However, I still have to say that these questions are not difficult, and you can't know any of them. Do it yourself in the future, think about it carefully, and ask someone else if you really can't think of it.
3. These four operations have no brackets. The algorithm is multiplication and division before addition and subtraction. If you want to get the maximum value, you must find ways to increase and decrease this number. Multiplication must be placed on the two largest numbers, and the result of division should be as small as possible and placed after the negative sign. This is 1-2 ÷ 3+4× 5 = 20 and 1/3.
4. Because diagonal addition requires equality, the upper and lower numbers of the middle number, the left and right numbers and the two numbers at both ends of the two diagonals must be equal. 3+ lower right corner =7+ upper right corner =4+★. That is, 3+ lower right corner +7+ upper right corner = 10+ upper right corner+lower right corner =2*(4+★)=8+2★. And the upper right corner +4+ lower right corner =3+★+7= 10+★. 10+ upper right corner+lower right corner-(upper right corner +4+ lower right corner) =6=8+2★-( 10+★)=-2+★. So ★=6+2=8
The price before the Chinese New Year is 100%, and a 20% discount during the Chinese New Year is a 20% discount. To restore the original price, it is necessary to raise the price (100%-80%)÷80%=0.25=25%.
6. big, increase. Comparing the bar charts in 2007 and 2003, we can see that the demand has increased more. The increase in oil demand is greater than the increase in oil supply, which depends on oil imports, so the dependence on oil imports also increases.
7. If Xiaoming missed three books, Xiaohong missed two, and Teacher Liu missed 20+3-2=2 1, accounting for1/4-40% = 35%. All the books have 2 1÷35%=60, so the number of books edited by Xiaohong and Xiaoming is 60-20=40.
8. The working efficiencies of Party A, Party B and Party C are110,115 and 1/20 respectively. Three people worked together for three days * * * to complete all 3 * (110+115+1/20) =13/20. The remaining 1- 13/20=7/20. Need (7/20) ÷ (115+1/20) = 3. The total number of days is 3+3=6.
9. No matter how long, the distance ratio between Party A and Party B is always 50: 40 = 5: 4. When they met, A walked 1/3 and 50 kilometers, and B walked 2/3 of the whole journey and 50 kilometers less. That is, 2/3-50km whole journey = (1/3+50km whole journey) *(4/5)= 4/ 15+40 whole journey. That is 2/3-50= 4/ 15+40. So the whole journey is (40+50) ÷ (2/3-4/ 15) = 225km.
10. The age gap between father and son remains unchanged. This year, the age difference between them is three times that of their sons (4- 1)/ 1 =, and after 1 1-5)/5 =. That is, 5/6 of the age difference minus 15 equals 1/3 of the age difference (that is, the age of the son minus 15 is the current age). Age difference = 15÷(5/6- 1/3)=30. So my son's age this year is 30÷3= 10.
1 1. Every 144 hours, A orbits the earth 144÷(9/5)=80 cycles. Satellite b orbits the earth 80-35=45 times. It takes 144÷45=3.2 hours for satellite B to orbit once. That is 3 1/5 hours.
12. Because p, p+ 1 is a prime number, it can only be 1 and 2 or 2 and 3. If p is large, at least one of p and p+ 1 is even, but not both prime. P= 1, p+3=4, not a prime number. If p=2, p+3=5 is a prime number. So the three prime numbers are 2, 3 and 5. Their reciprocal sum is1/2+1/3+1/5 = 31/30, and the reciprocal of the reciprocal sum is 30/3 1.
13. Two digits are A and one digit is B. The size of this two digit is 1 0a+B. The size of the new three digit is100a+B. Three digits plus1is eight times that of the original two digits. 1 00a+b+1-(10a+b) = 90a+1,and among multiples of 7, only 9 1 has two digits and1,so the original two digits are 9/kloc-.
14. 10/ 1 1, 1 1/ 12,… 18/ 19, 19/20。 Observe that the number of these columns is approximately equal to 1, and they are equal to11,112, …119,6544. The difference of 10 is less than 110, so the sum of the differences is less than1. That is to say, the sum of this number 10 plus a number less than 1 is equal to 10, so their sum must be between 9 and 10, more than 9 and less than 10.
15. Starting from home, because we are already at home, there is only one way, so write 1 here at home. Go to the next intersection, because there are only 1 roads here, so I still write 1. Go down or turn right from here, and the next 1 intersection is still only 1. All intersections are written as 1. At this time, if you go down from the intersection above or go right from the intersection below, you can reach the same intersection. There are two roads, so write 2 here. This intersection can only go down, so write 2 at the next intersection. ..... The rule is, if there is a road to the left and top of an intersection, add up the figures of the first two intersections and write them here. If an intersection can only be reached by the left or the upper road, continue to write the number of the intersection in front, indicating the number of routes from home to an intersection, so that the number on the whole picture is as follows (the picture displayed on the webpage will be deformed and copied to the notepad. -and | indicate roads, and dotted lines indicate blanks. )
1 - 1 - 1 - 1
|..........|..........|..........|
|.......... 1 - 2..........|
|..........|..........|..........|
1 - 2..........2 - 3
|..........|..........|..........|
|..........2 - 4..........|
|..........|..........|..........|
1 - 3 - 7 - 10
So there is 10 road to the school.
16. 10 shows 100000, and 10 shows 103000 at 0: 30.
During this period, the first two digits can only be 1, 0. The third number can't be 0, 1, and it can't be 3 at the same time, because when you take 3, it's only 10: 30, and the numbers after it are all 0, which is irrelevant, so the third number can only be 2. The first three numbers have been determined as 102, and only the possible values and arrangement of the last three numbers ABC need to be considered. The possible values of the last three numbers are as follows
A: Three, four, five, six, seven, eight and nine.
B: Three, four, five
C: Three, four, five, six, seven, eight, nine.
Group a and group c can be divided into two parts, namely a1= c1= b = (3,4,5); A2=C2=(6，7，8，9)
The possible values and ideal number of groups are as follows:
Choose one from A 1, b and C 1, and the number of possible combinations is 3*2* 1=6.
Choose one from A 1, b and C2 in turn, and the number of possible combinations is 3*2*4=24.
Choose one from A2, B, C 1, and the number of possible combinations is 4*3*2=24.
Choose one from A2, B and C2 in turn, and the number of possible combinations is 4*3*3=36.
The total number of combinations is 6+24+24+36=90.
Because there is only one combination of the first three numbers, there are 90 kinds of * * *.
17. The intersection of the diagonals of a semicircle and a square is O. Connect OB. It can be seen that the shadow area is 1/4- arch area of square area. A bow is a part surrounded by an arc OB and a line segment OB. And the area of arch = (area of semicircle-area of triangle OAB) ÷ 2 = (5 * 5 * 3.14 ÷ 2-10 *1/4) ÷ 2 = 7.65438+. The shadow area is10 *10 * (1/4)-7.125 =17.875cm2.
18. Both ends of the opening of the door are marked as AB, and the position of the cat is marked as O, which connects OA and OB and extends, and has an intersection point CD with the right lower wall, which is the line of sight of the cat. The rest of the room is the range of mice. Point C is the first brick on the right wall and point D is the second brick on the right wall. The area is 2 *1÷ 2+(3.5+1) * 2.5 ÷ 2 = 6.625 square meters = 66,250 square centimeters.
19. If the monthly expenditure in 800 yuan is less than 1000 yuan, that is, the monthly deposit is more than 200 yuan, then there will be 12800 yuan after two years (24 months). If you spend 1000 yuan and deposit it for two years, the deposit amount should be 200 * 24 = 4,800 yuan less than12,800 yuan. 12800-4800=8000, which means that the deposit after two years is still 8000 yuan. (8000-8000)÷6=0, that is, although you have saved for more than 6 months, the monthly new deposit is 0.
So Xiao Li's income is 1000 yuan, and his existing deposit is 8000 yuan.
20. Assuming that the total amount of brine after the first water addition is 100, the total amount of water is 100- 15=85, and the total amount of salt is 15. The amount of salt is unchanged, and it becomes 12% after the second water addition, so the total amount of salt after the second water addition is15 ÷12% =125. That is to say, the water added each time is 125- 100=25 parts. Then add 25 parts of water for the third time, and the salt content of brine is15/(125+25) = 0.1=10%.