of tossing an even number on the first toss and a prime number on the second toss?

Answer:

There are 36 total possible outcomes and 12 of them are favorable outcomes
P(E,P) = favorable outcomes/total outcomes
= 12/36
=1/3
Therefore, the probability of obtaining an even number and a prime number on the 2 tosses is 1/3.

Question 12.
The probability that Fiona wakes up before 8 A.M. when she does not need to set her alarm is \(\frac{4}{10}\). On any two consecutive days that Fiona does not need to set her alarm, what is the probability of her waking up before 8 A.M. for at least one of the days?
Answer:
The probability of waking up earlier than 8 AM at least once is 0.64.
Explanation:
P(E) + P(L) = 1
P(L) = 1 – P(E)
P(L) = 1 – 2/5
P(L) = 3/5
Therefore, the probability of that she will wakeup late is 3/5.
The probability of Fiona waking up late two days consecutively is 0.36
P(L,L) = P(L) x P(L)
= 3/5 x 3/5
= 9/25
= 0.36
The probability of waking up earlier than 8 AM at least once
A = 1 – P(L,L)
A = 1 – 0.36
A = 0.64

Question 13.
A globe is spinning on a globe stand. The globe surface is painted with 30% yellow, 10% green, and the rest is painted blue. Two times Danny randomly points to a spot on the globe while it spins. The color he points to each time is recorded.
a) What is the probability that he points to the same color on both spins?
Answer:
The probability of getting to the same color on both spins is 0.46
Explanation:
P(M) = P(Y,Y) + P(G,G) + P(B,B)
= P(Y) x P(Y) + P(G) x P(G) + P(B) x P(B)
= 0.3 x 0.3 + 0.1 x 0.1 + 0.6 x 0.6
= 0.09 + 0.01 + 0.36
= 0.46

b) What is the probability that he points to yellow at least one time?

Answer:
The probability of selecting yellow at least once is 0.51
Explanation:
P(NY) + P(Y) = 1
P(NY) = 1 – P(Y)
P(NY) = 1 – 0.3
P(NY) = 0.7
Therefore, the probability of not selecting yellow is 0.7
P(NY,NY) = P(NY) x P(NY)
P(NY,NY) = 0.7 x 0.7
P(NY,NY) = 0.49
Therefore, the probability of selecting twice is 0.49
P(A) + P(NY, NY) = 1
P(A) = 1 – 0.49
P(A) = 0.51
Therefore, the probability of selecting yellow at least once is 0.51.

Question 14.
Math Journal Sally thinks that for two independent events, because the occurrence of one event will not have any impact on the probability of the other event, they are also mutually exclusive. Do you agree with her? Explain your reasoning using an example.
Answer:
In an independent event, the existence of one does not affect the others. For example, eating chocolates and watching TV as they both are not mutually exclusive. Mutually exclusive events are two (or more) events that cannot be done at the same time. Thus independent events are not mutually exclusive.

Question 15.
A game is designed so that a player wins when the game piece lands on the letter A. The game piece begins on letter G. A fair six-sided number die is tossed. If the number tossed is odd, the game piece moves one step counterclockwise. If the number tossed is even, the game piece moves one step clockwise.
a) What is the probability that a player will win after tossing the number die once?
Answer:
Winning with one roll of die means getting an even number, say P(E). A number die has 6 possible results and 3 of these are even so the probability of getting even number die is the same as the probability of winning in one roll of the number die, which is 3/6 or 1/2.

b) What is the probability that a player will win after tossing the number die twice?

Answer:
The probability of arriving on A with two rolls of the dice is 1/4.
Explanation:
P(O,O) = P(O) x P(O)
P(O,O) = 1/2 x 1/2
P(O,O) = 1/4
Therefore, the probability of arriving on A with two rolls of the dice is 1/4.

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Lesson 11.2 Probability of Compound Events detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Lesson 11.2 Answer Key Probability of Compound Events

Math in Focus Grade 8 Chapter 11 Lesson 11.2 Guided Practice Answer Key

Use a possibility diagram to find each probability.

Question 1.
Two fair four-sided number dice, each numbered 1 to 4 are rolled together. The result recorded is the number facing down. Find the probability that the product of the two numbers is divisible by 2.
Answer:
Given,
Two fair four-sided number dice, each numbered 1 to 4 are rolled together.
The result recorded is the number facing down.

3/4

Question 2.
One colored disc is randomly drawn from each of two bags. Both bags each have 5 colored discs: 1 red, 1 green, 1 blue, 1 yellow, and 1 white. Find the probability of drawing a blue or yellow disc.
Answer:
Given,
One colored disc is randomly drawn from each of two bags. Both bags each have 5 colored discs: 1 red, 1 green, 1 blue, 1 yellow, and 1 white.

R represents red
G represents green
B represents blue
Y represents yellow
W represents white
16/25

Question 3.
A box has 1 black, 1 green, 1 red, and 1 yellow marble. Another box has 1 white, 1 green, and 1 red marble. A marble is taken at random from each box. Find the probability that a red marble is not drawn.
Answer:
Given,
A box has 1 black, 1 green, 1 red, and 1 yellow marble.
Another box has 1 white, 1 green, and 1 red marble.
Marble is taken at random from each box.

B represents black
R represents red
G represents green
Y represents yellow
W represents white
1/2

Solve. Show your work.

Question 4.
Three fair coins are tossed together.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

H represents Heads
T represents tails

b) Using your answer in a), find the probability of getting all heads.
Answer:
Total outcomes = 24
Number of getting all heads = 3
3/24 = 1/8
The probability of getting all heads = 1/8

c) Using your answer in a), find the probability of getting at least two tails.
Answer:
Total outcomes = 24
1/2
The probability of getting at least two tails = 1/2

Math in Focus Course 3B Practice 11.2 Answer Key

Solve. Show your work

Question 1.
A bag contains 2 blue balls and 1 red ball. Winnie randomly draws a ball from the bag and replaces it before she draws a second ball. Use a possibility diagram to find the probability that the balls drawn are different colors.
Answer:
Given,
A bag contains 2 blue balls and 1 red ball = 3
P(red) = 1/3
P(blue) = 2/3
possible outcomes = 4
Thus the probability that the balls drawn are different colors = 4/6

Question 2.
A letter is randomly chosen from the word FOOD, followed by randomly choosing a letter from the word DOG. Draw a tree diagram to find the probability that both letters chosen are the same.
Answer:

There are 12 possible outcomes and 3 among them are favorable outcomes, then
Probability = favorable outcomes/total outcomes
= 3/12
= 1/4
Therefore, the probability of getting two similar letters is 1/4.

Question 3.
Three pebbles are placed in a bag: 1 blue, 1 green, and 1 yellow. First a pebble is randomly drawn from the bag. Then a fair four-sided number die labeled from 1 to 4 is rolled. The result recorded is the number facing down. Use a possibility diagram to find the probability of drawing a yellow pebble and getting a 4.
Answer:

As there are 12 possible outcomes and 1 favorable outcome, then
Probability = favorable outcomes/total outcomes
= 1/12
Therefore, the probability of getting (4,y) is 1/12.

Question 4.
Thomas rolled a fair six-sided number die and a fair four-sided number die labeled 1 to 4 together. Use a possibility diagram to find the probability of rolling the number 3 on both.

Answer:

As there are total of 24 possible outcomes and 1 favorable outcome, then
Probability = favorable outcome/total outcomes
= 1/24
Therefore, the probability of getting (3,3) is 1/24.

Question 5.
At a bike shop, there are 3 bikes with 20-speed gears and 2 bikes with 18-speed gears. The bike shop also sells 1 blue helmet and 1 yellow helmet. Use a possibility diagram to find the probability of getting an 18-speed bicycle and a blue helmet if randomly selecting one bike and one helmet from among these.

Answer:

As there are 10 possible outcomes and 2 of them are favorable outcomes, then
Probability = favorable outcomes/total outcomes
= 2/10
= 1/5
Therefore, the probability of getting (B,18) is 1/5.

Question 6.
Susan randomly draws a card from three number cards: 1, 3, and 6. After replacing the card, Susan randomly draws another number card. The product of the two numbers drawn is recorded.

a) Use a possibility diagram to represent the possible outcomes.
Answer:

b) Using your answer in a), what is the probability of forming a number larger than 10 but less than 30?
Answer:

There are 9 possible outcomes and 2 of them are favorable outcomes,then
Probability(>10<30) = favorable outcomes/total outcomes
= 2/9
Therefore, the probability of getting greater than 10 and less than 30 is 2/9.

Question 7.
Jane and Jill watch television together for 2 hours. Jane selects the channel for the first hour, and Jill selects the channel for the second hour. Jane’s remote control randomly selects from Channels A, B, and C. Jill’s remote control randomly selects from Channels C, D, and E.
a) Use a possibility diagram to represent the possible outcomes for the channels they watch on television for 2 hours.
Answer:

b) Using your answer in a), what is the probability of watching the same channel both hours?
Answer:

As there are 9 possible outcomes and 1 favorable outcome, then
Probability = favorable outcomes/total outcomes
= 1/9

Question 8.
A color disc is randomly drawn from a bag that contains the following discs.

After a disc is drawn, a fair coin is tossed. Use a possibility diagram to find the probability of drawing a red disc and landing on heads.
Answer:

As there are 10 possible outcomes and 2 of them are favorable outcomes, then
Probability = favorable outcomes/total outcomes
= 2/10
= 1/5
Therefore, the probability of getting a red disc and head is 1/5.

Question 9.
Karen tosses a fair coin three times. Draw a tree diagram to find the probability of getting the same result in all three tosses.
Answer:

As there are total of 8 possible outcomes and 2 favorable outcomes, then
Probability = favorable outcomes/total outcomes
= 2/8
= 1/4
Therefore, the probability of acquiring a (H,H’,H”) and (T,T’,T”) is 1/4.

Question 10.
Mrs. Bridget’s recipes require her to put in some fine maize flour in bowl 1, followed by wheat flour in bowl 2, and rice flour in bowl 3. However, the jars of flour are not labeled, so she randomly guesses which flour to put in which bowl.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

b) Using your answer in a), find the probability of getting the correct flour in the correct order.

Answer:
There are total of 6 possible outcomes and 1 favorable outcomes, then
Probability = favorable outcomes/total outcomes
= 1/6
Therefore, the probability of selecting a cornmeal flour and low fat milk is 1/6.

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Lesson 11.1 Compound Events detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Lesson 11.1 Answer Key Compound Events

Math in Focus Grade 8 Chapter 11 Lesson 11.1 Guided Practice Answer Key

Tell whether the outcomes described are from a simple or compound event. If it is a compound event, identify the simple events that form the compound event.

Question 1.
Obtaining two heads when two coins are tossed
Answer:
There are two simple events each consisting of tossing a coin.
Thus it is a compound event

Question 2.
Winning a football game
Answer:
Simple event

Question 3.
Getting a number less than 4 or getting a number greater than 5 when a fair six-sided number die is rolled
Answer: Simple event

Question 4.
Rolling two fair six-sided number dice and obtaining a sum of 10 from the throws
Answer: There are two simple events each consisting of rolling a fair number die.

Represent and tell the number of possible outcomes for each compound event described.

Question 5.
Two fair coins are tossed together.
Answer:
The possibilities of tossing two coins are {(HH), (TT), (H,T), (T, H)}
Thus there are 4 outcomes of tossing two fair coins.

Question 6.
The results of rolling two fair six-sided number dice are multiplied.
Answer:
(1, 1), (1, 2) (1, 3), (1, 4) (1, 5) (1, 6)
(2, 1), (2, 2) (2, 3), (2, 4) (2, 5) (2, 6)
(3, 1), (3, 2) (3, 3), (3, 4) (3, 5) (3, 6)
(4, 1), (4, 2) (4, 3), (, 4) (4, 5) (4, 6)
(5, 1), (5, 2) (5, 3), (5, 4) (5, 5) (2, 6)
(6, 1), (6, 2) (6, 3), (6, 4) (6, 5) (6, 6)
The possible outcomes are 36.

Question 7.
A fair six-sided number die and a fair four-sided number die labeled 1 to 4 are rolled. The results that face down on both number dice are recorded.

Answer:

For each compound event, draw a tree diagram to represent the possible outcomes. Then tell the number of possible outcomes.

Question 8.
Joshua has two bags. In the first bag, there are 2 blue beads and 1 green bead. In the second bag, there are 3 lettered cards with the letters P, Q, and R. Joshua randomly takes an item from the first bag, and then from the second bag.
Answer:

The possible outcomes are 9
Where,
B represents blue
G represents Green
P represents letter P
Q represents letter Q
R represents letter R

Question 9.
A fair coin is tossed then a fair four-sided color die with faces painted yellow, green, blue, and black is rolled. The color facing down is the result recorded.
Answer:

The possible outcomes are 8.
Where,
H represents heads
T represents tails
Y represents yellow
G represents green
B represents blue
B represents black

Math in Focus Course 3B Practice 11.1 Answer Key

Tell whether each statement is True or False.

Question 1.
Selecting letter A from the word PROBABILITY is a compound event.
Answer:
P – 1
R – 1
O – 1
B – 2
A – 1
I – 2
L – 1
T – 1
Y – 1
So, the statement is false.

Question 2.
Selecting letter B from the word BASEBALL and ABLE is a simple event.
Answer:
B – 2
A – 2
S – 1
E – 1
L – 2
The statement is false
ABLE is a simple event

Question 3.
Tossing a fair six-sided number die to get either an even number or a five is a compound event.
Answer:
Rolling an even number: {2, 4, 6}
Rolling a 5: {5}
The statement “Tossing a fair six-sided number die to get either an even number or a five is a compound event” is false.

Question 4.
Umberto has 3 red cards and 4 blue cards. Drawing two red cards in a row, without replacing the first card before drawing the second card, is a compound event.
Answer: true

Tell whether the outcomes described are from a simple or compound event. If it is a compound event, identify the simple events that form the compound event.

Question 5.
Getting a 6 when a fair six-sided number die is rolled.
Answer:
Rolling a 6: {6}
A simple event is an event with only one outcome.

Question 6.
Rolling three fair six-sided number dice and obtaining a sum of 18 from the throws.
Answer:
There are 3 possible outcomes.
A compound event is an event with more than one outcome.

Question 7.
Getting an eighteen when a fair twenty-sided number die is rolled.

Answer:
There is one possible outcome thus it is a simple event.

Question 8.
Susan has 3 red cards and 4 blue cards. She first draws a blue card. Without replacing the first card, she then draws another blue card.
Answer: Compound event

Solve. Show your work.

Question 9.
In the top drawer, there are two battery operated flash lights, one red and one yellow. In the second drawer, there are three packages of batteries: small, medium, and large. A flashlight and a package of batteries are randomly selected.
a) Use a possibility diagram to represent the possible outcomes.
Answer:

R represents red
Y represents yellow
S represents small
M represents medium
L represents large

b) How many possible outcomes are there?
Answer: 6 outcomes

Question 10.
Two electronic spinners, A and B, are spun by pressing a button. Spinner A has four sectors labeled 1 to 4, while B has three sectors, labeled 1 to 3. Spinner B, due to technical error, will never land on number 2 if spinner A lands on a 4.
a) Use a possibility diagram to represent the possible outcomes.
Answer:
The target of this task is to analyze the events of turning two spinners labelled t 2, 3, and 4, and 1, 2, and 3 then make a diagram that shows all its possibLe outcomes then determine the number of its possible outcomes given a condition of not being able to get (4, 2).

Since there are two spinners A and B, marked from 1 – 4 and 1 – 3 then the illustration for the possible outcomes would be as depicted in the table.

b) How many possible outcomes are there?
Answer:
The possible outcomes are (1, 1), (2, 1), (3, 1), (4, 1), (1, 2), (2, 2), (3, 2), (1, 3), (2, 3), (3, 3). and (4, 3)Therefore, there are 11 possible outcomes.

Question 11.
Winston has two boxes. The first box has 3 black pens and 1 red pen. The second box has 1 green ball and 1 yellow ball. Use a tree diagram to represent the possible outcomes for randomly drawing a pen and a ball. Then tell the number of possible outcomes.

Answer:

8 outcomes
R represents red
Y represents yellow
G represents green
B represents black

Question 12.
Seraphina first tosses a fair six-sided number die. She then tosses a fair coin. Use a tree diagram to represent the possible outcomes.
Answer:
The objective of this task is to make a tree diagram for the possibLe outcomes for tossing a number die and flipping a coin.

Observe that rolling a die has 6 possible outcomes, then start the tree diagram with 6 branches. Now, at the end of these branches put the possible outcomes: 1, 2, 3, 4, 5 and 6. Next in tossing a coin, since there are 2 possible outcomes then for each number, make another 2 branches. At the end of these branches put the possibLe outcomes: H for head and T for tail. Finally, conclude the possible results. The tree diagram for this event would be as depicted below.

Question 13.
A game was designed such that a participant needs to accomplish 2 rounds to be considered the overall winner. The first round is to roll a 4 from a fair four-sided number die labeled 1 to 4. The result recorded is the number facing down. The second round is to randomly draw a red ball from a box of 2 differently colored balls.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

R represents red
R’ represents not red

b) How many possible outcomes are there?
Answer: 8 outcomes

c) Math Journal If the participant first draws the colored ball and then rolls the four-sided number die, will the number of possible outcomes be the same? Use a diagram to explain your reasoning.
Answer: Yes, there are still 8 outcomes

R represents red
R’ represents not red

Question 14.
Zoe first rolls a fair four-sided number die labeled 1 to 4. Then she rolls another fair four-sided number die labeled 2 to 5. The result recorded is the number facing down.

a) Use a possibility diagram to find the number of favorable outcomes for an odd sum.
Answer:
Note that there are 4 possible outcomes for born toss of the die, add these results, then the outcomes in tossing a die twice and adding their result would be as depicted in the table. Mark the sum that are odd.

Observe that there are 16 possib[e outcomes and 8 are odd then

Therefore, the probability of obtaining an odd sum is \(\frac{1}{2}\).

b) Use a possibility diagram to find the number of favorable outcomes for a difference greater than 2.
Answer:
Since there are 4 possible outcomes for both toss of the die, subtract these results, then the outcomes in tossing a die twice and subtracting their result would be as depicted in the table. Mark the difference that are bigger than 2.

Notice that there are 16 possible outcomes and only 1 is larger than 2 then

Therefore, the probability of obtaining a difference greater than 2 is \(\frac{1}{16}\).

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Probability detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Answer Key Probability

Math in Focus Grade 8 Chapter 11 Quick Check Answer Key

Solve. Show your work.

A box has 2 black balls, 7 red balls, and 3 green balls. A ball is randomly chosen from the box.

Question 1.
What is the probability of choosing a green ball?
Answer:
Given,
A box has 2 black balls, 7 red balls, and 3 green balls.
Total number of balls = 2 + 7 + 3 = 12
We have to find the probability of choosing a green ball.
Number of green balls = 3
Probability of choosing green balls = 3/12 = 1/4
Thus the Probability of choosing green balls is 1/4.

Question 2.
What is the probability of choosing a black ball?
Answer:
Given,
A box has 2 black balls, 7 red balls, and 3 green balls.
Total number of balls = 2 + 7 + 3 = 12
We have to find the probability of choosing a black ball.
Number of green balls = 2
Probability of choosing green balls = 2/12 = 1/6
Thus the Probability of choosing green balls is 1/6.

Question 3.
What is the probability of choosing a blue ball?
Answer:
Given,
A box has 2 black balls, 7 red balls, and 3 green balls.
Total number of balls = 2 + 7 + 3 = 12
We have to find the probability of choosing a black ball.
Number of blue balls = 0
Thus the Probability of choosing blue balls is 0.

Question 4.
What is the probability of choosing a ball that is not red?
Answer:
A box has 2 black balls, 7 red balls, and 3 green balls.
Total number of balls = 2 + 7 + 3 = 12
We have to find the probability of choosing a ball that is not red.
Number of red balls = 7
The number of balls that are not red is 5
So, the probability of choosing a ball that is not red is 5/12.

Question 5.
What is the probability of choosing a red or green ball?
Answer:
Given that,
A box has 2 black balls, 7 red balls, and 3 green balls.
Total number of balls = 2 + 7 + 3 = 12
Find the probability of choosing a red or green ball.
Number of green balls = 2
Probability of choosing green balls = 2/12 = 1/6
Thus the Probability of choosing green balls is 1/6.
Number of red balls = 7
Probability of choosing green balls = 7/12
The probability of choosing a red or green ball is 9/12 = 3/4.

Tell whether the events X and Y are mutually exclusive events.

Question 6.
A fair coin and a fair six-sided number die are tossed. X is the event that a head is obtained. Y is the event that a six is obtained.
Answer:
Given,
A fair coin and a fair six-sided number die are tossed.
X is the event that a head is obtained.
The probability of getting heads is p(A) = 1
Y is the event that a six is obtained.
The probability of getting six is p(B) = 1
Yes the events X and Y are not mutually exclusive events.

Question 7.
A fair six-sided number die is rolled. X is the event of obtaining a three. Y is the event of obtaining a five.
Answer:
A fair six-sided number die is rolled.
X is the event of obtaining a three.
Y is the event of obtaining a five.
Events 3 and 5 are mutually exclusive because we cannot get 3 and 5 at the same time.

Question 8.
Two fair six-sided number dice are tossed. X is the event that the sum of the score is six. Y is the event that the sum of the score is 10.
Answer:
Given,
Two fair six-sided number dice are tossed.
X is the event that the sum of the score is six.
Y is the event that the sum of the score is 10.
Total number of possible results from two six-sided dice is 6 × 6 = 36.
The possibility of the sum of the score is six is (1, 5) (2, 4) (3, 3) (4, 1) (5, 1) = 5/36
The sum of the score is 10 is (4, 6) (5, 5) (6, 4) = 3/36 = 1/12
Thus they are mutually exclusive events.

Question 9.
X is the event consisting of the factors of 24. Y is the event consisting of multiples of 6 less than 20.
Answer:
Factors of 24 is (1, 24), (2, 12) (3, 8) and (4, 6).
Multiples of 6 less than 20 is 6, 12, 18.
Thus they are mutually exclusive events.

Go through the Math in Focus Grade 4 Workbook Answer Key Chapter 1 Place Value of Whole Numbers to finish your assignments.

Math in Focus Grade 4 Chapter 1 Answer Key Place Value of Whole Numbers

Math Journal

Question 1.
Kim wrote these statements about the three numbers shown here.
Do you agree? Explain why or why not.
3,869 is less than 85,945.
85,691 is greater than 85,945.

Answer:
The above-given statements are:
1. 3,869 is less than 85,945.
2. 85,691 is greater than 85,945.
The first statement is correct.
The second statement is wrong.
The first statement is correct because we are comparing numbers with their place values.
In Mathematics, place value charts help us to make sure that the digits are in the correct places. To identify the positional values of numbers accurately, first, write the digits in the place value chart and then write the numbers in the usual and the standard form.
Rules for comparing numbers:
There are certain rules, based on which it becomes easier to compare numbers. These rules are:
* Numbers with more digits
* Numbers starting with a larger digit
Rule 1: Numbers with more digits
– When we compare numbers, then check if both the numbers are having the same number of digits or not. If a number has more digits, then it is greater than the other number.
Examples:
– 33 > 3
– 400 > 39
– 5555 > 555
– 10000 > 9999
Rule 2: Numbers starting with a larger digit
This rule is applicable when two numbers are having the same number of digits. In such cases, we need to check the digit at the leftmost place, whichever is greater. Therefore, the number with a greater digit at the leftmost place of the number is greater than the other number.
Examples:
* 323>232 [323 is greater than 232]
* 343<434 [343 is less than 434]
Thus, in the above examples, we can see that, when we compare the two numbers, though the number of digits is the same, one number is greater than/less than the other number.
Now according to the above rules, we check statement 1:
1. 3,869 is less than 85,945.
– In this statement, two numbers are given. The first number is having fewer digits than the second number so here rule 1 is applicable.
absolutely, 3,869 < 85,945
This can be written in another way:
85,945 > 3,869.
Now we check for statement 2:
2. 85,691 is greater than 85,945.
Here the same number of digits are having so rule 2 is applicable.
– 85,691 and 85,945 both the numbers have an equal number of digits, therefore, we will compare the left-most digit of both the numbers.
– As we can see the first two digits are the same for both the numbers, thus we need to compare the next left-most digit of both numbers.
6 < 9
Here it is a wrong statement because 6 is not greater than 9. So the second statement is wrong.

Question 2.
Sam continued this number pattern.
Do you agree? Explain why or why not.

Answer:

For number patterns there are certain rules:
To create a complete pattern, there are a set of rules to be considered. To apply the rule, we need to understand the nature of the sequence and the difference between the two successive numbers. It takes some amount of guesswork and checking to see whether the rule works throughout the whole series.
There are two basic divisions to find out the rules in number patterns:
* When the numbers in the given pattern get larger, they are said to be in ascending order. These patterns usually involve addition or multiplication.
* When the numbers in the given pattern get smaller, they are said to be in descending order. These patterns usually involve subtraction or division.
So the number pattern is correct.

Question 3.
Read the example. Then write your own 5-digit number and clues. Ask a friend or family member to solve your puzzle.

Example
45,870
1. The digit 5 is in the thousands place.
2. The value of the digit 7 is 70.
3. The digit in the hundreds place is 10 – 2.
4. The digit in the ten thousands place is 1 less than the digit in the thousands place.
5. The digit in the ones place is 0.

Number: _______
Clues: ______
Answer:
– According to the clues, the number is 15,870
– Now in line number 4, the digit in the ten thousand places is 1 less than hundreds of places. So, 1 < 5. 5 is in the thousand places so I placed 5 in the thousands place.
– In line 3 the hundreds place is 8. In the above-example 10-2 is given which is 8. So I placed 8 in the hundreds place.
– In line 2 the number 70 is given, 70 is nothing but 7 tens so I placed 7 in the tens place.
– In line 5 directly given that one’s place is 0.

Put on Your Thinking Cap!

Challenging Practice

Complete.

A 5-digit number is made up of different digits that are all odd numbers.

Question 1.
What is the greatest possible number? ______
Answer: 9,99,99
Explanation:
The largest 5 digit number that you can have when all digits are odd would be 99999.
that’s with no restrictions.
any other number you choose would have to be smaller.
the restrictions are:
the ten thousand digits are 3 times the one’s digit.
since the ten-thousands digit is fixed at 9, then the one’s digit has to be 3 because 3 * 3 = 9.
the thousands digit is 2 more than the hundreds digit.
since the thousands digit is fixed at 9, then the hundreds digit has to be 7 because 7 + 2 = 9.
that makes your number equal to 99793.
any other arrangement would force the ten-thousands digit to be something less than 9 or the thousands digit to be less than 9 which would result in a smaller number.

Question 2.
What is the value of the digit in the hundreds place? _____
Answer: 9
Explanation:
The largest 5 digit number that you can have when all digits are odd would be 99999.
The place values of 99999 are:
9 is in the hundred thousand place.
9 is in the ten thousand place
9 is in the thousand place
9 is in the hundreds place
9 is in the tens place
9 is in the one’s place.

Continue the pattern.

Question 3.

Answer: 487  502
Explanation:
A list of numbers that follow a certain sequence is known as patterns or number patterns. The different types of number patterns are algebraic or arithmetic patterns, geometric patterns, Fibonacci patterns and so on.
The above-given numbers are in the arithmetic pattern.
The arithmetic pattern is also known as the algebraic pattern. In an arithmetic pattern, the sequences are based on the addition or subtraction of the terms. If two or more terms in the sequence are given, we can use addition or subtraction to find the arithmetic pattern.
The numbers are 412  427  442  457  472  –  -. Now, we need to find the missing term in the sequence.
Here, we can use the addition process to figure out the missing terms in the patterns.
In the pattern, the rule used is “Add 15 to the previous term to get the next term”.
In the numbers given above, take the second term (427). If we add “15” to the second term (427), we get the third term 442.
Similarly, we can find the unknown terms in the sequence.
First missing term: The previous term is 472. Therefore, 472+15 = 487.
Second missing term: The previous term is 487. So, 487+15 = 502
Hence, the complete arithmetic pattern is 412  427   442   457   472   487   502.

Fill in the blanks.

Question 4.
What is 3 ten thousands + 14 tens + 16 ones? ______
Answer:30,156
Explanation:
In a number, the place (local) value of a non-zero digit is the value of this digit according to its position.
A number can have many digits, and each digit has a special place and value
ten thousand can be written as 10,000
tens can be written as 10
ones can be written as 1
3 x 10,000 = 30,000, 14 x 10 = 140 and 16 x 1 = 16
30 000 + 140 + 16 = 30 156.

Question 5.
7 thousands = _____ hundreds 10 tens
Answer:
7 thousand=7 hundreds 10 tens
This can be written as:
7000=700*10
7 thousands=70 hundreds
7 thousand=700 tens

Answer these questions.

Question 6.
what is the value of the digit 5 in the ? ____
Answer: 5000
Explanation:
In Mathematics, place value charts help us to make sure that the digits are in the correct places. To identify the positional values of numbers accurately, first, write the digits in the place value chart and then write the numbers in the usual and the standard form.
According to the above solution, 5 is in the thousands place.
5 thousands is nothing but 5000
So it can be written as 5000.

Question 7.
what is the value of the digit 5 in the ? ____
Answer:50
Explanation:
In Mathematics, place value charts help us to make sure that the digits are in the correct places. To identify the positional values of numbers accurately, first, write the digits in the place value chart and then write the numbers in the usual and the standard form.
According to the above solution, 5 is in the tens place.
5 tens are nothing but 50
So it can be written as 50.

Question 8.
what is the difference between the answers in Exercises 6 and 7? ____
Answer: Place values.
Explanation:
The place value is the position of each digit in a number. The place value of digits is determined as ones, tens, hundreds, thousands,ten-thousands and so on, based on their position in the number. For example, the place value of 1 in 1002 is thousands, i.e.1000.
The above-given number 75,859
7 is in the ten thousand place
5 is in the thousands of place
8 is in the hundreds place
5 is in the tens place
9 is in the one’s place.

Question 9.
, the difference between the values of the digits in the and the is 8,930. What is the digit in the ?
Answer:9
The above-given number 5 X,278
We need to find out the X.
The difference between and =8,930
Already we know the value in the square that is 7
X-7=8,930
If we subtract 59,278 and 50,348 then we get 8,930
so 9 is in the ten thousand places.
Check which number we subtract from the given number to get 8,930.
If we kept 9 in the X place and subtract with the assumed number that is 50,348 then we get the given number 8,930. Likewise we need to think and process the problem.

Put On Your Thinking cap!

Problem Solving

The button and the button do not work.
Justin wants to enter the number 82,365.
Explain what he can do to key in the number. Give two solutions.

Answer:
The above-given number is 82,365
But the 2 and 6 buttons are not working.
Here given clue we can add or subtract
First solution:
For example, if we add two 1’s means 1+1=2
if we add two 3’s or we can add 5+1=6 and 3+3=6.
Second solution:
If we subtract 5-3=2
if we subtract 7-1=6
These are the chances to enter the right values.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Systems of Linear Equations to score better marks in the exam.

Math in Focus Grade 8 Course 3 A Chapter 5 Answer Key Systems of Linear Equations

Math in Focus Grade 8 Chapter 5 Quick Check Answer Key

Complete the table of values and graph each linear equation.

Question 1.
y = 5x

Answer:
When x = 0, y=0,
x =1 , y = 5, x = 2, y = 10, when x = 3 y =15,

Explanation:

Question 2.
y = -x + 2

Answer:
x =0, y = 2, x= 1, y =1,
x= 2, y = 0, x =3,y =-1,

Explanation:

Solve. Show your work.

Question 3.
Samuel bought 30 books. The hardcover books cost $20 each while the rest,
which are paperbacks, cost $8 each. If he spent a total of $480, how many paperbacks did he buy?
Answer:
Number of paperbacks he buy are 10,

Let x be the no. of hard cover books and y be the no. of paper back books.
x + y = 30,
x = 30 – y,
20x + 8y = 480,
20(30 – y) + 8y = 480,
600 – 20y + 8y = 480,
-12y = – 120,
y = 120/12, y = 10,
x = 30-10 = 20,
No.of hard cover books = x = 20,
No.of paper back books = y = 10.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Lesson 5.1 Introduction to Systems of Linear Equations to score better marks in the exam.

Math in Focus Grade 7 Course 3 A Chapter 5 Lesson 5.1 Answer Key Introduction to Systems of Linear Equations

Math in Focus Grade 8 Chapter 5 Lesson 5.1 Guided Practice Answer Key

Solve the system of linear equations by copying and completing the tables of values.
The values x and y are positive integers.

Question 1.
A bottle of water and a taco cost $3. The cost of 3 bottles of water is $1 more than the cost of a taco.
Let x be the price of a bottle of water and y be the price of a taco in dollars.
The related system of equations and tables of values are:
3x – y = 1
x + y = 3

Answer:
The cost of a bottle of water is $2 and the cost of a taco is $1,

Explanation:

Solve each system of equations by making tables of values, x and y are positive integers.

Question 2.
x + y = 6
x + 2y = 8
Answer:

Explanation:
Solved each system of equations by making tables of values, x and y are positive integers.
For equation x+y = 6, when x =1 then y =5 and when x =2 then y=4,
For equation x +2y =8, when x =1 then y =7/2 and when x = 2 then x =3.

Question 3.
x + y = 8
x – 3y = -8
Answer:

Explanation:
Solved each system of equations by making tables of values, x and y are positive integers.
For equation x+y = 6, when x =1 then y =5 and when x =2 then y=4,
For equation x +2y =8, when x =1 then y =7/2 and when x = 2 then x =3.

For each linear equation, list in a table enough values for x and y to obtain a solution.
Remember that they must be positive integers.

Technology Activity

Materials:

• graphing calculator

Use Tables On A Graphing Calculator To Solve A System Of Equations

Work in pairs.

You can use a graphing calculator to create tables of values and solve systems of equations.
Use the steps below to solve this system:

8x + y = 38
x – 4y = 13

Step 1.
Solve each equation for y in terms of x. Input the two resulting expressions for y into the equation screen.

Caution
Use parentheses around fractional coefficients and the key for negative coefficients

Step 2.
Set the table function to use values of x starting at 0, with increments of 1.

Step 3.
Display the table. It will be in three columns as shown.

Step 4.
Find the row where the two y-values are the same. This y-value and the corresponding x-value will be the solutión to the equations.
The solution to the system of equations is given by x = and y = .

Math Journal How can you tell from the two columns of y-values that there is only one row where the y-values are the same?

Math in Focus Course 3A Practice 5.1 Answer Key

Solve each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 1.
2x + y = 5
x – y = -2
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 2.
x + 2y = 4
x = 2y
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 3.
3x + 2y = 10
5x – 2y = 6
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 4.
x – 2y = -5
x = y
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 5.
2y – x = -2
x + y = 2
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 6.
2x + y = 3
x + y = 1
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 7.
x + 2y = 1
x – 2y = 5
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 8.
2x – y = 5
2y + x = -1
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 9.
2x + y = -1
x + y = 1
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Solve by making a table of values. The values x and y are integers.

Question 10.
A shop sells a party hat at x dollars and a mask at y dollars.
On a particular morning, 10 hats and 20 masks were sold for $30.
In the afternoon, 8 hats and 10 masks were sold for $18. The related system of linear equations is:
10x + 20y =30
8x + 10y = 18
Solve the system of linear equations. Then find the cost of each hat and each mask.
Answer:
The cost of each hat is $1 and each mask is$1,

Explanation:
Given A shop sells a party hat at x dollars and a mask at y dollars.
On a particular morning, 10 hats and 20 masks were sold for $30.
In the afternoon, 8 hats and 10 masks were sold for $18.
The related system of linear equations is:
10x + 20y =30—(1)
8x + 10y = 18—-(2) dividing equation 1 by 10 we get
x +2y = 3,
x = 3-2y substituting in equation 2 we get
8(3-2y) +10y =18,
24 – 16y+10y =18,
24 -6y =18,
6y = 24 -18,
6y =6, y =1 we have x = 3-2X 1 = 3-2 = 1.

Question 11.
Alicia is x years old and her cousin is y years old.
Alicia is 2 times as old as her cousin.
Three years later, their combined age will be 27 years.
The related system of linear equations is:
x = 2y
x + y = 27
Solve the system of linear equations. Then find Alicia’s age and her cousin’s age.
Answer:
Alicia’s age is 18 and her cousin’s age is 9 years old,

Explanation:
GivenAlicia is x years old and her cousin is y years old.
Alicia is 2 times as old as her cousin.
Three years later, their combined age will be 27 years.
The related system of linear equations is: x = 2y,
x + y = 27 solving the equations  2y + y = 27,
3y = 27, y = 27/3 =9, so x = 2X 9 = 18, Alicia’s age is 18 and her
cousin’s age is 9 years old.

Question 12.
Steve and Alex start driving at the same time from Boston to Paterson.
The journey is d kilometers. Steve drives at 100 kilometers per hour and
takes t hours to complete the journey. Alex, who drives at 80 kilometers per hour is
60 kilometers away from Paterson when Steve reaches Paterson.
The related system of linear equations is:
100t = d
80t = d – 60

Solve the system of linear equations by making tables of values.
Then find the distance between Boston and Paterson.
Answer:
300 kilometers is the distance between Boston and paterson,

Explanation:
Given Steve and Alex start driving at the same time from Boston to Paterson.
The journey is d kilometers. Steve drives at 100 kilometers per hour and
takes t hours to complete the journey. Alex, who drives at 80 kilometers per hour is
60 kilometers away from Paterson when Steve reaches Paterson.
The related system of linear equations is:
100t = d
80t = d – 60 substituting 80t = 100t -60,
100t-80t = 60,
20t = 60, t = 60/20, t =3 so distance is 100 X 3= 300 kilometers.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Lesson 5.2 Solving Systems of Linear Equations Using Algebraic Methods to score better marks in the exam.

Math in Focus Grade 7 Course 3 A Chapter 5 Lesson 5.2 Answer Key Solving Systems of Linear Equations Using Algebraic Methods

Math in Focus Grade 8 Chapter 5 Lesson 5.2 Guided Practice Answer Key

Solve each system of linear equations using the elimination method.

Question 1.
2a+ 3b= 29 —Equation 1
2a — b = 17 — Equation 2
Subtract Equation 2 from Equation 1:

Answer:
The solution to the system of linear equations is a = 23/2 and b =6,

Explanation:

Question 2.
2x – y = 2
3x + y = 13
Answer:
The solution to the system of linear equations is  x = 3 and y =4.

Explanation:
Given 2x -y =2 -Equation 1,
3x + y = 13- Equation 2 Adding Equation 2 and 1,
2x – y =2
3x + y =13
5x = 15,
x = 15/5 = 3, substituting x =3 in Equation 1
y = 2x -2= 2 X 3 -2 = 6-2 =4.

Question 3.
x + 6 = 1
x + y = 6
Answer:
The solution to the system of linear equations is x = -5 and y = 11,

Explanation:
Given x + 6 =1 means x = 1-6 = -5,
substituting x = -5 in Equation 2 – x + y  = 6 as x = -5,
-5 + y = 6, y = 6 +5 = 11.
Therefore x = -5 and y = 11.

Solve each system of linear equations using the elimination method.

Question 4.
7m + 2n = -8
2m = 3n – 13
Answer:
The solution to the system of linear equations is m = -2 and n =3,

Explanation:
Given 7m +2n = -8 – Equation 1 and 2m = 3n – 13- Equation 2,
to make elmination we need common values we multiply
Equation 1 by 3 and Equation 2 by 2 so that n becomes common,
3 X(7m +2n )= 3 X -8 = 21m + 6n = -24 and
2 X 2m = 2 X (3n – 13)= 4m = 6n – 26 means 6n = 4m + 26,
so 21m + 4m + 26 = -24, 25m = -24-26,
25m = -50,
m = -50/25 = -2, so substituting m= -2 in 2m = 3n -13,
2(-2)+13 = 3n,
-4+13 = 3n,
3n = 9, n = 9/3 = 3.

Question 5.
3x – 2y = 24
5x + 4y = -4
Answer:
The solution to the system of linear equations is x = 4 and y = -6,

Explanation:
Given Equation 1 as 3x -2y = 24 and Equation 2 as 5x +4y = -4,
to make elmination we need common values we multiply
Equation 1 by 2  so that y can be elminated,
2X(3x-2y =24)= 6x -4y = 48 adding with Equation 2
6x -4y =48
(+)5x +4y = -4
11x = 44,
x = 44/11 = 4, we substitute x = 4 in Equation 1 ,
3 X 4 – 2y = 24,
12- 2y = 24,
12-24 = 2y,
2y = -12, y = -12/2 = -6.

Question 6.
2x + 7y = -32
4x – 5y = 12
Answer:
The solution to the system of linear equations is y = -4 and x = -2,

Explanation:
Given 2x +7y = -32 Equation 1 and 4x -5y = 12 Equation 2,
to make elmination we need common values we multiply
Equation 1 by -2  so that x can be elminated,
-2 X(2x +7y=-32) = -4x – 14y = 64 now we add eqaution 1 with
equation 2 as -4x – 14y = 64
(+) 4x- 5y =12
                                 -19y = 76, y = -76/19 =-4, substituting y =-4
4x -5(-4)=12, 4x +20 = 12, 4x = 12-20 = -8, 4x = -8 , x =-8/4 = -2.

Solve each system of linear equations by using the substitution method.

Question 7.
2x + y = 5 — Equation 1
y = 4x – 7 — Equation 2
Substitute Equation 2 into Equation 1:

Answer:
The solution to the system of linear equations is x =2 and y =1,

Explanation:

Question 8.
4x + 3y = 23 — Equation 1
5x + y = 15 — Equation 2
Use Equation 2 to express y in terms of x:
5x + y = 15

Answer:
The solution to the system of linear equations is x =2 and y = 5,

Explanation:

Question 9.
3x – y = 8
2x + 3y = 9
Answer:
The solution to the system of linear equations is x = 3 and y = 1,

Explanation:
Given 3x – y =8 Equation 1 and 2x +3y = 9 Equation 2,
we can write Equation 1 as y = 3x -8,
substituting y in Equation 2 as
2x +3 (3x -8) = 9,
2x + 9x – 24 =9,
11x = 9+ 24 = 33, x = 33/11 = 3,so y = 3 X 3 – 8 = 9-8 = 1.

Question 10.
7m + 2n = -8
2m = 3n – 13
Answer:
The solution to the system of linear equations is m = -2 and n =3,

Explanation:
Given 7m +2n = -8  Equation 1 and 2m = 3n – 13 Equation 2,
to make elmination we need common values we multiply
Equation 1 by 3 and Equation 2 by 2 so that n can be elminated as
3(7m +2n =-8) = 21m +6n = -24 – equation 3and
2(2m = 3n -13) = 4m = 6n – 26 means 6n = 4m +26 substituting 6n in eqaution 3
21m +4m + 26 = -24,
25m = -24 -26 = -50 , m = -50/25 = -2, now
6n = 4(-2)+ 26 = -8 +26 = 18,
6n = 18, n = 18/6 = 3.

Solve each system of linear equations using elimination method or
substitution method. Explain why you choose each method.

Question 11.
2x + 3y = 29
2x – 17 = y
Answer:
The solution to the system of linear equations is x = 10 and y = 3,
used substitution method as already y is given in Equation 2,

Explanation:
Given 2x+3y = 29 as equation 1 and y already there ,
so we use substitution method as y = 2x -17  in equation 1
2x + 3(2x – 17) = 29,
2x + 6x – 51= 29,
8x = 29 +51= 80,
x = 80/8 = 10,
y = 2 X 10 -17 = 20 -17 = 3.

Question 12.
3a – 2b = 5
2a – 5b = 51
Answer:
The solution to the system of linear equations is a= -7 and b = -13,
Used elmination method because of no common values of a and b,

Explanation:
Given 3a- 2b =5 Equation 1 and 2a – 5b = 51 in Equation 2 in both
equations a or b is not common so we use elmination and we multiply
equation 1 by 2 2(3a-2b =5) = 6a – 4b = 10 and equation 2 by 3
3(2a -5b = 51) = 6a – 15b = 153 subtracting eqaution 2 from eqaution 1
6a-4b =10
(-) -6a +15b =-153
11 b = -143,
b = – 143/11 = -13, substituting in eqaution 1
3a -2(-13)=5,
3a +26 =5,
3a = 5 -26 = -21,
a = -21/3 = -7.

Math in Focus Course 3A Practice 5.2 Answer Key

Solve each system of linear equations using the elimination method.

Question 1.
2j + k = 6
j – k = 8
Answer:
The solution to the system of linear equations is j = 14/3 and k = -10/3,

Explanation:
Given 2j +k =6 equation 1,
j – k = 8 equation 2 adding 1 and 2
2j+j = 6 +8,
3j = 14,
j = 14/3,
k = j – 8 = 14/3 – 8 = 14- 24/3 = -10/3,.

Question 2.
2j + 3k = 11
2j – 5k = 3
Answer:
The solution to the system of linear equations is j =4 and k =1,

Explanation:
Given 2j +3k = 11 – Equation -1, 2j – 5k =3- Equation -2
subtracting eqaution 2 from 1
2j +3k =11
-2j +5k =  -3
8k = 8, k = 8/8 = 1, substuting k =1 in equtation 2
2j – 5 = 3, 2j = 3+5 = 8, j = 8/2 = 4.

Question 3.
3m + n = 30
2m – n = 20
Answer:
The solution to the system of linear equations is m = 10, n =0.

Explanation:
Given 3m + n = 30 Equation 1, 2m -n = 20 Equation 2
adding 1 and 2 we get
3m +n =30
2m – n =20
5m = 50, m = 50/5 = 10, substituting m = 10 in equation2
2 X 10 – n =20,
n = 20 – 20 = 0.

Question 4.
3x – y = 9
2x – y = 7
Answer:
The solution to the system of linear equations is x =2 and y = -3,

Explanation:
Given 3x -y = 9 Equation 1 and 2x -y =7 Equation 2,
subtracting eqaution 2 from 1
3x -y =9
-2x +y =-7
x = 2, substituting x =2 in 2(2)-y = 7,
y = 4 – 7 = -3,

Question 5.
5s – t = 12
3s + t = 12
Answer:
The solution to the system of linear equations is s = 3 and t =3,

Explanation:
Given 5s -t =12 as equation 1 and 3s +t = 12 equation 2 adding1 & 2
5s -t =12,
(+)3s+t =12
8s = 24, s = 24/8 = 3, substituting s = 3 in eqaution 2
t = 12 – 3s,
t = 12 – 3 X 3 = 12-9 = 3.

Question 6.
2b + c = 10
2b – c = 6
Answer:
The solution to the system of linear equations is b = 4 and c = 2,
Explanation:
Given 2b + c = 10 as equation 1 and
2b – c =6 as equation 2 adding euqtion 1 & 2
2b +c =10
(+) 2b -c =6
4b = 16, b = 16/4 = 4, substituting b = 4 in equation 2
2X4 -6 = c,
c = 8-6 = 2.

Question 7.
3m – n = 7
21m + 6n = -29
Answer:
The solution to the system of linear equations is m = 1/3 and n = -6,

Explanation:
Given 3m – n = 7 as equation 1 means n = 3m -7,
21m +6n = -29 is equation 2 substituting n = 3m -7 in  equation 2
21m +6(3m -7) = -29,
21m +18m – 42 = -29,
39m = -29+42,
39m = 13,
m = 13/39 = 1/3, substituting n = 3(1/3)-7 = 1-7 = -6.

Question 8.
7a + b = 10
2a + 3b = -8
Answer:
The solution to the system of linear equations is

Explanation:
Given 7a + b = 10 as equation 1 we get b = 10-7a,
we have 2a + 3b = -8 as equation 2 substituting b = 10-7a in
equation 2 –2a +3(10-7a) =-8,
2a + 30 – 21a = -8,
-19a = -8 -30
a = 38/19= 2, substituting a=2 in b = 10 -7 X2 = 10 – 14 = -4.

Question 9.
2p + 5q = 4
7p + 15q = 9
Answer:
The solution to the system of linear equations is p= -3 and q = 2,

Explanation:
Given 2p +5q = 4 as eqaution 1 and 7p +15 q = 9 as equation 2,
if we multiply eqauation 1 with 3 we get 15 q so that we can easily
subtract from equation 2 so
3 X (2p +5q = 4)= equation 1 becomes
6p + 15q = 12 now we subtract equation 2
(-) 7p -15q = -9
-p = 3, p = -3 substituting p =-3 in 2p+5q = 4,
2(-3)+5q = 4,
5q = 4+6 = 10, q = 10/5 = 2.

Solve each system of linear equations using the substitution method.

Question 10.
2j + k = 3
k = j – 9
Answer:
The solution to the system of linear equations is j = 4 and k = -5,

Explanation:
Given 2j +k = 3 as equation 1 and wehave k = j -9 substituting in
equation 1 2j+(j -9) = 3,
3j = 3 + 9 = 12,
j = 12/3 = 4, therefore k = 4 – 9 = -5.

Question 11.
2h + 3k = 13
h = 2k – 4
Answer:
The solution to the system of linear equations is h =2 and k = 3,

Explanation:
Given 2h +3k = 13 as equation 1 and h = 2k-4
substituting h in equation 1 as
2 (2k-4) +3k = 13,
4k – 8 +3k = 13,
7k = 13+8= 21,
k = 21/7 = 3,So h = 2 X 3 – 4 = 6-4 =2.

Question 12.
3m + b = 23
m – b = 5
Answer:
The solution to the system of linear equations is b = 2 and m =7,

Explanation:
Given 3m+ b = 23 as equation 1 and as m-b = 5 ,
b = m-5 substituting in equation 1  3m + m -5 = 23,
4m = 23+5, 4m = 28, m= 28/4 = 7,
b = 7 – 5= 2.

Question 13.
3h – k = 10
h – k = 2
Answer:
The solution to the system of linear equations is h =4 and k =2,

Explanation:
Given 3h – k =10 as equation 1 and h -k = 2 ,
k = h -2 substituting in eqaution 1 k,
3h -h +2 = 10,
2h = 10-2 = 8,
h = 8/2 = 4 therefore k = h -2 = 4-2 =2.

Question 14.
3s – t = 5
s + 2t = 4
Answer:
The solution to the system of linear equations is s =2 and t = 1,

Explanation:
Given 3s – t = 5 as equation 1 and s +2t = 4 means
s = 4-2t substituting s in equation 1 we get
3(4-2t)- t =5,
12 – 6t – t =5,
12 -7t =5,
7t = 12-5 = 7,
t =7/7 = 1, therefore s = 4 – 2X 1 = 4-2 =2.

Question 15.
2x + y = 20
3x + 4y = 40
Answer:
The solution to the system of linear equations is x = 8 and y = 4,

Explanation:
Given 2x +y = 20 as equation 1 and 3x +4y = 40 equation2,
Equation 1 can be written as  y = 20-2x and
sustituted in equation 2 as 3x +4 (20-2x) = 40,
3x + 80 – 8x = 40,
-5x = 40 -80= -40,
5x = 40,
x = 40/5 = 8, so y = 20 -2 X 8 = 20 -16 =4.

Question 16.
3x + 2y = 0
5x – 2y = 32
Answer:
The solution to the system of linear equations is x = 4 and y = -6,

Explanation:
Given 3x+ 2y = 0 as equation 1 and 5x -2y = 32 as equation 2,
adding equation 1 and 2 we get
3x +2y =0,
5x -2y = 32
8x = 32,
x = 32/8 = 4 substituting x = 4 in equation 1
3(4) +2y = 0,
2y = -12, y = -12/2 = -6.

Question 17.
5x – y = 20
4x + 3y = 16
Answer:
The solution to the system of linear equations is x = 4 and y =0,

Explanation:
Given 5x – y = 20 as equation 1 and 4x +3y = 16 as equation 2
From equation 1 we have y = 5x -20,
substituting y in equation 2
4x + 3(5x -20) = 16,
4x + 15x – 60 = 16,
19x = 16 +60,
19 x = 76,
x = 76/19 = 4, so y = 5 x 4 – 20 = 0.

Question 18.
3p + 4q = 3
\(\frac{1}{2}\) + q = 3p
Answer:
The solution to the system of linear equations is p =\(\frac{1}{3}\) and
q = \(\frac{1}{2}\),

Explanation:
Given 3p +4q = 3 as equation 1 and \(\frac{1}{2}\) + q = 3p as equation 2
q = 3p – (1/2) substituting in equation 1
3p+4(3p – 1/2) =3,
3p +12 p – 2 = 3,
15p = 3+2 = 5,
p = 5/15 = 1/3, now q = 3 X 1/3  -1/2 = 1 – 1/2 = 1/2,
so p =\(\frac{1}{3}\) and q = \(\frac{1}{2}\).

Solve each system of linear equations using the elimination method or
substitution method. Explain why you choose each method.

Question 19.
2x + 7y = 32
4x – 5y = -12
Answer:
The solution to the system of linear equations is x= 2 and y =4,
used elimination method to elminate x,

Explanation:
Given 2x +7y = 32 as equation 1 and we have
4x – 5y = -12 as equation 2 if we multiply equation 1 by 2 and
subtarct equation 2 we can eliminate x
2X(2x +7y) = 64,
4x +14 y = 64
-4x +5y = 12
19y = 76, y = 76/19 = 4, now 4x -5 X 4= -12,
4x – 20 = -12,
4x = 8, x = 8/4 = 2.

Question 20.
3x + 3y = 22
3x – 2y = 7
Answer:
The solution to the system of linear equations is x = 13/3 and
y =3 used elmination method to elminate x,

Explanation:
Given 3x +3y = 22 as equation 1 and 3x – 2y = 7 as
equation 2 subtracting equation 2 from 1 we can eliminate x
3x + 3y = 22,
-3x +2y = -7
5y = 15,  y = 15/5 = 3, substituting y =3 in eqaution 2
3x – 2 X 3 = 7,
3x = 7+6 = 13,
x = 13/3.

Question 21.
7m + 2n = 20
2m = 3n – 5
Answer:
The solution to the system of linear equations is m = 2 and n =3,
used subtitution method,

Explanation:
Given 7m + 2n = 20 as equation 1 and 2m = 3n -5,
so using substitution m = (3n-5)/2 in equation 1,
7 (3n- 5)/2 + 2n = 20
7 (3n -5) + 4n = 40,
21n – 35 + 4n = 40,
25n = 40 +35 = 75,
25n = 75, n = 75/25 = 3,
n = 3, so m = (3 X3 -5)/2 = (9-5)/2 = 4/2 = 2.

Question 22.
3h – 4k = 35
k = 2h – 20
Answer:
The solution to the system of linear equations is h = 15 and
k =10 used substitution method,

Explanation:
Given 3h – 4k = 35 as equation 1 and k =2h – 20,
substituting k in equation 1
3h – 4(2h – 20) = 35,
3h – 6h + 80  = 35,
– 3h + 80 = 35,
3h = 80-35 = 45,
3h = 45, h = 45/3 = 15, so k = 2 X 15 – 20 = 30-20 = 10.

Question 23.
2h + 7k = 32
3h – 2k = -2
Answer:
The solution to the system of linear equations is h = 2 and k = 4,
used elmination method,

Explanation:
Given 2h +7k = 32 as equation 1 and 3h – 2k = -2 as equation 2
multiplying equation 1 by 3 and eqaution 2 by 2 to elminate h,
3(2h+ 7k= 32) becomes 6h + 21 k = 96 and
2(3h – 2k=-2) becomes 6h – 4k = -4 substracting 2 from 1,
6h +21k = 96
-6h +4k = 4
25 k = 100,
k = 100/25 = 4, substituting k =4 in eqaution 2
3h – 2 X 4= -2,
3h = -2 + 8 = 6,
h = 6/3 = 2.

Question 24.
2m+4=3n
5m – 3n = -1
Answer:
The solution to the system of linear equations m = -11 and n -6,
used substitution method,

Explanation:
Given 2m+ 4 = 3n as equation 1 and 5m – 3n = -1 as equation 2,
substituting 3n in equation 2 as
5m -3(2m +4) =-1,
5m – 6m -12 = -1,
-m = -1 + 12 = 11, m = -11, therefore 3n = 2 (-11) + 4,
3n = -22 + 4 = -18,
n = -18/3 = -6.

Solve.

Question 25.
Math Journal Sam solves the following system of linear equations by the elimination method,
without using calculator.

He can multiply the first equation by 3 and the second equation by 2 in order to eliminate x.
Or he can eliminate y by multiplying the first equation by 17 and the second equation by 3.
Which way should Sam choose? Explain.
Answer:
Sam can either choose any of the methods by
using method 1 he can elminate x or
by using method 2 he can elminate y and find solutions,

Explanation:
Given in method 1 sam can multiply the first equation by 3 and the
second equation by 2 in order to eliminate x  and in method 2
he can eliminate y by multiplying the first equation by 17 and
the second equation by 3 therefore Sam can either choose any of the method by
using method 1 he can elminate x or
by using method 2 he can elminate y and find solutions,
Method 1 :
3 X (2x + 3y = 1) and 2 X (3x – 17 y =23)
6x + 9y =3  and
(-)6x – 34y = 46
43 y = -43, y = -43/43, y = -1, substuting y = -1 in 2x +3y = 1,
2x + 3X -1 = 1,
2x – 3 = 1,
2x = 1+3 = 4, x = 4/2 = 2.
Method 2:
17 X (2X +3y = 1) and 3 ( 3x – 17y =23),
34x + 51 y = 17
9x – 51 y = 69
43x = 86, x = 86/43 = 2, x = 2 substituting  2 X 2 + 3y = 1,
4 +3y = 1, 3y = 1-4 = -3, 3y = -3, y = -3/3 = -1,
therefore by both methods we got x =2 and y = -1 ,
Sam can either choose any of the methods.