A relief organization donated 240 kg of maize and 150 kg of beans to needy families. Each family received…

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A relief organization donated 240 kg of maize and 150 kg of beans to needy families. Each family received exactly the same quantity by mass of either maize or beans. No family received both. Determine

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here are the solutions to questions 11 and 12.

11. A relief organization donated 240 kg of maize and 150 kg of beans to needy families. Each family received exactly the same quantity by mass of either maize or beans. No family received both. Determine the least possible number of needy families.

To find the least possible number of needy families, each family must receive the largest possible quantity of food.
This quantity must be a common divisor of 240 kg and 150 kg. Therefore, we need to find the Greatest Common Divisor (GCD) of 240 and 150.
Prime factorization of 240:
$240 = 24 \times 10 = (2^3 \times 3) \times (2 \times 5) = 2^4 \times 3 \times 5$
Prime factorization of 150:
$150 = 15 \times 10 = (3 \times 5) \times (2 \times 5) = 2 \times 3 \times 5^2$
Calculate the GCD by taking the lowest power of common prime factors.
$\text{GCD}(240, 150) = 2^1 \times 3^1 \times 5^1 = 2 \times 3 \times 5 = 30 \text{ kg}$
So, each family receives 30 kg of food.
Calculate the number of families that received maize.
$\text{Number of maize families} = \frac{240 \text{ kg}}{30 \text{ kg/family}} = 8 \text{ families}$
Calculate the number of families that received beans.
$\text{Number of bean families} = \frac{150 \text{ kg}}{30 \text{ kg/family}} = 5 \text{ families}$
Determine the total least possible number of needy families.
$\text{Total families} = 8 + 5 = 13 \text{ families}$
The least possible number of needy families is $\boxed{13}$ .

12. A worker completes a job in 12 hours while another takes 8 hours. How long will it take them to finish the job all together?
Determine the work rate for each worker.
Worker 1's rate ( $R_1$ ) = $\frac{1 \text{ job}}{12 \text{ hours}} = \frac{1}{12}$ job/hour.
Worker 2's rate ( $R_2$ ) = $\frac{1 \text{ job}}{8 \text{ hours}} = \frac{1}{8}$ job/hour.
Calculate their combined work rate when working together.
$R_{\text{total}} = R_1 + R_2 = \frac{1}{12} + \frac{1}{8}$
Find a common denominator for the fractions, which is 24.
$R_{\text{total}} = \frac{2}{24} + \frac{3}{24} = \frac{2+3}{24} = \frac{5}{24} \text{ job/hour}$
The time taken to complete the job together is the reciprocal of the combined rate.
$T_{\text{total}} = \frac{1}{R_{\text{total}}} = \frac{1}{\frac{5}{24}} = \frac{24}{5} \text{ hours}$
Convert the fraction to a mixed number or decimal.
$T_{\text{total}} = 4 \frac{4}{5} \text{ hours}$
To express this in hours and minutes:
$4 \frac{4}{5} \text{ hours} = 4 \text{ hours} + \left(\frac{4}{5} \times 60\right) \text{ minutes} = 4 \text{ hours} + 48 \text{ minutes}$
It will take them $\boxed{4 \frac{4}{5} \text{ hours}}$ (or 4 hours and 48 minutes) to finish the job all together.

A relief organization donated 240 kg of maize and 150 kg of beans to needy families. Each family received exactly the same quantity by mass of either maize or beans. No family received both. Determine

Answer

Step-by-step solution

To find the least possible number of needy families, each family must receive the largest possible quantity of food.

Calculate the GCD by taking the lowest power of common prime factors.

Calculate the number of families that received maize.

Calculate the number of families that received beans.

Determine the total least possible number of needy families.

Determine the work rate for each worker.

Calculate their combined work rate when working together.

Find a common denominator for the fractions, which is 24.

The time taken to complete the job together is the reciprocal of the combined rate.

Convert the fraction to a mixed number or decimal.

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