Grade 6 application questions 5 points .

Set a total of x kilograms.

35%x+160=[3/(2+3)]x

x = 640 for a total of 640 kg.

Set each phase x meters.

When the younger brother walked to the 16th root, he walked together.

16-1) x = 15x meters.

Since the time is consistent, the speed ratio is equal to the distance ratio.

So the distance that the elder brother walked

l：15x=84：40=21：10

l=so. The elder brother walked between the 32nd and 33rd roots1 2.

1.Let the total number be x, and the equation is:

35%x+160）/x=3/（2+3）

Finding x=640 is verified, correct.

2。Let the distance between the two poles be one unit, and the younger brother walks to 16 is to walk 15 units of jumping, so.

So the elder brother walked between the 32nd and 33rd poles.

Don't, everyone. Instead of showing your IQ level, you made the schoolboy lazy and delayed him.

Solution: Let's take this batch of fruit to a total of x kilograms.

x-35%x-160):(35%x+160)=2:3:(35%x+160)=2:3

x=640A: This batch of fruit is 640 kg in total.

16 roots).

Answer: Between 33 roots and 34 heels.

Call me if you have a problem, I'm also a primary school student, let's work together!

1.Let the total number be x (which is actually quite simple).

35%x+160=2/5x

160=2/5x-7/20x

160=1/20x

x=3200

Buy is 3 (2+3)=60%.

The next day it is 60%-35%=25%.

The total weight of the water is 160 25% = 640 (kg) 84 40 * 15 + 1 = 32 (this = is about the same).

Today's elementary school students are amazing.

2: Think of the total amount of work as the unit "1". Chase the problem, the engineering problem, i.e., move in the same direction and meet or (at the same time go against each other), find another factor:

According to the speed, the distance time = speed, that is, the unit "1": the speed difference and the time to chase = the distance difference, to find the law of solving the problem of A, the key to solving the problem is to find a single quantity (i.e., normalization) from a known corresponding quantity. (3) Find a simple problem that contains several other numbers in a number.

1) According to the subtractive sense. The basic quantitative relation is that the slower one comes first: (A B) B.

A is known to be a fraction (or less) of a fraction (hundredth) more than B, simple division problems.

Find the law of solving the problem by a fraction (or less) than B; Speed and (Encounter) Time = Total Distance, Distance Speed = Time, i.e. the product is unchanged.

Types of compound problems and their solutions.

Two of the working hours are used to find the third quantity.

4. The problem of opposing or moving in the same direction.

3. Calculate the opposite direction. Encounter Questions:

Productivity Working hours = total amount of work.

Total work efficiency = working hours.

Total amount of work worked hours = work efficiency.

5。Based on the total amount of work, the amount sought is calculated based on the total amount. The quantitative relationship is that the text narrative often contains words like "calculated as such":

This kind of problem implies that the total amount is constant, and the relationship between division, time and distance is used to find how much each part is: this kind of application problem implies that a single quantity is unchanged, and then it is used as the standard, the "normalization" problem, and the "summing" problem.

4。(5) Knowing how many times (or fractions) of a number is, multiplication calculation: refers to the application problem solved by one-step calculation. (3) Find the number that is a few less than a number; If the unit "1" is unknown. If the unit "1" is known.

5. Work efficiency is expressed as "fraction" of the total amount of work completed in a unit time.

3. Simple multiplication problems. (2) Finding the difference between two numbers is called a travel problem, a simple subtraction problem, and a type of simple application problem.

1. Find the sum of two numbers. (2) Divide a number into several equal parts, find this number, and find the remainder. (2) Find out how many multiples (fractions) of a number are.

The key to solving the problem is to first find the total number (i.e., the sum), the simple addition problem, the fraction problem, that is, at the same time in the same direction, to find the problem solving law of B, the fastest one is last: speed time = distance, travel problem. (1) Find the sum of several identical additions, and calculate the amount according to the requirements of the question.

2) Find the number that is more than a number. (4) Find how many times (or fractions) of one number is another.

1) The product of two factors and one of the factors is known.

1) According to the meaning of addition, work efficiency: the key is to find the standard quantity:

B (fractions of 1).

B (fractions of 1).

A is known to be a fraction (or less) of a fraction (or a hundredth) more than B

Let the landlord be confused, is it the middle point or the end?

If it is the midpoint: 150 (3 5- 1 2) = 150 1 10 = 1500 (km).

If it's the finish line: 150 (1-3 5) = 375 (km) you choose.

If it's the midpoint:

150 (3 5-1 2) = 150 1 10 = 1500 (meters) if the end point:

150 (1-3 5) = 150 2 5 = 375 (m).

Touch: Set the original boy x person, the original girl (6 7) x

6/7)x+6]：[6/7）x+6+x]=20：4120[（13/7)x+6]=41[(6/7)x+6}(260/7)x+120=(246/7)x+2462x=126

x = 63 original female students (6 7) * 63 = 54 people.

1-3/5 = 2/5

150 2/5 = 375 km.

It's gone, trouble to adopt. Although it doesn't make sense anymore......

The total length of the solution is x kilometers.

Three-fifths x + 150 = one-half x

1/2 - 1/5) x = 150

x=150 three-tenths.

x = 500 A and B are a total of 500 kilometers.

The midpoint, which is 1/2 of the total length, has now gone 3/5, with 1/2-3/5, it is the corresponding score of 150 kilometers, and it is OK to divide 150 by the corresponding score!

You made a mistake!

Let the whole journey be x(km).

The meaning of the title is: 3 5*x-1 2*x=150

The solution is x=1500 (km) and the answer is omitted.

3 5x+150=x solution x=375 Estimate yours is a typo, which is the end.

Let the full length be x, (1-3 5x)=150

x = 375 km.

Midpoint: Endpoint

An eraser is 2/5 of a dollar, which is 4/5 of a pencil. That is to say, the ** of the pencil multiplied by 4/5 is equal to 2/5 of 2 yuan, that is to say, 2/5 of the 5 yuan divided by 4/5 is equal to the ** of the pencil. So the ** of the pencil is equal to one-half of the yuan.

The actual aspect ratio is 3:1

Then on paper, the length is 5 cm, and the width should be 5 3 cm 60m: 5cm = 1200:1

So it is 5 cm long and 5 3 cm wide, and the scale is 1200:1

:1 or :3, so a scale of 120:1 should be used, with a length of 5 decimeters and a width of 20 120 = decimeters.

The length is 60m, in order to fill the drawings as much as possible, the length of the foundation is drawn on the edge of 5 decimeters, so it is: 1, obviously the side of 3 decimeters is completely enough to draw the width of 20 meters of the foundation at a scale of 120:1.

The recommended ratio is 200:1, because it is impossible to completely draw the line on the edge of the paper with a length of 5 demeters and a width of 3 decimeters, thank you!