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Math Expert
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Math Expert
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Bunuel wrote:
If $$x^2 y^3 = 200$$, what is the value of $$xy$$?

(1) $$y$$ is an integer

(2) $$\frac{x}{y} = 2.5$$

In this question if the first one would have said that both x and y are integers, would it then be sufficient?
Math Expert
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priyatomar wrote:
Bunuel wrote:
If $$x^2 y^3 = 200$$, what is the value of $$xy$$?

(1) $$y$$ is an integer

(2) $$\frac{x}{y} = 2.5$$

In this question if the first one would have said that both x and y are integers, would it then be sufficient?

No. Because x = ± 5 and y = 2 satisfy x^2 y^3 = 200, so xy can be 10 or -10. If we were told that x and y are positive integers, then yes, it would be sufficient.
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I think this is a high-quality question and I agree with explanation.
Math Expert
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Bunuel,

How should someone simplify (6.25) (y^5)=200 just to know that 200 is multiple of 6.25? Is there anyway to know that?

I know when i divide 200 by 6.25, the result is 32.

Math Expert
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Rebaz wrote:
Bunuel,

How should someone simplify (6.25) (y^5)=200 just to know that 200 is multiple of 6.25? Is there anyway to know that?

I know when i divide 200 by 6.25, the result is 32.

You can express 6.25 as $$6.25=6\frac{1}{4}=\frac{25}{4}$$. With this representation, it's evident that both the left-hand side and the right-hand side, which is 200, can be divided by 25. After this simplification, the calculation becomes straightforward.
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Bunuel wrote:
Rebaz wrote:
Bunuel,

How should someone simplify (6.25) (y^5)=200 just to know that 200 is multiple of 6.25? Is there anyway to know that?

I know when i divide 200 by 6.25, the result is 32.

You can express 6.25 as $$6.25=6\frac{1}{4}=\frac{25}{4}$$. With this representation, it's evident that both the left-hand side and the right-hand side, which is 200, can be divided by 25. After this simplification, the calculation becomes straightforward.

Ofcourse converting the decimal into fraction makes the calculation way more easier.

Brilliant explanation and very straightforward!

Thank you!
Math Expert
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