Mathematics problems in the seventh grade, math problems in the second volume of the seventh grade

When set to x hours, the cost is equal.

2+x=1000

Therefore, when the duration is less than 1000 hours, it is cost-effective to use incandescent lamps, and when the duration is 1000 hours, it is cost-effective to use energy-saving lamps when the duration is more than 1000 hours.

The original price cost, white light 2 yuan, energy-saving lamp 32, the price difference is 30 yuan, 30 yuan is equal to 60 kilowatt-hours of electricity.

1 kWh, 10 hours for white lights, 25 hours for energy-saving lights, so the life span is more than 600 hours, choose energy-saving lamps.

x=a+b

y=a-ba=（x+y）÷2

b=（x-y）÷2

Let these three digits be ABC

a+b+c=17

a+b=c+3

The number after the reversal is the bac

Expressed algebraically.

abc=100a+10b+c

bac=100b+10a+c

bac-abc=100b+10a-100a-10b=100（b-a）-10（b-a）

90(b-a)=495 (This is a wrong number!) ）

1.If the speed of the ship in still water is a and the speed of the water is b, let the speed of the ship sail with the current be x and the speed of the ship against the current is y, then there are x=(a+b), y=(a-b), a=((x+y) 2), b=((x-y) 2).

2.The sum of the digits of a three-digit number is 17, and the sum of the digits in the hundredth digit and the digits in the tenth digit is 3 greater than the number in the single digit, and if the digits in the hundredth digit are reversed with the digits in the tenth, the resulting new number is 495 larger than the original number, then the original number is (

*The second question seems to be a mistake.

If you reverse the number in the hundred place with the number in the ten place, the resulting new number is larger than the original number*** is an integer.

1. Solution: x=a+b

y=a-ba=(x+y)/2

b=(x-y)/2

2. Solution: Let the single digit be x, the ten digit be y, and the hundred digit be z, then x+y+z=17

y+z-x=3

100y+10z+x-495=100z+10y+xSolve the system of equations to obtain:

x=7,y= wrong question).

1. x=(a+b) y=(a-b) a=(x+y)/2 b=(x-y)/2

2.From the sum of the digits in the hundred digit and the number in the 100th digit is 3 greater than the number in the single digit, and the sum of the digits in a three-digit number is 17, we know that the single digit is 7, and the sum of the 100 digits plus the 100th digit is 10

The second question, "The number in the hundred digit is reversed with the digit in the tenth digit, and the new number obtained is 495 larger than the original number" is obviously wrong, because the single digit does not change, so the difference between them can only be 0You'd better read the question again and give you the answer.

a-b x-b a-y

2.If the second question is wrong, the single digit is 7

1. The absolute value of x=a+b y= a= 2 b= 2

2. Let the original number be 100x+10y+z,x+y+z=17,x+y=z+3,100y+10x+z=495+100x+10y+z,x=,y=,z=7,obviously your teacher's problem is not good, and the numbers are not made up.

1. Original formula = 2x 4y 8-5x 3y 6-xy = 2*(-2) 4-5(-2) 3-(-2) = 742, formula = x 4+(m-3) x 3+(n+8-3m) x 2+mnx-24x+8n

As can be seen from the title, m-3=0

n+8-3m=0②

m=3,n=1

Set the road width x meters, regardless of these three roads in the ** construction, you can imagine that the triangle road is moved to the edge of the vegetable garden, then the vegetable garden is less than the courtyard: x meters less in the east-west direction, and 2x meters less in the north-south direction, then the vegetable field area = (32-x) * (20-2x) = 558 can be found x

Let the width of the road be x m, according to the question: 32x * 2 + 20x = 32 * 20-558

Solution: Set the road width x meters.

32×20-32x×2-20x+2x²=5582x²-84x+640-558=0

2x²-84x+82=0

x²-42x+41=0（0≤x≤20）

The solution yields x1 = 41 (rounded), x2 = 1

A: The width of the road is 1 meter.

B has a diameter of 12 and a width of 2.

First of all, assuming that a total of 2n rings are interlocking, and the two ends of the two ends are also interlocking, that is, 2n small rings are interlocking to form a large ring, then the total length is (12+8)n-(1+2)n=17n

Then there are three cases:

1) Both ends are armor, that is, a certain place is interlocked by two armors, so the total length is 17n-2

2) Both ends are B, that is, untie a certain place where two Bs are interlocked, so the total length is 17n-4

3) The two ends are a A and a B, that is, a certain place where A and B are interlocked, so the total length is 17N-3

Is this a math problem under seven?

B has a diameter of 12 and a width of 2.

First of all, assuming that a total of 2n rings are interlocking, and the two ends of the two ends are also interlocking, that is, 2n small rings are interlocking to form a large ring, then the total length is (12+8)n-(1+2)n=17n

Then there are three cases:

1) Both ends are armor, that is, a certain place is interlocked by two armors, so the total length is 17n-2

2) Both ends are B, that is, untie a certain place where two Bs are interlocked, so the total length is 17n-4

3) The two ends are a A and a B, that is, a certain place where A and B are interlocked, so the total length is 17N-3

1) Both ends are armor, that is, a certain place is interlocked by two armors, so the total length is 17n-2

2) Both ends are B, that is, untie a certain place where two Bs are interlocked, so the total length is 17n-4

3) The two ends are a A and a B, that is, a certain place where A and B are interlocked, so the total length is 17N-3

Analysis: Represent the algebraic formulas of a and b respectively, make them equal, and see if there are corresponding values Answer: This statement is incorrect

The reason is as follows: although when n = 3, 20, 120, a b or a b, a = b can be made a = b, so that 60 n = 60 + 7 n + 7, i.e. 60 n = 67 n + 7

60n+420=67n, the solution is n=60, and n=60 is the root of the equation

When n = 60 and a = b, the value of n that does not fit this statement is 60

This question itself is a bit of a problem, you say: "with 45,000 yuan at the same time to buy ABC three lottery tickets 20 za". The number of ties in this question is certain, and the amount of money is also certain, according to what you said, there will be a problem with the chain leakage. The strict point should be a total of 20 ties.

Solution: a (a za): yuan).

b (one bundle): 2 * 1000 = 2000 (yuan).

c (a blind call Xunza): Yuan).

It can be obtained from the question. 1500a+2000b+2500c=45000a+b+c=20

Available from (1).

3a+4b+5c=90

(3)-(2)*3, get.

b+2c=30

4)-(2), got.

c=a+10

Let's start the analysis).

a+b+c=20, and c=a+10

c from this to determine what are the types of c).

then c = 11, 12, 13, 14

then a = 1, 2, 3, 4

Substituting the two values into the original equation will give us b).

Substitute the values of a,c into the original equation.

Solution. b=8,6,4,2

Answer: There are four schemes, namely Zha A, 8 Zha B, 11 Zha C;

Zha A, 6 Zha B, 12 Zha C;

Zha A, 4 Zha B, 13 Zha C;

Zha A, 2 Zha B, 14 Zha C.