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Sat Dec 28, 2019 12:48 am
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
50% (02:22) correct
50%
(02:15)
wrong
based on 90
sessions
History
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Time
Result
Not Attempted Yet
Mike the mechanic has a machine which has four cog wheels in connection. The largest wheel has 242 teeth and the others have 66, 48 and 26, respectively. How many rotations must the largest wheel make before each of the wheels is back in its starting position?
A. 660
B. 512
C. 484
D. 312
E. None of these
A. 660
B. 512
C. 484
D. 312
E. None of these
Sat Dec 28, 2019 2:38 am
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After starting to rotate together, each of these wheels will complete rotations at different times.
When the smallest cog with 26 teeth will already have completed its first rotation (think of this as rotating by 26 teeth), the other three will still be on the way. (for example, the one with 66 teeth will still have to rotate by 66-26=40 more teeth to complete its rotation.)
The first time that they will all reach the starting positions together will be when they will have rotated by the same number of teeth, say, x.
Then, x = least common multiple of their respective number of teeth = LCM {242, 66,48,26}=75504.
Now, for a rotation by 75504 teeth, the largest wheel will have to rotate 312 times (75504/242).
[The second largest wheel will have to rotate by 75504/66 times, and so on.]
ANSWER: (D) 312
When the smallest cog with 26 teeth will already have completed its first rotation (think of this as rotating by 26 teeth), the other three will still be on the way. (for example, the one with 66 teeth will still have to rotate by 66-26=40 more teeth to complete its rotation.)
The first time that they will all reach the starting positions together will be when they will have rotated by the same number of teeth, say, x.
Then, x = least common multiple of their respective number of teeth = LCM {242, 66,48,26}=75504.
Now, for a rotation by 75504 teeth, the largest wheel will have to rotate 312 times (75504/242).
[The second largest wheel will have to rotate by 75504/66 times, and so on.]
ANSWER: (D) 312
Sun Dec 29, 2019 3:08 am
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Solution
Given
- • Mike the mechanic has a machine which has four cog wheels in connection.
• The largest wheel has 242 teeth and the others have 66, 48 and 26, respectively.
To find
- • The number of rotations the largest wheel must make before each of the wheels is back in its starting position
Approach and Working out
LCM of the given number of teeth of all wheels = LCM (242, 66, 48, 26) = 75504
- • Hence, number of rotations the largest wheel will make = \(\frac{75504}{242}\) = 312
Thus, option D is the correct answer.
Correct Answer: Option D
Sun Jan 05, 2020 1:56 pm
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Bunuel
I came up with the following approach:
With what number do I have to multiply 242 in order to evenly divide it by all other given number of teeth?
The least common multiplier (LCM) is the value we are looking for.
We use factorization to do so:
242 = (2*11*11)
66 = (2*3*11)
48 = (2*2*2*2*3)
26 = (2*13)
Multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs. This gives you the LCM.
LCM = (2*2*2*2*3*11*11*13)
Divide by 242 = (2*11*11)
Leaves you with (2*2*2*3*13) = 312
IMO: D
