Chapter 1.2, Problem 18PSB

a.

To determine

To find: Which number would the clock point.

Answer to Problem 18PSB

  $3$ O’clock.

Explanation of Solution

Given information:

A figure is given as below.

Hand is rotated by ${90}^0$ clockwise.

  

Calculation:

A clock is ${360}^0$ divided into $12$ equal parts with each part covering ${30}^0$ .

So, when hand is rotated ${90}^0$ clockwise. Then hand would point at $\frac{{90}^0}{{30}^0}=3$ .

Hence, hand will point at $3$ O’clock.

b.

To determine

To find: Which number would the clock point.

Answer to Problem 18PSB

  $4$ O’clock.

Explanation of Solution

Given information:

Hand is rotated by ${150}^0$ clockwise and ${30}^0$ anti-clockwise.

Calculation:

A clock is ${360}^0$ divided into $12$ equal parts with each part covering ${30}^0$ .

So, when hand is ${150}^0$ clockwise and ${30}^0$ anti-clockwise. Final rotation was of ${150}^0-{30}^0={120}^0$ .

And hand would point $\frac{{120}^0}{{30}^0}=4$ .

Hence, hand will point at $4$ O’clock.