Timed Pacing Drill

Set a 12-minute timer for the 10 questions below. Question 4 is deliberately more expensive than the questions around it.

The purpose is not to prove that Question 4 can be solved. The purpose is to notice when it stops buying useful information. If a route is not clear after a serious first look, mark it, choose the best current answer, and continue. When Question 5 begins, solve it at normal quality. Do not try to “win back” the time by rushing.

After the set, record three times: the time when Question 4 first became expensive; the time when you actually left it; and the time remaining after Question 10.

If the first and second times are far apart, the problem was not mathematical speed. It was exit delay. Redo the set later with the same questions in a different order. The goal is a faster decision, not faster arithmetic.

▸ Drill: 10 questions, 12 minutes

1. Solve for x: 7x − 4 = 24.

A) 4 B) −4 C) 2.86 D) 3.43

2. Find 15% of 260.

F) 39 G) 41 H) 26 J) 45

3. Simplify √144.

A) 12 B) 72 C) 24 D) 14.4

4. A triangle has sides 9 and 12 with an included angle of 55°. Find the third side, rounded to the nearest hundredth.

F) 15.00 G) 9.85 H) 21.00 J) 10.06

5. Find the mean of 12, 15, 18, 21, 24.

A) 20 B) 90 C) 18 D) 17

6. Solve the system 3x + 2y = 16 and x = 4 for y.

F) 4 G) 2 H) 14 J) −2

7. Find the area of a circle with radius 6.

A) 12π B) 36π C) 6π D) 72π

8. A value increases 40% from 50. Find the new value.

F) 54 G) 20 H) 70 J) 90

9. Find the slope of the line through (2, 5) and (6, 13).

A) 4 B) 8 C) 0.5 D) 2

10. How many ways can a committee of 3 be chosen from 8 people?

F) 56 G) 336 H) 24 J) 21

▸ Answers and Explanations

1: A. 7x = 28, so x = 4. Choice D reports 24/7 without first adding 4 to isolate the x-term.

2: F. 0.15 × 260 = 39. Choice H comes from an arithmetic slip while computing 0.15×260.

3: A. √144 = 12. Choice B halves 144 instead of taking its square root.

4: J. By the Law of Cosines: c² = 9² + 12² − 2(9)(12)cos(55°) ≈ 101.11, so c ≈ 10.06. Choice F omits the −2ab·cos C term entirely, computing √(9²+12²)=15 as if this were a right triangle.

5: C. (12+15+18+21+24)/5 = 90/5 = 18. Choice A reports the sum of the values without dividing by the count.

6: G. Substituting x=4: 3(4)+2y=16, so 12+2y=16, 2y=4, y=2. Choice F comes from a sign slip while isolating the 2y term.

7: B. Area = πr² = π(6)² = 36π. Choice A omits the squaring in the area formula, using πr instead of πr².

8: H. 50 × 1.40 = 70. Choice G reports only the amount of increase (50×0.40=20) instead of the new total value.

9: D. Slope = (13−5)/(6−2) = 8/4 = 2. Choice C inverts the slope ratio, computing the change in x divided by the change in y instead of the reverse.

10: F. Order does not matter when choosing a committee, so this is a combination: ₈C₃ = 56. Choice G treats the selection as if order mattered, computing a permutation instead of a combination.