Calculating right time in clock angle problem

Question

A clock is set right at 5 a.m. The clock loses 16 minutes in 24 hours.What will be the true time when the clock indicates 10 p.m. on 4th day?

I am completely messed up in this question. Can anyone please explain it with each step.

Answer 1

The clock loses 16 minutes every 24 hours, which means that it loses $$ \frac{16\text{ minutes}}{24 \text{ hour}} = \frac{2}{3} \frac{\text{min}}{\text{hr}}$$ minutes every hour. In order to figure out what time it really is when the clock says 10 PM on the 4th day, we need to figure out how much time has really past since 5 AM on the first day, and compare that to how much time the clock indicates has passed. Let us call this actual number of hours $T$ (or the Time that has passed), and let $S$ be the number of hours that the clock Shows have passed. Note that $$ 60 \frac{\text{min}}{\text{hr}}\cdot S\text{ hr} = 60\frac{\text{min}}{\text{hr}}\cdot T \text{ hr} - \frac{2}{3}\frac{\text{min}}{\text{hr}}\cdot T \text{ hr}. $$ That is, the time shown is the actual time, minus $\frac{2}{3}$ of a minute for every hour that has actually passed. Now, we actually know what $S$ is: the clock indicates that 3 days and 17 hours have passed, which means that (dropping the units, because I am tired of editing them in) $$ S = 3\cdot 24 + 17 = 89. $$ Solving for $T$, we get $$ 60 S = \left( 60 - \frac{2}{3} \right)T = \frac{180-2}{3} T = \frac{178}{3} T \implies T = \frac{60\cdot 2}{178}S = \frac{180}{178}S = \frac{90}{89}(80) = 90. $$ Thus the actual amount of time that has elapsed is 90 hours. That is, 3 days (72 hours) and 18 hours have elapsed. Thus the actual time is 18 hours past 5 AM, or 11 PM on the 4th day.

Answer 2

When we say a clock "loses" 16 minutes each day. What does that precisely mean? At the end of the day, the clock reads $11:44$. Thus, when $1$ hour has elapsed, the clock has "lost" $\frac23$ of a minute, or $40$ seconds. Let's use the word "cl-minutes" to denote what this clock calls a minute. Our conversion factor is this:

$$59\frac13\text{ cl-minutes} = 60\text{ minutes}$$

Now, from $5$am on day $1$ to $10$pm on day $4$ is a duration of $19+24+24+22$ hours, or $89$ cl-hours total. That's $5340$ cl-minutes.

$$\begin{align} 5340\text{ cl-min}\times\frac{60\text{ min}}{59\frac13\text{ cl-min}} &= \left(5340\times60\div\frac{178}{3}\right)\text{ min}\\ &= 5400\text{ min}\\ &= 90\text{ hours} \end{align}$$

It is actually $11:00$pm

Answer 3

Let us interpret what "losing $16$ minutes in $24$ hours" means: when $24$ true hours pass, the clock will show that $23$ hours and $44$ minutes have passed (thus, it "lost" $16$ minutes). I will denote with $h$ true hour, and with $fh$ fake hour that the clock shows, so we can write the sentence as $$24\, h = (24-\frac{16}{60})\, fh$$

(division by $60$ is because $1$ hour = $60$ minutes)

Dividing by $24$ we get

$$1\, h = (1 - \frac{16}{24\cdot 60})\, fh = (1-\frac{2}{3\cdot 60})\,fh =(1- \frac 1{90})\,fh = \frac{89}{90}\, fh,$$

and if we multiply by $90$, we get that $90\,h = 89\,fh$.

So, let us say that we started at $5\,\mathrm{am}$ Monday and ask ourselves how much (fake) time had passed when the clock shows $10\,\mathrm{pm}$ Thursday. It is $3$ (fake) days to get to $5\,\mathrm{am}$ Thursday plus $17$ (fake) hours to get to $10\,\mathrm{pm}$ Thursday, and the total is $(3\cdot 24 + 17)\, fh = 89 fh$, but $89\, fh = 90\,h$, so to get true time, we need to add $90\,h$ to $5\,\mathrm{am}$ Monday. Luckily, we can use what we know, that if we add $89$ hours, we will get to $10\,\mathrm{pm}$ Thursday, and adding another hour gives us $11\,\mathrm{pm}$ Thursday of true time.