So subtract half of the numerator from the denominator of 2/7 to get a form of the original fraction.
2/(7- 1)=2/6
If you look up the first sentence, subtract 2 from the denominator, and the subtraction is equal to 0 of 65438+2 points.
2/(6-2)=2/4= 1/2。 Meet the conditions.
So the original score is 2/6, and the simplest score is 1/3.
2. The rising distance of the water surface = the extra volume in the cylinder; The bottom region of a cylinder.
Excess volume = iron volume =12×12×12 =1728 cubic centimeters.
So the rising distance = 1728÷60÷50=0.576 cm.
Up to 20+0.576 = 20.576cm.
3. Three people add up to = boss+(boss× 3+1)+(boss× 0.5-1) = 4.5× boss =180.
So the number of stamps of Boss: 180/4.5=40.
Third: 40×0.5- 1= 19 pieces.
4. Let three numbers be A, B and C,
Then 35 is half a plus half b plus half c, 27 is half b plus half c plus half a, and 25 is half c plus half a plus half b.
So, add up the three numbers 35, 27 and 25,
This result is equal to two A's plus two B's plus two C's.
Therefore, A+B+C =(35+27+25)/2=43.5.
The final result has two decimal places.
So the decimal point must be added before the ten digits.
It becomes ABCD+AB. laser record
List vertical type
CD=98
AB= 19
The original four digits are 1998.
6. There are 3600 ÷ (1-40%) ÷ (1-25%) ÷ (1-20%) =10000.
Ethylene propylene * * * processing10000× (1-20%)-3600 = 4400 pieces.
7.1000+1000× 0.7 %×12 =1084 yuan.
8. After adding salt, the brine concentration is 1%, so the salt: water ratio is 1: 100.
If salt is 1, water is 100.
After adding 1 serving of salt, the concentration = [(1+1)/(100+1)] ×100% =1.98%.
9. The greatest common denominator is 15.
These two numbers must be selected from 15, 30, 45.
Any two of these three numbers add up to 50.
Therefore, 0 groups are eligible.
10、326025=3×3×3×3×5×5×7×23
One of the odd numbers must be 23,
The odd number before 23 is 2 1=3×7.
Odd number 25 after 23 = 5× 5
The odd number after 25 is 27 = 3×3×3.
So these four odd numbers are 2 1, 23, 25, 27.
1 1, backward from last time:
After the third oil pour from C to A and B, A, B and C are all 16kg.
Then before the third oil pouring, that is, after the second oil pouring from B to A and C, A and B each have 16/2 = 8kg of oil.
C poured 8+8 = 16kg to A and B, so C has 16+ 16 = 32kg.
Before the second oil pouring, that is, after the first oil pouring from A to B and C, A has 8/2 = 4kg of oil and C has 32/2 = 16kg of oil.
B poured 4+ 16 = 20kg for A and C, so B has 8+20 = 28kg.
Before the first oil pouring, that is, when the oil is not poured, B has 28/2= 14 kg, and C has 16/2=8 kg.
A reaches 14+8 = 22kg in B and C, so A has 4+22 = 26kg.
That is, a barrel of crude oil is 26kg, b barrel 14kg and c barrel is 8kg.
12, let the side length of a cube be a, after the cube becomes a cuboid, the areas of two vertices have not changed, but the areas of four sides have increased.
So the surface area of a cuboid is larger than that of a cube =3×a×4=60.
So, a=5.
The surface area of the cube =5×5×6= 150.
13. The master processes more pieces per hour than the apprentice 12 pieces.
Then it will take 4.5 hours to process 12×4.5=54 pieces.
That's 4.5 hours, and 54 disciples haven't finished.
At this time, I did it with the master. The master did 30, and the apprentice did 54-30=24.
The master earns 30-24 = 6 more than the apprentice.
Then these 54 parts were made by master and apprentice in 6÷ 12=0.5 hours.
That is, the apprentice made 24 in 0.5 hours.
So apprentices do 24÷0.5=48 per hour.
The least common multiple of 14, 3, 5 and 7 = 105
a/3 = 35a/ 105； b/5 = 2 1b/ 105； c/7= 15c/ 105
1.16×105 =121.8, so1.16 ≈122//kloc-.
35a+2 1b+ 15c = 122
Integers that meet the conditions: a= 1, b=2, c=3.
So the three scores are 1/3, 2/5 and 3/7.
() Fill in 1+2+3=6.
15. If you walk 80M per minute, you can get to school 6 minutes earlier.
At this time, if you walk for another 6 minutes, you can leave the school 6×80=480 meters at the specified time;
If you walk 50 meters per minute, you will be three minutes late.
Arriving at the school, it is 50×3= 150 meters away from the school.
That is, the first case can walk 480+150 = 630m from the second case within the specified time.
Specified time = 630(80-50)= 2 1 min.
Distance from home to school = 80× (2 1-6) = 1200m.
16, the first encounter can be 1 the whole journey.
The second encounter can go through three whole courses, and the number of encounters = (3+1)/2 = 2;
The third encounter can take five whole courses, and the number of encounters = (5+1)/2 = 3;
……
10 minute = 10×60=600 seconds.
At this time, Party A * * * walked 600× (3+2) = 3000m.
3000÷90=33……30
At this point, Party A and Party B walked 33 laps and added 30 meters.
Then they met (33+ 1)/2= 17 times.