Clock, Variation, Sequences, Miscellaneous

Example

Find the time after 4 o’clock when the hands of a clock overlap for the first time.

Solution

At 4 o’clock, hour \(H = 4\), so \(5H = 20\ \text{minutes}\). Set up the hand movement equation:

Answer: 4:21.82

Example

In how many minutes after 2 o’clock will the hands of the clock extend in opposite directions for the first time?

Solution

For \(H = 2\), since \(H < 6\), the hands are opposite when separated by 30 minute spaces:

Answer: 43.64 minutes

Example

How many minutes after 3 o’clock will the hands of a clock be perpendicular for the second time?

Solution

For \(H = 3\), use the second perpendicular position formula:

Answer: 32.73 minutes

Example

At what time after 12:00 noon will the hour hand and minute hand first form an angle of \(120^\circ\)?

Solution

Convert angle \(\theta = 120^\circ\) to minute spaces: \(M_\theta = \dfrac{120^\circ}{6^\circ} = 20\ \text{minutes}\). With \(H = 12\) (\(0\ \text{minutes}\) initial offset):

Answer: 12:21.82 PM

Example

A man leaves home past 3:00 PM and returns between 2 and 3 hours later, finding the clock hands interchanged. At what time did he leave?

Solution

Departure hour \(H_1 = 3\) (\(5H_1 = 15\)). Return hour \(H_2 = 6\) (\(5H_2 = 30\)). Set up system of equations:

Substitute \(y\) into \(x\) equation:

Answer: 3:31.47 PM

Example

It is now between 3 and 4 o’clock, and in 20 minutes the minute hand will be as far ahead of the hour hand as it is now behind it. Find the time now.

Solution

For \(H = 3\), initial angle behind is \(\theta = 15 + \dfrac{x}{12} - x = 15 - \dfrac{11x}{12}\). After \(T = 20\ \text{minutes}\), angle ahead is \(\beta = (x + 20) - \left(15 + \dfrac{x + 20}{12}\right) = \dfrac{11x}{12} + \dfrac{10}{3}\). Set \(\beta = \theta\):

Answer: 3:06.36

Example

From 6:15 PM to 7:45 PM of the same day, determine the arc described by the minute hand.

Solution

Elapsed time \(\Delta t = 465 - 375 = 90\ \text{minutes}\). Calculate rotation angle:

Answer: \(540^\circ\)

Example

A clock has minute hand 9 in long and hour hand 6 in long, with minute hand rotating 2 in above hour hand plane. Find distance between tips at 5:40 AM.

Solution

At 5:40 AM, hour hand angle is \(5(30^\circ) + 40(0.5^\circ) = 170^\circ\). Minute hand angle is \(40(6^\circ) = 240^\circ\). Angle between hands \(\theta = 240^\circ - 170^\circ = 70^\circ\). Planar distance squared:

Include vertical offset \(z = 2\ \text{in}\):

Answer: 9.17 inches

Example

A grandfather clock strikes once at 1 o’clock, twice at 2 o’clock, thrice at 3 o’clock, and so on. How many times will it strike in one day?

Solution

Calculate sum for 12 hours and double for 24 hours:

Answer: 156 times

Example

Given that \(w\) varies directly as the product of \(x\) and \(y\) and inversely as the square of \(z\). If \(w = 4\) when \(x = 2\), \(y = 6\), and \(z = 3\), find the value of \(w\) when \(x = 1\), \(y = 4\), and \(z = 2\).

Solution

Set up the variation equation with constant \(k\):

Substitute initial values to find \(k\):

Substitute \(k = 3\), \(x = 1\), \(y = 4\), and \(z = 2\) to solve for \(w\):

Answer: 3

Example

If \(x\) varies directly as \(y\) and inversely as \(z\), and \(x = 14\) when \(y = 7\) and \(z = 2\), find \(x\) when \(y = 16\) and \(z = 4\).

Solution

Set up the proportion formula:

Find constant \(k\):

Calculate \(x\) for \(y = 16\) and \(z = 4\):

Answer: 16

Example

Electric power transmitted by a line is proportional to design voltage and current capacity, and inversely to transmission distance. A \(115\ \text{kV}\) line rated at \(100\ \text{A}\) transmits \(150\ \text{MW}\) over \(150\ \text{km}\). How much power can a \(230\ \text{kV}\) line rated at \(150\ \text{A}\) transmit over \(100\ \text{km}\)?

Solution

Express power as a joint variation equation:

Solve for proportionality constant \(k\):

Calculate power \(P\) for \(V = 230\ \text{kV}\), \(I = 150\ \text{A}\), and \(d = 100\ \text{km}\):

Answer: 675 MW

Example

The time required for an elevator to lift a weight varies directly with weight and distance, and inversely as motor power. If a \(10\ \text{hp}\) motor lifts \(100\ \text{lbs}\) through \(50\ \text{ft}\) in \(30\ \text{seconds}\), what motor size is required to lift \(800\ \text{lbs}\) through \(40\ \text{ft}\) in \(40\ \text{seconds}\)?

Solution

Set up the formula for time \(t\):

Determine constant \(k\):

Solve for required power \(P\) when \(W = 800\ \text{lbs}\), \(S = 40\ \text{ft}\), and \(t = 40\ \text{seconds}\):

Answer: 48 hp

Example

The electrical resistance of a cable varies directly as its length and inversely as the square of its diameter. If a cable \(600\ \text{meters}\) long and \(25\ \text{mm}\) in diameter has a resistance of \(0.1\ \Omega\), find the length of a cable of the same material \(75\ \text{mm}\) in diameter with resistance of \(\dfrac{1}{6}\ \Omega\).

Solution

Set up the variation equation for resistance:

Find constant \(k\):

Find length \(L\) when \(D = 75\ \text{mm}\) and \(R = \dfrac{1}{6}\ \Omega\):

Answer: 9000 meters

Example

Compare the electrical resistance offered by two pieces of wire of the same material, one being \(100\ \text{m}\) long and \(5\ \text{mm}\) in diameter, and the other \(50\ \text{m}\) long and \(3\ \text{mm}\) in diameter.

Solution

Form the ratio of resistances:

Substitute dimensions \(L_1 = 100\), \(D_1 = 5\), \(L_2 = 50\), \(D_2 = 3\):

Express \(R_1\) in terms of \(R_2\):

Answer: \(R_1 = 0.72 R_2\)

Example

Find the 30th term and the sum of the first 30 terms of the sequence 4, 7, 10, 13, …

Solution

Identify first term \(a_1 = 4\) and common difference \(d = 7 - 4 = 3\).

Answer: 30th term is 91; Sum is 1425

Example

Find the sum of 7 arithmetic means inserted between 3 and 35.

Solution

With \(k = 7\) means, total terms \(n = 7 + 2 = 9\). First term \(a_1 = 3\) and last term \(a_9 = 35\). Total sum of all 9 terms:

Sum of the 7 inserted means equals total sum minus end terms:

Answer: 133

Example

The 4th term of a geometric progression is 216 and the 6th term is 1944. Find the 8th term.

Solution

Set up term equations \(a_4 = a_1 r^3 = 216\) and \(a_6 = a_1 r^5 = 1944\). Divide \(a_6\) by \(a_4\):

Find 8th term using \(a_8 = a_6 r^2\):

Answer: 17496

Example

Find the sum of the infinite geometric series 6, -2, 2/3, -2/9, …

Solution

First term \(a_1 = 6\). Calculate common ratio \(r = \dfrac{-2}{6} = -\dfrac{1}{3}\). Since \(|r| < 1\):

Answer: 9/2

Example

A rubber ball is dropped from a height of 50 feet and rebounds 2/3 of the distance it falls. Find total distance traveled before coming to rest.

Solution

Initial drop \(h = 50\ \text{ft}\). First rebound height \(a_1 = 50 \times \dfrac{2}{3} = \dfrac{100}{3}\ \text{ft}\) with ratio \(r = \dfrac{2}{3}\). Apply formula:

Answer: 250 feet

Example

The 3rd term of a harmonic progression is 15 and the 9th term is 6. Find the 11th term.

Solution

Convert to arithmetic progression terms: \(a_3 = \dfrac{1}{15}\) and \(a_9 = \dfrac{1}{6}\). Set up linear equations:

Subtract \(a_3\) from \(a_9\):

Find \(a_{11}\) in arithmetic progression:

Take reciprocal to find 11th harmonic term \(h_{11}\):

Answer: 5

Example

The geometric mean and harmonic mean of two numbers are 12 and 36/5 respectively. Find the two numbers.

Solution

Given \(G = 12\) and \(H = \dfrac{36}{5}\). Find arithmetic mean \(A\) using \(G^2 = A \times H\):

Set up equations for two numbers \(x\) and \(y\):

Solve quadratic equation \(t^2 - 40t + 144 = 0\):

Answer: 36 and 4

Example

Find the total number of spheres in a complete triangular pyramid pile if each side of the base contains 4 spheres.

Solution

A pile of spheres with equilateral triangle base forms a tetrahedral figurate number with \(n = 4\) layers:

Answer: 20 spheres

Example

Find the positive integer solutions for \(x\) and \(y\) in the equation \(5x + 3y = 31\).

Solution

Express \(y\) in terms of \(x\):

For \(y\) to be a positive integer, \(31 - 5x\) must be positive and divisible by \(3\). Test positive integer values for \(x\):

If \(x = 2\), \(y = \dfrac{31 - 10}{3} = 7\) (valid integer solution).

If \(x = 5\), \(y = \dfrac{31 - 25}{3} = 2\) (valid integer solution).

If \(x \ge 8\), \(y \le 0\) (not a positive integer).

Answer: \((x, y) = (2, 7)\) and \((5, 2)\)

Example

A merchant has three items on sale: a radio for \(P 50\), a clock for \(P 30\), and a flashlight for \(P 1\). At the end of the day, he sold a total of 100 items for exactly \(P 1000\). How many radios were sold?

Solution

Let \(x = \text{radios}\), \(y = \text{clocks}\), and \(z = \text{flashlights}\). Set up the system of equations:

Substitute \(z\) into the total revenue equation:

Solve for \(y\) in terms of \(x\):

For \(y\) to be an integer, \(900 - 49x\) must be divisible by \(29\). Since \(900 \equiv 1\) and \(49 \equiv 20 \pmod{29}\), this requires \(20x \equiv 1 \pmod{29}\), which gives \(x \equiv 16 \pmod{29}\). Because \(y \ge 0\) requires \(x \le 18\), the only choice is \(x = 16\):

Find \(z\) using \(z = 100 - 16 - 4 = 80\). Check total cost: \(50(16) + 30(4) + 80 = 800 + 120 + 80 = 1000\).

Answer: 16 radios