0
$\begingroup$

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.

enter image description here

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I got the same answer as you do, with a different way of calculation. $\endgroup$ – LaBird Mar 19 '15 at 16:16
  • $\begingroup$ 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 $x = 37\frac{109}{143}$. $\endgroup$ – LaBird Mar 19 '15 at 16:22
  • $\begingroup$ @LaBird, many thanks for the confirmation. You might want to post your calculation as a separate answer. $\endgroup$ – Barry Cipra Mar 19 '15 at 16:38
  • $\begingroup$ $\quad~~~$thnks! $\endgroup$ – R K Mar 19 '15 at 18:28
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thnks,wow ! ur solution is pretty simple. $\endgroup$ – R K Mar 19 '15 at 18:28
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.