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Difficulty:
85%
(hard)
Question Stats:
54% (02:26) correct
46%
(02:34)
wrong
based on 2688
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History
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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
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2019-09-21_1425.png [ 19.34 KiB | Viewed 40251 times ]
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Posts: 1,348
Updated on: 07 Oct 2022, 01:58
Originally posted by GMATGuruNY on 21 Sep 2019, 03:00.
Last edited by GMATGuruNY on 07 Oct 2022, 01:58, edited 1 time in total.
Last edited by GMATGuruNY on 07 Oct 2022, 01:58, edited 1 time in total.
gmatt1476
Width of the nut:
\(1 \frac{5}{16}\) inches = \(\frac{21}{16}\) inches
Conversion note:
\(25.4\) millimeters per inch \(= 25\frac{2}{5}\) millimeters per inch \(= \frac{127}{5}\) millimeters per inch
Converting the width of the nut from inches to millimeters, we get:
\(\frac{21}{16}\) inches * \(\frac{127}{5}\) millimeters per inch = \(\frac{21}{16} * \frac{127}{5} = \frac{21}{5} * \frac{127}{16}\) = (4.2)(almost 8) = more than 33 millimeters.
Of the five answer choices, only E will accommodate a width greater than 33 millimeters.
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General Discussion
Archit3110
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Updated on: 21 Sep 2019, 05:00
Originally posted by Archit3110 on 21 Sep 2019, 04:32.
Last edited by Archit3110 on 21 Sep 2019, 05:00, edited 1 time in total.
Last edited by Archit3110 on 21 Sep 2019, 05:00, edited 1 time in total.
given fraction;
\(1 \frac{5}{16}\) inches ; 21/16
and 1 inch = 25.4 mm
so at least closest measurement ; 25.4*21/16 = ~33.3 = 34
IMO E
\(1 \frac{5}{16}\) inches ; 21/16
and 1 inch = 25.4 mm
so at least closest measurement ; 25.4*21/16 = ~33.3 = 34
IMO E
gmatt1476
Posts: 1,348
Archit3110
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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Santoshk31
Joined: 01 Jun 2017
Last visit: 14 Nov 2025
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Posts: 2
this is a pretty straight fwd calculation. convert inches to mm that comes around > 33 (~33.337).
Thus, ANS -> E
Thus, ANS -> E
Mbahousegmat
Joined: 03 Jan 2020
Last visit: 10 Jan 2020
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Location: United States
Schools: Kellogg EMBA"18 (A$)
GMAT 1: 720 Q56 V40
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
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
Schools: Insead '14
Posts: 6
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
during conversion use closest estimates..
21/16 inch x 25.4
20 / 15 x 25 = 33.33.. u know now answer is E.34
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
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
\(1 \frac{5}{16}\) inches = \(\frac{21}{16 }\) inches
\(21/16 * 25.4\) = \(\frac{21}{16} * 25 \frac{4}{10}\)
\(\frac{21}{16 }* \frac{254}{10}\)
\(\frac{21}{16} * \frac{127}{5} = \frac{2667}{80}\)
\(= approximately 33 \\
\)
The only answer that's greater is 34. Answer is E.
\(21/16 * 25.4\) = \(\frac{21}{16} * 25 \frac{4}{10}\)
\(\frac{21}{16 }* \frac{254}{10}\)
\(\frac{21}{16} * \frac{127}{5} = \frac{2667}{80}\)
\(= approximately 33 \\
\)
The only answer that's greater is 34. Answer is E.
Pritishd
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Fixed. Thank you!
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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.
gmatt1476
\(1 \frac{5}{16}\) \(= \frac{21}{16} ≈ \frac{21}{15}\)
1 inch ≈ 25.4 millimeters ≈ 25 millimeters
Now, \(\frac{21}{15}*25=1.4*25=35≈34\)
The answer is E
Posts: 259
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
Add the given and your solution above:
1 inch + 5/16th inch = 25.4mm+8.5mm = 33.9mm which is about 34mm.
E
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
Add the given and your solution above:
1 inch + 5/16th inch = 25.4mm+8.5mm = 33.9mm which is about 34mm.
E
Posts: 2
1 5/16= 21/16
25.4 * 21/16= 33.33 ~ 34
Posted from my mobile device
25.4 * 21/16= 33.33 ~ 34
Posted from my mobile device
Posts: 259
Arnobdas09
Did you use a calculator? 25.4 * 21/16 seems like a ridiculously tedious calculation that can be avoided with some logic.
\(1\frac{5}{16}\) * 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
IMO E. 34
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Products:
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?
\(1 \frac{5}{16} = 25 + 25*5/16 + .4 + .4*5/16 = 25 + 7.8 + .4 + .125 = 33.325 mm\)
Only wrench with width of 34 mm will fit most closely.
IMO E
gmatftw777
Yes, this is a unit-conversion problem, not a geometry question, so it is still relevant to the GMAT Focus Edition.
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