AlgebraHardConsecutive integers under a limit
Algebra practice question
The sum of three consecutive integers is at most 48. What is the greatest possible value of the largest of the three integers?
Answer: 17
Let the integers be n, n + 1, n + 2: 3n + 3 ≤ 48, so 3n ≤ 45 and n ≤ 15. The largest is n + 2, at most 17. Check: 15 + 16 + 17 = 48.
Common wrong answers
- If you answered 15
- 15 is the SMALLEST of the three integers, not the largest.
- If you answered 16
- 16 is the middle integer (48/3), or the largest if n were 14.
- If you answered 18
- 18 would give 16 + 17 + 18 = 51, more than 48.
More Algebra questions
- Hard
For how many integer values of x is −2 < (x − 5)/3 ≤ 1 true? - Hard
The system 2x + 3y = 6 and 4x + 6y = k has infinitely many solutions. What is the value of k? - Hard
Which of the following is the complete solution set to |x − 4| < 6? - Hard
A student has $40 to spend on notebooks costing $4 each and pens costing $2 each, and wants at least 12 items … - Hard
What is the least whole number of minutes of international calling in a month for which the total monthly cost… - Hard
How many liters of the 30% acid solution did the technician use?
Not affiliated with the College Board. SAT is a trademark of the College Board.