"Application of linear equations of one variable - discount sale" PPT courseware download of linear equations of one variable

Description

"Application of linear equations of one variable - discount sale" PPT courseware download of linear equations of one variable

Part One Content: Basic Knowledge Key Points

Knowledge point 1: Profit and loss issues in sales

1. A store sells a batch of clothing. Each piece sells for 150 yuan and can make a profit of 20%. Find the cost price of this kind of clothing. Assuming that the cost price of this kind of clothing is x yuan, the equation can be obtained as (C)

A.x=15×20% B.20%×150x=150

C.150-x=20%x D.150-x=20%

2. A store is liquidating its inventory and selling the last two down jackets at a special price. Type A down jacket sells for 1,200 yuan, with a profit of 20%, and type B down jacket also sells for 1,200 yuan, but with a loss of 20%. The store has two transactions (D)

A. Profit of 100 yuan B. Loss of 125 yuan

C. No loss, no profit D. Loss of 100 yuan

3. Xiao Ming and Xiao Li need to buy the same classic book. Xiao Ming gets a 10% discount from the bookstore, and Xiaoli gets a 20% discount from the online store, but needs to pay an additional 10 yuan for express delivery. As a result, Xiaoli spends less than Xiao Ming. 2 yuan, what is the price of this classic book? If the price of this classic book is x yuan, the equation that can be formulated is 0.9x-2=0.8x+10.

PPT on linear equations of one variable, part 2: improvement of comprehensive capabilities

7. A certain brand of clothing in a shopping mall is priced at 1,000 yuan per piece. In order to participate in the market competition, the shopping mall sells it at a discount of 15% of the listed price (i.e. 85% of the listed price) and an additional 40 yuan in profit. As a result, each piece of clothing can still make a profit of 20%. .If the purchase price of each piece of this kind of clothing is x yuan, please list the equation about x (D)

A.1000×85%-40=20%x

B.(1000-40)×85%-x=20%x

C.1000×85%-40-x=20%×1000

D.1000×85%-40=(1+20%)x

8. A bookstore holds a discount event for book purchases:

① If the one-time purchase of books does not exceed 100 yuan, you will not enjoy the discount;

② If the one-time purchase of books exceeds 100 yuan but does not exceed 200 yuan, a 10% discount will be applied to the original price;

③If you purchase books for more than 200 yuan at one time, you will get a 30% discount on the original price.

In this event, Xiaoli paid a total of 229.4 yuan for two book purchases. The original price of the second book purchase was three times that of the first book purchase, so the total original price of Xiaoli's two book purchases was 248 or 296 yuan.

9. (Adapted) A store sells a certain type of clothing at a discount due to the change of seasons. If each piece of clothing is sold at 50% off the listed price, it will lose 20 yuan, and if it is sold at 20% off the listed price, it will make 40 yuan. In order to ensure that you do not lose money, at least 60% off.

10. Lan Lan went to a stationery store to buy exercise books, and the salesperson told her that if she purchased more than 10 exercise books, she would get a 30% discount on the portion exceeding 10. Lan Lan bought 20 copies, and it turned out to be 1.8 yuan cheaper. You know what happened? What's the price per copy?

Solution: Assume that the original price of each copy is x yuan. According to the meaning of the question, we get

20x-10x-0.7×10x=1.8, the solution is x=0.6.

Answer: The original price per copy was 0.6 yuan.

PPT on linear equations of one variable, the third part: Expanding exploration and breakthroughs

14. This year, an online shopping mall is conducting promotional activities during the "Double 11 Shopping Festival". The rules of the activities are as follows:

① No discount will be given if the purchase does not exceed 100 yuan; ② If the purchase exceeds 100 yuan but is less than 500 yuan, a 10% discount will be given; ③ If the purchase exceeds 500 yuan, a 10% discount will be provided for the portion over 500 yuan, and a 20% discount will be provided for the portion exceeding 500 yuan.

(1) The total price of the goods Xiaoli purchased for the first time (list price and ) was 200 yuan. How much yuan did she actually pay according to the event regulations?

(2) Xiaoli spent 490 yuan on her second shopping trip. Compared with no promotion, how much money did she save on her second trip? (Please use a linear equation of one dollar to answer)

(3) If Xiaoli combines these two purchases into one purchase, will it save more money? Why?

Solution: (1)200×0.9=180(yuan).

(2) Because 500×0.9=450 (yuan) and 490>450, the second purchase exceeds 500 yuan.

Assume that the total price of the second shopping product is x yuan. According to the meaning of the question, we get

500×0.9+0.8(x-500)=490,

The solution is x=550, then 550-490=60.

Answer: I saved 60 yuan on my second purchase.

(3) Save money by combining two purchases into one purchase.

Reason: 200+550=750 (yuan),

500×0.9+(750-500)×0.8=450+200=650(yuan),

Because 180+490=670>650,

Therefore, Xiaoli combined the two purchases into one purchase to save money.

Keywords: Beijing Normal University Edition seventh-grade mathematics PPT courseware free download, linear equation of one variable PPT download, application of linear equation of one variable discount sales PPT download, .PPT format;

For more information about the PPT courseware "Linear Equations of One Variable and Applying the Linear Equations of One Variable to Discounted Sales", please click the "Linear Equations of One Variable PPT Applying the Linear Equations of One Variable to Discounted Sales PPT" tag.

"Application of linear equations of one variable - discount sales" PPT download of linear equations of one variable:

"Application of the linear equation of one variable for discount sales" The first part of the linear equation PPT download content: Learning objectives 1. Accurately understand the relationship between profit (profit rate), cost, and sales price in the discount sales problem. (Difficulty) 2. Be able to use Solving simple linear equations of one variable...

"Application of linear equations of one variable - discount sales" linear equation of one variable PPT:

"Application of linear equations of one variable on sale" PPT of linear equations of one variable Part 1 content: Learning objectives 1. Analyze the quantitative relationship in the problem, establish an equation to solve the problem 2. Find the equivalent relationship and list the equation; accurately solve the equation... .. . ...apply the power of one variable..

File Info

Update Time: 2024-09-27

This template belongs to Mathematics courseware Beijing Normal University Edition Seventh Grade Mathematics Volume 1 industry PPT template

Preview