### All ACT Math Resources

## Example Questions

### Example Question #1 : How To Find The Probability Of An Outcome

If two regular six-sided dice are rolled at the same time, what is the probability that the sum of the their numbers will be prime?

**Possible Answers:**

1/2

1/4

5/12

7/18

**Correct answer:**

5/12

There number of possible outcomes is equal to six times six, or thirty-six. The sum of the two dice must be either 2, 3, 5, 7, or 11. There are 15 out of the 36 outcomes that would result in a sum that is a prime number:

[1,1], [1,2], [1,4], [1,6], [2,1], [2,3], [2,5], [3,2], [3,4], [4,1], [4,3], [5,2], [5,6], [6,1], [6,5]

### Example Question #2 : How To Find The Probability Of An Outcome

Maria is planning the seating for the head table at a college gala. There are eight speakers that will be seated along one side of the table. Richard wants to sit beside Hang, and Maria knows that Thomas and Lily should not be seated together. In how many ways can Maria make up the seating plan?

**Possible Answers:**

**Correct answer:**7200

The simplest way to solve this is to find the number of seating arrangements in which Richard and Hang are seated together and then subtract those in which Thomas and Lily are also seated together. Consider Richard and Hang as a unit. This pair can be arranged with the other six speakers in _{7}*P*_{7} ways. For each of these ways, Hang could be either on Richard’s left or his right. Thus, there are _{7}*P*_{7} × 2 = 10 080 arrangements in which Richard and Hang are seated together. Now also consider Lily and Thomas as a unit. The two pairs can be arranged with the remaining four speakers in _{6}*P*_{6} ways, and the total number of arrangements with each of the pairs together is _{6}*P*_{6} × 2 × 2 = 2880.

Therefore, the number of seating arrangements in which Richard and Hang are adjacent but Thomas and Lily are not is 10 080 − 2880 = 7 200.

### Example Question #3 : How To Find The Probability Of An Outcome

{10, 12, 24, 50, 60, 100, 260, 480, 606, 1000}

What is the probability that a number selected randomly from the set will be divisible by both 4 and 6?

**Possible Answers:**

3/10

3/5

2/5

1/10

1/2

**Correct answer:**

2/5

First, find the numbers that in the set that are divisible by 4. 12, 24, 60, 100, 260, 480, and 1000 are all divisible by 4. Now find the numbers that are divisible by 6. 12, 24, 60, 480, 606 are all divisible by 6. The numbers that are divisible by both 4 and 6 are 12, 24, 60, and 480, or 4 total numbers from the set. So 4 out of the 10 numbers are divisible by 4 and 6. The probablility is 4/10, which reduces to 2/5. The correct answer is 2/5.

### Example Question #4 : How To Find The Probability Of An Outcome

Jacob was 27 years old when his son Mike was born. Mike was 23 years old when his son Sam was born. Sam will celebrate his seventh birthday in 2014. What year was Jacob born?

**Possible Answers:**

1984

2007

1957

1987

1964

**Correct answer:**

1957

If Sam will celebrate his seventh birthday in 2014, then he was born in 2007. So Mike was born 23 years before in 1984 and Jacob was born 27 years before that in 1957.

If you answered 1964, then you did not factor that Sam was 7 years old at the time of calculation.

If you answered 1984, then you found the year that Mike was born, not Jacob.

If you answered 2007, then you found the year that Sam was born, not Jacob.

If you answered 1987, then you just subtracted 27 years old from the day of the party in 2014.

### Example Question #1 : Probability

What is the probability of choosing two consecutive red cards from a standard deck of cards, if replacement is not allowed?

**Possible Answers:**

25/102

51/100

23/25

17/25

61/204

**Correct answer:**

25/102

Probability = what you want ÷ total number

A standard deck of playing cards has 52 cards, with 4 suits and 13 cards in each suit

Choosing two red cards = 26 * 25 = 650

Choosing two cards = 52 * 51 = 2652

So the probabiulity of choosing 2 red cards is 650/2652 = 25/102

If replacement is allowed, then the probability of choosing 2 red cards becomes 676/2704 = 1/4

### Example Question #6 : How To Find The Probability Of An Outcome

If *x* percent of 300 is 120, what is 175% of *x*?

**Possible Answers:**

50

310

70

320

**Correct answer:**

70

120 is 40% of 300, so *x* = 40.

175% of 40 is equal to 70.

### Example Question #7 : How To Find The Probability Of An Outcome

A spinner has 3 equally distributed colors on it: blue, red, and yellow. There is an equal chance for the spinner to land on each color (1/3). What's the probability it will land on blue twice after 2 spins?

**Possible Answers:**

1/2

1/9

1/4

1/3

1/6

**Correct answer:**

1/9

There are 9 possible ways for this outcome to be determined. They are contained in this set

{BB, BR, BY, RB, RR, RY, YB, YR, YY}.

Only one of these outcomes is desired, so the value is 1/9

Another way to solve this is to multiply the probability by the number of instances: thus two consecutive instances of "blue" will have the probability of one instance (1/3) multiplied by one instance (1/3), or 1/9.

### Example Question #8 : How To Find The Probability Of An Outcome

Two dice are rolled. What is the probability of getting two odd numbers on two dice?

**Possible Answers:**

1/8

1/2

1/6

2/3

1/4

**Correct answer:**

1/4

Two independent events that each have a probability of 1/2 of occuring.

1/2 * 1/2 = 1/4

### Example Question #9 : How To Find The Probability Of An Outcome

You are at the drawing for a raffle to win a new car. You bought 8 tickets for the raffle. If a total of 526 tickets were sold, what is the probability that your ticket is drawn?

**Possible Answers:**

0.0152

0.025

0.018

0.008

0.022

**Correct answer:**

0.0152

Probability = # of Tickets You Bought / # of Tickets Sold

Probability = 8/526 = 0.0152

### Example Question #10 : How To Find The Probability Of An Outcome

You roll a six sided die three times. What are the chances that all three rolls are 2?

**Possible Answers:**

1/216

1/12

1/36

1/18

1/6

**Correct answer:**

1/216

Probability of each event = 1 side of die / # of sides = 1/6

Probability for multiple events = P1 * P2 * P3

1/6 x 1/6 x 1/6 = 1/216