Quizzes & Puzzles55 mins ago
please can you help 2
8 Answers
Question 4 - Trouble at Hubble
In 1990, the Hubble Space Telescope was launched. Twenty-one years on, it was in need of repair, and a rocket was sent into space to repair it. The rocket made good progress on the way up to the telescope, travelling at an average speed pf 3,000 mph. Whilst at the telescope, the rocket developed a fault and the rocket's speed on the return jurney was reduced to 1,000 mph. What was the overall average speed of the rocket?
Question 5 - Line Dancing
Mercury orbits the sun in 89 days, Earth orbits in 365.25 days. If, at midnight on 21st December 2010, the panets are in alignment on the same side of the sun, on which dates will Mercury pass between the Earth and the Sun during 2011?
Question 6 - Ma's Boys
Four astronaut brothers, Neil, John, Buzz, and Eugene, all do tours of duty on a space station. Neil returns to Earth every 9 weeks; Jhn returns every 10 weeks; Buzz goes home every 11 weeks; and Eugene every 12 weeks. Each visuit home is on a friday for 1 day, and is included in the number of weeks above. They are al home for their mother's 50th burthday party. How old wuil she be when she next has all 4 sons home together?
Question 7 - Space Men
Toy space men of different shapes are each formed from five 1cm cubes, with adjacent cubes firmly welded together face to face. To qualify for space flight, each space man has to be able to fit into a 3cm x 3xm x 3cm escape pod. How many such space men are there, each distinct from the other?
In 1990, the Hubble Space Telescope was launched. Twenty-one years on, it was in need of repair, and a rocket was sent into space to repair it. The rocket made good progress on the way up to the telescope, travelling at an average speed pf 3,000 mph. Whilst at the telescope, the rocket developed a fault and the rocket's speed on the return jurney was reduced to 1,000 mph. What was the overall average speed of the rocket?
Question 5 - Line Dancing
Mercury orbits the sun in 89 days, Earth orbits in 365.25 days. If, at midnight on 21st December 2010, the panets are in alignment on the same side of the sun, on which dates will Mercury pass between the Earth and the Sun during 2011?
Question 6 - Ma's Boys
Four astronaut brothers, Neil, John, Buzz, and Eugene, all do tours of duty on a space station. Neil returns to Earth every 9 weeks; Jhn returns every 10 weeks; Buzz goes home every 11 weeks; and Eugene every 12 weeks. Each visuit home is on a friday for 1 day, and is included in the number of weeks above. They are al home for their mother's 50th burthday party. How old wuil she be when she next has all 4 sons home together?
Question 7 - Space Men
Toy space men of different shapes are each formed from five 1cm cubes, with adjacent cubes firmly welded together face to face. To qualify for space flight, each space man has to be able to fit into a 3cm x 3xm x 3cm escape pod. How many such space men are there, each distinct from the other?
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Question 6:
she will be 88 years and 28 days old.
the least common multiple (LCM) of 9, 10, 11, 12.
...one way to get the least common multiple is to factor each number into primes:
9= 3∙3
10 = 2∙5
11 = 11
12 = 2∙2∙3
now create the shortest list that contains all the numbers needed to get any of the original numbers
LCD = 2∙2∙3∙3∙5∙11 = 1980
1980 weeks divided by 52 (weeks in a year) ≈ 38.077 years ≈ 38 years and 28 days
she will be 88 years and 28 days old.
the least common multiple (LCM) of 9, 10, 11, 12.
...one way to get the least common multiple is to factor each number into primes:
9= 3∙3
10 = 2∙5
11 = 11
12 = 2∙2∙3
now create the shortest list that contains all the numbers needed to get any of the original numbers
LCD = 2∙2∙3∙3∙5∙11 = 1980
1980 weeks divided by 52 (weeks in a year) ≈ 38.077 years ≈ 38 years and 28 days
Here's the maths:
Let the distance to the space telescope be D miles. Then the outbound time, in hours, T1 = D/3000
Similarly, the return time T2 = D/1000.
The overall average speed is the total distance travelled (2 x D), divided by (T1 + T2)
i.e Average speed = (2 x D) / (D/3000 + D/1000)
A fraction remains unchanged in value if we multiply or divide the top and bottom terms by the same amount. So we can get rid of D by dividing top and bottom of that fraction by it.
We've now got Average speed = 2 / (1/3000 + 1/1000)
In order to carry out the addition in that equation we must make both the numerators the same, giving
Average speed = 2 / (1/3000 + 3/3000) = 2 / (4/3000)
Remembering that dividing by a fraction is accomplished by 'turning upside down and multiplying', gives us
Average speed = (2 x 3000) / 4 = 6000/4 = 1500 mph
Chris
Let the distance to the space telescope be D miles. Then the outbound time, in hours, T1 = D/3000
Similarly, the return time T2 = D/1000.
The overall average speed is the total distance travelled (2 x D), divided by (T1 + T2)
i.e Average speed = (2 x D) / (D/3000 + D/1000)
A fraction remains unchanged in value if we multiply or divide the top and bottom terms by the same amount. So we can get rid of D by dividing top and bottom of that fraction by it.
We've now got Average speed = 2 / (1/3000 + 1/1000)
In order to carry out the addition in that equation we must make both the numerators the same, giving
Average speed = 2 / (1/3000 + 3/3000) = 2 / (4/3000)
Remembering that dividing by a fraction is accomplished by 'turning upside down and multiplying', gives us
Average speed = (2 x 3000) / 4 = 6000/4 = 1500 mph
Chris
An easier way to do the sums is to pretend that the distance to the telescope is 3000 miles.
That gives the outward time as 1 hour, and the return time as 3 hours.
So the average speed is the total distance (6000 miles) divided by the total time (4 hours) = 1500mph.
To make that a more rigid proof, the distance to the telescope should be written as 3000K, where K is a constant. The working is basically the same, with the 'K's cancelling out, so that they don't appear in the answer.
Chris
That gives the outward time as 1 hour, and the return time as 3 hours.
So the average speed is the total distance (6000 miles) divided by the total time (4 hours) = 1500mph.
To make that a more rigid proof, the distance to the telescope should be written as 3000K, where K is a constant. The working is basically the same, with the 'K's cancelling out, so that they don't appear in the answer.
Chris
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