Urgent! Two Mathematics Olympiad questions for the sixth grade of primary school

1. The picture is not well drawn, the blank part of the midpoint, the three intersection points are the center of three circles, and it is not difficult to see that a small shadow can be calculated according to an equilateral triangle with a side length of 5 cm corresponding to the fan, that is to say, the area of the shadow part can actually be seen as three equilateral fans with a side length of 5 cm.

2. Assuming that a is A's one-day work progress and B is B's one-day work progress, A+B=1 18 6A+10B=1-7 12 can be listed as 1 48, and B can be concluded to be 48 days alone.

1. Connect the three points of one of the shaded parts to form an equilateral triangle, from the figure you will find: the area of each shaded part = the area of the regular triangle + two bow areas - one bow area = sector area So we can find the fan area based on this small shadow part = 60 * 25 360, so the total area of the shadow is = 3 60 * 25 360 = 25 2

2. Item A A (A + B) = 18 A = A 18-B 6 A + 10 B = A5 12 6 (A 18-B) + 10 B = A5 12 A = 48 B!

Problem 2: To get the smallest surface area of the object, stick as many surfaces as possible and as large as possible.

First, the two small cubes are put together into a large cuboid (4 long, 2 wide, 2 high), which covers the two sides, and the surface area is reduced by 2*2*2=8 square centimeters.

Now the largest surface of the cuboid is a 4*2 face, and this side is glued to a cube with an edge length of 5 cm, and two more sides are covered, and the surface area is reduced by 4*2*2=16 square centimeters.

Calculation: The original surface area of three. 2*2*6*2+5*5*6=198 square centimeters.

Surface area of a new object: 198-8-16=174 square centimeters.

It's too troublesome to draw, so I have to express it in words, hoping to understand it.

Combining the two into one, imagine a and b as a large container (but with the same bottom area), and the water depth in the two containers is 300 (8 6 + 4 3) = 5 dm.

A container poured 8 * 6 * 5 = 240 cubic decimeters = 240 liters.

Pour 4 * 3 * 5 = 60 cubic decimeters = 60 liters in container.

1. Factor 1764.

Because of the difference between A and B is 4 rings.

So it's a combination of the following two.

2. A 120, B 100, C 70

Find the least common multiple of 4200, which is the total mileage required, divided by the slowest person, 4200 70 60 (minutes).

(2)300÷120=5/2

Find the least common multiple of several fractions: that is, find the least common multiple of the numerator of these fractions and divide it by the greatest common factor of the denominator, and the quotient obtained is the least common multiple of these fractions. ）

Salt content of 5% of salt water: 800 5% = 40 (grams) 400 grams of 20% salt content: 400 20% = 80 (grams) If you mix these 400 grams of brine and the 800 grams of water first, keep 20% of the 400 grams of salt water, and turn 800 grams of water into 5% salt water.

Then the salt in the 400 grams of brine should be: 40 + 80 = 120 (grams) So: the concentration of the third alcohol is:

120 400 = 2, when meeting the passenger car more than the truck: 3 2 = 6 (km), go 1-9 10 = 1 10, the bus go: 6 1 10 = 60 (km) the truck goes:

60 9 10 = 54 (km), after meeting, the bus walks another 54 km to reach the end point, at which point the truck goes:

54 9 10 = km). Trucks are still 60 km from station B).

1.Set the container with x grams of alcohol.

x+800*5℅)/(x+800)=20℅x+40=

x = 150 (grams).

Let the concentration of the third type of alcohol be one percent y, then.

150+400*y℅)/(150+800+400)=20℅150+400*y℅=270

y=30A: The concentration of the third type of alcohol is 30 percent

2.At the time of the encounter, the bus walked 3 [(1-9 10)*2] = 60 (km) and the truck walked 60*9 10 = 54 (km).

After the encounter, the bus walked 54 kilometers to station A, and the truck walked 54 * 9 10 = kilometers) The distance from station B is kilometers).