Find time, when angle between the minute and hour hands are given

Question

The question which I was trying to solve is:

Find the time between four and five o' clock when the angle between the hour hand and the minute hand is $78^\circ$.

My approach:

At four o' clock, the hour hand and the minute hand of the clock will be at 4 and 12 respectively, hence the angle they make will be $120^\circ$. Let us assume some 'x' minutes have passed since then when the hands of the clock finally acquired the position described in the question. In 'x' minutes, the minute hand would have moved by $6x^\circ$, at the same time the hour hand would have moved by $0.5x^\circ$. At that position, the separation between them as given by the question is $78^\circ$. Hence the equation we can derive from here is:

$120^\circ + 0.5x^\circ - 6x^\circ = 78^\circ$

Solving for 'x' I get $x = \left(\frac{84}{11}\right)$ minutes. Hence, as per my calculation since $\left(\frac{84}{11}\right)$ minutes have elapsed since four o' clock, the time in question should be around 7~8 minutes past 4.

The answer to the exercise given in the book: At $\left(\frac{84}{11}\right)$ and 36 minutes past 4.

Can someone explain me, where did I go wrong?

Answer 1

Reverse the signs and calculate the other answer. One assumes the minute hand is ahead of the hour hand, the other assumes the minute hand is behind.

Answer 2

HINT, here is how much the hour hand travels per hour, minute, second:

• Per Hour: $$\text{H}=\frac{360^{\circ}}{12\space\text{hours}}=30^{\circ}\text{/}\space\text{hour}\tag1$$
• Per Minute: $$\text{M}=\frac{\text{H}^{\circ}}{60\space\text{minutes}}=\left(\frac{1}{2}\right)^{\circ}\text{/}\space\text{minute}\tag2$$
• Per Second: $$\text{S}=\frac{\text{M}^{\circ}}{60\space\text{seconds}}=\left(\frac{1}{120}\right)^{\circ}\text{/}\space\text{second}\tag3$$