Primary Olympics Formula 1

Basic mathematical formula

 

1. Number of Parts X number of parts = Total number of parts = number of parts

2. Multiples of 1 x multiples = multiples of a few values; multiples of 1 = multiples of a certain number of values; multiples of a given number = multiples of 1

3. speed X time = distance between minutes speed = time distance between minutes time = speed

4. unit price × quantity = total price ÷ unit price = Total Quantity ÷ quantity = unit price

5. Work Efficiency × work time = total work volume worker work efficiency = total work volume during work hours worker work time = Work Efficiency

6. Addition + addition = and-one addition = another addition

7. subtrahend-subtrahend = difference subtrahend-Difference = subtrahend + subtrahend = subtrahend

8. factor X Factor = product accumulation one factor = another factor

9. divisor = quotient divisor operator = divisor × divisor = Divisor

1. Square

C perimeter S area a side length perimeter = side length × 4 C = 4a

Area = side length x side length S = a ×

Surface area = Rib length × 6 S table = a × 6 volume = Rib length × Rib length V = a ×

3. rectangle

C perimeter S area a side long perimeter = (Length + width) × 2 C = 2 (a + B)

Area = long x wide S = AB

4. Cube

V: volume S: Area A: Long B: wide H: high (1) surface area (long x Width + long x height + width x height) x 2

S = 2 (AB + AH + bH)

(2) volume = length x Width x height v = ABH

5. triangles

S area a bottom h high area = bottom X high Region 2 S = Ah middle 2

Triangle Height = Area × 2 base triangle Bottom = Area × 2 height

6. Parallelogram

S area a bottom h high area = bottom X high S = ah

7. Trapezoid

S area a upper bottom B lower bottom h high area = (upper bottom + lower bottom) X high bottom 2 S = (a + B) x H bottom 2

8. Circle

S area C perimeter d = Diameter R = radius

(1) circumference = diameter x Diameter = 2 x radius X radius c = diameter d = 2 kg/r

(2) Area = radius X region

9. Cylinder

V: volume H: High S; bottom area R: bottom radius C: Bottom perimeter

(1) side area = bottom circumference × high (2) Surface Area = side area + bottom area × 2 (3) volume = bottom area × high (4) volume = side area ÷ 2 × radius

10. Cone

V: volume H: High S; bottom area R: bottom radius

Volume = base area × high cost 3 Total number of parts = average

And difference formula

(And + difference) limit 2 = large number (and-difference) limit 2 = decimal

And times Problems

And decimal places x multiples = large numbers (or-decimal places = large numbers)

Difference

Decimal decimal number x multiple = large number (or decimal number + Difference = large number)

Tree planting

1. Tree Planting on non-closed lines can be divided into the following three situations:

(1) If tree planting is required at both ends of a non-closed line, then: number of plants = number of segments + 1 = total length of plants-1 total length = plant distance × (number of plants-1) plant distance = full-length growth (number of plants-1)

(2) If planting trees at one end of a non-closed line and at the other end do not, then:

Number of plants = number of segments = total length of the two plants = total length of the four plants

(3) If no trees are planted at both ends of a non-closed line, then:

Number of plants = number of segments-1 = total length of plants-1 total length = plant distance × (number of plants + 1) total length of plants = total length of plants (number of plants + 1)

2 The relationship between the number of tree planting problems on closed lines is as follows:

Number of plants = number of segments = total length of the two plants = total length of the four plants

Profit and Loss Problems

(Profit + loss) the difference between two distributions = number of participating distributions (large profit-small profit) The difference between the two distributions = number of participating distributions (large loss-small loss) the difference between two assignments = number of participating assignments

Encounter problems

Encounter distance = speed and × encounter time = encounter distance encounter speed and = encounter distance encounter time

Tracking Problems

Tracking distance = speed difference x tracking and time = chasing and distance tracing speed difference = tracing and distance tracing time

Concentration problems

The weight of the solutes + the weight of the solvent = the weight of the solution × 100% = the concentration

Weight of solution × concentration = weight of the solutes × weight concentration = weight of the Solution

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Octa weekly special training (IV)

1. Vehicle A and vehicle B start from a and vehicle B respectively. When starting, the speed ratio of A and B is. After the meeting, the speed of a is reduced by 20%, and that of B is increased by 20%. In this way, when a reaches B, B is 10 km away from location. Then A and B are ___ kilometers apart.

[Solution] the original speed ratio of A and B is, and the speed ratio after the meeting is

5 × (1-20%): 4 × (1 + 20%) =. 8 =.

When they met, Jia and went through the entire process and.

A and B are 10 km apart (-×) = 450 (km)

2. At a.m., two cars left the fertilizer factory and drove to Xingfu village. The speed of both vehicles is 60 km per hour. At 08:32, the first car was three times the distance from the fertilizer plant. By 08:39, the distance between the first car and the fertilizer plant was twice that of the second one. So the first car left the fertilizer factory at a.m?

[Solution] 39-32 = 7. The distance of each vehicle in the seven minutes is exactly equal to 1 (= 3-2) times the distance of the second car passing by 08:32, so the first car had been running 7 × 3 = 21 (points) at 08:32, and it left the fertilizer plant at 08:11 (32-21 = 11)

Note: The conclusion of this question is irrelevant to the speed of the two vehicles. The answer is 08:11 as long as they are at the same speed.

3. Both vehicles a and B depart from location a and travel to Location C. The distance between locations A and B is equal to the distance between locations B and C. The speed of Car B is 80% of that of car. It is known that car B leaves 11 minutes earlier than car A, but has stayed in B for 7 minutes. Car A cannot stop sailing to C. Finally, B arrived at Location C four minutes later than. Then, the number of minutes after B's departure will exceed that of B's.

[Solution] from location a to Location C, do not consider the stopover. It takes 8 minutes for Car B to stay in the middle of the road. It takes four minutes for Car B to stay in the end,

Therefore, chelaile A is over B in the middle of the line B. The time used by chelaile A is 80% of the time used by B.

8 bytes (1-80%) = 40 (minutes). The total duration of Row A is 40-8 = 32 (minutes)

Use 32 minutes 2 = 16 (minutes) for line A to B)

That is, after departure from B, 11 + 16 = 27 (minutes). Car A exceeds B.

4. On an equal road next to the railway, a pedestrian and a cyclists are traveling south at the same time. The pedestrian speed is 3.6 km/hour, and the cyclists speed is 10.8 km/hour. At this time, there was a train coming from behind them. It took 22 seconds for a train to pass through a pedestrian, and 26 seconds for a passenger to pass through the cyclists. The total length of the Train Body is ____ (① 22 meters ② 56 Meters ③ 781 meters ④ 286 meters ⑤ 308 meters)

[Solution] set the train speed to X meters/Second, the pedestrian speed to 1 meter/second, And the cyclists speed to 3 meters/second. The length of the Train Body is (x-1) × 22 = (X-3) × 26 reduced to 4 x = 56, that is, x = 14 (meter/second) therefore, the total length of the Train Body is (14-1) × 22 = 286 (meters), so select ④.

6. A driver drove from City A to City B. If you move forward at the original speed, you can arrive on time. When the distance is halfway through, the driver finds that the actual average speed of the first half can only reach 11/13 of the original speed. Now the driver wants to arrive at City B on time. In the next half of the trip, the actual average speed is _______ compared with the original speed _______.

[Solution] the first half of the journey was originally scheduled and used-1 =. To arrive on time, the last half of the journey can only be 1-= of the original time, so the speed of the last half of the trip is the original speed, that is

7. A and B depart from both Station A and Station B at the same time. The first encounter was at 28 km from Station A. After the match, the two cars continued to travel, after arriving at Station B and Station A, they immediately returned along the original road. The second encounter was at 60 km from Station. The distance between A and B is ___ km.

[Solution] The first encounter between A and B is in place C. At this time, the sum of the routes between A and B is equal to the distance between A and B.

A and B met at D for the second time, and B went from C to A to D in the opposite direction. A total of 60 + 28 = 88 (km ), from C to B and then from the opposite direction to D. At this time, the sum of the routes of A and B is two times the distance between A and B, so the second and the second are two times the distance between A and B, the distance between Party A and Party B is twice that of their own journey when they first met each other. In this way, the first time B met, the path BC = 88 km 2 = 44 (km ). Thus AB = 28 + 44 = 72 (km)