Practice Test

8.EE.8c systems in real-world problems

Question 1 of 3

2026 Practice Test A Math, question 8

Question

Liam and Kevin each collect stamps. Liam has twice as many stamps as Kevin. Liam gives 8 of his stamps to Kevin. Liam now has 12 more stamps than Kevin.

How many stamps did Liam start with?

Select the correct answer from each drop-down to complete the sentences.

An equation that relates the number of stamps originally in Liam’s and Kevin’s collections is $L =$ $K$ .

After Liam gives 8 stamps to Kevin, the equation becomes $L - 8 =$ .

By solving the equation, it can be determined that Liam started with stamps.

Choose an answer from each menu, then check. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Show the answer

Correct answer: blank 1: 2; blank 2: K + 20; blank 3: 56

Liam originally has twice as many stamps as Kevin, so the equation is $L = 2K$ .

After Liam gives 8 stamps to Kevin, the equation becomes $L - 8 = (K + 8) + 12$ , which simplifies to $L - 8 = K + 20$ .

Substitute $L = 2K$ into the equation: $2K - 8 = K + 20$ .

Solve for K:

$2K - K = 20 + 8$

$K = 28$

Substitute to find L: $L = 2K = 2 \times 28 = 56$ .

So, Liam started with 56 stamps.

Question analysis

This word problem about two stamp collections is answered with three drop-down menus: an equation for the starting counts, an equation for after Liam gives Kevin 8 stamps, and the number Liam started with. Students turn "twice as many" and "12 more" into algebra and solve. Tracking both boys' counts before and after the gift is the tricky part, and every menu must be right.

NYS standards: 8.EE.8c systems in real-world problems
7.EE.4 equations and inequalities from problems
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve one equation in one unknown. The menus lay out the equations, but the problem still ties two unknowns together before and after a gift, a grade 8 system, and all three menus must be right.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Much to keep track of moderate
Solution: 4 steps
Reading level: Question and choices: about grade 5, 86 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The menus do much of the setting up, since each sentence already shows the shape of its equation. The second menu is the hard one: the 8 stamps leave Liam and join Kevin before the gap of 12 is measured. Solving gives Kevin's 28 first, and K + 20 then equals 48, Liam's count after the gift, which waits on the last menu beside his starting 56. Testing each starting count against the story is a quick check on the last menu.

Turning words into math moderate Understanding the question
The gift changes both counts, so "Liam now has 12 more stamps than Kevin" compares Liam's L − 8 with Kevin's K + 8, and the second menu's K + 20 holds a 20 that appears nowhere in the stem. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of moderate Working it out
Each boy has a count before and after the gift, four numbers in all, and all three menus must be right; the last asks for Liam's start, not the 48 he has after giving 8 away. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let K be Kevin's starting count, write each boy's count after the gift in terms of K, and set Liam's 12 above Kevin's; testing each number in the last menu is a fast check.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

BCB Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Sets up both equations and solves for Kevin's 28, then takes the value of K + 20, 48, which is Liam's count after the gift, not at the start.
BBC Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Follows the words "12 more" to L + 12, adding 12 to Liam's own count instead of Kevin's, then finds 56 by testing the three counts against the story.
ACC Equation written backwards (Clement, Lochhead & Monk, 1981)
Reveals a misconception Expected pull: weak
Writes "twice as many stamps as Kevin" backwards, as 2L = K, and picks 0.5, giving Liam half of Kevin's stamps, while the other two menus are right.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Twice as many as: $L = 2K$; L is the number of stamps Liam starts with, and K is the number Kevin starts with. Twice as many means 2 times as many, so Liam’s count is 2 times Kevin’s. Because L and 2K are equal, you can put 2K in place of L in any equation.
Doing the same to both sides: $\text{if } a = b \text{, then } a + c = b + c$; a and b are the two sides of an equation, and c is any number or term. An equation is like a balanced scale. Add the same thing to both sides, or take the same thing away from both, and the scale stays level. This is how you move a term to the other side.

Sort out the question
Let L be Liam’s stamps at the start and K be Kevin’s. You know three facts:
- Liam has twice as many stamps as Kevin.
- Liam gives 8 of his stamps to Kevin.
- After that, Liam has 12 more stamps than Kevin.
The first menu finishes the starting equation. The second finishes the equation after the gift. The third is Liam’s count at the start.
Fill in the first menu
Twice as many means 2 times as many. Liam has twice as many as Kevin, so:
$L = 2K$
Pick 2 in the first menu.
Watch out: the 0.5 in the menu turns the fact around. $L = 0.5K$ would give Liam half as many as Kevin. Liam has more, so his count must be the bigger one.
Write both counts after the gift
The gift changes both counts. Liam loses 8 stamps, and Kevin gains those same 8.
- Liam now has $L - 8$ stamps.
- Kevin now has $K + 8$ stamps.
Liam has 12 more than Kevin. So Liam’s count equals Kevin’s count plus 12:
$L - 8 = (K + 8) + 12 = K + 20$
Pick K + 20 in the second menu.
Watch out: the 12 goes on Kevin’s count, not Liam’s. Picking L + 12 adds it to the wrong boy.
Solve for K
Now use the first equation. Since $L = 2K$ , put 2K in place of L:
$2K - 8 = K + 20$
Subtract K from both sides. Then add 8 to both sides. Each move keeps the scale level.
$K - 8 = 20$
$K = 28$
Kevin started with 28 stamps.
Find Liam’s starting count
Liam started with twice as many as Kevin:
$L = 2K = 2 \times 28 = 56$
Liam started with 56 stamps.
Watch out: $K + 20 = 48$ is Liam’s count after the gift. The menu offers 48, but the question asks about the start.
So the menus read 2, K + 20 and 56.

Check your answer

At the start, Liam has 56 stamps and Kevin has 28, and 56 is twice 28. After the gift, Liam has 48 and Kevin has 36. $48 - 36 = 12$ , so Liam has 12 more. Every fact fits.

Question 2 of 3

2018 Test A Math, question 9

Question

Jenny starts a game with twice as many marbles as Keiko. Jenny gives Keiko 5 marbles, but she still has 10 more than Keiko. How many marbles did Jenny have to start with?

Show the answer

Correct answer: D

Set up some equations. Jenny (J) has twice as many marbles as Keiko (K):

$J=2K$

Jenny gives Keiko 5 marbles, so now they each have: J − 5 and K + 5 marbles. Jenny still has 10 more than Keiko:

$J-5=(K+5)+10$

To find how many marbles Jenny had to start with, solve J = 2K for K and substitute that into the second equation:

In equation J = 2K, solve for K:

$K=\frac{J}{2}$

Substitute $\frac{J}{2}$ in for K.

$J-5=(K+5)+10$

$J-5=(\frac{J}{2}+5)+10$

$J-5=\frac{J}{2}+15$

$J=\frac{J}{2}+20$

$\frac{J}{2}=20$

$J=40$ marbles

Question analysis

A two-variable word problem about marbles: Jenny starts with twice as many as Keiko, gives Keiko 5, and still has 10 more. Students set up equations for before and after the trade and solve for Jenny's starting number. The trade changes both players' counts, and the question asks about the start, not the end.

NYS standards: 8.EE.8c systems in real-world problems
7.EE.4 equations and inequalities from problems
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve one equation in one unknown. This needs two related unknowns and a before-and-after change, which leads to a system of equations or a variable on both sides, both grade 8 skills.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math major
Much to keep track of moderate
Solution: 4 steps
Reading level: Question and choices: about grade 4, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Most of the difficulty is in setting up the equations, not solving them. The gift of 5 marbles changes both counts, so the gap between the girls shrinks by 10, not 5, and students who take the marbles from Jenny without giving them to Keiko reach a consistent but wrong answer. Two other choices are true counts from after the trade, so finishing the algebra is not the same as answering the question. Testing the choices backward is a fast, safe alternative.

Turning words into math major Understanding the question
Three relations hide in two sentences: "twice as many", the gift of 5 that lowers Jenny's count and raises Keiko's, and "still has 10 more", which compares the counts after the gift. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of moderate Working it out
Each girl has a count before and after the trade, four quantities in all, and the question asks for only one of them, Jenny's at the start. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Work backward from the choices: halve each one for Keiko's start, move 5 marbles from Jenny to Keiko, and check whether Jenny ends with exactly 10 more.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Solves correctly for Keiko's start, 20, then reports her count after receiving the 5 marbles, 25, instead of Jenny's starting count.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Subtracts the 5 marbles from Jenny but forgets that Keiko gains them, solving 2K − 5 = K + 10 to get K = 15 and J = 30.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Gives Jenny's count after she hands over 5 marbles, 40 − 5 = 35, instead of the number she had at the start.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Twice as many: $J = 2K$; J is the number of marbles Jenny starts with, and K is the number Keiko starts with. Twice as many means 2 times as many, so J equals 2 times K. Because J and 2K are equal, you can write 2K anywhere you see J.
Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c$; a and b are the two sides of an equation, and c is any number or term. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. The same is true if you add the same thing, or divide both sides by the same number that is not zero.

Sort out the question
Let K be Keiko’s marbles at the start. Jenny starts with twice as many, so she has 2K. Then Jenny gives Keiko 5 marbles. After that, “she still has 10 more than Keiko”. You need Jenny’s count at the start, before the gift.
Write the counts after the gift
The gift changes both counts. Jenny loses 5, and Keiko gains 5.
- Jenny now has $2K - 5$ .
- Keiko now has $K + 5$ .
Jenny has 10 more than Keiko. So Jenny’s count equals Keiko’s count plus 10.
$2K - 5 = (K + 5) + 10$
Watch out: Keiko gains the 5 marbles too. Taking 5 from Jenny but not adding 5 to Keiko leads to 30, choice B.
Solve for K
Add the numbers on the right side first.
$2K - 5 = K + 15$
Subtract K from both sides. Then add 5 to both sides. Each move keeps the equation balanced.
$K - 5 = 15$
$K = 20$
Keiko started with 20 marbles.
Find Jenny’s starting count
Jenny started with twice as many as Keiko.
$J = 2K = 2 \times 20 = 40$
Watch out: 25 is Keiko’s count after the gift, choice A. 35 is Jenny’s count after the gift, choice C. The question asks about the start. So the answer is D, 40.

Check your answer

Start: Jenny has 40 and Keiko has 20, and 40 is twice 20. After the gift, Jenny has 35 and Keiko has 25. $35 - 25 = 10$ , so Jenny still has 10 more. Every fact fits.

Question 3 of 3

2018 Test A Math, question 30

Question

Today, Tien’s age is $\frac{1}{4}$ of Jordan’s age. In 2 years, Tien’s age will be $\frac{1}{3}$ of Jordan’s age. How old is Jordan today?

A4 years old
B6 years old
C12 years old
D16 years old

Show the answer

Correct answer: D

First, set up an equation to express Tien’s age ( $T$ ) and Jordan’s age ( $J$ ) today:

$T=\frac{1}{4}J$

Two years from now, Tien’s age will be $T+2$ , and Jordan’s age will be $J+2$ .

Set up an equation about the relationship between Tien’s age and Jordan’s age in two years:

$T+2=\frac{1}{3}(J+2)$

Solve the above equation for $T$ :

$T=\frac{1}{3}(J+2)-2$

Now set the two equations equal to each other and solve for $J$ :

$\frac{1}{4}J=\frac{1}{3}(J+2)-2$

$\frac{1}{4}J=\frac{1}{3}J-\frac{4}{3}$

$-\frac{1}{12}J=-\frac{4}{3}$

$J=-\frac{4}{3}\left(-\frac{12}{1}\right)$

$J=16$

Question analysis

This age word problem says Tien is one fourth of Jordan's age today and will be one third of Jordan's age in two years, then asks for Jordan's current age. Students must turn two fractional relationships into equations and solve them together. Both people age by the same two years, and it is easy to solve for the wrong person.

NYS standards: 8.EE.8c systems in real-world problems
8.EE.7b linear equations with rational coefficients
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 equations have one unknown on one side, like 3x + 5 = 20. This one links two unknown ages with fractional coefficients and leads to a variable on both sides, which is grade 8 algebra.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math major
Error-prone arithmetic moderate
Solution: 4 steps
Reading level: Question and choices: about grade 1, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The difficulty is in the setup: two fractional relationships at two different times, with both people aging. Once the equations are right, the algebra is short, but it ends with two ages and only one is asked for. Testing the choices against both conditions is a reliable alternative that avoids the algebra entirely.

Turning words into math major Understanding the question
The second sentence must become T + 2 = (J + 2)/3; writing J/3, or forgetting that Jordan also ages two years, breaks the equation. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic moderate Working it out
Substituting J/4 for T leaves J/4 + 2 = (J + 2)/3, an equation with fractions on both sides that must be cleared, usually by multiplying by 12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Let Jordan's age be J, write Tien's age as J/4, and set J/4 + 2 equal to one third of (J + 2); or test each choice against both conditions.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Solves the equations correctly but reports Tien's age today, 4, which is one fourth of Jordan's 16, instead of Jordan's age.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Gives Tien's age two years from now, 4 + 2 = 6, answering for the wrong person and the wrong time.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Picks 12 because it is a multiple of both 4 and 3, but then Tien would be 3 and later 5, not a third of 14.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property: $a \times (b + c) = a \times b + a \times c$; a, b and c are any numbers. To multiply a sum by a number, multiply each part of the sum, then add. It works because a groups of (b + c) is the same as a groups of b plus a groups of c.

Sort out what you know
Today, Tien’s age is $\frac{1}{4}$ of Jordan’s age. That means Jordan is 4 times as old as Tien.
In 2 years, both of them will be 2 years older. Then Tien’s age will be $\frac{1}{3}$ of Jordan’s. So Jordan will be 3 times as old as Tien. You need Jordan’s age today.
Name the ages with a letter
Let t stand for Tien’s age today. Jordan is 4 times as old, so his age is 4t, which means 4 times t. Add 2 years to each age:
Person Today In 2 years
Tien $t$ $t + 2$
Jordan $4t$ $4t + 2$
Watch out: Jordan gets 2 years older too. Do not forget to add 2 to his age.
Write an equation for later
In 2 years, Jordan’s age will be 3 times Tien’s age. Write that as an equation:
$4t + 2 = 3 \times (t + 2)$
Use the distributive property on the right side. Multiply both t and 2 by 3:
$4t + 2 = 3t + 6$
Solve for Tien’s age
Take 3t away from both sides. Doing the same thing to both sides keeps them equal:
$t + 2 = 6$
Now take 2 away from both sides:
$t = 4$
So Tien is 4 years old today. Watch out: 4 is Tien’s age, not Jordan’s. Choice A, 4 years old, is a trap.
Find Jordan’s age
Jordan is 4 times as old as Tien:
$4 \times 4 = 16$
So the answer is D, 16 years old.

Person	Today	In 2 years
Tien	$t$	$t + 2$
Jordan	$4t$	$4t + 2$

Check your answer

Put 16 back into the story. Today, Tien is 16 ÷ 4 = 4. In 2 years, Tien is 6 and Jordan is 18. Is 6 one third of 18? Yes, because 18 ÷ 3 = 6. Choice C fails: Tien would be 3, then 5, and 5 is not a third of 14.

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