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 Lower Sec Maths - Help me please!
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By: geniuskids (offline)  Thursday, September 25 2014 @ 10:42 PM CDT (Read 2025 times)  
geniuskids

Hi there, kindly refer to the attachments and help provide me with solutions.

The answers are as follows:

Q9a) 165 cm3

Q9b) 217 cm2

Q10) 90cm

Q11a) 149cm2

Q11b (I) 4460cm3 b(ii) 2340cm2 b (iii) 22.2cm

Q12a(I) 553cm2 a(ii) 737cm3 b) 335g

Q13a) 2827.8cm3 b) 12871 c) 1040cm3

Q14(a) 33.5cm b) 27.5cm2 c) 592cm2 d)440cm3 e)$11.83

Q15(a) 3 3/7cm2 b) 66 2/7cm2 c) $0.33


Thank you so much!
Regards





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By: jo sarah (offline)  Saturday, September 27 2014 @ 01:59 AM CDT  
jo sarah

Hi, would have liked to supply more detailed solutions, but couldn't login to reply yesterday. But my time is very tight now, so hope the following indicators can be helpful for you to proceed.

Qn. 9 is a case of a solid with uniform cross-section (one such familiar solid is the prism).
Generally, in such cases,
Volume = Area of uniform cross-section x height
[now, one calls it height if the solid stands on the uniform cross section as its base; calls it length if the uniform cross section is on either end; or calls it depth if it is at the front and back. So be not constrained by the 'name' of it]
And,
total surface area = perimeter of uniform cross-section x height + (0, 1 or 2) x area of cross-section, depending on whether it's a solid or a hollow with or without cover

(a) Area of uniform cross-section = [ 2 x 7 + .5 (pi) (3.5)^2 - (pi) x 2^2 ] cm^2 = [ 14 + 2.125 (pi) ] cm^2
.....[you can easily compute the numerical values with a calculator, so pardon if I made some computation mistakes]
.....And so the volume of wood needed = above answer x 8 cm^3

(b) Total area of painted surfaces = [ 7 + 2 x 2 + .5 x 2 (pi) x 3.5 ] x 8 cm^2 + 2 x answer in (a)
.....[I assume the inside hole is not counted as "exposed" face that is painted; else add 4 (pi) x 8 cm^2 to above answer.]


Qn. 10
Let the height to be found be h cm
Total area of the entire figure
= area of parallelogram + area of semicircle
= (140 x h) + .5 (pi) x (70^2) cm^2..........................................[corrected typo here; 2 Oct]
= 2.03 m^2 + 2.03 x 100^2 cm^2

[now, solve the algebraic equation for h. and this is a simple linear equation, and you will get the answer.
the key here is to remember to convert the given area in m^2 to cm^2, or convert your dimensions to m before using them.]

Qn. 11
(a) Required area = area of trapezium - area of semicircle
= [ .5 x 12 (7 + 21) - .5 (pi) (3.5^2) ] cm^2....................................[key: know formula for area of trapezium]

(b) (i) Volume = answer found in (a) x 30 cm^3........................[answer in (a) is the cross-sectional area]
(ii) ... [trust this is not a problem now Smile ]

(c) Since solid is melted and recast, new solid has volume equal to this melted down one.
So, answer of (b) (i) = (pi) x 8^2 x height of cylinder
and solve this equation.


Qn. 12
(a) (i) Total surface area of trophy = surface area of hemisphere + surface area of cylinder
= top circular face + .5 curved face of sphere + 2 flat faces of cylinder + curved face of cylinder
= [ (pi) x 4^2 + .5 x 4 (pi) x 4^2 + 2 x (pi) x 4^2 + 2 (pi) x 4 x 12 ] cm^2.......................[to know: surface area of sphere = 4 (pi) r^2]

[Note: many students simply punch each of the above into the calculator and waste lots of precious time in keypunching. It is good if you are able to factorise common factors and save on keystrokes on calculator.
So, the above is simplified as, (16 + 32 + 32 + 96) (pi), or if you are better at mental work, it can be simplified further...]

(ii) Volume of trophy = volume of hemisphere + volume of cylinder
= [ (4/3) (pi) x 4^3 + (pi) x 4^2 x 12 ] cm^3

(b) Since density = mass / volume............................[as you learnt even in Science]
2.5 = mass / (answer in (a) (ii) above)
and solve the equation.


Qn. 13
(a) should not be a problem.

(b) Let the number of cubes be n.
answer of (a) x 10 = n (1.3)^3..............................[13 mm = 1.3 cm, you got to convert to same unit of measure before you can equate]
when you solve this, you are not likely to get an integer answer.
So the maximum no. of cubes that can be formed must be an integer smaller than this answer.

(c) Simply, cut the remaining solid into a tall and a short half cylinder, and you should be able to handle it, right?


Sorry, i have to stop here for now. If i can find time, will return to do the rest.
Hope the above are sufficient to help you solve the questions, hopefully, even the remaining two.

All the best! Smile

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By: geniuskids (offline)  Tuesday, September 30 2014 @ 05:19 PM CDT  
geniuskids

Hi Jo-Sarah,

Thank you! Sorry was busy with my exams. I'll take a look soon.

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