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.
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