Question 1

Calculate the values of x and y in the following table.   $\begin{array} { | l | l | l | l | } 
\hline F _ { 1 } & d _ { 1 } & F _ { 2 } & d _ { 2 } \
\hline 16 \mathrm { lb } & 4 \text { in. } & 8 \mathrm { lb } & x \text { in. } \
\hline 110 \mathrm { lb } & y \text { in. } & 60 \mathrm { lb } & 11 \text { in. } \
\hline
\end{array}$
A) x = 16, y = 10
B) x = 8, y = 6
C) x = 16, y = 5
D) x = 6, y = 5
E) x = 6, y = 10

Accepted Answer

The answer of Calculate the values of x and y...

Question 2

Calculate the values of x and y in the following table. $\begin{array} { | l | l | l | l | } 
\hline \text { Gear A } && \text { Gear B } \
\hline \begin{array} { l } 
\text { Number of } \
\text { teeth }
\end{array} & \mathrm { rpm } & \begin{array} { l } 
\text { Number of } \
\text { teeth }
\end{array} & \mathrm { rpm } \
\hline 40 & 700 & 25 & x \
\hline 50 & 56 & y & 80 \
\hline
\end{array}$
A) x = 1,120, y = 35
B) x = 1,125, y = 72
C) x = 560, y = 35
D) x = 2,240, y = 72
E) x = 1,125, y = 36

Accepted Answer

The gear ratio is determined by the number of teeth and rpm. For Gear A and Gear B, the ratio of teeth to rpm should be equal. For the first row: (40 teeth / 700 rpm) = (25 teeth / x rpm), solving gives x = 1,120. For the second row: (50 teeth / 56 rpm) = (y teeth / 80 rpm), solving gives y = 35.

Question 3

Draw a sketch for the problem and solve it. A piece of machinery weighs 5 tons. It is to be balanced by two men whose combined weight is 400 lb. The piece of machinery is placed 8 in. from the fulcrum. How far from the fulcrum must the two men exert their weight in order to balance it?
A) 23 in.
B) 460 in.
C) 230 in.
D) 200 in.
E) 198 in.

Accepted Answer

The answer of Draw a sketch for the problem and...

Question 4

A large pulley turns at 60 rpm. A smaller pulley 6 in. in diameter turns at 400 rpm. What is the diameter of the larger pulley?
A) 21.5 in.
B) 100 in.
C) 43 in.
D) 50 in.
E) 40 in.

Accepted Answer

The relationship between the diameters and rotational speeds of two pulleys in a belt drive system can be described by the formula $D_1 \times RPM_1 = D_2 \times RPM_2$, where $D_1$ and $D_2$ are the diameters of the pulleys, and $RPM_1$ and $RPM_2$ are their rotational speeds in revolutions per minute. Given that one pulley is 6 inches in diameter and turns at 400 rpm, and the other turns at 60 rpm, we can solve for the diameter of the larger pulley: $6 \times 400 = D_2 \times 60$. Solving for $D_2$ gives $D_2 = \frac{6 \times 400}{60} = 40$ inches.

Question 5

A large pulley turns at 105 rpm. A smaller pulley 7 in. in diameter turns at 300 rpm. What is the diameter of the larger pulley?
__________ inches

Accepted Answer

The answer of A large pulley turns at 105 rpm....

Question 6

A small pulley is 9 in. in diameter, and a larger one is 17 in. in diameter. How many rpm does the larger pulley make if the smaller one revolves at 850 rpm?
__________ rpm

Accepted Answer

The answer of A small pulley is 9 in. in...

Question 7

Calculate the values of x and y in the following table? $$\begin{array} { | l | l | l | l | } 
\hline \multicolumn{2}{|l|} { \text { Pulley A } } & \text { Pulley B } \
\hline \text { Diameter } & \mathrm { rpm } & \text { Diameter } & \mathrm { rpm } \
\hline 34 \mathrm {~cm} & 54 & 68 \mathrm {~cm} & x \
\hline 18 \mathrm {~cm} & y & 12 \mathrm {~cm} & 96 \
\hline
\end{array}$$
A) x = 108, y = 32
B) x = 27, y = 64
C) x = 13.5, y = 64
D) x = 108, y = 64
E) x = 13.5, y = 32

Accepted Answer

The relationship between the diameters and rpm of pulleys in a belt drive system is given by the formula $D_1 \times N_1 = D_2 \times N_2$, where $D_1$ and $D_2$ are the diameters of the pulleys, and $N_1$ and $N_2$ are their respective rpms. For Pulley A and Pulley B with diameters 34 cm and 68 cm, respectively, and given that Pulley A has an rpm of 54, we can find $x$ by setting up the equation $34 \times 54 = 68 \times x$, which simplifies to $x = 27$. For the second set of pulleys with diameters 18 cm and 12 cm, and given that Pulley B has an rpm of 96, we can find $y$ by setting up the equation $18 \times y = 12 \times 96$, which simplifies to $y = 64$.

Question 8

A large gear with 100 teeth turning at 210 rpm turns a small gear with 12 teeth. At how many rpm does the small gear turn?
A) 1,749 rpm
B) 1,790 rpm
C) 895 rpm
D) 3,580 rpm
E) 1,750 rpm

Accepted Answer

The speed ratio of the gears is equal to the ratio of their number of teeth. So, the small gear turns at 210 * 100 / 12 = 1750 rpm.

Question 9

Calculate the values of x and y in the following chart.

$\begin{array}{l}
\hline { \text { Pulley A } }\quad\quad\quad\quad\quad\quad \text { Pulley B } \
\begin{array}{ | l | l | l | l | }

\hline \text { Diameter } & \mathrm { rpm } & \text { Diameter } & \mathrm { rpm } \
\hline 36 \mathrm {~cm} & 400 & x \mathrm {~cm} & 720 \
\hline 15 \mathrm {~cm} & 300 & 24 \mathrm {~cm} & y \
\hline
\end{array}
\end{array}$
A) x = 240, y = 1,885
B) x = 24, y = 188.5
C) x = 48, y = 375
D) x = 24, y = 375
E) x = 20, y = 187.5

Accepted Answer

The answer of Calculate the values of x and y...

Question 10

Draw a sketch for the problem and solve it. A rocker arm raises oil from an oil well. On each stroke, it lifts a weight of 3 tons on a weight arm 1 ft long. What force is needed to lift the oil on a force arm 2 ft long?
A) 0.75 ton
B) 3 tons
C) 6 tons
D) 12 tons
E) 1.5 tons

Accepted Answer

The answer of Draw a sketch for the problem and...

Question 11

One pulley is twice as large in diameter as a second pulley. If the larger pulley turns at 259 rpm, what is the rpm of the smaller?
A) 518 rpm
B) 129.5 rpm
C) 538 rpm
D) 1,076 rpm
E) 522 rpm

Accepted Answer

The speed of the smaller pulley will be twice that of the larger pulley because it has half the diameter. Therefore, if the larger pulley turns at 259 rpm, the smaller pulley will turn at 259 * 2 = 518 rpm.

Question 12

Calculate the values of x and y in the following table.

$\begin{array}{l}
\hline { \text { Pulley A } }\quad\quad\quad\quad\quad\quad \text { Pulley B } \
\begin{array}{ | l | l | l | l | }

\hline \text { Diameter } & \text { rpm } & \text { Diameter } & \text { rpm } \
\hline x \text { in. } & 625 & 25 \text { in. } & 975 \
\hline 96 \mathrm {~cm} & 240 & y \mathrm {~cm} & 360 \
\hline
\end{array}
\end{array}$
A) x = 78, y = 122
B) x = 42, y = 61
C) x = 39, y = 64
D) x = 42, y = 122
E) x = 21, y = 61

Accepted Answer

The relationship between the diameters and rpm of pulleys is given by the formula: $ \text{Diameter}_A \times \text{rpm}_A = \text{Diameter}_B \times \text{rpm}_B $. Solving for x and y using this formula gives x = 39 in and y = 64 cm.

Question 13

Calculate the values of x and y in the following table.   $\begin{array}{|l|l|l|l|}
\hline \text { Gear A } & &\text { Gear B } \
\hline \begin{array}{l}
\text { Number of } \
\text { teeth }
\end{array} & \text { rpm } & \begin{array}{l}
\text { Number of } \
\text { teeth }
\end{array} & \mathrm{rpm} \
\hline 10 & 100 & x & 125 \
\hline 260 & y & 50 & 520 \
\hline
\end{array}$
A) x = 8, y = 25
B) x = 2, y = 200
C) x = 8, y = 100
D) x = 16, y = 200
E) x = 16, y = 25

Accepted Answer

To find the ratio of the number of teeth between gears A and B, we can set up the proportion: $\frac{10}{x} = \frac{100}{125}$. Solving for $x$, we get $x=8$. We can also set up the proportion $\frac{260}{50} = \frac{y}{520}$ and solve for $y$ to get $y=1040$. However, we need to convert the units of $y$ to rpm, which can be done by dividing by the number of teeth on Gear B: $\frac{1040}{50} = 20.8$ rpm. Since $y$ must be a whole number of rpm, we round 20.8 to 21 rpm. Therefore, the values of $x$ and $y$ are $8$ and $21$, respectively, which corresponds to choice C.

Question 14

Draw a sketch for the problem and solve it. An object is 4 ft from the fulcrum and balances a second object 8 ft from the fulcrum. The first object weighs 64 lb. How much does the second object weighs?
A) 32 lb
B) 37 lb
C) 128 lb
D) 64 lb
E) 42 lb

Accepted Answer

In a balance, the weight on one side is equal to the weight on the other side. We can use the formula:

Weight of object 1 × distance from fulcrum to object 1 = Weight of object 2 × distance from fulcrum to object 2

Substituting the given values, we get:

64 × 4 = Weight of object 2 × 8
Weight of object 2 = (64 × 4)/8 = 32 lb

Therefore, the second object weighs 32 lb. Choice A) is the correct answer.

Question 15

A large gear with 300 teeth turning at 11 rpm turns a small gear at 150 rpm. How many teeth does the small gear have?
A) 5.5
B) 36
C) 27
D) 22
E) 72

Accepted Answer

The answer of A large gear with 300 teeth turning...

Question 16

Calculate the values of x and y in the following table.   $\begin{array} { | l | l | l | l | } 
\hline F _ { 1 } & d _ { 1 } & F _ { 2 } & d _ { 2 } \
\hline 27 \mathrm { lb } & 6 \text { in. } & x \mathrm { lb } & 2 \mathrm { in. } \
\hline y \mathrm { lb } & 4.2 \mathrm { in } . & 512.6 \mathrm { lb } & 4.4 \mathrm { in } . \
\hline
\end{array}$
A) x = 240, y = 537.0
B) x = 8.1, y = 5,370.1
C) x = 81, y = 537.0
D) x = 80, y = 537.0
E) x = 240, y = 1,074.0

Accepted Answer

The principle of moments (or the law of the lever) states that for a system in equilibrium, the sum of clockwise moments about a pivot is equal to the sum of anticlockwise moments about the same pivot. For the first row, using $F_1d_1 = F_2d_2$, we have $27 \times 6 = x \times 2$, solving for $x$ gives $x = 81$. For the second row, using the same principle, $y \times 4.2 = 512.6 \times 4.4$, solving for $y$ gives $y = 537.0$.

Question 17

A small pulley is 13 in. in diameter, and a larger one is 19 in. in diameter. How many rpm does the larger pulley make if the smaller one revolves at 570 rpm?
A) 820 rpm
B) 833.0769230769 rpm
C) 410 rpm
D) 390 rpm
E) 380 rpm

Accepted Answer

The speed of the pulleys is inversely proportional to their diameters. Therefore, if the smaller pulley (13 in.) revolves at 570 rpm, the larger one (19 in.) would revolve at (13/19) * 570 = 390 rpm.

Question 18

A large gear with 405 teeth running at 300 rpm turns a small gear 450 rpm. How many teeth does the small gear have?
A) 180
B) 540
C) 320
D) 267
E) 270

Accepted Answer

The number of teeth on the gears is inversely proportional to their speeds. So, $405 \text{ teeth} \times 300 \text{ rpm} = x \text{ teeth} \times 450 \text{ rpm}$. Solving for $x$ gives $x = \frac{405 \times 300}{450} = 270$ teeth.

Question 19

Calculate the values of x and y in the following table.

$\begin{array}{l}
\hline { \text { Pulley A } }\quad\quad\quad\quad\quad\quad \text { Pulley B } \
\begin{array}{|l|l|}
\hline \text { Diameter } & \mathrm{rpm} \
\hline 30 \mathrm{~cm} & 150 \
\hline y \text { in. } & 40 \
\hline
\end{array}
\begin{array}{l|l|}

\hline \text { Diameter } & \mathrm{rpm} \
\hline 45 \mathrm{~cm} & x \
\hline 24 \mathrm{in} . & 20 \
\hline
\end{array}
\end{array}$
A) x = 200, y = 24
B) x = 100, y = 12
C) x = 50, y = 6
D) x = 300, y = 18
E) x = 50, y = 18

Accepted Answer

First, let's use the formula for pulley speed, which is:

pulley speed = (diameter of other pulley * speed of other pulley) / diameter of pulley in question

For the first row, we have:

30 * 150 = y * 40
y = 112.5

For the second row, we have:

45 * x = 24 * 20
x = 100

Therefore, the answer is B: x = 100, y = 12.

Question 20

A hydraulic pump is driven with an electric motor(see figure below). The pump must rotate at 1,000 rpm. The pump is equipped with a 6.0-in.-diameter belt pulley. The motor runs at 1,500 rpm. What diameter pulley is required on the motor? 
A) 3 in.
B) 2 in.
C) 4 in.
D) 5 in.
E) 10 in.

Accepted Answer

To determine the required diameter of the motor pulley, we can use the formula:

motor speed x motor pulley diameter = pump speed x pump pulley diameter

Plugging in the given values, we get:

1,500 rpm x motor pulley diameter = 1,000 rpm x 6.0 in.

Solving for motor pulley diameter, we get:

motor pulley diameter = (1,000 rpm x 6.0 in.) / 1,500 rpm

motor pulley diameter = 4 in.

Therefore, the required diameter of the motor pulley is 4 in. (Choice C).

Quiz 7: Ratio and Proportion

Draw a sketch for the problem and solve it. A piece of machinery weighs 5 tons. It is to be balanced by two men whose combined weight is 400 lb. The piece of machinery is placed 8 in. from the fulcrum. How far from the fulcrum must the two men exert their weight in order to balance it?

A large pulley turns at 60 rpm. A smaller pulley 6 in. in diameter turns at 400 rpm. What is the diameter of the larger pulley?

A large pulley turns at 105 rpm. A smaller pulley 7 in. in diameter turns at 300 rpm. What is the diameter of the larger pulley? __________ inches

A small pulley is 9 in. in diameter, and a larger one is 17 in. in diameter. How many rpm does the larger pulley make if the smaller one revolves at 850 rpm? __________ rpm

A large gear with 100 teeth turning at 210 rpm turns a small gear with 12 teeth. At how many rpm does the small gear turn?

Draw a sketch for the problem and solve it. A rocker arm raises oil from an oil well. On each stroke, it lifts a weight of 3 tons on a weight arm 1 ft long. What force is needed to lift the oil on a force arm 2 ft long?

One pulley is twice as large in diameter as a second pulley. If the larger pulley turns at 259 rpm, what is the rpm of the smaller?

Draw a sketch for the problem and solve it. An object is 4 ft from the fulcrum and balances a second object 8 ft from the fulcrum. The first object weighs 64 lb. How much does the second object weighs?

A large gear with 300 teeth turning at 11 rpm turns a small gear at 150 rpm. How many teeth does the small gear have?

A small pulley is 13 in. in diameter, and a larger one is 19 in. in diameter. How many rpm does the larger pulley make if the smaller one revolves at 570 rpm?

A large gear with 405 teeth running at 300 rpm turns a small gear 450 rpm. How many teeth does the small gear have?

A hydraulic pump is driven with an electric motor(see figure below) . The pump must rotate at 1,000 rpm. The pump is equipped with a 6.0-in.-diameter belt pulley. The motor runs at 1,500 rpm. What diameter pulley is required on the motor?