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A rectangular solid P has height 2c and a base of width a and length

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Bunuel
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A rectangular solid P has height 2c and a base of width a and length [#permalink] New post   27 May 2018, 12:28
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A rectangular solid P has height 2c and a base of width a and length b. Two other rectangular solids, Q and R, each have height c and bases of width a and length b. Which of the following represents the amount by which the sum of the surface areas of Q and R exceeds the surface areas of P?

A. 2ab

B. 4ab

C. 2ab + 2bc

D. 2ab + 4ac

E. 2ab +4ac +4bc
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zishu912
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Re: A rectangular solid P has height 2c and a base of width a and length [#permalink] New post   27 May 2018, 12:56
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Bunuel wrote:
A rectangular solid P has height 2c and a base of width a and length b. Two other rectangular solids, Q and R, each have height c and bases of width a and length b. Which of the following represents the amount by which the sum of the surface areas of Q and R exceeds the surface areas of P?

A. 2ab

B. 4ab

C. 2ab + 2bc

D. 2ab + 4ac

E. 2ab +4ac +4bc



Given bigger box
ht. = 2c
width = a
length = b

surface area of the box = 2ab + 2*b*2c + 2*a*2c = 2ab + 4bc + 4ac

Now for the smaller boxes with height = c width = a and length = b

surface area of the each of the box = 2 a* b + 2 b*c + 2 c*a= 2ab + 2bc + 2ca
so their sum = 2 (2ab + 2bc + 2ca) = 4ab + 4bc + 4ca

so the difference between sum of smaller boxes and the larger boxes = 4ab + 4bc + 4ca - 2ab - 4bc - 4ca = 2ab

Hence A is the correct choice
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Re: A rectangular solid P has height 2c and a base of width a and length [#permalink] New post   27 May 2018, 23:54
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Solution



Given:
    • Rectangular solid P has length b, width a, and height 2c
    • Rectangular solid Q has length b, width a, and height c
    • Rectangular solid R has length b, width a, and height c

To find:
    • Amount by which the sum of the surface areas of Q and R exceeds the surface areas of P

Approach and Working:
For any rectangular solid of length l, width b, and height h, the total surface area is 2 (lb + bh + lh)
    • Surface area of solid P = 2 (ab + 2ac + 2bc)
    • Surface area of solid Q = 2 (ab + ac + bc)
    • Surface area of solid R = 2 (ab + ac + bc)


Therefore, surface area of Q + surface area of R – surface area of P = 2 (ab + ac + bc) + 2 (ab + ac + bc) - 2 (ab + 2ac + 2bc) = 2ab

Hence, the correct answer is option A.

Answer: A
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Re: A rectangular solid P has height 2c and a base of width a and length [#permalink] New post   27 May 2018, 12:36
Lets say a=b=c=1. The bigger rectangular has a surface of 10 and the two smaller combined a surface of 12. Difference is 2. Answer choice a providers the 2 we are looking for.
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Re: A rectangular solid P has height 2c and a base of width a and length [#permalink] New post   27 May 2018, 14:31
Surface area of rectangular solid - 2(lb*bh*ha)
Surface area of P = 2(ab+2cb+2ca)
Surface area of Q= 2(ab+bc+ca)
Surface area of R = 2(ab+bc+ca)

Difference in area = Sum of surface Area of R and Q - Surface Area of P
= 4ab+4bc+4ca- 2ab-4bc-4ac
= 2ab Ans
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Re: A rectangular solid P has height 2c and a base of width a and length [#permalink] New post   28 May 2018, 02:45
Surface area of P = 2ab + 4bc + 4ca
Surface area of Q = Surface area of R = 2ab + 2bc + 2ca
Hence, sum of the surface area of Q and R = 4ab + 4bc + 4ca

Therefore, the difference between the sum of the surface areas of Q and R and the surface area of P = 2ab
Hence, answer is A.
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Re: A rectangular solid P has height 2c and a base of width a and length [#permalink] New post   20 Sep 2023, 09:33
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Re: A rectangular solid P has height 2c and a base of width a and length [#permalink]
20 Sep 2023, 09:33
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