A small table has a length of 12 inches and a breadth of b inches. Cub

Question

A small table has a length of 12 inches and a breadth of b inches. Cubes are placed on the surface of the table so as to cover the entire surface. The maximum side of such cubes is found to be 4 inches. Also, a few such tables are arranged to form a square. The minimum length of side possible for such a square is 48 inches. Find b.

A. 8
B. 16
C. 24
D. 32
E. 48

nick1816, Archit3110

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A. 8
B. 16
C. 24
D. 32
E. 48

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nick1816 · Accepted Answer

Given-
HCF(12,b)=4
LCM(12,b)=48

Hence, 12*b=48*4
b=16

Archit3110 · Answer

for the given length of 12 inches the side b inches is found to be 4inches
so now to form a square ie 12x12 ; we need total 3 such tables 
and so as to make table with min length 48 inches we need 48/3 ; 16 such tables
IMO B

axelr · Answer

What does it mean "The maximum side of such cubes is found to be 4 inches"? I guessed it means that every cube has a side of 4 inches. Is it right?
While "The minimum length of side possible for such a square is 48 inches" means that the square obtained with the few tables has a side lenght of 48?

Thanks, I am not english and am afraid I might not have understood the text right, because I find 24 as the answer.

nick1816 · Answer

You've correctly understood it.

If b=24, then use can use cubes of length 12 cm to cover the table; hence, your solution contradicts the statement given.

Archit3110 · Answer

axelr  
See attached image

Posted from my mobile device

nick1816 · Answer

This is wrong...though you get right answer

Table is not a square first of all.

Cubes are placed on the surface of the table so as to cover the entire surface. The maximum side of such cubes is found to be 4 inches .

This statement implies that b is a multiple of 4. b=4k, where k is not multiple of 3.

a few such tables are arranged to form a square. The minimum length of side possible for such a square is 48 inches

This statement implies that LCM of 12 and b is 48.

12*b/4=48
12*4k/HCF(12,4k)=48

k=4
b=16

Archit3110 · Answer

nick1816  ; okay i got your point what I did was that I actually formed square shaped table considering 3 equal '4inches' sides i.e 12 each .. and so as to get min length side of 48 ; 16 such table arrangements would be required...

satya2029 · Answer

A small table has a length of 12 inches and a breadth of b inches. Cubes are placed on the surface of the table so as to cover the entire surface. The maximum side of such cubes is found to be 4 inches. Also, a few such tables are arranged to form a square. The minimum length of side possible for such a square is 48 inches. Find b.

A. 8
B. 16
C. 24
D. 32
E. 48

nick1816, Archit3110

Posted from my mobile device

Attachments

IMG_20191106_093716__01.jpg [ 1.58 MiB | Viewed 7256 times ]

A. 8
B. 16
C. 24
D. 32
E. 48

nick1816 ,  Archit3110

Posted from my mobile device

dc2880 · Answer

Hi, can you please explain where you obtained the equation: 12*4k/HCF(12,4k)=48 ? Thank you!

nick1816 · Answer

dc2880

Product of a and b= LCM(a,b) * HCF(a,b) ........{ true for any 2 positive integers}

12*b = LCM(12,b) * HCF(12*b)

12*b/  = LCM(12,b)

Fdambro294 · Answer

To understand what the question is asking and the underlying concepts at play, you could start with a couple of simple numbers and see what the conditions ultimately mean for the unknown side B.

Suppose we had a Rectangle with a Length of 6 and a Width of 4.

"cubes are placed on the surface of the table so as to COVER THE ENTIRE surface"

A cube, with each side equal, must be placed side-by-side on the rectangle such that the entire surface area of the rectangle is covered.

then we are told the MAXIMUM SIDE that such a cube could have.

in our made up example (L = 6 and W = 4) what length could we make the cubes such that they would cover the entire surface area?

1: we could have 1-by-1 and have 6 across the length and 4 across the width --- total of 24

2: we could have 2-by-2 and have 3 across the length and 2 across the width --- total of 6

3. could we use 3-by-3 cubes?
we would be able to put TWO 3-by-3 cubes across the length. However, when we put one row of two cubes down across the length, there would only be 1 inch remaining across the width. we could NOT cover the entire surface area with just 3-by-3 cubes.

1 and 2 are COMMON FACTOR of the Length and Width. The MAXIMUM side of such cubes would be 2 inches, where 2 inches is the GREATEST Common Factor of the Length (6) and the Width (4)

Therefore, the statement "the maximum side of such cubes is found to be 4 inches" is essentially telling us the following:

the GREATEST COMMON FACTOR of (12 and b) = 4 (i)

(Part 2 of the question stem)

we then take the 12 - by - B rectangles and lay them down side by side, creating a square. The MINIMUM Length possible of a Square that could be created is 48 inches.

For the Square to be 48 - by - 48 we would have to lay FOUR tables (12 inches) along the Length Side in the 1st Row.

how many we would have to place in each of the columns would depend on the value of B. The B inch width of each table MUST be evenly divisible into 48 inches.

and since 48 inches is the MINIMUM Length possible of such a Square, the statement is essentially telling us the following:

LOWEST COMMON MULTIPLE of (12 ; b) = 48 (ii)

at this point we can use the following Property:

for any two positive integers (X and Y) the following equation will always hold:

LCM (X ; Y) * GCF (X ; Y) = X * Y

at this point we just need to insert (i) and (ii) for the LCM and GCF and the side lengths of the rectangle.

LCM(12 ; b) * GCF (12 ; b) = 12 * b
48 * 4 = 12 * b

b = 16

Answer *B*

kk47 · Answer

This question is one of the BEST question to test application of HCF and LCM.  nick1816  method is the best one to solve this question in my opinion.

Another take -

Assuming 'n' number of cubes. The square formed would have the side as 4n. Therefore, Area = 16n^2.

Area is also = 12 x b x n

=> 12bn = 16n^2
=> 12b = 16n
=> b = 4n/3

Also, for min. sq. side = 48, n = 12.

=> b = 4 x 12/3 = 16.

magnfyhr · Answer

Why is E not a valid answer? Four 12x48 tables form a 48x48 square which can be covered in such cubes, and no smaller table can be formed with said table. B and E seem like the only 2 valid answers given this principle, but I think I am not understanding something from the question.

sanjitscorps18 · Answer

Although some might disagree, but I find the wording of the question to be ambiguous. This statement for example

"The maximum side of such cubes is found to be 4 inches"

The only role of this statement is to push the test taker towards thinking this is somehow related to LCM.

However, in doing so it creates another problem, an ambiguity. Does this mean there are cubes with less than 4-inch sides?

A question appears to try hard to test the concept of LCM and HCF by trading a little bit of clarity.

A small table has a length of 12 inches and a breadth of b inches. Cub

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A small table has a length of 12 inches and a breadth of b inches. Cub

Prep Toolkit

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