Question:
Stan glued a large cube A, a smaller cube B and 3 1-cm cubes together to form a solid as shown in Figure 1. He painted the solid including its base.
He then glued a rectangular block C to the solid in Figure 1 to form another solid as shown in Figure 2. Cube A and block C have the same base area.
(a) What fraction of the length of an edge of cube A was the length of an edge of cube B?
(b) Find the painted area of the solid in Figure 1.
(c) Stan glued some more 1-cm cubes to the solid in Figure 2 to form a larger rectangular solid. What was the least number of 1-cm cubes used?
Solution:
![P6 / Methodist Girls / 2019 / SA2 / Paper 2 – Q08 [3 marks]](https://i0.wp.com/sgprimaths.com/wp-content/uploads/2020/07/P6_MGS_2019_Prelims_Paper2_Q08-Qns.jpg?resize=400%2C260&ssl=1)
![P6 / Catholic High / 2019 / SA1 / Paper 2 – Q12 [4 marks]](https://i0.wp.com/sgprimaths.com/wp-content/uploads/2020/03/P6_Catholic_2019_SA1_Paper2_Q12-Qns.jpg?resize=400%2C260&ssl=1)
Please explain (b). Painted Area of A = 4x4x5+1×2. Where is this “1×2” comes from? And Painted Area of C = 1x1x10? Where does “10” comes from?
Hi! Let me try to explain as concretely as I can.
(b) Painted area of A = Area of 4 vertical sides + Bottom + Top
Area of 4 Vertical sides + Bottom = 4×4 x 5
Area of Top refers to the area not blocked by Cube B and the column of 3 small cubes = 1 x 2
Therefore, Painted area of A = 4x4x5 + 1×2
(c) Painted area of C.
I think I made a mistake here. Instead of “C”, it should be “Painted area of 3 1-cm cubes”
In Figure 1, the 3 1-cm cubes are stacked on top of each other.
I shall refer to them as Top, Middle, Bottom.
The Top has 4 sides that could be painted.
The Middle and Bottom have 3 sides each that could be painted.
No. of sides to be painted = 4 + 3 + 3 = 10 sides.
Therefore, Painted area of 3 1-cm cubes = 1x1x10.
Does this explanation helps?
Thanks for replying. I have figured out the 3 small 1 cm cube. But for area A, i would think the Top A is being blocked by 3 by 3 and 1 by 1, so the remaining that is unblock should be 6. That means the total area of top A is 4 by 4, then less top B 3 by 3 then less top 1 by 1 small cube => (4×4)-(3×3)-(1×1) = 6. So total of area of painted A should be 86 instead of 82.
Oh ya, you’re right! I missed out some area on the Top. Sorry about the mistake there!
Thanks for pointing it out! I would correct the solution as soon as I could and upload a corrected one.
Hope you like this website so far. =)
If you find other mistakes, please let me know in the comments!
Hi no problem. This website is helpful! Thanks for the effort in posting the solutions. Once again, thanks!
Glad it’s helpful!
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I still dont understand part (c)
Hi! Here’s a more detailed explanation for part (c).
Figure 2 is currently made up of:
– Cube A
– Cube B
– Block C
– 3 1-cm cubes
The question said that Stan want to add more 1-cm cubes to Figure 2 to form a larger rectangular solid.
Vol of 1-cm cubes needed = Vol of Larger Rectangular Solid – Vol of current cubes & block
Volume of Larger Rectangular Solid
This larger rectangular solid will have the following dimensions:
– Length : 4 + 4 = 8 cm [length of A + length of C]
– Breadth : 4 cm [breadth of C]
– Height : 4 + 3 = 7 cm [height of A + height of B]
Vol of Larger Rect Solid = 8 x 4 x 7 = 224cm^3
Volume of current cubes & block
Vol of A = 4 x 4 x 4 = 64cm^3
Vol of B = 3 x 3 x 3 = 27cm^3
Vol of C = 4 x 4 x 6 = 96cm^3
Vol of 3 1-cm cubes = 3cm^3
Total vol of current cubes & block = 64+27+96+3 = 190cm^3
Vol of 1-cm cubes needed = 224 – 190 = 34cm^3
No. of 1-cm cubes needed = 34cm^3 / 1cm^3 = 34 cubes
Hope this clarifies!