Today, Tien’s age is 1/4 of Jordan’s age. In 2 years, Tien’s age will be 1/3 of Jordan’s age. How old is Jordan today?

Today, Tien’s age is 1/4 of Jordan’s age

2018 Handbook

2018 Test A Math · Question 30 of 57

Also in 2026 Practice Test A Math, question 26; 2025 Practice 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.

The original DOE question

Question 87 as printed on page 87 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 87. Open page 87 of the handbook (PDF)