What number is 80% of 50% of 20% of 400?
Translate to an equation and then solve.
Solution: “What number” translates into a variable, I’ll use “n”. The word “is” translates into an equals sign. The three percents get translated into their decimal equivalents: .8, .5, and .2. The “of’s” gets translated into multiplication signs. And 400 is an integer which needs no translation. So the equation looks like this:
n = .8 x .5 x .2 x 400
n = 32
It is 6:20.
a) What is the acute angle formed by the hands?
b) What will that angle be when the time is 6:54?
a) As we established (see Problem 1), the big hand moves at a rate of 6? per minute, while the small hand moves at .5? per minute. Right now the big hand is on the 4 and the small hand is somewhere between the 6 and 7. So, the angle they form is: the angle from 4 to 5, plus the angle from 5 to 6, plus the sliver beyond 6 that the small hand has moved. The wedges between the numbers on an analog clock are all 30? (360?/12 = 30?). So from 4-5 is 30?, and from 5-6 is 30?; we just need to add to this whatever ground was covered by the small hand in the 20 minutes since it was 6:00 (and the small hand was exactly on the 6). If the small hand covers .5?/minute, in 20 minutes it sweeps out an angle of .5?(20) which is 10?. So, adding up our sectors, 30 + 30 + 10 = 70. The current angle formed by the hands is 70?.
b) At 6:54, the big hand will be just below the 11, and the small hand will be just before the 7. Counting up the full sectors, we have three of them: from 7-8, from 8-9, and from 9-10, all 30 each?. So the sector from the 7 to the 10 is 90?.
We have to add to this the angle between the 10 and the big hand, and the small angle between the 7 and the small hand. If the big hand moves at 6?/minute, then at 54 minutes past the hour it has covered 6?(54) = 324?. We know the gap from the 10 to the 12 is 60?. So think of the 10 as being 300? past the hour (360?-60?). That’s a gap of 324-300 = 24?. Adding that to 90 we are now at 114?. Lastly, if the hour hand moves at .5?/minute, then at 54 minutes past the hour it has moved .5?(54) or 27? past the 6, and is only 3? away from the 7. So the little sliver between the 7 and the hour hand is 3?; adding that to 114 we get 117.
At 6:54, the obtuse angle between the hour hand and the minute hand is 117?.
The time is 5:20. The acute angle formed by the clock hands is 40?. In exactly how many minutes will the big hand be right on top of the little hand? Show your work. Express your answer as a mixed number or as a decimal rounded to the nearest hundredth. Let N stand for “New angle”, let S stand for “Starting angle”, let m stand for time in minutes.
SOLUTION: When the big hand is right on top of the small hand with no space between them, they create an angle of 0?. So N, the new angle (i.e. the final angle) will be 0.?The time that will elapse in minutes is unknown. The starting angle is 40. The speeds of the two hands (see Problem 1) are: big hand, 6?/minute; small hand, .5?/minute. The big hand in this case is shrinking the angle, so we’ll subtract its movement from 40. The small hand will slightly grow the angle, so we’ll add its movement to 40. The general layout of the equation will be:
New angle = Starting angle – (big-hand-movement) + (small-hand-movement)
Plugging in our values:
0 = 40 – 6m + .5m . . . From here it’s just equation solving.
0 = 40 – 5.5m
5.5m = 40
m = 7 3/11 or 7.27 minutes
It’s 4:30. The acute angle formed by the big hand and the little hand is 45?. To the nearest tenth, what will that angle be when the time is 4:44? Solve by writing a formula which uses the following variables:
N = new angle
S = starting angle
SOLUTION: In the 14 minutes that transpire between 4:30 and 4:44, the big hand will greatly enlarge the angle as it moves away from the small hand; while the small hand will slightly shrink the angle as it “chases after” the big hand. So the general layout of our equation will look like this:
New angle = Starting angle *plus* Big-hand-movement *minus* Small-hand-movement
As established in Problem 1 (https://classsizeofone.com/clock-hands-math-problem-part-1/), the big hand moves at a rate of 6? per minute, while the small hand moves at .5? per minute. The time is known, and is the same for both hands: 14 minutes. So,
N = S + 6(14) – .5(14)
N = 45 + 84 – 7 = 122. . . . At 4:44, the acute angle formed by the hands will be 122?.
It’s 3:00 on an analog clock – the big hand is on the “12” and the little hand is on the “3”. The two hands form a 90 degree angle. What size angle will the hands form when the time is 3:10? Write a formula using the following variables, and then solve: S = starting angle, N = new angle.
SOLUTION: When the big hand moves from its current position, it approaches the small hand, shrinking the angle. When the small hand moves, it moves away from the big hand, enlarging the angle. So the general layout of the equation would go like this:
New angle = Starting angle *Minus* Movement-of-big-hand *Plus* Movement-of-small-hand
The hands move at different rates, so let’s figure out each of those rates. Imagine the big hand sitting on the “12” moves 5 minutes so it’s sitting on the “1”. That’s obviously a change of 5 minutes but it’s also a movement of 30?, since the original angle was 90? and moving to the “1” represents 1/3 of the distance between the “12” and the “3”. So, if the ratio is 30? per 5 minutes, that’s a unit rate of 6? per minute. (Another way to get there is this: There are 360? in a circle and 60 minutes on a clock face. 360/60 reduces to a ratio of 6:1.) So if the big hand moves 6? in 1 minute, then in 10 minutes it moves is 6?(10) or 60?. Meanwhile the small hand would take a full hour to go from its current position, “3”, to the next number, “4”, which also represents a movement of 30?. So if something moves 30? in 60 minutes, that means its unit rate would be 30/60 or .5 degrees per minute. If something moves .5 degrees in 1 minute, then in 10 minutes it moves .5?(10), which is 5?. So getting back to the general equation above,
N = S – (big-hand-movement) + (small-hand-movement)
let’s plug in the values for the degrees covered by the two hands, plus 90 for the starting angle which was given:
N = 90 – 60 + 5 … So at 3:10, the angle between the two hands is 35?.