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Sixteen rectangles are formed by choosing all possible combinations of

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Sixteen rectangles are formed by choosing all possible combinations of  [#permalink]

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New post 11 Sep 2015, 02:09
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Sixteen rectangles are formed by choosing all possible combinations of integer length l and integer width w such that 1 ≤ l ≤ 4 and 1 ≤ w ≤ 4. The probability that a rectangle chosen at random from this set of rectangles has perimeter p and area a such that both 10 ≤ p ≤ 12 and 4 ≤ a ≤ 8 is

(A) 9/256
(B) 49/256
(C) 3/8
(D) 7/16
(E) 7/8

Kudos for a correct solution.

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Sixteen rectangles are formed by choosing all possible combinations of  [#permalink]

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New post 11 Sep 2015, 06:07
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Bunuel wrote:
Bunuel wrote:
Sixteen rectangles are formed by choosing all possible combinations of integer length l and integer width w such that 1 ≤ l ≤ 4 and 1 ≤ w ≤ 4. The probability that a rectangle chosen at random from this set of rectangles has perimeter p and area a such that both 10 ≤ p ≤ 12 and 4 ≤ a ≤ 8 is

(A) 9/256
(B) 49/256
(C) 3/8
(D) 7/16
(E) 7/8

Kudos for a correct solution.


Check other Probability and Geometry questions in our Special Questions Directory.


Probabilty = number of rectangles that have 10 ≤ p ≤ 12 and 4 ≤ a ≤ 8 / total rectangles

About perimeter:
10 ≤ p ≤ 12 => 5 ≤ l+w ≤ 6

l and w are integers. So

4+1=5
1+4=5
3+2=5
2+3=5
4+2=6
2+4=6
3+3=6
5+1=6 is out because l and w are between 1 and 4

About area:

4 ≤ a ≤ 8, areas as 5 and 7 are out because l*w must be integer and can not be 5*1 and 7*1

4*1=4
1*4=4
2*3=6
3*2=6
4*2=8
2*4=8

Then, If we compare perimeters and areas l and w as 3 and 3 is out because 3*3=9. Remained combinations are

4 and 1
1 and 4
2 and 3
3 and 2
4 and 2
2 and 4

Total of 6 combinations.

As given total rectangles are 16. So the probability is 6/16=3/8.

Answer C.
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Re: Sixteen rectangles are formed by choosing all possible combinations of  [#permalink]

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New post 11 Sep 2015, 02:10
Bunuel wrote:
Sixteen rectangles are formed by choosing all possible combinations of integer length l and integer width w such that 1 ≤ l ≤ 4 and 1 ≤ w ≤ 4. The probability that a rectangle chosen at random from this set of rectangles has perimeter p and area a such that both 10 ≤ p ≤ 12 and 4 ≤ a ≤ 8 is

(A) 9/256
(B) 49/256
(C) 3/8
(D) 7/16
(E) 7/8

Kudos for a correct solution.


Check other Probability and Geometry questions in our Special Questions Directory.
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Sixteen rectangles are formed by choosing all possible combinations of  [#permalink]

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New post 11 Sep 2015, 04:42
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L = 1 or 2 or 3 or 4 .
B = 1 or 2 or 3 or 4.

Remember square is also a kind of rectangle.

Total combinations of area (a) i.e L X B are as following.

1X1 , 1X2 , 1X3 , 1X4..............1 , 2 , 3 , 4
2X1 , 2X2 , 2X3 , 2X4..............2 , 4 , 6 , 8
3X1 , 3X2 , 3X3 , 3X4..............3 , 6 , 9 , 12
4X1 , 4X2 , 4X3 , 4X4..............4 , 8 , 12 , 16

So total 16 combinations for area (a) can be made and 7 out of these combinations satisfy our condition i.e 4<= a <=8

So the prob is 7 / 16 ........Eq 1.

Total combinations of perimeter (p) i.e 2(L + B) are as following.

4 , 6 , 8 , 10................................. 2(1+1) , 2(1+2) , 2(1+3) , 2(1+4)
6 , 8 , 10 , 12............................... 2(2+1) , .... so on......................
8 , 10 , 12 , 14...............................2(3+1) , 2(3+2) , ....so on..........
10 , 12 , 14 , 16..............................2 (4+1) , 2 (4+2) , ....so on........

So total 16 combinations of perimeter (p) can be made & 7 out of these combinations satisfy our condition i.e 10<= p <=12

There for prob is 7 / 16 ..................Eq 2.

Multiply Eq 1 & Eq 2.

7 / 16 * 7 / 16 = 49 / 256.

Ans is B.

Hope the answer is right and the approach is correct.
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Re: Sixteen rectangles are formed by choosing all possible combinations of  [#permalink]

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New post 11 Sep 2015, 06:40
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Bunuel wrote:
Sixteen rectangles are formed by choosing all possible combinations of integer length l and integer width w such that 1 ≤ l ≤ 4 and 1 ≤ w ≤ 4. The probability that a rectangle chosen at random from this set of rectangles has perimeter p and area a such that both 10 ≤ p ≤ 12 and 4 ≤ a ≤ 8 is

(A) 9/256
(B) 49/256
(C) 3/8
(D) 7/16
(E) 7/8

Kudos for a correct solution.


Total 16 combinations.

When,
l=1, w=4...........1 combination

When,
l=2, w=3,4.......2 combinations

When,
l=3; w=2....1 combination

When,
l=4; w=1,2...2 combinations

1+2+1+2=6 combinations

6/16=3/8

Answer: C
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Re: Sixteen rectangles are formed by choosing all possible combinations of  [#permalink]

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New post 13 Sep 2015, 08:59
Bunuel wrote:
Sixteen rectangles are formed by choosing all possible combinations of integer length l and integer width w such that 1 ≤ l ≤ 4 and 1 ≤ w ≤ 4. The probability that a rectangle chosen at random from this set of rectangles has perimeter p and area a such that both 10 ≤ p ≤ 12 and 4 ≤ a ≤ 8 is

(A) 9/256
(B) 49/256
(C) 3/8
(D) 7/16
(E) 7/8

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

First, get a sense of the sixteen rectangles. Here are some of the possibilities:
1×1
1×2
1×3
1×4
2×1 (notice that this is a different rectangle from the 1×2)
2×2
etc.

To compute the desired probability, count the rectangles that meet both conditions. Note that some rectangles will meet one condition but not both. If necessary, make separate lists, then cross-reference.

Perimeter between 10 and 12, inclusive:
2×3 and 3×2
1×4 and 4×1
2×4 and 4×2
3×3
Total: 7 rectangles

Area between 4 and 8, inclusive:
2×2
2×3 and 3×2
1×4 and 4×1
2×4 and 4×2
Total: 7 rectangles

There are only 6 triangles on both lists:
2×3 and 3×2
1×4 and 4×1
2×4 and 4×2

Thus, the probability of choosing one of these triangles is 6/16, or 3/8.

One trap in this problem is that you might think that you should calculate the separate probabilities of meeting either condition and then multiply those probabilities together, because there is an and condition at work. However, these probabilities are not independent. The perimeter and the area of the rectangles in question are not independently determined. Instead, what you should do is simply count the rectangles that meet both conditions, then divide by the total number of rectangles.

The correct answer is C.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Sixteen rectangles are formed by choosing all possible combinations of  [#permalink]

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