# Between $6$ and $8$ pm, the minute hand the hour hand interchange positions

A man enters his home from some time between $6$ to $7$ pm . when he leaves his home sometime between $7$ to $8$ pm, he observes that the minute hand the hour hand have interchanged position.At what time did the man entered his house.

Options

$\color{green}{a.)\quad 38\cfrac{82}{121}\quad \text{minutes past} \quad 6}.\\ b.)\quad 37\cfrac{42}{121}\quad \text{minutes past}\quad 6.\\ a.)\quad 37\cfrac{82}{121} \quad\text{minutes past}\quad 6.\\ d.)\quad 37\cfrac{62}{121}\quad \text{minutes past}\quad 6.\\$

i made the $4$ figures and formed the equation for the figure from $6$ am to $7$ am. assuming the circumference of clock is $360$. time taken as $t$ . and $x$ be the specified distance between $2$ hands.

$6t-\frac{t}{2}=180+x$

for the second i am confused on forming equation and stucked.

I don't get any of the given options as an answer. Instead, I find the man entered his house at $37{109\over143}$ minutes after $6$.

Here's my thinking. Let $t$ denote the time in minutes starting at $6$ o'clock, let $M(t)$ denote the minute number (between $0$ and $60$) that the minute hand is pointing at, and let $H(t)$ denote the minute number that the hour hand is pointing at. For the time range of interest ($0\lt t\lt 120$), the key formula is

$$H(t)=30+{t\over12}$$

Now if $t_1$ denotes the time (after $6$) that the man enters his house and $t_2$ denotes the time (between $7$ and $8$) when he exits, we want

$$M(t_2)=H(t_1)\quad\text{and}\quad H(t_2)=M(t_1)$$

But since $0\lt t_1\lt 60$, we have $M(t_1)=t_1$, while $60\lt t_2\lt 120$ implies $M(t_2)=t_2-60$. The two equations are thus

$$t_2-60=30+{t_1\over12}\quad\text{and}\quad 30+{t_2\over12}=t_1$$

When I eliminate $t_2$ and solve for $t_1$, I get $t_1=5400/143=37{109\over143}$.

A final remark: The OP posted an earlier clock-type question, presumably from the same source, for which the book's solution was wrong. I hope someone will check my work as well.

• I got the same answer as you do, with a different way of calculation. – LaBird Mar 19 '15 at 16:16
• My working: Assume the man entered the house at $x$ minutes past $6$, and left at $y$ minutes past $7$. When he entered the house, the hour hand was at $(180 + 30 \times \frac{x}{60})$ degrees clockwise from the $12$ mark $(1)$, the minute hand was at $(6x)$ degrees clockwise from the $12$ mark $(2)$. When he left, the hour hand was at $(210 + 30 \times \frac{y}{60})$ degrees clockwise from the $12$ mark $(3)$, the minute hand was at $(6y)$ degrees clockwise from the $12$ mark $(4)$. The question requires $(1)=(4)$ and $(2)=(3)$. Solving the equations give $x = 37\frac{109}{143}$. – LaBird Mar 19 '15 at 16:22
• @LaBird, many thanks for the confirmation. You might want to post your calculation as a separate answer. – Barry Cipra Mar 19 '15 at 16:38
• $\quad~~~$thnks! – R K Mar 19 '15 at 18:28

An alternative approach:

Assume the man entered the house at $x$ minutes past $6$, and left at $y$ minutes past $7$.

When he entered the house, the hour hand was at $(180+30\times \frac{x}{60})$ degrees clockwise from the $12$ mark ...... $(1)$,

and the minute hand was at $(6x)$ degrees clockwise from the $12$ mark ...... $(2)$.

When he left, the hour hand was at $(210+30\times \frac{y}{60})$ degrees clockwise from the $12$ mark ...... $(3)$,

and the minute hand was at $(6y)$ degrees clockwise from the $12$ mark ...... $(4)$.

The question requires $(1)=(4)$ and $(2)=(3)$. Hence,

$180 + 0.5x = 6y$ ...... $(5)$

$210 + 0.5y = 6x$ ...... $(6)$

$(6) \times 12 + (5)$ gives: $2700 + 0.5x = 72x$

Solving the equation gives $x = 2700 / 71.5 = 5400 / 143 = 37\frac{109}{143}$, same as Barry's answer.

• Thnks,wow ! ur solution is pretty simple. – R K Mar 19 '15 at 18:28

Use hours, resp., full turns of hands, as units. The man enters at $6+x$ and leaves at $7+y$. When he enters we have $${\rm sh}={6+x\over12}, \quad{\rm lh}=x\ ,$$ and when he leaves we have $${\rm sh}={7+y\over12}, \quad{\rm lh}=y\ .$$ The interchanging of hands leads to the two equations $$x={7+y\over12},\quad y={6+x\over12}$$ with the solution $x={90\over143}$, $\>y=\ldots\$. Converting $x$ to minutes gives $37{109\over143}$ minutes.