AlgebraVery hardConsecutive odd integers with a condition on the largest
Algebra practice question
The sum of three consecutive odd integers is 3 more than twice the largest of them. What is the smallest of the three integers?
Answer: 5
With n, n + 2, n + 4: 3n + 6 = 2(n + 4) + 3, so 3n + 6 = 2n + 11 and n = 5.
Common wrong answers
- If you answered 3
- 3 comes from 3n + 6 = 2n + 9, forgetting the extra 3.
- If you answered 7
- 7 is the middle one.
- If you answered 9
- 9 is the largest of the three.
More Algebra questions
- Very hard
For what value of c does the system 3x − 2y = 8 and 9x + cy = 5 have no solution? - Very hard
In the system 5x + 4y = 32 and 3x − 2y = 6, what is the value of 8x + 2y? - Very hard
The solution set of the inequality |ax − b| ≤ 6, where a and b are constants and a > 0, is 3 ≤ x ≤ 9. What is … - Very hard
The function f is linear, and for every value of x, f(x + 2) − f(x) = 6. If f(5) = 4, what is the value of f(0… - Very hard
In the system of equations 2x + 3y = 12 and 5x − 2y = k, where k is a constant, the solution (x, y) satisfies … - Very hard
In the xy-plane, the system of equations kx + 6y = 12 and 2x + (k − 1)y = 6, where k is a constant, has infini…
Not affiliated with the College Board. SAT is a trademark of the College Board.