1. Questions and Answers on Olympiad Mathematics in the Sixth Grade of Primary School
Bind exercise books with a batch of paper. If 120 volumes have been bound, the remaining paper is 40% of this batch of paper; If 185 copies are bound, 1350 copies remain. How many sheets of paper are there in this batch? Answer and analysis: Method 1: 120 books correspond to 60% of the total number (1-40%=), so the total number is 120÷60%=200 books. When 185 books are bound, there are still 200- 185: 15 books unbound, corresponding to 1350, so each book needs 1350÷ 15=90. That is, there are 18000 sheets of paper.
Method 2: Bind 120 copies, leaving 40% of the paper, that is, using 60% of the paper. Then to bind 185 copies, you need185× (60% ÷120) = 92.5% of paper, that is, the remaining 1-92.5%=7.5% of paper is/kloc-. So this batch of paper has 1350÷7. 5%= 18000 sheets.
2. Test questions and answers of Olympiad Mathematics in the sixth grade of primary school
A and B are going to explore the desert. They go deep into the desert for 20 kilometers every day. It is known that each person can carry one person's food and water for up to 24 days. If some food is not allowed to be stored on the way, how many kilometers can one of them go deep into the desert (the last two have to return to the starting point)? What if some food can be stored on the way back? Answer and analysis:
It can go 360 kilometers into the desert.
Suppose A goes back after X days, A leaves the food he needs when he goes back, and the rest is transferred to B. At this time, B*** has (48-3X) days of food, so X=8. Of the remaining 24 days of food, B can only go forward for 8 days, leaving 16 days of food for him to return, so B can go deep into the desert.
If the conditions are changed, the crux of the problem is the food left over in B24 days when A returns. Because 24 days of food can make B go deep into the desert alone 12 days, and the other 24 days of food will provide A and B with a walk back and forth, that is, 24÷4=6 days, so B can go deep into the desert 18 days, that is, one day.
3. Sixth-grade olympiad questions and answers.
Sixth grade students take part in the school math competition. There are 50 test questions. The grading criteria are: 3 points for a correct answer, 1 point for a wrong answer and 1 point for a wrong answer. Please note: the total score of students in this class must be even. Answer and analysis: If all 50 questions are correctly answered, * * * will get 150 points, which is an even number. Every time you answer a wrong question, you will miss 4 points. No matter how many questions you answer wrong, the multiple of 4 is always even. 150 MINUS even number, difference or even number. Similarly, every time you don't answer a question, you will be 2 points short. No matter how many questions you answer, the multiple of 2 is always even, and the sum of even and even is even. So every student's score in the class is even. Then the sum of the class scores must be even.
4. Sixth grade Olympic math test questions and answers
It is known that the number of students in A school is 40% of that in B school, the number of girls in A school is 30% of that in A school, and the number of boys in B school is 42% of that in B school. Then, the number of girls in the two schools accounts for () of the total number of students in the two schools. Test center: the practical application of percentage.
Analysis: 40% and 42% of the unit "1" are students in school B, so the number of students in school A is 40%, and the number of girls in school B is1-42%; The number of girls in a school is 30% of the number of students in a school, so the number of girls in a school is 40% × 30%; Then divide the number of girls in the two schools by the total number of students in the two schools.
Answer: Solution: The number of girls in a school is 40%×30%= 12%.
Number of girls in school B:1-42% = 58%;
( 12%+58%)÷( 1+40%),
=70%÷ 140%,
=50%;
A: The number of female students in the two schools accounts for 50% of the total number of students in the two schools.
So the answer is 50%.
Comments: The key to solve this problem is to distinguish the difference between the two units "1", find out who each takes as the standard, and then solve the problem according to the quantitative relationship.
5. Question and answer of the sixth grade Olympic Mathematics in primary school
It is known that the number of students in A school is 40% of that in B school, the number of girls in A school is 30% of that in A school, and the number of boys in B school is 42% of that in B school. Then, the number of girls in the two schools accounts for () of the total number of students in the two schools. Answer and analysis:
Test center: the practical application of percentage.
Analysis: 40% and 42% of the unit "1" are students in school B, so the number of students in school A is 40%, and the number of girls in school B is1-42%; The number of girls in a school is 30% of the number of students in a school, so the number of girls in a school is 40% × 30%; Then divide the number of girls in the two schools by the total number of students in the two schools.
Answer: Solution: The number of girls in a school is 40%×30%= 12%.
Number of girls in school B:1-42% = 58%;
( 12%+58%)÷( 1+40%),
=70%÷ 140%,
=50%;
A: The number of female students in the two schools accounts for 50% of the total number of students in the two schools.
So the answer is 50%.
6. Question and answer of the sixth grade Olympic Mathematics in primary school
Three people, A, B and C, each shot three times. The product of the number of rings in their respective targets is that according to the total number of rings in a single target, from high to low, they are A, B and C. Who shot the fourth ring on the target? (The number of rings is a natural number that does not exceed) The analysis shows that the product of the number of rings of three people and three gun targets is 60, that is, the number of rings of each gun target is a divisor of 60. The prime factor of 60 decomposition is 60=22×3×5, and because the number of rings per gun does not exceed 10, there are only four situations in which 60 is written as the product of three natural numbers that do not exceed 10:
60=3×4×5; 60=2×6×5; 60=2×3× 10; 60= 1×6× 10。
Among them, the total number of rings is 12, 13, 15 and 17 respectively. In the case of four rings, ① the total number of rings is the least, so the four rings are played by C.