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The figures above show a hexagonal nut that has a width of 1 5/16 : Problem Solving (PS)
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# The figures above show a hexagonal nut that has a width of 1 5/16

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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
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Archit3110 wrote:
given fraction;
$$1 \frac{5}{16}$$ inches ; 21/16
and 1 inch = 25.4 mm
so closest measurement ; 25.4*21/16 = 33.3 ~ 33
IMO D

Careful!

From the prompt:
a wrench that, in order to fit the nut, must have a width of at least $$1 \frac{5}{16}$$ inches
Implication:
$$w ≥ \frac{21}{16}$$ inches

In your solution, the value in red indicates that $$\frac{21}{16}$$ inches > 33 millimeters.
Thus:
$$w ≥ \frac{21}{16}$$ inches > 33 millimeters
$$w > 33$$ millimeters

Only E satisfies this condition.
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
this is a pretty straight fwd calculation. convert inches to mm that comes around > 33 (~33.337).
Thus, ANS -> E
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
MBA HOUSE KEY CONCEPT: Mixed Numbers and Rule of 3(directly proportional).

1 5/16 = (1 x 16 + 5) / 16 = 21/16

Inches 21/16 x 25.4 = 33...

So, the next integer greater than 33 and something is 34.

E
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
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avoid calculating sophisticated numbers. simply do this

during conversion use closest estimates..

21/16 inch x 25.4

20 / 15 x 25 = 33.33.. u know now answer is E.34
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
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The figures above show a hexagonal nut that has a width of 55669.png inches and a wrench that, in order to fit the nut, must have a width of at least 1+5/16inches.

What 55669.png means?
If the nut has 1+5/16 inches too, why they give the information about inches of nut and of wrench separately using “and”?
And the image of the question indicates that the nut has 1+5/16 inches. But at the same time the question says “a wrench that, in order to fit the nut, must have a width of at least 1+5/16inches”??

This question is annoying geez can someone explain me why this question lends infos in this way please?

Posted from my mobile device
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
$$1 \frac{5}{16}$$ inches = $$\frac{21}{16 }$$ inches

$$21/16 * 25.4$$ = $$\frac{21}{16} * 25 \frac{4}{10}$$

$$\frac{21}{16 }* \frac{254}{10}$$

$$\frac{21}{16} * \frac{127}{5} = \frac{2667}{80}$$

$$= approximately 33 \\$$

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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
1
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suminha wrote:
The figures above show a hexagonal nut that has a width of 55669.png inches and a wrench that, in order to fit the nut, must have a width of at least 1+5/16inches.

What 55669.png means?
If the nut has 1+5/16 inches too, why they give the information about inches of nut and of wrench separately using “and”?
And the image of the question indicates that the nut has 1+5/16 inches. But at the same time the question says “a wrench that, in order to fit the nut, must have a width of at least 1+5/16inches”??

This question is annoying geez can someone explain me why this question lends infos in this way please?
Posted from my mobile device

Even I got confused here. Adding Bunuel to make this correction
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
Pritishd wrote:
suminha wrote:
The figures above show a hexagonal nut that has a width of 55669.png inches and a wrench that, in order to fit the nut, must have a width of at least 1+5/16inches.

What 55669.png means?
If the nut has 1+5/16 inches too, why they give the information about inches of nut and of wrench separately using “and”?
And the image of the question indicates that the nut has 1+5/16 inches. But at the same time the question says “a wrench that, in order to fit the nut, must have a width of at least 1+5/16inches”??

This question is annoying geez can someone explain me why this question lends infos in this way please?
Posted from my mobile device

Even I got confused here. Adding Bunuel to make this correction

_________________________
Fixed. Thank you!
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
1
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Why is this question rated hard 95%? I think it's fairly simple: 1+5/16 so approximately 25.4+8=33.4, so 33 wouldn't fit at all, only option is 34.
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
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gmatt1476 wrote:

The figures above show a hexagonal nut that has a width of $$1 \frac{5}{16}$$ inches and a wrench that, in order to fit the nut, must have a width of at least $$1 \frac{5}{16}$$ inches. Of all the wrenches that fit the nut and have widths that are whole numbers of millimeters, the wrench that fits the nut most closely has a width of how many millimeters?

(Note: 1 inch ≈ 25.4 millimeters)

A. 30
B. 31
C. 32
D. 33
E. 34

PS92751.01

Attachment:
2019-09-21_1425.png

$$1 \frac{5}{16}$$ $$= \frac{21}{16} ≈ \frac{21}{15}$$

1 inch ≈ 25.4 millimeters ≈ 25 millimeters

Now, $$\frac{21}{15}*25=1.4*25=35≈34$$

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The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
the key is the 5/16th. They aready give you the value of 1 inch. Other solutions involve significant calcuations with unkind numbers - highly recommend the below approach:

Approximate values:
5/16 is about 5/15 = 1/3
25.4 is roughly 25.5

Calculate the mm of 5/16ths of a inch:
25.5*1/3 = 8.5 --> 5/16th of an inch is approximately in mm

1 inch + 5/16th inch = 25.4mm+8.5mm = 33.9mm which is about 34mm.
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
1 5/16= 21/16
25.4 * 21/16= 33.33 ~ 34

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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
Arnobdas09 wrote:
1 5/16= 21/16
25.4 * 21/16= 33.33 ~ 34

Posted from my mobile device

Did you use a calculator? 25.4 * 21/16 seems like a ridiculously tedious calculation that can be avoided with some logic.
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
$$1\frac{5}{16}$$ * 25.4 = 25.4 + $$\frac{5}{16}$$*25.4 = 25.4 + 5*1.6 (because 254 ~ 256 = $$16^2$$) = 25.4 + 8 = 33.4

IMO E. 34
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
This is how I solved it. Difference between 5/16 and 5/15 is extremely minimal. So I took

1/3 and divided 25.4. Easily got 8.46.

25 Plus 8 equals to 33. Include the .4s it has to be greater than 33.
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
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Re: The figures above show a hexagonal nut that has a width of 1 5/16 [#permalink]
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