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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
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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 equal-sized intervals. At which [#permalink]
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IMO E.

On dividing by 8, the remainder is 6 which brings the pointer to E.
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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
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Bunuel wrote:

The dial shown above is divided into equal-sized 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:
2015-10-16_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 equal-sized intervals. At which [#permalink]
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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 equal-sized intervals. At which [#permalink]
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Attached is a visual that should help.
Attachments

Screen Shot 2016-05-10 at 5.49.54 PM.png
Screen Shot 2016-05-10 at 5.49.54 PM.png [ 67.15 KiB | Viewed 59376 times ]

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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
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 equal-sized intervals. At which [#permalink]
I am still not able to understand the logic. please help!
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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
[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.
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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
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INSEADIESE wrote:
I am still not able to understand the logic. please help!


Divide it by 8 and check the reminder, the remainder will be the answer in clockeise direction from start.

Posted from my mobile device
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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
Bunuel wrote:

The dial shown above is divided into equal-sized 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:
2015-10-16_0859.png


This question is simply asking us to find out the remainder when 1174 is divided by 8. If we notice carefully, the clock is divided into 8 intervals. Therefore, the pointer stops at 'S' every 8 intervals. Hence, all we need to do is find out how many more intervals will the pointer go after it shifts 146 intervals.

\( \frac{1174}{8} \) leaves a quotient of 146 and a remainder of 6.

This means, that the pointer strikes S in the 146th interval then moves 6 intervals more to complete the rotation through 1174 intervals.

Answer : E
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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
Correct Option E
1174 Cycle / 8 equal parts = 146 Cycle + 6 Reminder, and 6 position is allocated to option E
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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
Instead of doing the actual division to find out the reminder, can we just not apply the "units digits" theory as:

1174 -> 4 ( units digits) -> the 2nd power so 4,6,4,6,4,6,4,6,4 etc... . Because its divided by 8 -> We count to 8, which gives us de reminder of 6 ...

Is this approach correct?

thank you
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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
Skywalker18 wrote:
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



Bunuel
Why can't we reduce 1174/8 to its lowest form i.e. 587/4 and then find remainder = 3 ?
What is wrong in my approach?
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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
When we divide 1174/8 Remainder is 6

But when we divide 587/4 remainder is 3 why?

Posted from my mobile device
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The dial shown above is divided into equal-sized intervals. At which [#permalink]
Let point S be the starting point.
Since the pointer travelled 1174 intervals, it must be on point S at the completion of 1168 intervals(since 1168 is divided by 8).
Now@, after travelling E intervals from S, the pointer will reach point E.
Thus, the correct option will be E.
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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: The dial shown above is divided into equal-sized intervals. At which [#permalink]
Am I the only one who assumed A,B,C,D,E= 5 equal spaced intervals :/
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