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 / Nan Hua / 2021 / SA2 /
Paper 2 – Q14 [4 marks]](https://i0.wp.com/sgprimaths.com/wp-content/uploads/2022/08/P6_SA2_2021_Nan_Hua_Paper2_Q14-Qns.jpg?resize=400%2C260&ssl=1)
![P6 / Nanyang / 2019 / CA1 / Paper 2 – Q12 [4 marks]](https://i0.wp.com/sgprimaths.com/wp-content/uploads/2022/04/P6_CA1_2019_Nanyang_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.
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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!
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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!