Homework 21 Description More Info Core Math 7 Mathematics
Homework 21descriptionmoreinfocore Math 7 Mathematicshomework 17 G
Solve the following problems involving various mathematical concepts including exponential growth, work rate, proportions, geometry, algebra, and number patterns. The problems are drawn from a homework assignment for Grade 7 students, covering practical applications of math in real-life scenarios and problem-solving exercises.
Paper For Above instruction
Mathematics plays a vital role in understanding and solving real-world problems, especially in middle school, where foundational concepts are reinforced and applied. The following solutions address a diverse set of problems, including exponential growth, work collaboration, proportions, geometry, algebra, and number patterns, reflecting the curriculum for Grade 7 students.
1. Exponential Growth: Bacteria in a Dish
The problem involves bacteria doubling their covered area each day, with the entire dish covered after 16 days. The question asks on which day only one-quarter of the dish was covered.
Given that bacteria double their coverage daily, the coverage grows exponentially. If the dish is fully covered after 16 days, the coverage on previous days can be traced backwards by halving the area since doubling occurs daily:
- Day 16: fully covered (100%)
- Day 15: half covered (50%)
- Day 14: quarter covered (25%)
Therefore, the dish was only one-quarter covered on Day 14.
2. Work Rate: Benny and Frank
Benny takes 2 hours to wash 500 dishes, while Frank takes 3 hours for the same task. The problem asks how long they would take together to wash 1000 dishes.
First, find each person's rate per hour:
- Benny: 500 dishes / 2 hours = 250 dishes per hour
- Frank: 500 dishes / 3 hours ≈ 166.67 dishes per hour
Combined rate:
- 250 + 166.67 ≈ 416.67 dishes per hour
Time to wash 1000 dishes together:
Time = Total dishes / Combined rate = 1000 / 416.67 ≈ 2.4 hours
Thus, they will take approximately 2 hours and 24 minutes.
3. Weekly Earnings: Richard, Stephan, Rachel, and Stephanie
Richard and Stephan earn $5.15/hour. Rachel works 13 hours weekly; Stephanie works 20 hours weekly. The total earnings for all four are calculated.
Richard and Stephan combined weekly earnings:
- 2 persons × $5.15/hour × total hours (unknown) — but since only their hourly rate is provided, assume they work a standard week (say 40 hours each):
However, since the problem states their hourly rate without specifying their weekly hours, the total amount earned in a week can be computed as:
- Richard: $5.15 × 40 = $206
- Stephan: $5.15 × 40 = $206
- Rachel: $5.15 × 13 ≈ $66.95
- Stephanie: $5.15 × 20 ≈ $103
The total of their earnings in one week:
$206 + $206 + $66.95 + $103 ≈ $581.95
4. Work and Job Duration
Six men can complete a job in 14 days. How many men are needed to do the same job in 21 days?
Using inverse proportionality of men and days:
Number of men × days = constant
Initial: 6 men × 14 days = 84
Desired: x men × 21 days = 84
x = 84 / 21 = 4
Therefore, 4 men are required.
5. Salt-water Solution Proportion
The solution contains 15 parts salt in 50 ounces of water. The question asks, what is the proportion of water in the solution?
Total parts: 15 salt + water = total parts (assuming total parts total up to the sum)
If salt is 15 parts in 50 ounces, the proportion of salt:
15 / (15 + water parts) = (amount of salt)/(total solution)
But the question explicitly asks for the proportion of water in the solution, which is:
Water: 50 ounces, and total solution includes salt and water. So, if only water is 50 ounces and salt is 15 parts, then the total solution is 50 ounces plus the salt component.
Alternatively, since the problem specifies salt in 50 ounces, the proportion of water is:
Proportion of water = (total water) / (total solution)
If the total solution is water plus salt, and salt proportion is known, the ratio of water is proportional to the total weight, and the proportion of water in the solution is:
Water proportion = (Total water / Total solution). Since the problem gives water in ounces, assuming the total solution's weight is the sum of salt and water, the proportion of water in the solution is:
Water / (Salt + Water) = 50 / (15 + x), but since the total parts are not specified for salt or water, more information is needed. However, if only the proportion of salt (15 in 50 ounces), then:
Proportion of water = (50 ounces of water) / (50 + amount of salt).
In typical mix, the salt proportion is 15/50 = 0.3, so water proportion would be 0.7.
Therefore, the proportion of water in the solution is approximately 70%.
6. Theatre Seating Arrangement
There are 800 seats divided into 3 theatres: Theatre 1 has 270 seats, Theatre 2 has 150 more seats than Theatre 3. Find the number of seats in Theatre 2.
Let the number of seats in Theatre 3 be x. Then, Theatre 2 has x + 150 seats.
Total seats:
270 + (x + 150) + x = 800
Simplify:
270 + x + 150 + x = 800
2x + 420 = 800
2x = 380
x = 190
Theatre 2 has x + 150 = 190 + 150 = 340 seats.
7. Geometry: Rectangular Chalkboard
The length of a chalkboard is 3 times its width. When 3 meters are subtracted from its length and 3 meters are added to its width, will it become a square?
Let width = w, then length = 3w.
Original dimensions:
- Length = 3w
- Width = w
After adjustments:
- New length = 3w - 3
- New width = w + 3
For it to be a square:
3w - 3 = w + 3
3w - w = 6
2w = 6
w = 3
Original width = 3 meters; length = 3 × 3 = 9 meters.
Adjusted dimensions: length = 9 - 3 = 6 meters, width = 3 + 3 = 6 meters. Yes, the new dimensions form a square of 6 meters on each side.
8. Algebraic Problems and Number Patterns
(a) Sum of two numbers is 60; larger is twice the smaller
Let smaller number = x, larger = 2x
Sum: x + 2x = 60
3x = 60
x = 20
Thus, the two numbers are 20 and 40.
(b) Ages of three people where sum is 120
Second person's age = 2 × first person's age
Third person's age = 3 × first person's age
Sum: x + 2x + 3x = 120
6x = 120
x = 20
First: 20, second: 40, third: 60
(c) Sum of consecutive numbers is 153
Let first number = n, next = n + 1, etc. Assuming two consecutive numbers:
n + (n + 1) = 153
2n + 1 = 153
2n = 152
n = 76
Numbers are 76 and 77.
(d) Number of boys and girls in a class
Boys = 2 × girls, total students = 75
Boys + Girls = 75
2g + g = 75
3g = 75
g = 25
Girls = 25, Boys = 50
(e) Age problem involving grandfather and Miguel
Let Miguel's current age = M, grandfather's = G.
In three years, G + 3 = 6(M + 3)
G + 3 = 6M + 18
G = 6M + 15
Sum of current ages: G + M = 68
Substitute G:
6M + 15 + M = 68
7M + 15 = 68
7M = 53
M = 53 / 7 ≈ 7.57 years
G = 6(7.57) + 15 ≈ 45.43 years
Thus, Miguel is approximately 7.6 years old, and grandfather is about 45.4 years old.
9. Number Patterns and Equations
(a) Consecutive odd integers summing to 88
Let first odd integer = n, next = n + 2
n + n + 2 = 88
2n + 2 = 88
2n = 86
n = 43
Numbers: 43 and 45
(b) Sum of four consecutive integers is 106
Let first = n, next = n + 1, etc.
n + (n + 1) + (n + 2) + (n + 3) = 106
4n + 6 = 106
4n = 100
n = 25
Numbers: 25, 26, 27, 28
(c) Sum of two consecutive even integers is 26
Let first = 2n, next = 2n + 2
2n + (2n + 2) = 26
4n + 2 = 26
4n = 24
n = 6
The integers: 12 and 14
10. Number Theory Problems
Two consecutive odd integers with sum 88: 43 and 45.
Sum of 4 consecutive integers: 25, 26, 27, 28.
Sum of two consecutive even integers: 12 and 14.
These problems collectively develop algebraic reasoning, proportional reasoning, geometric understanding, and number pattern recognition, equipping Grade 7 learners with essential problem-solving skills for their academic and everyday applications.
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