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The left pulley points at 4 and the right pulley points at 3, which means we get 4+3=7. Hence, the correct answer is E.
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There are three pulleys connected to one another. The one on the left has a radius of 8, has number 1,2,3,4 marked clock-wise (arrow is pointing at 1). The one on the right has radius of 12 and has 1,2,3,4 marked clock-wise (again, arrow pointing to 0). The one at the bottom is a pulley with a radius of 4. If the bottom pulley turns 1&1/2 times anti-clockwise, what is the sum of the numbers that the arrows of the two pulleys on the upper part of the figure would be pointing at after 1&1/2 turns?

A. 3
B. 4
C. 5
D. 6
E. 7

*An answer will be posted in 2 days.

Can it be solved via ratio problem. Since radius of bottom is 1/2 of left side circle and 1/3 of right side circle.

I anti clock wise rotation of 3/2 revolution of bottom circle will make 3/4th revolution of left circle and 3/6 or 1/2 revolution of right triangle. which make 3 on arrow in case of right and 4 in case left circle making it 7.

Since radius of circle are in proportion. we can calculate reduction ratio.
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There are three pulleys connected to one another. The one on the left has a radius of 8, has number 1,2,3,4 marked clock-wise (arrow is pointing at 1). The one on the right has radius of 12 and has 1,2,3,4 marked clock-wise (again, arrow pointing to 0). The one at the bottom is a pulley with a radius of 4. If the bottom pulley turns 1&1/2 times anti-clockwise, what is the sum of the numbers that the arrows of the two pulleys on the upper part of the figure would be pointing at after 1&1/2 turns?

A. 3
B. 4
C. 5
D. 6
E. 7

*An answer will be posted in 2 days.

Responding to a pm:

How much each pulley moves depends on the circumference of the pulley. More the circumference, less it will move.
Since the radii of the pulleys are 4, 8 and 12, their circumferences will be in the ratio 1:2:3
So the pulleys will move in the ratio 1 : 1/2 : 1/3 = 6:3:2

When the smallest moves 1.5 rotations anticlockwise, it moves 6 steps on the number wheel 1-2-3-4-1-2-3.
So the middle pulley will move 3 steps 1-2-3-4
And the largest pulley will move 2 steps 1-2-3

Sum of the numbers on the arrows = 4+3 = 7
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chetan2u
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There are three pulleys connected to one another. The one on the left has a radius of 8, has number 1,2,3,4 marked clock-wise (arrow is pointing at 0). The one on the right has radius of 12 and has 1,2,3,4 marked clock-wise (again, arrow pointing to 0). The one at the bottom is a pulley with a radius of 4. If the bottom pulley turns 1&1/2 times anti-clockwise, what is the sum of the numbers that the arrows of the two pulleys on the upper part of the figure would be pointing at after 1&1/2 turns?

A. 3
B. 4
C. 5
D. 6
E. 7

*An answer will be posted in 2 days.

since there is NO 0, i believe you mean 1, when you right arrow pointing at 1...
The Q depends on circumference...

the lower pulley has a radius 4 and will travel \(1.5 ( 2pi*4)=12*pi\) in 1\(\frac{1}{2}\)rotations...
lets see the effect on the other two pulleys as the all will travel same distance during this interval...

1)The circumference of pulley of radius 8 is \(2*pi*8 = 16*pi..\)
so the pulley will move \(\frac{12*pi}{16*pi} = \frac{3}{4}\)of the rotation, so the arrow will cover 1 to 2, 2 to 3 and 3 to 4... AND will point at 4..

1)The circumference of pulley of radius 12 is \(2*pi*12 = 24*pi..\)
so the pulley will move \(\frac{12*pi}{24*pi}= \frac{1}{2}\) of the rotation, so the arrow will cover 1 to 2 and 2 to 3... AND will point at 3..

\(SUM = 4+3 = 7\)
E

Hi, thanks for the explanation. Can you also clarify why the upper two pulleys move clockwise if the bottom pulley spins counter-clockwise? I presumed that all pulleys would move counter clockwise and hence got 5 ==> Left pulley goes from 1-4-3-2 and right pulley goes 1-4-3, the sum of both being five.
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There are three pulleys connected to one another. The one on the left has a radius of 8, has number 1,2,3,4 marked clock-wise (arrow is pointing at 0). The one on the right has radius of 12 and has 1,2,3,4 marked clock-wise (again, arrow pointing to 0). The one at the bottom is a pulley with a radius of 4. If the bottom pulley turns 1&1/2 times anti-clockwise, what is the sum of the numbers that the arrows of the two pulleys on the upper part of the figure would be pointing at after 1&1/2 turns?

A. 3
B. 4
C. 5
D. 6
E. 7

*An answer will be posted in 2 days.

since there is NO 0, i believe you mean 1, when you right arrow pointing at 1...
The Q depends on circumference...

the lower pulley has a radius 4 and will travel \(1.5 ( 2pi*4)=12*pi\) in 1\(\frac{1}{2}\)rotations...
lets see the effect on the other two pulleys as the all will travel same distance during this interval...

1)The circumference of pulley of radius 8 is \(2*pi*8 = 16*pi..\)
so the pulley will move \(\frac{12*pi}{16*pi} = \frac{3}{4}\)of the rotation, so the arrow will cover 1 to 2, 2 to 3 and 3 to 4... AND will point at 4..

1)The circumference of pulley of radius 12 is \(2*pi*12 = 24*pi..\)
so the pulley will move \(\frac{12*pi}{24*pi}= \frac{1}{2}\) of the rotation, so the arrow will cover 1 to 2 and 2 to 3... AND will point at 3..

\(SUM = 4+3 = 7\)
E

Hi, thanks for the explanation. Can you also clarify why the upper two pulleys move clockwise if the bottom pulley spins counter-clockwise? I presumed that all pulleys would move counter clockwise and hence got 5 ==> Left pulley goes from 1-4-3-2 and right pulley goes 1-4-3, the sum of both being five.

Initially, I thought the same but then I realized, that the arrow is stationary so yes the upper pulleys will indeed move anti clockwise and so the numbers move and not the arrow. Therefore the arrows will point to 4 & 3.

Hit Kudos if this helped!
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E[/quote]

Hi, thanks for the explanation. Can you also clarify why the upper two pulleys move clockwise if the bottom pulley spins counter-clockwise? I presumed that all pulleys would move counter clockwise and hence got 5 ==> Left pulley goes from 1-4-3-2 and right pulley goes 1-4-3, the sum of both being five.[/quote]

Initially, I thought the same but then I realized, that the arrow is stationary so yes the upper pulleys will indeed move anti clockwise and so the numbers move and not the arrow. Therefore the arrows will point to 4 & 3.

Hit Kudos if this helped![/quote]

Hey, thanks for replying. Figured it as soon as I replied. Visualized it in my head and then face-palmed.
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MathRevolution
There are three pulleys connected to one another. The one on the left has a radius of 8, has number 1,2,3,4 marked clock-wise (arrow is pointing at 1). The one on the right has radius of 12 and has 1,2,3,4 marked clock-wise (again, arrow pointing to 0). The one at the bottom is a pulley with a radius of 4. If the bottom pulley turns 1&1/2 times anti-clockwise, what is the sum of the numbers that the arrows of the two pulleys on the upper part of the figure would be pointing at after 1&1/2 turns?

A. 3
B. 4
C. 5
D. 6
E. 7

circumferences:
L=12*2π=24π
M=8*2π=16π
S=4*2π=8π

S turned 1.5 times: 8π*3/2=12π
M turned: 12π/16π=3/4, arrow=4
L turned: 12π/24π=1/2, arrow=3
M+L=7

Ans (E)
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This involves the concept of speed remains constant or the gear ratio(as we read in physics). To keep the speed constant, the smaller pulley will rotate more than the bigger pulley. Once we identify that, we can establish a relation as: n1r1 = n2r2. Solving this two times, we can get the answer.
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chetan2u
MathRevolution
There are three pulleys connected to one another. The one on the left has a radius of 8, has number 1,2,3,4 marked clock-wise (arrow is pointing at 0). The one on the right has radius of 12 and has 1,2,3,4 marked clock-wise (again, arrow pointing to 0). The one at the bottom is a pulley with a radius of 4. If the bottom pulley turns 1&1/2 times anti-clockwise, what is the sum of the numbers that the arrows of the two pulleys on the upper part of the figure would be pointing at after 1&1/2 turns?

A. 3
B. 4
C. 5
D. 6
E. 7

*An answer will be posted in 2 days.

since there is NO 0, i believe you mean 1, when you right arrow pointing at 1...
The Q depends on circumference...

the lower pulley has a radius 4 and will travel \(1.5 ( 2pi*4)=12*pi\) in 1\(\frac{1}{2}\)rotations...
lets see the effect on the other two pulleys as the all will travel same distance during this interval...

1)The circumference of pulley of radius 8 is \(2*pi*8 = 16*pi..\)
so the pulley will move \(\frac{12*pi}{16*pi} = \frac{3}{4}\)of the rotation, so the arrow will cover 1 to 2, 2 to 3 and 3 to 4... AND will point at 4..

1)The circumference of pulley of radius 12 is \(2*pi*12 = 24*pi..\)
so the pulley will move \(\frac{12*pi}{24*pi}= \frac{1}{2}\) of the rotation, so the arrow will cover 1 to 2 and 2 to 3... AND will point at 3..

\(SUM = 4+3 = 7\)
E

Hi Chetan,

As the bottom pulley moves anti-clockwise, shouldn't the total be 2 (left) + 3 (right) = 5?
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Pully 2 is twice as big as Pully 1, and Pully 3 is three times as big a Pully 1
This means pully 2 will rotate numbers half as much as P1, and P3 rotates 1/3 as much

Pully 1 = radius 4
Pully 2 = radius 8
Pully 3 = radius 12

Pully 1 = 1.5 full rotations = 6 spots
Pully 2 = 6/2 = 3 spots --> lands on 4
Pully 3 = 6/3 = 2 spots --> lands on 3

4+3 = 7
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