An analogue clock is gaining three minutes every hour. If you set the correct time at 6p. m on Saturday. When will it show the correct time again?

6PM on Tuesday

1 week and 3 days after the initial Saturday

Explanation:

The clock gains 3 minutes every 1 hour.

We can represent that as the ratio below

3 min : 1 hour

There are 12 hours on a round analogue clock face. I'm assuming that there's no AM or PM indicators shown. In other words, something like 6 AM would look identical to 6 PM.

There are 12*60 = 720 minutes needed so that things wrap back around. If the clock is 720 minutes ahead, then there's no change and the person wouldn't know something is wrong (that the clock is 12 hours ahead).

We need to transform that "3 minutes ahead" into "720 minutes ahead".

720/3 = 240

We need to multiply both sides of that ratio by 240

So,

3 min : 1 hour

240*(3 min) : 240*(1 hour)

720 minutes : 240 hours

The clock will be 720 minutes ahead when 240 real time hours have passed.

240 hours = 240/24 = 10 days

10/7 = 1 remainder 3

10 days = (1 full week) + 3 extra days

The remainder 3 tells us how far to move ahead in the days of the week. Move 3 days from Saturday and you should arrive on Tuesday of the next week.

Therefore, it will show 6 PM correctly on Tuesday. More specifically, it will be 1 full week and then another 3 days to move from Saturday to Tuesday.

d 192

step-by-step explanation:

we find our area to find area we multiply thoses numbers to get our answer so our area is 192

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°