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The dial shown above is divided into

Expert replies
by boomgoesthegmat » Tue May 03, 2016 2:47 pm
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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

OA: E
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Source: — Problem Solving |

by OptimusPrep » Tue May 03, 2016 7:22 pm
boomgoesthegmat wrote:Image

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

OA: E
Number of intervals in one circle = 8
Multiple of 8 closest to 1174 = 1168

Hence the pointer will cover 6 additional intervals
Therefore it will stop at E

Correct Option: E
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by [email protected] » Mon Mar 12, 2018 5:00 pm
Hi All,

We're told that the dial shown above is divided into equal-sized intervals. We're asked which of the following letters the pointer will stop at if it is rotated CLOCKWISE from S through 1,174 intervals. This question is ultimately about 'remainders.' The clock face has 8 intervals, so we can calculate the number of complete rotations - and the 'remainder' (the additional intervals that the clock dial would move through):

1174/8 = 146 r 6 --> that remainder 6 means that the dial would move through the first dash, then A, B, C and D before finally stopping on E.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by Scott@TargetTestPrep » Wed Mar 14, 2018 3:33 pm
boomgoesthegmat wrote:Image

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
We are given a diagram with 8 intervals of equal length. Let’s start by sketching the diagram.

Image

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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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