A math teacher has 30 cards, each of which is in the shape

Question

A math teacher has 30 cards, each of which is in the shape of a geometric figure. Half of the cards are rectangles, and a third of the cards are rhombuses. If 8 cards are squares, what is the maximum possible number of cards that re circles.

A. 9
B. 10
C. 11
D. 12
E. 13

A. 9
B. 10
C. 11
D. 12
E. 13

arthuro69 · Accepted Answer

Remember:
a square is a special kind of rhombus (sides are perpendicular)
a square is a special kind of rectangles (sides with same length)

Among the 30 cards with have:
15 rectangles
10 rhombus
8 squares

Among the 15 rectangles, there could be 8 special ones (with sides of same length) that are squares. That lets at least 7 rectangles that are not square.
Among the 10 rectangles, there could be 8 special ones (with sides perpendicular) that are squares. That lets at least 2 rhombus that are not square.
We have 8 squares.
So the minimum different cards that represent a square, a rhombus or a rectangle is 2 + 7 + 8 = 17
Which means that the maximum number of circles that you could have is 30 - 17 = 13

Answer (E)

EvaJager · Answer

Squares are both rectangles and rhombuses. Circles are neither rectangles, nor rhombuses. 
There are 15 rectangles and 10 rhombuses.
Using a Venn diagram or a 2 X 2 table, we can deduce that the maximum number of cards that re circles is 30 - (15 + 10 - 8) = 13.

Answer: E

ellexuxin · Answer

can anyone actually get me 13 non-rectangular AND non-rhombus shape cards out of 30 cards with the specifications clearly stated half of the 30 cards (15 cards) are rectangular (including some square cards) and one third of the 30 cards (10 cards) are rhombus (including some square cards)??????

clearly the question stem stated there are 25 cards that is the combination of rectangular shape and rhombus shape.. there are 8 cards within the 25 cards so 25-8=13 cards are non square but still either rectangular or rhombus.. so the non-rectangular AND non-rhombus shape only has 5 spots left in a hand of 30 cards..

i know van-diagram or matrix is what the test or the study of the test want people to go with.. but that is not reasoning and in fact in real life that can never happen.. if anyone disagrees and can make me a hand of 30 cards with exactly what the question stem asked and have 13 circle (non-rectangular AND non-rhombus) cards in real life, let me know!

madn800 · Answer

Ellexuxin GMAT does not tests real world practicality; rather it tests practicality within the constraints given in the problem.

roadrunner · Answer

Key to remember here is that square is also a rhombus. After that its plug-n-play, 15 rectangles , all squares are also rhombus ==> only 2 rhombuses are not square.

Total 17 items so far, thus, max 13 circles are possible.

Ans. E

stne · Answer

Can anyone clarify, what is the meaning of " re circles ",as given in the question or is this a typo?

pandeyashwin · Answer

it's a typo. It should be "are circles"

ScottTargetTestPrep · Answer

We see that we have ½ x 30 = 15 rectangles and ⅓ x 30 = 10 rhombuses. Since there are 8 squares (recall that squares are both rectangles and rhombuses), we have 15 + 10 - 8 = 17 rectangles or rhombuses. If the remaining cards are all circles, we have a maximum number of 30 - 17 = 13 circles.

Answer: E

ummekazi · Answer

i think there is

A 10
B 20
C 30

RastogiSarthak99 · Answer

Bunuel  possible to share official Manhattan social for this?

ThatDudeKnows · Answer

RastogiSarthak99 
I have no affiliation with Manhattan GMAT, but I can't imagine their solution is all that different from this.

See attached.

Answer choice E.

Rainman91 · Answer

What if the question asked the "Minimum number of circles?" Then what would we do?

luisdicampo · Answer

Deconstructing the Question  
Total Cards = 30.
Rectangles (R) =  \frac{1}{2} 	imes 30 = 15 .
Rhombuses (H) =  \frac{1}{3} 	imes 30 = 10 .
Squares (S) = 8.

Step 1: Geometric Relationship  
By definition, a Square is a quadrilateral that is both a Rectangle and a Rhombus.
Therefore, the set of Squares is the  intersection  of the set of Rectangles and the set of Rhombuses.
 S = R \cap H = 8 .

Step 2: Calculate the Union of Polygons  
We want to find how many cards are "taken" by these polygon shapes. We use the set union formula:
 |R \cup H| = |R| + |H| - |R \cap H| 
 |R \cup H| = 15 + 10 - 8 
 |R \cup H| = 17 .

So, 17 cards are either Rectangles, Rhombuses, or Squares.

Step 3: Maximize Circles  
The remaining cards are the ones that are not in the union calculated above.
 	ext{Remaining} = 	ext{Total} - 17 
 	ext{Remaining} = 30 - 17 = 13 .

To maximize the number of circles, we assume all remaining cards are circles (no other shapes like triangles exist).
Max Circles = 13.

Answer:  E

A math teacher has 30 cards, each of which is in the shape

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