Wiki

Bài 28, 29, 30 trang 22 SGK Toán 9 tập 2 – Giải bài toán bằng cách

H2 Heading: Solving Problems 28, 29, and 30 on Page 22 of the 9th-Grade Mathematics textbook

Problem 28: Finding Two Natural Numbers with Specific Properties

Question:

Find two natural numbers such that their sum is 1006, and if the larger number is divided by the smaller number, the quotient is 2 and the remainder is 124.

Solution:

Let’s assume the larger number as (x) and the smaller number as (y). (Condition: x > y; x, y ∈ ℕ*)

According to the given condition that the sum of the two numbers is 1006, we have: (x + y = 1006).

Since the larger number divided by the smaller number gives a quotient of 2 and a remainder of 124, we can write: (x = 2y + 124) (with y > 124).

We have the following system of equations:

``````⇔ {x + y = 1006
x - 2y = 124}``````

Simplifying further:

``````⇔ {x + y = 1006
3y = 882}``````

Solving for the variables:

``````⇔ {x = 1006 - y
y = 294}``````
``````⇔ {x = 1006 - 294
y = 294}``````
``````⇔ {x = 712
y = 294} (satisfies the condition)``````

Therefore, the two natural numbers we are looking for are 712 and 294.

Problem 29: Solving a Classic Problem

Question:

Solve the following classic problem:

Seventeen fresh fruits, consisting of oranges and tangerines, are divided amongst a hundred people. If each orange is divided into three equal parts and each tangerine is divided into ten equal parts, how many oranges and tangerines are there in total?

Solution:

Let’s assume the number of tangerines as (x) and the number of oranges as (y). The condition is that x and y should be positive integers.

Since there are seventeen fruits in total, we can write the equation: (x + y = 17) (1)

Since each orange is divided into three equal parts and each tangerine is divided into ten equal parts, we can write the equations:

For oranges: (3y parts)

For tangerines: (10x parts)

Also, given that there are a hundred parts in total, we can write the equation: (10x + 3y = 100) (2)

From equations (1) and (2), we have the following system of equations:

``````⇔ {x + y = 17
10x + 3y = 100}``````

Simplifying further:

``````⇔ {3x + 3y = 51
10x + 3y = 100}``````
``````⇔ {3x + 3y - (10x + 3y) = 51 - 100
10x + 3y = 100}``````
``````⇔ {-7x = -49
10x + 3y = 100}``````

Solving for x and y:

``````⇔ {x = 7
3y = 100 - 10x}``````
``````⇔ {x = 7
y = 10} (satisfies the condition)``````

Therefore, there are 10 tangerines and 7 oranges in total.

Problem 30: Calculating Distance and Departure Time of a Car Journey

Question:

A car travels from point A to point B, intending to arrive at B at 12:00 PM. If the car travels at a speed of 35 km/h, it arrives 2 hours late. If it travels at a speed of 50 km/h, it arrives 1 hour early. Find the distance of the journey (AB) and the departure time of the car from A.

Solution:

Let’s assume the distance of the journey as (x) km and the estimated travel time from A to B as (y) hours. The condition is that x > 0 and y > 1 (as the car arrives 1 hour early).

+) Case 1:

When the car travels at a speed of 35 km/h:

Since the car arrives 2 hours late, the total travel time is (y + 2) hours.

The distance traveled can be calculated using the formula (S = v * t), where S is distance (in km), v is speed (in km/h), and t is time (in hours). So, we have: (35(y + 2)) km.

Since the distance remains constant, we can set up the equation: (x = 35(y + 2)) (1).

+) Case 2:

When the car travels at a speed of 50 km/h:

Since the car arrives 1 hour early, the total travel time is (y – 1) hours.

The distance traveled can be calculated using the formula (S = v * t), where S is distance (in km), v is speed (in km/h), and t is time (in hours). So, we have: (50(y – 1)) km.

Since the distance remains constant, we can set up the equation: (x = 50(y – 1)) (2).

From equations (1) and (2), we have the following system of equations:

``````⇔ {x = 35(y + 2)
x = 50(y - 1)}``````

Simplifying further:

``````⇔ {x = 35y + 70
x = 50y - 50}``````
``````⇔ {x - 35y = 70  (1)
x - 50y = -50  (2)}``````

Subtracting equation (2) from equation (1), we get:

``````⇔ {15y = 120
x - 50y = -50}``````

Solving for y:

``````⇔ {y = 8
x = -50 + 50y}``````
``````⇔ {y = 8
x = -50 + 50 * 8}``````
``````⇔ {y = 8
x = 350} (satisfies the condition)``````

Therefore, the distance of the journey (AB) is 350 km, and the car departs from A at 4:00 AM.

Visit Kienthucykhoa.com for more educational content.

Next Article:

Kiến Thức Y Khoa

Xin chào các bạn, tôi là người sở hữu website Kiến Thức Y Khoa. Tôi sử dụng content AI và đã chỉnh sửa đề phù hợp với người đọc nhằm cung cấp thông tin lên website https://kienthucykhoa.edu.vn/.

Check Also
Close