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- bob bundy
- 2013-03-18 19:31:34
hi Bourne88
That's great news! 
Thanks for the update.
Bob
- Bourne88
- 2013-03-18 09:18:20
Hi Bob,
Just wanted to let you know that I passed my Royal Marines Recruit test with such a high score that I can apply to join as an officer.
Thank you for all your help,
Bourne
- Bourne88
- 2013-01-30 07:09:11
Hi Bob,
Thanks very much for your help. I might have a few other questions for you. I'm glad you enjoyed helping me out.
Will post them here when they most likely pop up.
Cheers!
- bob bundy
- 2013-01-30 05:24:21
hi Bourne88
That's it! All answers correct!
What you have done for Q20 is what I intended. I'll put in some more steps
If you make 40 components it will take 25metres of steel. Cost is $200.
But you can get some money back for the wastage as scrap.
56% is waste and you only get half the value for the scrap.
So value of scrap = 200 x 0.5
How much ? 56% so do
56% of 200 x 0.5
So 40 components has cost less than $200 by this amount
new cost = 200 -(200x56%x0.5)
I should have put the bracket there last post shouldn't I?
But that's for 40, so divide by 40 to get the true cost of one.
Bob
ps. I've enjoyed helping so post again if you have another problem.
- Bourne88
- 2013-01-30 01:19:33
Hi Bob,
Thanks for the help. I've so far got the following;
18. Percentage wastage of steel = 70 x 62.5
= 4375 cm squared.
Then minus the area of the remaining shape = 4375 - 1926
= 2449cm squared
Then 2449 / 4375 x 100%
= 55.9
= 56%
19. Length of coil = 25m or 2500cm
= 2500 x 70
= 175000 cm squared
Number of Units = Area of coil/ Area to make one component
= 175000 / 4375
= 40 Components.
I'm having some trouble with what you said for question 20, If $200 is required for a 25m Length where is the 200 - 200 coming from and the 0.5(Is that the 50% of the cost part)? The answer I'm currently getting is not what it should be.
Cost of individual component = 200 - 200 x 56/100 x 0.5 (Divided by Ans to Q19)
= 200 - 11200/100 x 0.5
= 200 - 112 x 0.5
= 200 - 56
= 144 / 40
= $3.6
Again thanks very much with your help on this.
- bob bundy
- 2013-01-29 10:40:43
hi Bourne88
Q17. 1926 is the answer I got too.
I never thought about drawing it like that with the information given.
I think making that diagram may well be a part of the assessment. They've given you a part of the picture to get you started but they want to see if you can convert the rest of the information into a clear diagram. Even if that's not a requirement it's certainly the way I'd go with it because I find maths problems are easier for me once I have a picture of what I'm doing. I'd recommend it every time.
So for Q18.
(i) Calculate one unit of the strip ie. the yellow part.
(ii) Subtract your answer for Q17, so you get the wastage.
(iii) Make this a fraction and times by 100%
Q19. How many yellow units will fit in 25 metres ? (ie. 2500 cm)
Q20. I worked out for 25 metres worth and then divided by the answer to Q19.
Then divide by answer to Q19.
Give those a try and post back answers or ask for more help.
Bob
- Bourne88
- 2013-01-29 07:29:25
Hi Bob,
Thanks very much for your help.
For question 17 I used:
Area of a circle = Pi x r squared
= 3.14 x (30 x 30)
= 3.14 x 900
= 2826cm squared.
Then minus the area of the square
= L x B
=30 x 30
= 900cm squared
So 2826 - 900 = 1926cm squared?
The diagram makes sense to me, I never thought about drawing it like that with the information given.
- bob bundy
- 2013-01-28 23:55:37
OK. Now I have a diagram. I have NOT attempted to make it accurately to scale ... just tried to make it look like the problem and put in red all the measurements. For the purposes of calculating wastage I've shaded in yellow one unit of steel from which a stamping can be made.
You say you've done Q17. If you post your answer, I can check it and make sure you are ok with that before we go on to the harder bits.
Study the diagram (click to enlarge) and make sure you agree with it as a correct representation of the problem. Any questions, just ask.
Bob
- bob bundy
- 2013-01-28 23:38:43
hi Bourne88
Welcome to the forum.
Looks like a diagram for a length of steel is needed for Q18 onwards so I'll do that now and post again.
Bob
- Bourne88
- 2013-01-28 19:42:40
Hi there,
I am really struggling with the following questions regarding working out the area and percentages with the following questions 17 - 20. The site has been extremely useful for practicing for my recruit test for the Royal Marines but I constantly get stuck on these type of questions and cant work out how to complete them. Because I'm a new member I cannot post a direct link to the paper.
But if you google - 'Psychometric success mechanical comprehension Practice test 1' a link to the paper will come up.
I get the right answer for question 17 but have no idea where to start for 18-20.
If anyone could aid me with working out or point me in the right direction I would be extremely grateful.
http://www.psychometric-success.com/pra … st%201.pdf
Link added for you.
Thank you