Word Problem Practice and Tutorial

Word Problem Practice

Word Problems are common on these types of exam:

Nursing Entrance Exams ( NET, DET, PSB, HOBET, HESI, TEAS)

Armed Services Entrance including ASVAB and AFQT

High School Exams ( CAHSEE, COOPSHSAT, TACHS, BC Provincial, ISEE, HSPT)

 

Example Word Problem and extended solution – see below for more

For a certain board game, two dice are thrown to determine the number of spaces to move.   One player throws the two dice and the same number comes up on each of the dice.  What is the probability that the sum of the two numbers is 9?

A. 0

B. 1/6

C. 2/9

D. 1/2

E. 1/3

Let’s look at a few different methods and steps to solving this problem.

1. Create an Algebra Problem

While you might think that creating an algebra problem is the last thing that you would want to do, it actually can make the problem extremely simple.

Consider what you know about the problem.   You know that both dice are going to roll the same number, but you don’t know what that number is.  Therefore, make the number “x” the unknown variable that you will need to solve for.

Since you have two dice that both would roll the same number, then you have “2x” or “two times x”.  Since the sum of the two dice needs to equal nine, that gives you “2x = 9”.

Solving for x, you should first divide both sides by 2.   This creates 2x/2 = 9/2.  The twos cancel out on the left side and you are have x = 9/2 or x = 4.5

You know that a dice can only roll an integer: 1, 2, 3, 4, 5, or 6, therefore 4.5 is an impossible roll.   An impossible roll means that there is a zero possibility it would occur, making choice A, zero, correct.

2. Run Through the Possibilities for Doubles

You know that you have to have the same number on both dice that you roll.   There are only so many combinations, so quickly run through them all.

You could roll:

Double 1’s = 1 + 1 = 2

Double 2’s = 2 + 2 = 4

Double 3’s = 3 + 3 = 6

Double 4’s = 4 + 4 = 8

Double 5’s = 5 + 5 = 10

Double 6’s = 6 + 6 = 12

Now go through and see which, if any, combinations give you a sum of 9.   As you can see here, there aren’t any.  No combination of doubles gives you a sum of 9, making it a zero probability, and choice A correct.

3. Run Through the Possibilities for Nine

Just as there are only so many possibilities for rolling doubles, there are also only so many possibilities to roll a sum of nine. Quickly calculate all the possibilities, starting with the first die.

If you rolled a 1 with the first die, then the highest you could roll with the second is a 6.   Since 1 + 6 = 7, there is no way that you can roll a sum of 9 if your first die rolls a 1.

If you rolled a 2 with the first die, then the highest you could roll with the second is a 6.   Since 2 + 6 = 8, there is no way that you can roll a sum of 9 if your first die rolls a 2.

If you rolled a 3 with the first die, then you could roll a 6 with your other die and have a sum of 9.   Since 3 + 6 = 9, this is a valid possibility.

If you rolled a 4 with the first die, then you could roll a 5 with your other die and have a sum of 9.   Since 4 + 5 = 9, this is a valid possibility.

If you rolled a 5 with the first die, then you could roll a 4 with your other die and have a sum of 9.   Since 5 + 4 = 9, this is a valid possibility.

If you rolled a 6 with the first die, then you could roll a 3 with your other die and have a sum of 9.   Since 6 + 3 = 9, this is a valid possibility.

Now review all the possibilities that give you a combination of 9.   You have:  3 + 6, 4 + 5, 5 + 4, and 6 + 3.  These are the only combinations that will give you a sum of 9, and none of them are doubles.  Therefore, there is a zero probability that doubles could give you a sum of 9, and choice A is correct.

4. Calculate the Odds

Quickly calculate the odds for just rolling a 9, without setting any restrictions that it has to be through doubles or anything else.   You’ve seen in Method 3 that there are 4 ways that you can roll a sum of 9.  Since you have two dice, each with 6 sides, there are a total of 36 different combinations that you could roll (6*6 = 36).  Four of those thirty-six possibilities give you a sum of 9.  Four possibilities of rolling a 9 out of thirty-six total possibilities = 4/36 = 1/9.  So that means there is a 1/9 chance that would roll a 9, without any restrictions.  Once you add restrictions, such as having to roll doubles, then your odds are guaranteed to go down and be less than 1/9.  Since the odds have to be less than 1/9, the only answer choice that satisfies that requirement, is choice A, which is zero, making choice A correct.

 

More Word Problems Practice

1. Two trains start at the same time, from the same station, in different directions. One travels at an average speed of 72 km/hr., and the other at 52 km/hr. After 20 minutes how far apart are they?

a. 6.67 km
b. 17.33 km
c. 24.3 km
d. 41.33 km

2. Brian jogged 7 times around a circular track of 75 meter diameter. How much linear distance did he cover?
a. 1250 meters
b.1450 meters
c. 1649 meters
d.1725 meters

3. If 70 workers build a wall of 150 meters in 12 days then how many workers are required to build a 600 meter wall in 30 days?

a. 100
b. 108
c. 112
d. 116>

4. The length a rectangle is twice of its width and its area is equal to the area of a square with sides of 12 cm. What will be the perimeter of the rectangle near to the nearest whole number?

a. 36 cm
b. 46 cm
c. 51 cm
d. 56 cm

5. A farmer wants to plant trees at the outside boundaries of his rectangular field of dimensions 650 meters × 780 meters. Each tree requires 5 meter free space all around it from is stem. How much free area will be left?

a. 478,800
b. 492,800
c. 507,625
d. 518,256

6. Sarah wants to paint her 17m × 21m bedroom wall. It has one 3m × 4m window and one 0.5m diameter exhaust outlet which are not to be painted. She employed a painter who can paint 1 square meter area in 5 minutes. How much time he requires then to paint the wall?

a. 29 hours
b. 20 hours
c. 24 hours
d. 32 hours

7. A coin and a dice are rolled, a person wins if the coin come up heads, or the dice with a number greater than 4.  In 20 games how many times will a player win?

a. 13
b. 8
c. 11
d. 15

8. Tony bought 15 dozen of eggs for $80. From which, 16 eggs were broken during loading and unloading. Remaining he sold at $0.54 each. What will be his percentage profit? Provide answer in 2 significant digits.

a. 11%
b. 11.2%
c. 11.5%
d. 12%

9. A driver traveled from city A to city B in 1 hour and 13 minutes. On the way he had to stop at 5 traffic signals with an average time of 80 seconds. If the distance between the cities is 65 kilometers then what was the average driving speed?

a. 56.42
b. 58.77
c. 60.34
d. 63.25

10.  In a weaving factory it is estimated that if 10 machines run on 100% efficiency for 8 hours, they will produce 1450 meters of cloth. But due to some technical problems, 4 machines run of 95% efficiency and the remaining 6 at 90% efficiency. How many meters of cloth can these machines will produce in 8 hours?

a. 1334 meters
b. 1310 meters
c. 1330 meters
d. 285 meters

 

Answer Key and Solutions

 1. D

Distance traveled by 1st train in 20 minutes = (72 km/hr × 20 minutes) /60 minutes = 24 m. Distance traveled by 2nd train in 20 minutes = (52 km/hr × 20 minutes)/60 minutes =
17.33 km. Since the trains are travelling in opposite directions, add the distances for the difference apart, 24 + 17.33 = 41.33 km.

2.  C

In one round trip he covers the distance equal to the circumference of the circular path. Circumference/Diameter = ?.  75/X = 3.14159.   75 X 3.14159 = X.  Circumference of the path=X=235.61 meters.  Distance covered 7 times around = 235.65 × 7=1649.33 meters, or about 1649.

3.  C

This is a compound proportionality problem. There is inverse proportionality between workers and days and direct proportionality between workers and length of the wall.

Number of workers

Working days

Length of wall

70

12

150

X

30

600

So the equation will be x/70 = 12/30 X 600/150
x/70 = 7200/4500
x/70 = 72/45
45/72 * x/70
45x = 5040
X = 112

4. C

Area of the square = 12×12=144cm 2. Let x be the width so 2x  will be the length of rectangle.  The area will be 2x2 and the perimeter will be 2(2x+x)=6x.  According to the condition   2x2 = 144 then x=8.48cm.   The perimeter will be  6×8.48  =50.88  =51 cm.

5. A

As one tree requires 10 meter diametric space, or, a 10 meter space on all four sides will be left. So the dimensions left are 630 × 760 = 478,800 meters2.

6. A First find the total paintable area – Area of wall=17m×21m=357m2. Area of window=3m×4m=12m 2. Area of round exhaust fan outlet=? (0.5/2) 2 = 0.2m2. Total paintable area=357-12-0.2=344.8m2. Time will be=344.8×5=1724 minutes=28 hours 45 minutes, or 29 hours.

7.  A

The sample space of this event will be S = { (H,1),(H,2),(H,3),(H,4),(H,5),(H,6)  (T,1),(T,2),(T,3),(T,4),(T,5),(T,6) }  So there are a total of 12 outcomes and 8 winning outcomes.  The probability of a win in a single event is P (W) =8/12=2/3.  In 20 games the probability of a win = 2/3 × 20 = 13

8. A

Let us first mention the money Tony spent: $80

Now we need to find the money Tony earned:

He had 15 dozen eggs = 15 * 12 = 180 eggs. 16 eggs were broken. So,

Remaining number of eggs that Tony sold = 180 – 16 = 164.

Total amount he earned for selling 164 eggs = 164 * 0.54 = $88.56.

As a summary, he spent $80 and earned $88.56.

The profit is the difference: 88.56 – 80 = $8.56

Percentage profit is found by proportioning the profit to the money he spent:

8.56 * 100/80 = 10.7%

Checking the answers, we round 10.7 to the nearest whole number: 11%

9. B

Time taken to travel from A to B in seconds = 3600 + (13 X 60) = 3600 + 780 = 4380 seconds.

Total time spent at traffic signals = 80 X 5 = 400 seconds.

The remaining driving time = 4380 – 400 = 3980 seconds = 3980/3600 = 1.106 hours

The speed will be 65/1.106 = 58.77 km/hr

10.  A

At 100% efficiency 1 machine produces 1450/10 = 145 m of cloth.

At 95% efficiency, 4 machines produce 4•145•95/100 = 551 m of cloth.

At 90% efficiency, 6 machines produce 6•145•90/100 = 783 m of cloth.

Total cloth produced by all 10 machines = 551 + 783 = 1334 m

Since the information provided and the question are based on 8 hours, we did not need to use time to reach the answer.

 

 

11 thoughts on “Word Problem Practice and Tutorial”

  1. Hey, thanks for this! It’s really even helping me for my test coming up this November! I was wondering if you could post more word problem videos whenever you get a chance, thanks!

  2. Hey there, Thanks for posting these mind benders!! 🙂 It’s great practice!
    I’m sure I am wrong but could you kindly help me understand in question 8 how the 164 remaining whole eggs became 165? It throws the percentage off.

    164 remaining eggs x 0.54 cents = $88.56 – $80 investment = $8.56 profit.
    Percentage of profit is 8.56 x 100/80= 10.7%
    Rounded up to two whole numbers = 11% Return On Investment

    I recognize that we arrived at the same answer but I find it confusing that our remaining eggs are different. THanks for any clarification. 🙂

  3. To all the comments re. 2 trains problem – #1. Yes the question was unclear and has been updated! thanks!

    1. Most standardized tests at this level, including the CFAT have word problems – they won’t be exactly the same as posted here – the problems and tutorials posted here are meant to give you practice solving this general type of problem.

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