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2 years ago

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A swimming pool that holds 1200 gallons of water can be completely drained by a two pumps. The first pump is small and can only

remove 300 gallons from the pool. If the second pump is used and it removes 20 gallons every minute, how many minutes will it take to completely drain the pool?

1 answer:

sp2606 [1]2 years ago

6 0

Answer: 45 minutes

Step-by-step explanation:

the first pump takes care of 300 gallons so we need to worry about the remaining 900 gallons. the second pump removes 20 gallons per minute so all u have to do to find how many minutes it would take is divide 900 gallons by 20 gallons a minute which gives you 45

it will take 45 minutes to completely drain the pool

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When the members of a family discussed where their annual reunion should take place, they found that out of all the family mem

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Answer:

The total number of family members is 21.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of these sets.

I am going to say that:

-The set A represents those that would not go to a park.

-The set B represents those who would not go to a beach.

-The set C represents those who would not go to the family cottage.

The value d represents those who would go to all three places.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a are those that would only not go to a park, $A \cap B$ are those who would not got to a park or to the beach, $A \cap C$ are those who would not go to a park or to the famili cottage. And $A \cap B \cap C$ are those that would not go to any of these places.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

This diagram has the following values:

$a,b,c,d,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

The total number of family members is the sum of all these values:

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

We start finding the values from the intersection of the three sets

5 would not go to a park or a beach or to the family cottage.

This means that $A \cap B \cap C = 5$

1 would go to all three places. This means that $d = 1$ .

8 would go to neither a park nor the family cottage

This means that:

$A \cap C + (A \cap B \cap C) = 8$

$A \cap C = 3$

8 would go to neither a beach nor the family cottage

$B \cap C + (A \cap B \cap C) = 8$

$B \cap C = 3$

7 would go to neither a park nor a beach

$A \cap B + (A \cap B \cap C) = 7$

$A \cap B = 2$

15 would not go to the family cottage

$C = 15$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$15 = c + 3 + 3 + 5$

$c = 4$

12 would not go to a beach

$B = 12$

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$12 = b + 3 + 2 + 5$

$b = 2$

11 would not go to a park

$A = 11$

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$11 = a + 2 + 3 + 5$

$a = 1$

Now, we can find the total number of family members.

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$T = 1 + 2 + 4 + 1 + 2 + 3 + 3 + 5$

$T = 21$

The total number of family members is 21.

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