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The dial shown above is divided into equalsized intervals. At which
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15 Oct 2015, 21:59
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The dial shown above is divided into equalsized intervals. At which of the following letters will the pointer stop if it is rotated clockwise from S through 1,174 intervals? (A) A (B) B (C) C (D) D (E) E Kudos for a correct solution.Attachment:
20151016_0859.png [ 6.5 KiB  Viewed 19021 times ]
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Re: The dial shown above is divided into equalsized intervals. At which
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15 Oct 2015, 23:51
Bunuel wrote: The dial shown above is divided into equalsized intervals. At which of the following letters will the pointer stop if it is rotated clockwise from S through 1,174 intervals? (A) A (B) B (C) C (D) D (E) E Kudos for a correct solution.Attachment: 20151016_0859.png We have 8 Intervals bit 1174 is not divisible by 8: the nearest number divisble by 8 is 1168 and we are left with 6 as a remainder > Answer (E)
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The dial shown above is divided into equalsized intervals. At which
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16 Oct 2015, 02:32
One complete rotation starting from S contains 8 equally spaced intervals Position at which Pointer will stop will be Remainder of 1174/8 i.e 6 The pointer will stop at E Answer E
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Re: The dial shown above is divided into equalsized intervals. At which
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16 Oct 2015, 02:38
IMO E.
On dividing by 8, the remainder is 6 which brings the pointer to E.



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The dial shown above is divided into equalsized intervals. At which
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16 Oct 2015, 02:42
Bunuel wrote: The dial shown above is divided into equalsized intervals. At which of the following letters will the pointer stop if it is rotated clockwise from S through 1,174 intervals? (A) A (B) B (C) C (D) D (E) E Kudos for a correct solution.Attachment: 20151016_0859.png My solution:
As there are 8 intervals we can divide 1174 by 8,
we get 146 as quotient and 6 as remainder. Therefore, 146th interval will be the point S, then count 6 more and we will reach at point E.
Answer choice E
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Re: The dial shown above is divided into equalsized intervals. At which
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16 Oct 2015, 07:57
My solution:
divide 1174 by 2 we get 587; now divide 587 by 4 we get 3 as a reminder. I'm counting 4 intervals as A(12 o clock), 3 o clock, 6 o clock and 9 o clock. So with 3 as the reminder needle will stop at 9 o clock. Hence at E it will stop. E is the solution.



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Re: The dial shown above is divided into equalsized intervals. At which
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10 May 2016, 18:08
Attached is a visual that should help.
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Screen Shot 20160510 at 5.49.54 PM.png [ 67.15 KiB  Viewed 15820 times ]
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Re: The dial shown above is divided into equalsized intervals. At which
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11 May 2016, 07:42
Bunuel wrote: The dial shown above is divided into equalsized intervals. At which of the following letters will the pointer stop if it is rotated clockwise from S through 1,174 intervals? (A) A (B) B (C) C (D) D (E) E We are given a diagram with 8 intervals of equal length. Let’s start by sketching the diagram. Notice we also assigned a number to each letter marking between the intervals. Begin with S = 0 and moving clockwise, we have A = 2, B = 3, C = 4, D = 5 and E = 6. Let's first study the pattern of the spinner's movement. If the spinner were to go through just 8 intervals, it would stop where it started, at letter S. If, instead, it were to go through 10 intervals, it would go all the way around the circle one time (8 intervals) plus 2 more, stopping at letter A. One more example: if it were to go through 35 intervals it would make four complete revolutions, plus 3 more intervals, stopping at letter B. We can see that if we divide the total number of intervals it travels by 8, the whole number part of the quotient will be the number of revolutions it has made, and the remainder will tell us how many "extra" intervals it has gone past S before stopping. Therefore, to determine at which letter the pointer will stop when it is rotated clockwise 1,174 intervals, we need to divide 1,174 by 8. 1,174/8 = 146 R 6. The 146 indicates that the spinner has gone through 146 complete revolutions. The remainder 6 indicates that the spinner has stopped 6 intervals past S, which is location E. Answer: E
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Re: The dial shown above is divided into equalsized intervals. At which
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04 Dec 2016, 01:26
Great Official Question. Here is what i did in this one => 1174 rotations Since there are 8 intervals => 1174=> 8k+6 so after 1168 rotations the pointer will be back at S and we just have to now rotate it 6 more times clockwise=> It will land up at E
Hence E
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Re: The dial shown above is divided into equalsized intervals. At which
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22 Nov 2018, 13:31
I am still not able to understand the logic. please help!



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Re: The dial shown above is divided into equalsized intervals. At which
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21 Mar 2019, 15:07
[1] Divide 1,174 by Number of Intervals
# Intervals = 8
\(1,174 \div 8 = 146 R 6\)
[2] Use Remainder to Count Number of Intervals for Final Rotation
Since the remainder is 6, count 6 intervals around the circle.
Answer E.




Re: The dial shown above is divided into equalsized intervals. At which
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