# ATPL theory questions

Join Date: Apr 2009

Location: down south

Age: 75

Posts: 13,226

Likes: 0

Received 0 Likes
on
0 Posts

The answer to your first question is incorrtect - the second correct.

My method answers any questions - trust me - I have been teaching this stuff for twenty years.

BEWARE commercial "real questions".

My method answers any questions - trust me - I have been teaching this stuff for twenty years.

BEWARE commercial "real questions".

Join Date: Jun 2007

Location: Cochin VOCI , India

Age: 34

Posts: 1,605

Likes: 0

Received 0 Likes
on
0 Posts

Thanks

I am not doubting your methods , infact you explained it perfectly and I understand that commercial question banks have quite few errors in them as well

But there some non EASA CAA's around the world that follow the EASA question bank word for word and naturally incorporate these errors in their question banks as well

I am not doubting your methods , infact you explained it perfectly and I understand that commercial question banks have quite few errors in them as well

But there some non EASA CAA's around the world that follow the EASA question bank word for word and naturally incorporate these errors in their question banks as well

Join Date: Jun 2007

Location: Cochin VOCI , India

Age: 34

Posts: 1,605

Likes: 0

Received 0 Likes
on
0 Posts

Help Please

Q. A turn indicator is built around gyroscope with ?

1) 1 Degree of Freedom

2) 3 Degrees of Freedom

3) 2 Degrees of Freedom

4) 0 Degree of Freedom

The correct answer should be 1 Degree of Freedom . This is also given and explained in the Keith Williams questions banks as well

However in two questions banks that I have seen Professional Volare and JAR Databank Presentation the correct answer is given

If you were to see this question for an airline written exam what would you choose ?

Q. A turn indicator is built around gyroscope with ?

1) 1 Degree of Freedom

2) 3 Degrees of Freedom

3) 2 Degrees of Freedom

4) 0 Degree of Freedom

The correct answer should be 1 Degree of Freedom . This is also given and explained in the Keith Williams questions banks as well

However in two questions banks that I have seen Professional Volare and JAR Databank Presentation the correct answer is given

**2 Degrees of Freedom**If you were to see this question for an airline written exam what would you choose ?

*Last edited by cyrilroy21; 27th Mar 2013 at 05:57.*

Join Date: Nov 2000

Location: White Waltham, Prestwick & Calgary

Age: 70

Posts: 4,027

Likes: 0

Received 2 Likes
on
1 Post

First of all, Volare was discredited years ago. Having said that, there are many questions that are just plain wrong, even now. If you feel that they are looking for a wrong answer, or one different from what it should be, then query the question at exam time - only don't mention any databases!

Here is what it should be:

An older method of classifying gyroscopes uses the

numbers of axes

freedom is the ability to move around an axis - for

example, a fuselage can pitch, roll and yaw. A turn

indicator has 1 degree of freedom. An airborne

instrument, with a gyro that has 2 degrees of freedom and

a horizontal spin axis could be a DGI.

Nudge, nudge, wink, wink

Here is what it should be:

An older method of classifying gyroscopes uses the

numbers of axes

*not*including the spin axis, so one with 3*planes*of freedom has 2*degrees*of freedom. A degree offreedom is the ability to move around an axis - for

example, a fuselage can pitch, roll and yaw. A turn

indicator has 1 degree of freedom. An airborne

instrument, with a gyro that has 2 degrees of freedom and

a horizontal spin axis could be a DGI.

Nudge, nudge, wink, wink

Join Date: Jun 2007

Location: Cochin VOCI , India

Age: 34

Posts: 1,605

Likes: 0

Received 0 Likes
on
0 Posts

@paco

Thanks

I re read the question again and I think the question is asking about the gyroscope used rather than the turn indicator itself

Here is an answer I got courtesy of the theflyingengineer.com

And while we are at it , can someone please tell me what the Volare Question bank is about ?

Is it some official JAA question bank ? Does it use CQB 14 or 15 ?

Thanks

I re read the question again and I think the question is asking about the gyroscope used rather than the turn indicator itself

Here is an answer I got courtesy of the theflyingengineer.com

Definition: "A turn indicator is based on the ability of a gyroscope with two degrees of freedom to superpose the angular velocity vector of the gyroscopic rotor’s own spin and the angular velocity vector of the rotation of the device’s case.". The rotation of the gyro about its own axis constitutes one degree of freedom. And the rotation about the longitudinal axis is the second.

Is it some official JAA question bank ? Does it use CQB 14 or 15 ?

*Last edited by cyrilroy21; 27th Mar 2013 at 07:13.*

Join Date: Apr 2009

Location: down south

Age: 75

Posts: 13,226

Likes: 0

Received 0 Likes
on
0 Posts

Is it some official JAA question bank ?

The computer is stand-alone, cannot be accessed from outside, and is multiple password protected. I know - I've worked on it!

edit to say paco is correct - the turn indicator gyro has one degree of freedom.

*Last edited by Lightning Mate; 27th Mar 2013 at 09:09.*

electriciantraining.tpub.com/14187/css/14187_139.htm

A gyro can have different degrees of freedom, depending on the number of gimbals in which it is supported and the way the gimbals are arranged. Do not confuse the term "degrees of freedom" with an angular value such as degrees of a circle. The term, as it applies to gyros, is an indication of the number of axes about which the rotor is free to precess. A gyro mounted in two gimbals has two degrees of freedom.

When two gimbals are used, the gyro is said to be UNIVERSALLY MOUNTED. This arrangement provides two axes about which the gyro can precess. These two axes and the spin axis intersect at the center of gravity of the entire system (excluding the support). Because of this arrangement, the force of gravity does not exert a torque to cause precession. The rotor, inner gimbal, and outer gimbal are balanced about the three principal axes.

Elmer A. Sperry - Degrees of Freedom

A gyro has two basic mountings: one with three degrees of freedom, and one with only two degrees of freedom. A gyro having three degrees of freedom is able to rotate about three axes including its own axis of rotation. Foucault used such a model to illustrate the rotation of the earth, crafting a model with a gyro mounted in two rings. One ring rotated on a horizontal axis within the other, which in turn rotated on a vertical axis.

http://www.navymars.org/national/tra.../14187_ch3.pdf

DEGREES OF FREEDOM

A gyro can have different degrees of freedom, depending on the number of gimbals in which it is supported and the way the gimbals are arranged. Do not confuse the term "degrees of freedom" with an angular value such as degrees of a circle. The term, as it applies to gyros, is an indication of the number of axes about which the rotor is free to precess.

A gyro mounted in two gimbals has two degrees of freedom. When two gimbals are used, the gyro is said to be UNIVERSALLY MOUNTED. This arrangement provides two axes about which the gyro can precess. These two axes and the spin axis intersect at the center of gravity of the entire system (excluding the support). Because of this arrangement, the force of gravity does not exert a torque to cause precession. The rotor, inner gimbal, and outer gimbal are balanced about the three principal axes.

TWO DEGREES-OF-FREEDOM GYROS

The two-degrees-of-freedom (free) gyros can be divided into two groups. In the first group, the gyro's spin axis is perpendicular to the surface of the Earth. Thus the gyro's rotor will spin in a horizontal plane. These gyros are used to establish vertical and horizontal planes to be used where stabilized reference planes are needed.

In the second group, the gyro's spin axis is either parallel to the surface of the Earth or at some angle other than perpendicular. The spin axis of the gyro in the gyrocompass, for example, is maintained in a plane parallel to the surface of the Earth It is aligned in a plane of the north-south meridian. Once set, it will continue to point north as long as no disturbing force causes it to precess out of the plane of the meridian.

en.wikipedia.org/wiki/Gyroscope

Within mechanical systems or devices, a conventional gyroscope is a mechanism comprising a rotor journaled to spin about one axis, the journals of the rotor being mounted in an inner gimbal or ring; the inner gimbal is journaled for oscillation in an outer gimbal for a total of two gimbals.

The outer gimbal or ring, which is the gyroscope frame, is mounted so as to pivot about an axis in its own plane determined by the support. This outer gimbal possesses one degree of rotational freedom and its axis possesses none. The next inner gimbal is mounted in the gyroscope frame (outer gimbal) so as to pivot about an axis in its own plane that is always perpendicular to the pivotal axis of the gyroscope frame (outer gimbal). This inner gimbal has two degrees of rotational freedom.

The axle of the spinning wheel defines the spin axis. The rotor is journaled to spin about an axis, which is always perpendicular to the axis of the inner gimbal. So the rotor possesses three degrees of rotational freedom and its axis possesses two. The wheel responds to a force applied about the input axis by a reaction force about the output axis.

en.wikipedia.org/wiki/Turn_and_balance_indicator

Turn indicator

The turn indicator is a gyroscope instrument that works on the principle of precession. The gyro is mounted in a gimbal. The gyro's rotational axis is in-line with the lateral (pitch) axis of the aircraft, while the gimbal has limited freedom around the longitudinal (roll) axis of the aircraft.

MY COMMENTS

The existence of the spin axis itself does not provide a degree of freedom because the spin axis cannot precess. But each gimbal enables the spin axis to precess in one additional direction. So the number of degrees of freedom is equal to the number of gimbals.

Join Date: Jan 2006

Location: Europe

Posts: 404

Likes: 0

Received 0 Likes
on
0 Posts

**Flight Planning - MJRT (CAP 697)**

Hi all,

would need some help with the following question:

How do I get this figured out and which charts, beside the one mentioned (I used the one on page 26, Section 4, MJRT - 28.000 ft PA) do I have to refer to get the answer? Thanks in advance!

would need some help with the following question:

*(Flight Planning MJRT refer to Fig. 4.5.3.1)*

Given: flight time from TOC to enrout point in FL280 is 48 min. Cruise procedure is long range (LRC).

Temp. ISA -5° C

TOM 56.000 kg

Climb fuel 1.100 kg

Find: distance in nautical air miles (NAM) for this leg and fuel consumption.

Answer would be: 345 NM / 2000 kg

Given: flight time from TOC to enrout point in FL280 is 48 min. Cruise procedure is long range (LRC).

Temp. ISA -5° C

TOM 56.000 kg

Climb fuel 1.100 kg

Find: distance in nautical air miles (NAM) for this leg and fuel consumption.

Answer would be: 345 NM / 2000 kg

*Last edited by Transsonic2000; 29th Mar 2013 at 23:16.*

Subtracting 1100 kg climb fuel from the 56000 kg take-off mass gives a mass of 56000 kg – 1100 kg = 54900 kg at the start of the cruise.

In Figure 4.5.3.1 the second cells in the 55000 kg row indicates that the TAS = 437 kts.

Note 2D at the base of the page states “decrease TAS by 1 kt for each 1 deg C below ISA.

So for the existing ISA deviation of -5 deg C the required correction is -5 kts.

So the corrected TAS = 437 – 5 = 432 kts.

The question states that the leg time is 48 min.

In 48 minutes at 432 kts TAS the distance flown = 432 x 48 / 60 = 345.6 NAM.

In Figure 4.5.3.1 the cell at the intersection of the 54000 kg row and the 900 kg column (for the mass of 54900 kg) indicates an initial cruise range of 3736 NAM.

Subtracting the 345.6 NAM flown in this leg gives 3736 – 345.6 = 3390.4 NAM remaining.

The closest figure to this in the table is 3395, which appears at the intersection of the 52000 kg row and the 900 kg column. This indicates a final mass of 52900 kg.

Subtracting this final mass from the initial mass gives a mass change of (54900 – 52900) = 2000 kg caused by the burning of fuel.

Note 2B at the base of the page states “decrease fuel by 0.6% for each 10 deg C below ISA.

So for the existing -5 deg C ISA deviation, the fuel must be decreased by 0.6% x 5/10 = 0.3%.

0.3% of 2000 kg = 6 kg. So the corrected fuel = 2000 kg – 6 kg = 1994 kg.

So the fuel consumed in the cruise is 1994 kg.

So for the cruise leg of this flight the distance flown is approximately 345 Nm and the fuel burned is approximately 1994 kg.

In Figure 4.5.3.1 the second cells in the 55000 kg row indicates that the TAS = 437 kts.

Note 2D at the base of the page states “decrease TAS by 1 kt for each 1 deg C below ISA.

So for the existing ISA deviation of -5 deg C the required correction is -5 kts.

So the corrected TAS = 437 – 5 = 432 kts.

The question states that the leg time is 48 min.

In 48 minutes at 432 kts TAS the distance flown = 432 x 48 / 60 = 345.6 NAM.

In Figure 4.5.3.1 the cell at the intersection of the 54000 kg row and the 900 kg column (for the mass of 54900 kg) indicates an initial cruise range of 3736 NAM.

Subtracting the 345.6 NAM flown in this leg gives 3736 – 345.6 = 3390.4 NAM remaining.

The closest figure to this in the table is 3395, which appears at the intersection of the 52000 kg row and the 900 kg column. This indicates a final mass of 52900 kg.

Subtracting this final mass from the initial mass gives a mass change of (54900 – 52900) = 2000 kg caused by the burning of fuel.

Note 2B at the base of the page states “decrease fuel by 0.6% for each 10 deg C below ISA.

So for the existing -5 deg C ISA deviation, the fuel must be decreased by 0.6% x 5/10 = 0.3%.

0.3% of 2000 kg = 6 kg. So the corrected fuel = 2000 kg – 6 kg = 1994 kg.

So the fuel consumed in the cruise is 1994 kg.

So for the cruise leg of this flight the distance flown is approximately 345 Nm and the fuel burned is approximately 1994 kg.

*Last edited by keith williams; 30th Mar 2013 at 12:40.*

Join Date: Jun 2007

Location: Cochin VOCI , India

Age: 34

Posts: 1,605

Likes: 0

Received 0 Likes
on
0 Posts

Hey guys help with this question please

The approximate range of a 10 KW NDB over the sea is

a) 500nm

b) 1000nm

c) 50nm

Option d 100nm is supposed to be the correct answer

I used the formula 3 (over the sea ) X Sqrt ( 10,000 Watts ) = 300nm

Anybody know how 100nm is the supposed to be the correct answer ?

The approximate range of a 10 KW NDB over the sea is

a) 500nm

b) 1000nm

c) 50nm

__d) 100nm__Option d 100nm is supposed to be the correct answer

I used the formula 3 (over the sea ) X Sqrt ( 10,000 Watts ) = 300nm

Anybody know how 100nm is the supposed to be the correct answer ?

Join Date: Apr 2009

Location: down south

Age: 75

Posts: 13,226

Likes: 0

Received 0 Likes
on
0 Posts

The range of an NDB depends to a large extent on the frequency used.

The ICAO band allocated is 190 kHz to 1750 kHz.

Without the frequency stated the question is impossible to answer and is yet another "invented" question I suspect.

The original CAA/JAA question was not the one as stated.

The ICAO band allocated is 190 kHz to 1750 kHz.

Without the frequency stated the question is impossible to answer and is yet another "invented" question I suspect.

The original CAA/JAA question was not the one as stated.

Join Date: Dec 2008

Location: UK

Posts: 21

Likes: 0

Received 0 Likes
on
0 Posts

Doing some Aircraft Performance revision I've come across a question in my school's 'exam preparation' questions (so not necesserily previous exam questions, just revision tools):

"What is the effect on V1 if mass is increased?"

The answer given is that it increases. Indeed, looking at the MRJT in CAP698 shows an increase in V1 with mass. This is a general theory question, however.

I don't understand how this can be a hard and fast rule - Will there not be situations where an increase in mass will slow the acceleration and deceleration such that ASDA will be insufficient should V1 be increased or even left unchanged after adding mass? I'm thinking, in this case, a lower V1 would be necessary in order to guarantee a stop within ASDA should it be required. Obviously, this requires better OEI performance, as we will have to accelerate further on only one engine, but if we know we can still reach screen height at V2, what's the issue?

I'm just thinking about taking off from a short runway, with a clearway, with a FLLTOM dictated by ASDA, and we increase the a/c mass up to the FLLTOM. Would V1 really go up?

Probably a question for my instructor, but as its a holiday weekend and this was on my mind, I thought I might see if some other ppruner was feeling simlilarly pensive...

"What is the effect on V1 if mass is increased?"

The answer given is that it increases. Indeed, looking at the MRJT in CAP698 shows an increase in V1 with mass. This is a general theory question, however.

I don't understand how this can be a hard and fast rule - Will there not be situations where an increase in mass will slow the acceleration and deceleration such that ASDA will be insufficient should V1 be increased or even left unchanged after adding mass? I'm thinking, in this case, a lower V1 would be necessary in order to guarantee a stop within ASDA should it be required. Obviously, this requires better OEI performance, as we will have to accelerate further on only one engine, but if we know we can still reach screen height at V2, what's the issue?

I'm just thinking about taking off from a short runway, with a clearway, with a FLLTOM dictated by ASDA, and we increase the a/c mass up to the FLLTOM. Would V1 really go up?

Probably a question for my instructor, but as its a holiday weekend and this was on my mind, I thought I might see if some other ppruner was feeling simlilarly pensive...

Join Date: Dec 2008

Location: UK

Posts: 21

Likes: 0

Received 0 Likes
on
0 Posts

So hopefully getting towards answering my own question....

In the case above, before adding mass, we have excess performance in terms of reaching screen height OEI, right? So would we have set V1 at the minimum speed that we can be assured to reach screen height/V2 after engine failure, rather than the maximum speed we can be assured of stopping within ASDA?

If the former is the case, I can see how an increase in mass would require an increase in V1. So there will be some speed range beyond V1 where in theory the aircraft will come to a stop within the (factored) ASDA, but we ignore this, because as long as we hit screen height on speed, we're happy? I suppose this fits in with a 'go-minded' philosophy..

Am I on the right lines?

In the case above, before adding mass, we have excess performance in terms of reaching screen height OEI, right? So would we have set V1 at the minimum speed that we can be assured to reach screen height/V2 after engine failure, rather than the maximum speed we can be assured of stopping within ASDA?

If the former is the case, I can see how an increase in mass would require an increase in V1. So there will be some speed range beyond V1 where in theory the aircraft will come to a stop within the (factored) ASDA, but we ignore this, because as long as we hit screen height on speed, we're happy? I suppose this fits in with a 'go-minded' philosophy..

Am I on the right lines?

You are correct in saying that if the aircraft is ASDA limited you cannot increase V1 to accommodate an increase in mass. This is because the increase in mass and the increase in V1 would both increase the ASD.

But if the aircraft is limited by its ability to get to V2 and screen height within the TODA, then increasing V1 will reduce the distance to reach screen height.

In using the CAP 698 you need to take care to read the small print. Para 2.1 in Section 4 MRJT1 page 7, states that "The field length used in the graph is based on minimum V1 being equal to VMCG". This is true throughout the CAP698, so the speeds in tables 4.8 and 4.9 are also based on minimum V1.

As a general rule for the exams, unless the question contains some additional restrictions such as "the aircraft is climb limited", then the effects listed in the CAP are the ones that they are looking for in the questions.

The thread below contains material on this subject.

http://www.pprune.org/professional-p...=increasing+V1

But if the aircraft is limited by its ability to get to V2 and screen height within the TODA, then increasing V1 will reduce the distance to reach screen height.

In using the CAP 698 you need to take care to read the small print. Para 2.1 in Section 4 MRJT1 page 7, states that "The field length used in the graph is based on minimum V1 being equal to VMCG". This is true throughout the CAP698, so the speeds in tables 4.8 and 4.9 are also based on minimum V1.

As a general rule for the exams, unless the question contains some additional restrictions such as "the aircraft is climb limited", then the effects listed in the CAP are the ones that they are looking for in the questions.

The thread below contains material on this subject.

http://www.pprune.org/professional-p...=increasing+V1

*Last edited by keith williams; 21st Apr 2013 at 18:22.*

Join Date: Dec 2008

Location: UK

Posts: 21

Likes: 0

Received 0 Likes
on
0 Posts

Thanks Keith. I shall have a read of the link.

Would your assessment be a fair representation of real life? That is to say the aim will tend to be to use as low a V1 as possible whilst guaranteeing screen height and directional control, despite the fact that ASDA may permit an abort at higher speed? (Before we start worrying about/considering reduced/debated thrust for the sake of engine life).

Thanks again.

Would your assessment be a fair representation of real life? That is to say the aim will tend to be to use as low a V1 as possible whilst guaranteeing screen height and directional control, despite the fact that ASDA may permit an abort at higher speed? (Before we start worrying about/considering reduced/debated thrust for the sake of engine life).

Thanks again.

Join Date: Jan 2012

Location: cardiff uk

Posts: 6

Likes: 0

Received 0 Likes
on
0 Posts

**hekp with this question**

hi this question cropped up in last mass & balance exam Given a crate at 1 ft 9 inches X 1 ft 7 inches X 1 ft 2 inches, what's the maximum load in it if RL is ____and load intensity is _____, what the method to tackle this question thanks sorry no figures i copied it of atp forum the answer was 257kg

Example 1.

An aircraft has a floor load limit of 4500 N/m squared. What is the maximum mass that can be loaded in this hold on a 1.4m x 0.4m pallet? (g = 9.81 m/s/s).

(a) 464 Kg.

(b) 122 Kg.

(c) 177 Kg.

(d) 256 Kg.

The floor loading limit is 4500 n/m squared and the floor area of the pallet is 1.4m x 0.4m = 0.56m squared.

Floor loading = load / area.

So load required to exert a given floor loading = floor loading x area.

This means that the pallet load required to exert the limiting floor loading of 4500 N/m squared, is equal to 4500 N/m squared x 0.56 m squared = 2520 N.

The force exerted by a mass = F = mass x g

Rearranging this equation gives mass = F/g

Using g = 9.81, the mass required on the pallet must be 2520 N/9.81 = 256.9 Kg.

Example 2.

What is the maximum mass that can be loaded onto a 0.8m square 25 Kg pallet, if the running load limit is 800 Kg/m?

(a) 644 Kg.

(b) 615 Kg.

(c) 475 Kg.

(d) 519 Kg.

The running load exerted by a load is equal to the total mass of the load divided by the length of its base.

In order to obtain the maximum allowable mass, it is necessary to orient the pallet such that its longest possible base length is used.

If the mass loaded on the pallet = M, then the total mass is the load (M Kg) plus the pallet mass (25 Kg) which gives a total mass of (M + 25) Kg.

Both of the edges of the pallet are 0.8 m long.

So the running load exerted by the loaded pallet is:

Running Load = Total mass/ base length = (M + 25) Kg / 0.8m

But the maximum allowable running load in this question is 800 Kg/m, so the maximum mass that can be loaded on the pallet is that which gives the condition:

(M + 25) Kg / 0.8 m = 800Kg / m.

This equation can be simplified by multiplying both sides by 0.8 m, to give:

(M + 25) Kg = (0.8 x 800) Kg Which is 640 Kg

Subtracting 25 Kg from each side gives M = 615 Kg

So the maximum mass that can be loaded on the pallet is 615 Kg.

General factors to remember:

1. The static load intensity for any given load will be a minimum when the load is sitting on its largest side. So to maximize the permitted load, it must be placed on its largest side.

2. The running load imposed by any given load will be minimum when the load is moving along its longest side. So to maximize the permitted load, it must be moved along its longest side.

Join Date: Jan 2012

Location: cardiff uk

Posts: 6

Likes: 0

Received 0 Likes
on
0 Posts

hi thanks for the reply this is a valid question any help will be appreciated.

what is the maxium load that can be placed in a baggage compartment with a running load of 75 kg/inch and a load intensity of 300 kg/sq ft if a package dimensions of 18x24x15 inches.

A.350kg

B.900kg

C.1200kg

D.1800kg

what is the maxium load that can be placed in a baggage compartment with a running load of 75 kg/inch and a load intensity of 300 kg/sq ft if a package dimensions of 18x24x15 inches.

A.350kg

B.900kg

C.1200kg

D.1800kg

You will learning nothing from simply having me or any other member giving you the answer. You must learn to do the calculations for yourself.

The examples in my previous post illustrate how to answer this type of question.

You need to do one calculation for the static load and another for the running load.

For the maximum load that does not exceed the static load limit use:

Max load = Static load limit x floor area of load

For the maximum load that does not exceed the running load limit use:

Max load = Running load limit x longest side of of the load.

Note that the running load limit is in kg/inch and the static load limit is in kg/sq ft. So for the static load limit you need to convert inches into feet (there are 12 inches in a foot).

The maximum load is the smaller of the two figures that you will derive from these calculations.

If you then post your step-by-step solutions I or someone else will comment on them.

The examples in my previous post illustrate how to answer this type of question.

You need to do one calculation for the static load and another for the running load.

For the maximum load that does not exceed the static load limit use:

Max load = Static load limit x floor area of load

For the maximum load that does not exceed the running load limit use:

Max load = Running load limit x longest side of of the load.

Note that the running load limit is in kg/inch and the static load limit is in kg/sq ft. So for the static load limit you need to convert inches into feet (there are 12 inches in a foot).

The maximum load is the smaller of the two figures that you will derive from these calculations.

If you then post your step-by-step solutions I or someone else will comment on them.

*Last edited by keith williams; 27th Apr 2013 at 22:33. Reason: To correct typing errors and to add a little bit more information.*