Math Puzzles & Problem Solving

For quadratic ax²+bx+c=0, sum of roots = -b/a = -2/4 = -1/2, product = c/a = 1/4. Their sum is -1/2+1/4 = -1/4. This uses Vieta's formulas for root relationships.

This is identical to question 115619. For 4x²+2x+1=0: sum of roots = -2/4 = -1/2, product = 1/4. Adding gives -1/4, consistent with Vieta's formulas for quadratic equations.

With sides 3, 4, 5, this satisfies 3²+4²=5², so it's a right triangle by the converse of Pythagoras theorem. The angle opposite the longest side (5) is 90°.

Multiplying the second equation by 2: 4x + 16y = 20. Subtracting the first (4x + 6y = 12) gives 10y = 8, so y = 0.8. Substituting y into 2x + 8(0.8) = 10 gives 2x + 6.4 = 10, 2x = 3.6, x = 1.8.

Expanding (sinA+cosA)²+(sinA-cosA)² gives sin²A+2sinAcosA+cos²A+sin²A-2sinAcosA+cos²A = 2(sin²A+cos²A) = 2(1) = 2, using the fundamental identity sin²θ+cos²θ=1.

A triangle with angles 45°, 45°, 90° has two equal angles, making it isosceles. Since one angle is 90°, it's also right-angled. Combining these properties gives a right-angled isosceles triangle.

Substitute the given angles: sin45° = 1/√2, cos45° = 1/√2, tan45° = 1, sec45° = √2, sin90° = 1, cos90° = 0. The expression becomes (1/√2) × (1/√2) × 1 × √2 × 1 × 0 = 0 because cos90° = 0. Any expression multiplied by zero equals zero.

A person born on February 29 (leap day) has their birthday once every 4 years, but the gap between consecutive celebrations is longest at 8 years when a leap year is skipped. This happens from age 64-68 (1984→1992: 8 years, skipping 1988) or age 68-72 (1992→2000: 8 years, skipping 1996).

The minute and hour hands meet when the minute hand gains 360° on the hour hand - this takes 12/11 hours (about 65.45 minutes). In 24 hours, this happens 22 times (11 times per 12-hour period). The 22nd meeting is at 12:00 exactly, completing the cycle.

Using the formula n(n-1)/2 = 66, we get n^2 - n - 132 = 0. Factoring gives (n-12)(n+11) = 0. Since the number of people must be positive, n = 12.

Let the youngest be x. The ages are x, x+3, x+6, x+9, x+12. Sum = 5x + 30 = 50. 5x = 20, so x = 4. The youngest child is 4 years old.

While there are 64 individual unit squares, the total number of squares of all sizes (1x1, 2x2... 8x8) is calculated as the sum of squares from 1 to 8: 1+4+9+16+25+36+49+64 = 204.

Let distance = d km. Forward speed = d/7 km/hr. Return speed = d/5 km/hr (12 km/hr faster). So: d/5 - d/7 = 12. Multiply by 35: 7d - 5d = 420. 2d = 420. d = 210 km. This checks out: forward speed = 210/7 = 30 km/hr, return speed = 210/5 = 42 km/hr (exactly 12 km/hr faster).

Classic exponential growth puzzle. If the lily doubles each day and completely covers the pool on day 20, it must have covered HALF the pool on day 19 (because 2 × half = full). This is not day 10 (that would be halving, not doubling). The pool dimensions don't matter - the doubling logic works regardless of size.

Step-by-step profit calculation: 1) Bought at $60, sold at $70 = +$10 profit. 2) Bought back at $80 (regret), sold at $90 = +$10 profit. Total = $10 + $10 = $20. The $80 purchase doesn't create a loss - it's the new cost basis. The -$10 mentioned in the problem is misleading intermediate accounting, not actual profit.

Classic conditional probability (Bayesian reasoning). You drew a white marble. This could come from Bag A (2 white) or Bag C (1 white, 1 black), not Bag B (2 black). Given white marble, probability you picked Bag A is 2/3 (2 white marbles in A vs 1 in C). If from Bag A, remaining marble is white. So answer = 2/3. This is not 1/3 (that's the prior probability of picking Bag A), not 1/2 (ignoring the evidence).

In a row of connected gears, the direction alternates with each gear. Gear 1 (CW), Gear 2 (ACW), Gear 3 (CW), Gear 4 (ACW), Gear 5 (CW). Therefore, the fifth gear turns in the same direction as the first.

In a train of gears, the speed of the last gear relative to the first depends only on the ratio of the first gear's teeth to the last gear's teeth. Intermediate gears (idlers) do not affect the final output speed regardless of their size.