Chapter 12.4, Problem 35E

(a)

To determine

To find the fraction of the distance between 4 and 5 that the hour hand have travelled.

Answer to Problem 35E

  $\frac{1}{3}$

Explanation of Solution

Given:

The hour and minute hands of a clock travel around its face at different speeds.

Calculation:

When the minute hand points to $4$ total it has travelled $20$ minutes.

Therefore, fraction of the distance between 4 and 5 that the hour hand have travelled is $\frac{20}{60}=\frac{1}{3}$

(b)

To determine

To find the fraction of the distance travelled by the hour hand if the minute hand reaches its position.

Answer to Problem 35E

  $\frac{1}{12}$

Explanation of Solution

Given:

The hour and minute hands of a clock travel around its face at different speeds.

Calculation:

The hour hand will travel only $5$ minutes when again minute hand will reach hour hand’s position.

Therefore, fraction of the distance, the hour hand have travelled is $\frac{5}{60}=\frac{1}{12}$

(c)

To determine

To list the next two terms of the series representing the distance travelled by the hour hand as the minute hand chases its position.

Answer to Problem 35E

  $\frac{1}{3},\frac{1}{3}+\frac{1}{20},\frac{1}{3}+\frac{1}{10},\frac{1}{3}+\frac{3}{20}$

Explanation of Solution

Given:

The hour and minute hands of a clock travel around its face at different speeds.

Calculation:

The terms of the series will be-

  $\frac{1}{3},\frac{1}{3}+\frac{1}{20},\frac{1}{3}+\frac{1}{10},\frac{1}{3}+\frac{3}{20}$

(d)

To determine

To find at what time between the two hands will coincide between $4$ o’ clock and $5$ o’ clock.

Answer to Problem 35E

  $21.81$ Minutes.

Explanation of Solution

Given:

The hour and minute hands of a clock travel around its face at different speeds.

Calculation:

At 4:00 the minute hand has covered  $0$ degrees and hour hand has covered  $120$ degrees

Now let time after which these two coincide be  $x$  min.

So hour hand covers  $120+\frac{x}{2}$  up to that time and minute hand covers $6x$ degrees up to that time when they coincide the angles should be same

So,  $120+\frac{x}{2}\text{}=6x$

Solving we get  $6x-\frac{x}{2}\text{}=120$

  $\begin{array}{l}\Rightarrow \frac{12x-x}{2}\text{}=120\\ \Rightarrow 11x=240\\ \Rightarrow x=21.81\text{ mins}\end{array}$

So after $21.81$ minutes the hands will coincide.