Answer to Olympiad Questions!?? Olympiad Mathematics Question Answer!

10 15 5) (7 2) 20 5 4 (years old) The age of the son 10 years ago.

4 7 28 (years old) father's age 10 years ago.

4 10 15 = 29 (years) The current age of the son.

28 10 15 = 53 (years old) The father's current age is the first time that they meet closer to point B, A walks 120 * 2 = 240 meters more than B, A is faster than B (80-60) = 20 meters per minute, then A and B both walk 240 20 = 12 minutes, and the distance between A and B is (80 + 60) * 12 = 1680 meters.

The second time is to meet closer to point A. It shows that B walked 120 * 2 = 240 meters more than A, B walked a total distance of 960 meters, A walked a total distance of 720 meters, B took 960 60 = 16 minutes, A took 720 80 = 9 minutes, A took 16-9 = 7 minutes less than B, indicating that A rested for 7 minutes.

1. The son is x years old, the father is 7 (x-10), and after 15 years, it is 2 (x+15)-5=7(x-10)+15, and the father is 42;

2. Because the speed of A is fast, the first meeting point is close to point B, because A is 20 meters faster than B per minute, so A walks 120 * 2 = 240 meters more than B when they meet for the first time, then the first encounter takes 240 20 = 12 minutes, then the meeting point is 12 * 80 = 960 meters from place A, and the distance from place B is 12 * 60 = 720 meters, and A needs (720 + 720) 80 = 18 minutes from the meeting point to point B and then turn back to the meeting point, It takes (960+960) 60=32 minutes for B to walk from the meeting point to point A and then back to the meeting point, because the second meeting point is the same as the first, so A rests for 32-18=14 minutes.

4 (years old) son 10 years ago.

4 7 28 (years old) father's age 10 years ago.

4 10 15 = 29 (years) The current age of the son.

28 10 15 = 53 (years) Father's current age.

1. Set the age of the current father and son to be x and y respectively

x-10=7(y-10)

x+15=2(y+15)-5

x = 14 y = 38 minutes.

1. Suppose the age of the son is x years old, then the age of the father ten years ago is 7x, and the age of the son after fifteen years is x+25 years old, then the age of the father after fifteen years is 2x+50-5 years old, and the age of the father ten years ago is 7x+25 years old, after 25 years it becomes 7x+25 years old, and the equation has 7x+25=2x+45, and the solution is x=4Then now the age of the son is 29 years old, and the age of the father is 43 years old.

2. If two people meet for the first time at 120 meters away from the midpoint, and then A rests and meets, it proves that B walks 120 * 2 = 240 meters, and B walks 60 meters per minute, then B is equal to walking 240 more divided by 60 equals 4 minutes, then A is resting for 4 minutes!

1. Now: Father: 38 years old, son 14 years old.

2. Because: 80t-60t=120m

So: t=6

AB distance: 840m

Midpoint o = 420m

A walks 420-120=300m, it takes minutes;

Then the two meet at a distance of 120 meters from the midpoint, and B takes time: (420+240) 60=11 minutes.

A rested: minutes.

Don't want equations, go straight to the answers, or do you want processes?

1. Calculation questions. (Off-form calculations, there must be a process!) ）

1. Solution: original =

2. (2 2 5 + 1 3x5) 3 2 5 + 342 3 7x7 9

Solution: Original (2 1 2 5 3 5) 17 5 (342 3 7)x7 9

3＋5/5）x5/17＋342x7/9＋3/7x7/9

4x5/17＋38x7＋1/3

267 and 26 51

3. (8 and 2 7 + 8 and 2 9) (5 7 + 5 9).

Solution: Original (8 8 2 7 2 9) 5x(1 7 1 9).

16＋2x16/63）÷〔5x16/63〕

16＋2x16/63）x63/5x16

16x63/5x16＋（2x16/63）x（63/5x16）

134, 2007 -2006 and 1 3 + 2005 -2004 and 1 3 + ....+

Original (2007, 2006, 2005, 2004......＋1－0）＋（1/2－1/3＋1/2－1/3……＋1/2）

1003x1＋1004x1/2－1003x1/3

1003x2/3＋502

668 and 2 3 502

1170 and 2 3

2. Multiple choice questions. (No process required, but make sure it's right!) Otherwise, ......No points are given).

1. Produce a batch of parts, after innovating the technology, produce the same number of parts, save 20% of time, and improve the efficiency of selection (a).

a、25% b、80% c、20% d、125%

2. A student used a computer to design a calculation program, when the input data is 10, the output data is (c).

Enter ......1 2 3 4 5 ……

Output ......2/5 3/10 4/17 5/26 ……

a、10/97 b、10/99 c、10/101 d、10/103

3. Use a batch of tiles with a length of 45 cm and a width of 30 cm to pave a square, and at least the number of tiles required is (a).

a、6 b、8 c、12 d、16

1. Solution: original =

2. (2 2 5 + 1 3x5) 3 2 5 + 342 3 7x7 9

Solution: Original (2 1 2 5 3 5) 17 5 (342 3 7)x7 9

3＋5/5）x5/17＋342x7/9＋3/7x7/9

4x5/17＋38x7＋1/3

267 and 26 51

3. (8 and 2 7 + 8 and 2 9) (5 7 + 5 9).

Solution: Original (8 8 2 7 2 9) 5x(1 7 1 9).

16＋2x16/63）÷〔5x16/63〕

16＋2x16/63）x63/5x16

16x63/5x16＋（2x16/63）x（63/5x16）

134, 2007 -2006 and 1 3 + 2005 -2004 and 1 3 + ....+

Original (2007, 2006, 2005, 2004......＋1－0）＋（1/2－1/3＋1/2－1/3……＋1/2）

1003x1＋1004x1/2－1003x1/3

1003x2/3＋502

668 and 2 3 502

1170 and 2 3aca

There are 58 basketballs and volleyballs, 45 volleyball and soccer balls, 77 soccer balls and basketballs, and how many basketballs, volleyballs, and soccer balls are there?

Solution: Adding the 3 numbers, we can find that each type of ball appears 2 times, so basketball + volleyball + football = (58 + 45 + 77) 2 = 90.

So football = 90-58 = 32.

Basketball = 90-45 = 45.

Volleyball = 90-77 = 13.

I'm a teacher, thank you.

There are 58-45 = 13 more basketball than football, and there are 77 more basketball and football, so basketball has (77+13) 2=45, and football has 77-45=32.

There are 45-32 = 13 in volleyball.

This is probably a question in the sixth grade or the first year of junior high school.

1.Take the square of (1000100101) as decimal as ( ).

2.The ten digits of a certain four-digit number plus 1 is equal to a single digit, and the single digit plus 1 is equal to a two-digit number, and the sum of the four digits in reverse order is equal to 10769, and the sum of the four-digit numbers is ( ).

3.The sum of a four-digit number and the digits on each digit is equal to 2001, and the four-digit number is ( )4If a four-digit number is equal to 2 times the sum of the first and last digits, then the four-digit number can be ( ).

5.If a six-digit abcdef is abcdef=1 6defabc, find the six-digit abcdef

6.For a certain four-digit number, the number on the ten digits is equal to the number on the single digit plus 2, and the number on the single digit plus 2 is equal to the number on the hundred digits, and the sum of the number formed by the four digits in reverse order is 9988, and the four digits are found.

7.A password is a two-digit number, and the sum and product of the two digits are equal to the two-digit number, and all the two-digit numbers are found.

8.Let a, b, and c be three consecutive positive integers, the square of a is equal to 14884, and the square of c is equal to 15376, find the square of b.

9.Let n be a positive integer, and the quotient 9 and remainder r obtained by dividing n(n+1) by 302 are positive integers, then the sum of the maximum and minimum values of r is ( ).

10.Let a724b be a multiple of 12 and find the maximum value of ab.

11.Find the largest integer ( ) that is divisible by the sum of any 3 consecutive certificates

a. 1 b. 2 c. 3 d. 6

12.Let abc+bac+bca+cab+cba=3194 find abc

13.Remove 63, 91, 130 with n, and the sum of the 3 remainders is 26, find n

14.The sum of any two digits in a three-digit number is divisible by the third digit, then there are ( ) such three-digit numbers.

1. Set A1 and A2 ,..a2002 is a positive real number, and 1 (2+a1)+1 (2+a2)+1/(2+a2002)=1/2

then a1*a2*.The minimum value of a2002 is:

2. If you put 4 small balls with a radius of 1 into one ball, the minimum radius of the big ball is

3. In the acute angle abc, a 2 b, then the value range of bc ac is

4. There are 52 people in a class, half male and half female. Men and women are divided into two groups equally, and four people are selected from this class to participate in an activity.

The probability of being from a different group is

5. A straight line is separated from the circle, there are 6 points on the straight line, there are 4 points on the circumference, and the minimum can be a straight line through two points.

6. The diameter of the bottom surface of the cylinder is 4R, the height is 42R, and the number of balls with a radius of R is at most.

40 salt water and 20 salt water are mixed to make 25 salt water 900 grams, ask how many grams of salt water are needed for 40 and 20 salt water?

This year's grandma is 70 years old, Xiao Ming is 10 years old, and how many young grandmothers are 5 times that of Xiao Ming?

9.75, the five digits of the five digits are 7 5 5 3 1

5 and 10 are not pictured.