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Gwar [14]
1 year ago
12

At a family reunion, each of Sana aunts and uncles is getting photographed once, The aunts are taking pictures in groups of5 and

the uncles are taking pictures in groups of 10. If Sana has the same total number of aunts and uncles, what is the minimum number of aunts that Sana must have
Mathematics
2 answers:
Ludmilka [50]1 year ago
7 0

Answer:

The minimum number of aunts that Sana must have is 10.

Step-by-step explanation:

Consider the provided information.

It is given that the aunts are taking pictures in groups of 5.

Uncles are taking pictures in group of 10.

We need to find minimum number of aunts. So find the LCM of 5 and 10.

5 = 5 × 1

10 = 5 × 2

LCM = 5 × 2 = 10

The LCM of 5 and 10 is 10.

Therefore, the minimum number of aunts that Sana must have is 10.

yaroslaw [1]1 year ago
3 0

This question is based on least common multiple method.

As given in the question,

Aunts are taking pictures in group of = 5

Uncles are taking pictures in group of = 10

So we have been asked what is the minimum number of aunts Sana have.

As we have been given that the number of aunts and uncles is equal so to find the minimum number of aunts we will apply the least common multiple method.

So we get LCM of 5,10 as 10

Hence there are minimum 10 aunts.


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Step-by-step explanation:

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You have two circles, one with radius r and the other with radius R. You wish for the difference in the areas of these two circl
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Answer:

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Step-by-step explanation:

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\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}

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