A rectangular field had a length

2024 Handbook

2024 Test B Math · Question 56 of 57

Question

A rectangular field had a length of 80 meters and a width of 50 meters. The field was recently enlarged, with the length increased by 15% and the width increased by 10 meters. By what percentage was the area of the field increased?

A26.5%
B35%
C38%
D42.5%

Show the answer

Correct answer: C

The original area of the field was 4,000 square meters.

$80\times 50=4{,}000$

When the field was enlarged, the length increased by 15%, so the new length is 92 meters.

$80\times 15\%$

$80\times 0.15=12$

$80+12=92$

The width increased by 10 meters, so the new width is 60 meters.

$50+10=60$

The new area of the field is 5,520 square meters.

$92\times 60=5{,}520$

The percent increase is found by finding the difference between the two areas and dividing the increase by the original area.

$5{,}520-4{,}000=1{,}520$

$1{,}520\div 4{,}000=0.38=38\%$

Question analysis

A rectangular field 80 meters long and 50 meters wide has its length increased by 15 percent and its width by 10 meters. Students find the percent by which the field's area grew. It tests percent increase with rectangle area, and the tricky part is that one side changes by a percent and the other by a fixed length.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.G.1 area of polygons
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually apply one percent change to one quantity. This changes a length by a percent and a width by a fixed amount, then asks for the percent change in area, a five-step chain with two kinds of change.
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: Combines ideas moderate
Many steps moderate
A very close wrong answer minor
Solution: 5 steps
Reading level: Question and choices: about grade 6, 39 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

Each step is ordinary, but the two changes are stated in different units, one a percent and one in meters, which tempts students to force them into the same form. Adding the two percents is the most natural shortcut and lands close to the right value, because it misses only the small extra area where both changes overlap. Computing the old and new areas directly is slower but leaves nothing to chance.

Combines ideas moderate Working it out
Percent change has to be combined with area: the length grows by a percent and the width by meters, and the question asks for the percent change of their product. (Embretson, 1998; Daniel & Embretson, 2010)
Many steps moderate Working it out
Five computations are chained: the new length 92, the new width 60, the areas 4,000 and 5,520, and the increase 1,520 as a share of 4,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The 35% from adding 15% and 20% sits only 3 points below 38%, so a rough estimate of the new area cannot rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Compute the new length and the new width, find the old and new areas, and divide the increase in area by the original area.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Units or scale (King et al., 2004)
Reveals a slip Expected pull: moderate
Treats the 10-meter widening as a 10% increase, so multiplies 1.15 × 1.10 = 1.265 and reports a 26.5% rise.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Adds the length's 15% to the width's 20% increase, as if percent increases in two dimensions simply add, missing the 3% where they overlap.
D Units or scale (King et al., 2004)
Reveals a slip Expected pull: weak
Treats the 15% as 15 meters, making the field 95 by 60 meters, an area of 5,700 square meters and a 42.5% increase.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle: $A = l \times w$; A is the area, the flat space inside. l is the length and w is the width. Multiplying counts how many unit squares fit, row by row.
Percent increase: $\text{percent increase} = \frac{\text{new} - \text{old}}{\text{old}} \times 100$; old is the starting amount, and new is the amount after. new − old is how much it grew. Dividing by old compares the growth with the start. Multiplying by 100 turns that into a percent.

Sort out the question
The field was 80 meters long and 50 meters wide. The length grew by 15%, a percent. The width grew by 10 meters, a number of meters. These are two different kinds of increase.
You need the percent by which the area grew.
Find the old area
You need the old area to compare with the new one.
$A = 80 \times 50 = 4{,}000$
The old field covered 4,000 square meters.
Find the new length and width
15% of 80 is $0.15 \times 80 = 12$ . So the new length is $80 + 12 = 92$ meters.
The new width is $50 + 10 = 60$ meters.
Watch out: 10 meters is not 10%. Treating it as 10% leads to 26.5%, choice A. Treating 15% as 15 meters leads to 42.5%, choice D.
Find the new area
Multiply the new length by the new width.
$A = 92 \times 60 = 5{,}520$
The new field covers 5,520 square meters. That is $5{,}520 - 4{,}000 = 1{,}520$ square meters more.
Find the percent increase
$\text{percent increase} = \frac{5{,}520 - 4{,}000}{4{,}000} \times 100 = \frac{1{,}520}{4{,}000} \times 100 = 38$
The area grew by 38%. Watch out: The width grew by 20%, since 10 is 20% of 50. Adding 15% + 20% = 35% gives choice B. That misses the extra corner piece where both new strips meet.
So the answer is C, 38%.

Check your answer

Use growth factors. The length is multiplied by 1.15. The width goes from 50 to 60, so it is multiplied by 1.2. The area is multiplied by $1.15 \times 1.2 = 1.38$ , which is 38% more.

The original DOE question

Question 113 as printed on page 215 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 113. Open page 215 of the handbook (PDF)