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FinnZ [79.3K]
2 years ago
14

Which of the following problems would NOT have a solution?

Mathematics
2 answers:
Vika [28.1K]2 years ago
6 0
The last one I think
Natasha2012 [34]2 years ago
6 0

Answer:

The correct option is D) Two pizzas are shared equally among zero people, and you want to know how much each person gets.

Step-by-step explanation:

Consider the provided information,

We need to identify the option which has no solution,

Consider the option A)

Six pizzas are shared equally among three people, and you want to know how much each person gets.

We can find how much each person gets by dividing the number or pizza with number of people:

Each person gets = 6/3 = 2

That means each person gets 2 pizza.

The problem has a solution.

Consider the option B)

Three pizzas are shared equally among two people, and you want to know how much each person gets.

We can find how much each person gets by dividing the number or pizza with number of people:

Each person gets = 3/2 = 1.5

That means each person gets 1.2 pizza.

The problem has a solution.

Consider the option C)

Zero pizzas are shared equally among three people, and you want to know how much each person gets.

We can find how much each person gets by dividing the number or pizza with number of people:

Each person gets = 0/3 = 0

That means each person gets 0 pizza.

The problem has a solution.

Consider the option D)

Two pizzas are shared equally among zero people, and you want to know how much each person gets.

We can find how much each person gets by dividing the number or pizza with number of people:

Each person gets = 2/0 = No solution

As we know any number divided by 0 has no solution.

Thus, the problem has no solution.

Hence, the correct option is D) Two pizzas are shared equally among zero people, and you want to know how much each person gets.

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Divide 3m 80cm of ribbon in the ratio of 1/3 : 3/4 : 1/2 what is the result?​
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Step-by-step explanation:

the smallest common multiple of 3, 4 and 2 is 12.

so, to be able to truly compare these fractions we need to bring all denominators to .../12.

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in total we have therefore 19 (4+9+6) equally large parts of the ribbon that are then combined :

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