把他们做成动画片让孩子懂---**工程问题**
【口诀】:
工程总量设为1,
1除以时间就是工作效率。
单独做时工作效率是自己的，
一齐做时工作效率是众人的效率和。
1减去已经做的便是没有做的，
没有做的除以工作效率就是结果。
例: 一项工程，甲单独做4天完成，乙单独做6天完成。甲乙同时做2天后，由乙单独做，几天完成?
[1-(1/6+1/4)×2]/(1/6)=1(天)
**植树问题**
【口诀】:
植树多少棵，要问路如何? 直的减去1，圆的是结果。
例 1: 在一条长为120米的马路上植树，间距为4米，植树多少棵?
路是直的。 所以植树 120/4-1=29(棵)。
例 2: 在一条长为120米的圆形花坛边植树，间距为4米，植树多少棵?
路是圆的， 所以植树 120/4=30(棵)。
牛吃草问题
[公式]
设定一头牛一天吃草量为 "1"
1) 草的生长速度=(对应的牛头数x吃的较多天数-相应的牛头数x吃的较少天数)/(吃的较多天数一吃的较少天数);
2) 原有草量=牛头数x吃的天数-草的生长速度x吃的天数;
3) 吃的 天数=原有草量/(牛头数-草的生长速度);
4) 牛头数=原有草量/吃的天数+草的生长速度
例题:
有一片草地上面的草每天生长的速度相同, 现在已知16头牛可以吃50天或者21头牛可以吃25天, 那么36头牛可以吃几天?
设每头牛每天吃的草量为1, 公式为:Y=(N-X)xT, 其中, Y表示原有草量, N表示牛的数量, X表示小草生长量, T表示所用时间。根据题意, 可得两个等量关系式:
Y=(16-X)x50, Y=(21-X)x25
解得X=11, Y=250
所以, 套入公式:250=(N-11)xT
当N=36时, 式子为:250=(36-11)xT
解得 T=10
**盈亏问题 (Surplus and Deficit Problems)**
**[口诀] (Rhyme/Formula):**
全盈全亏, 大的减去小的; 一盈一亏, 盈亏加在一起。
除以分配的差, 结果就是分配的东西或者是人。
**例 1: 士兵背子弹。**
每人 45 发则多 680 发;
每人 50 发则多 200 发,
多少士兵多少子弹?
全盈问题。大的减去小的。 则公式为:
(680-200)/(50-45)=96(人)
则子弹为 96×50+200=5000(发)。
**例 2: 学生发书。**
每人 10 本则差 90 本;
每人 8 本则差 8 本,
多少学生多少书?
全亏问题。大的减去小的。 则公式为:
(90-8)/(10-8)=41(人)
相应书为 41×10-90=320(本)
**例 3: 小朋友分桃子,**
每人 10 个少 9 个;
每人 8 个多 7 个。
求有多少小朋友多少桃子?
一盈一亏, 则公式为:
(9+7)/(10-8)=8(人)
相应桃子为 8×10-9=71(个)
Sum and Ratio Problems
Given the total, find the parts.
【Mnemonic Rhyme】:
The family wants to merge, dividing the property has principles.
The denominator is the sum of ratios, the numerator is one's own.
Multiply the sum by the ratio, and that's what you get.
Example: The sum of Jia, Yi, and Bing is 27, and Jia : Yi : Bing = 2 : 3 : 4. Find Jia, Yi, and Bing.
The denominator is the sum of ratios, which means the denominator is 2+3+4=9;
The numerator is one's own, so the proportions of Jia, Yi, and Bing relative to the sum are 2/9, 3/9, 4/9, respectively.
Multiply the sum by the ratio, so Jia is 27×2/9=6, Yi is 27×3/9=9, Bing is 27×4/9=12.
Difference and Ratio Problems (Difference and Multiple Problems)
【Mnemonic Rhyme】:
Mine is more than yours, the multiple is the reason. The numerator is the actual difference, the denominator is the difference in multiples.
The quotient is the value of one multiple, multiply by their respective multiples, and the two numbers can be found.
Example: Number Jia is 12 greater than Number Yi, and Jia : Yi = 7 : 4. Find the two numbers.
First find the value of one multiple: 12/(7-4)=4,
So Number Jia is: 4×7=28, Number Yi is: 4×4=16.
浓度问题 (Concentration Problems)
(1) 加水稀释 (Adding Water Dilution)
【口诀】(Mnemonic)
加水先求糖，糖完求糖水。糖水减糖水，便是加糖量。(Should be 加水量, adding water amount, not 加糖量, adding sugar amount, based on the title and example)
例: 有20千克浓度为15%的糖水，加水多少千克后，浓度变为10%?
(Example: There are 20 kilograms of sugar water with a concentration of 15%. After adding how many kilograms of water, will the concentration become 10%?)
加水先求糖，原来含糖为: 20×15%=3 (千克)
(Adding water first find sugar, the original sugar content is: 20 * 15% = 3 (kilograms))
糖完求糖水，含3千克糖在10%浓度下应有多少糖水,
(Sugar finished find sugar water, how much sugar water should there be with 3 kilograms of sugar at a 10% concentration,)
3/10%=30 (千克)
(3 / 10% = 30 (kilograms))
糖水减糖水，后的糖水量减去原来的糖水量，30-20=10 (千克)
(Sugar water minus sugar water, the later sugar water amount minus the original sugar water amount, 30 - 20 = 10 (kilograms))
(2) 加糖浓化 (Adding Sugar Concentration)
【口诀】(Mnemonic)
加糖先求水，水完求糖水。糖水减糖水，求出便解题。
(Adding sugar first find water, water finished find sugar water. Sugar water minus sugar water, the solution is found.)
例: 有 20 千克浓度为 15%的糖水, 加糖多少千克后, 浓度变为 20%?
(Example: There are 20 kilograms of sugar water with a concentration of 15%. After adding how many kilograms of sugar, will the concentration become 20%?)
加糖先求水, 原来含水为: 20×(1-15%)=17 (千克)
(Adding sugar first find water, the original water content is: 20 * (1 - 15%) = 17 (kilograms))
水完求糖水, 含 17 千克水在 20%浓度下应有多少糖水,
(Water finished find sugar water, how much sugar water should there be with 17 kilograms of water at a 20% concentration,)
17/(1-20%)=21.25 (千克)
(17 / (1 - 20%) = 21.25 (kilograms))
糖水减糖水, 后的糖水量减去原来的糖水量, 21.25-20=1.25 (千克)
(Sugar water minus sugar water, the later sugar water amount minus the original sugar water amount, 21.25 - 20 = 1.25 (kilograms))
**行程问题 (Distance Problems)**
**(1)相遇问题 (Meeting Problem)**
【口诀】: 相遇那一刻，路程全走过，除以速度和，就把时间得。
例:甲乙从相距120千米的两地相向而行，甲的速度为40千米/小时，乙的速度为20千米/小时，多少时间相遇?
相遇那一刻，路程全走过。即甲乙走过的路程恰好是两地的距离120千米。
除以速度和，就把时间得。即甲乙两人的总速度为两人的速度之和40+20=60(千米/小时)，所以相遇的时间就为120/60=2(小时)。
**(2)追及问题 (Catch-up Problem)**
【口诀】: 慢鸟要先飞，快的随后追。先走的的路程，除以速度差，时间就求对。
例:姐弟二人从家里去镇上，姐姐步行速度为3千米/小时，先走2小时后，弟弟骑自行车出发速度6千米/小时，几时追上?
先走的的路程，为3×2=6(千米)。
速度的差，为6-3=3(千米/小时)。
所以追上的时间为:6/3=2(小时)。
**Title:**
小学奥数经典口诀
(Primary School Olympiad Math Classic Jingles)
**Section 1: 和差问题 (Sum and Difference Problem)**
已知两数的和与差，求这两个数。
(Given the sum and difference of two numbers, find these two numbers.)
【口诀】:
和加上差，越加越大；除以2，便是大的；
(Sum plus difference, the more you add the bigger it gets; divide by 2, that is the larger number;)
和减去差，越减越小；除以2，便是小的。
(Sum minus difference, the more you subtract the smaller it gets; divide by 2, that is the smaller number.)
例:已知两数和是10，差是2，求这两个数。
(Example: Given the sum of two numbers is 10, and the difference is 2, find these two numbers.)
按口诀，则大数=(10+2)/2=6, 小数=(10-2)/2=4。
(According to the jingle, the larger number = (10+2)/2 = 6, the smaller number = (10-2)/2 = 4.)
**Section 2: 鸡兔同笼问题 (Chicken and Rabbit in the Same Cage Problem)**
【口诀】:
假设全是鸡，假设全是兔。
(Assume all are chickens, assume all are rabbits.)
多了几只脚，少了几只足？
(How many legs are too many, how many legs are too few?)
除以脚的差，便是鸡兔数。
(Divide by the difference in legs per animal, that is the number of chickens and rabbits.)
例:鸡兔同笼，有头36，有脚120，求鸡兔数。
(Example: Chickens and rabbits in the same cage, there are 36 heads, and 120 legs, find the number of chickens and rabbits.)
求兔时，假设全是鸡，则兔子数=(120-36×2)/(4-2)=24
(When finding rabbits, assume all are chickens, then the number of rabbits = (120 - 36 * 2) / (4 - 2) = 24)
求鸡时，假设全是兔，则鸡数=(4×36-120)/(4-2)=12
(When finding chickens, assume all are rabbits, then the number of chickens = (4 * 36 - 120) / (4 - 2) = 12)
**Diagram Description:**
* **Type:** Illustration
* **Main Elements:**
* A stylized open book is depicted at the bottom. It appears to be lying flat, with the spine facing forward and the pages slightly fanned out. The book is colored green.
* A sharpened pencil is shown positioned diagonally, resting on the open book, pointing towards the top right. The pencil is yellow with an orange/pink eraser and a grey lead tip.
* **Relative Position:** The book is at the bottom of the image, serving as a base. The pencil is on top of the book.
视频信息
答案文本
视频字幕
Welcome to engineering problems! These problems help us understand work efficiency. The key idea is to set the total work as one unit, and work efficiency equals one divided by the time needed. When people work together, we add their individual efficiencies.
Let's solve this example step by step. Jia's efficiency is one fourth per day, Yi's efficiency is one sixth per day. Working together for two days, they complete five sixths of the work. The remaining one sixth takes Yi exactly one more day to finish.
Tree planting problems have simple rules. For straight paths, we subtract one from the total positions because trees go at both ends but we count spaces between them. For circular paths, every position gets a tree with no subtraction needed.
Let's see both examples. The straight road needs twenty nine trees because we divide one hundred twenty by four and subtract one. The circular garden needs exactly thirty trees because we just divide one hundred twenty by four with no subtraction.
Welcome to our magical math adventure! Today we will learn classic formulas and tricks for primary school math olympiad problems. These special rhymes and formulas will help you solve math problems quickly and easily!
Engineering problems are about work and efficiency. We always set the total work as one unit. When we divide one by the time needed, we get the work efficiency rate. When working together, we add up everyone's efficiency rates.
In our example, worker A finishes in four days, so A's efficiency is one fourth. Worker B finishes in six days, so B's efficiency is one sixth. Working together for two days, they complete five sixths of the work. The remaining one sixth is finished by worker B alone in exactly one day.
Cow eating grass problems involve growing grass and multiple scenarios. We set one cow eating one unit per day as our standard. The key is finding the grass growth rate using two different scenarios with different numbers of cows and days.
In our example, we calculate that grass grows at rate eleven per day, and original grass amount is two hundred fifty units. With thirty six cows eating, they consume twenty five units more than growth rate, so the grass lasts exactly ten days.
Tree planting problems depend on the shape of the path. For straight paths, we subtract one from the total segments because trees are placed at endpoints and intervals. For circular paths, every interval gets a tree with no endpoints.
For the circular flower bed, every four meter interval gets a tree. Since it's a circle with no endpoints, we divide one hundred twenty by four to get thirty trees total.
The chicken and rabbit cage problem uses the assumption method. We assume all animals are one type, then calculate the difference in feet to find the real numbers. Each chicken has two feet, each rabbit has four feet.
If we assume all thirty six animals are chickens, we get seventy two feet, but we have one hundred twenty feet. The extra forty eight feet means twenty four rabbits. If we assume all are rabbits, we get one hundred forty four feet, which is twenty four feet too many, meaning twelve chickens.
To summarize what we have learned today: Engineering problems use total work as one unit and add efficiency rates. Cow eating grass problems require finding the growth rate first. Tree planting depends on path shape - subtract one for straight paths. Chicken and rabbit problems use the assumption method with foot count differences. Practice these formulas and you will master them quickly!
Surplus and deficit problems involve distribution scenarios. When all cases show surplus or all show deficit, we subtract the smaller amount from the larger amount. When one case shows surplus and another shows deficit, we add the surplus and deficit amounts together.
In the first example, both distributions have surplus bullets. We subtract two hundred from six hundred eighty and divide by the difference in bullets per soldier. In the second example, one distribution lacks peaches while another has extra, so we add nine and seven together.
To summarize what we have learned: Engineering problems use total work as one unit and add efficiency rates. Tree planting depends on path shape with different counting rules. Cow eating grass problems require finding growth rate first. Surplus and deficit problems use assumption methods. Practice these formulas and you will master them quickly!