How many primary school math Olympiad questions can I ask? To step

Question 1: According to the original price**, the profit margin per product is 125% 90% 1 The original profit is 25 pieces 1

Now the profit is 2 25 pieces, 1 125 pieces, 1 so the sales volume after the price reduction is double the original.

Question 2: The question of "Chinese Remainder Theorem".

Look at the first two first, and meet the first two with a minimum of 30 2 28.

So the number of apples is 28 30 4

So there are 118 of them.

The original price is 1 + 25% = 125% of the cost

Later, the selling price is 125%*90%=

Profit per piece is:

The total profit is 25%*(1+25%)=

The sales volume is the original: double.

The number of apples, divided by 10 by 8, by 15 by 13, by 17, by 16 plus 2, divisible by 10 and 15, by 17 The least common multiple of 110 and 15 is 30

The minimum number that is divisible by 30 and divides by 17 is 120, so the least number of apples is: 120-2=118.

1. If the cost is A, the sales volume before the price reduction is X, and the sales volume after the price reduction is Y, there is the following relationship:

Original Pricing: Pricing after Price Reduction:

Original profit: p1 =

Now profit: P2 = (a)*y

Then the title is: p2 =

In turn, we can get the proportional relationship between x and y.

2. According to the title:

10a+8 = 15b+13 = 17c+16 where a, b, and c are all natural numbers.

From 10a + 8 = 15b + 13 we get 2a = 3b + 1 ...1)

From 10a + 8 = 17c + 16, we get 10a = 17c + 8 ...2)

From (2), we can see that c is at least 6, where a = 11, b = 7

So there are at least 118 apples.

I don't know if there are specific solutions such as unknowns in the Primary School Olympiad, but the idea is like this, and you can change it appropriately according to your own actual situation.

2. According to the meaning of the topic, you may wish to set up a total of m apples, of which every 10 are counted x times, every 15 are counted y times, and every 17 are counted z times, and the equation is listed.

10x+8=m 15y+13=m 17z+16=m simplified to 10x=15y+5=17z+8

Analysis: y must be odd, 17z, the single digit must be 2, then when z=6, the substitution test is in line with the topic.

So m=118 has at least 118 apples.

Don't do two questions. Excuse me.

There's no easy way for you to have a problem.

=(3791+

Isn't it supposed to be in this formula?

1.Original =

2.Original =

3.Original = 12 * 12 12 13 * 13 = 124Original =Wrong title, right?

1.Find the greatest common number of .

Find the length, width, and height of 6 times each.

Find the total number of blocks. 6 4 3 72 (block).

2.Solution 1: (10,12,15) 60, set a wooden stick of 60 cm.

60 10 6 cm, 60 12 5 cm, 60 15 4 cm.

The 10 equal divisions are the first type of tick marks, a total of 10 1 9 bars.

The 12 equal divisions are the second type of tick marks, a total of 12 1 11.

15 equal divisions are the third type of tick mark, over 15 1 14.

The first type coincides with the second type of tick mark (6, 5) 30, 60 30 1 2 1 1.

(6,4) 12,60 12 1 5 1 4 that coincide with the third tick mark.

The second and third tick marks coincide with (5, 4) 20, 60 20 1 3 1 2.

The three tick marks coincide without ) 60

Thus, there are 9 11 14 1 4 2 27 tick marks, and the sticks are sawn into a total of 27 1 28 segments.

Solution 2: The total length is divided into segments as unit 1. The least common multiple of 1 10 and 1 12 is 1 2, 1 10 is the least common multiple of 1 15 1 5, 1 12 and 1 15 The least common multiple of 1 3, 1 10, 1 12 and 1 15 is 1, there is 10 + 12 + 15 - (2 + 5 + 3) + 1 = 28

The least common multiple of solution 3 is 60, assuming that the stick is 60 long, 1, then, divided into 10 equal parts of each 6, the scale is.

2. Each part divided into 12 equal parts is 5, 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60

3. Each part divided into 15 equal parts is 4, 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60

4. Merge the same scale, that is, there are scales as follows:

Therefore, there are 28 paragraphs at the end.

3.Year Year Year (Year 11) (Year 11) (Year 11) 121 Year Old Flower is similar.

i.e. flowers similar to 121 years old. Flower resemblance is a three-digit number, 121 (1 9) to exclude: first year and year cannot be equal to 1, so:

It can be ruled out that they can only be broken down into 1 times themselves, such as 2:

Year 11 22 121, with repetition.

9 doesn't work, 121 9 1089, not a three-digit number.

Only, one by one, it can be seen that the flowers are similar to 121 8 968. Age 8.

Years and years are different from people to know the years of the year. Longitudinal: Year 4, age 2.

People of different ages 2 4

0 9 remains, apparently people are different 5 10.

If the original purchase price x is the current purchase price.

If the selling price is y, then the original profit margin (y-x) x

The new profit rate ( = 6 100

The solution is y= i.e., the original profit rate is 14%, then the current profit rate is 20%, and y= is substituted into the new profit rate (

Verify the original formula =

1. Set the purchase price to A sales price B

2. The original profit margin is (b-a) a

3. The current profit margin is (

4. Profit increased by 6%.

Then (b-a) a+6%=(

b-a+6%a=(

1/19b= b=

5. Therefore, the original profit margin is (b-a) a = ( =14%, so the current profit rate is 14% + 6% = 20%.

Let the selling price be P and the purchase price be C, which can be obtained from the question:

p ( 1 ) (p c 1) 6 100 so, (p c) (1 1).

p c So, now the profit from selling this dress is.

p/c －1＋

Let the original purchase price be x and the selling price be y, then (profit =