Sixteen rectangles are formed by choosing all possible combinations of

Question

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.

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

Kudos for a correct  solution .

Bambaruush · Accepted Answer

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.

Bunuel · Answer

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

goldfinchmonster · Answer

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.

AmoyV · Answer

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

Bunuel · Answer

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.

exc4libur · Answer

total combinations: 16 rectangles = 4 choices length • 4 choices of width = 16 combinations
4 ≤ area ≤ 8 and 10 ≤ perimeter ≤ 12 are made of integers 1 ≤ length • width ≤ 4
if area = 4, then lw={4,1}{2,2}{1,4}, but lw={2,2} gives perimeter=2(2+2)=8 outside range, lw={4,1}{1,4}=2 options
if area = 6, then lw={2,3}{3,2}{6,1}{1,6}, but {6} outside range, lw={2,3}{3,2}=2 options
if area = 8, then lw={4,2}{2,4}{8,1}{1,8}, but {8} outside range, lw={4,2}{2,4}=2 options
if area = 5, then lw={5,1}{1,5}, but {5} is outside range, 0 options (notice this happens for all "prime" areas)
if area = 7, then 0 options;
favorable combinations: {2,3}{3,2}{4,2}{2,4}{4,1}{1,4}=2+2+2=6 
probability(fav/total): 6/16=3/8

Answer (C)

TestPrepUnlimited · Answer

Notice we only have 16 rectangles to choose from so we can already eliminate A and B. Next, L = 4 and W = 1 is the same as W = 4 and L = 1, so we can save time by listing effective rectangles only with W ≤ L and double-counting any results with W != L. Finally, we just need to find the dimensions that satisfy 10 ≤ p ≤ 12 and 4 ≤ a ≤ 8. Instead of using 10 ≤ p ≤ 12, it's easier to use 5 ≤ L + W ≤ 6. With this problem laid out properly we only need to consider:
L = 4, W = 2. A = 8, 2 rectangles satisfy.
L = 4, W = 1. A = 4, 2 rectangles satisfy.
L = 3, W = 3. A = 9. does not satisfy
L = 3, W = 2. A = 6, 2 rectangles satisfy.

Out of the 16 rectangles, 6 satisfy. Therefore the answer is 3/8.

TestPrepUnlimited · Answer

First, notice we have only 16 rectangles to choose from so we can already eliminate A and B as there are only 16 possibilities. 
Next, for this problem L = 4 and W = 1 is the same as W = 4 and L = 1, so we can save time by listing effective rectangles only with W ≤ L and double-counting any results with W ≠ L. 
Finally, we just need to find the dimensions that satisfy 10 ≤ p ≤ 12 and 4 ≤ a ≤ 8. Instead of using 10 ≤ p ≤ 12, it's easier to use 5 ≤ L + W ≤ 6. 
With this problem laid out properly we only need to consider:
L = 4, W = 2. A = 8, 2 rectangles satisfy.
L = 4, W = 1. A = 4, 2 rectangles satisfy.
L = 3, W = 3. A = 9.
L = 3, W = 2. A = 6, 2 rectangles satisfy.

Out of the 16 rectangles, 6 satisfy. Therefore the answer is 3/8.

MathRevolution · Answer

1 ≤ l ≤ 4 => possible values of 'l' = {1, 2, 3, 4}

1 ≤ w ≤ 4 => possible values of 'w' = {1, 2, 3, 4}

10 ≤ p ≤ 12 => possible values of 'p' = {10, 11, 12}

4 ≤ w ≤ 8 => possible values of 'a' = {4, 5, 6, 7, 8}

The Perimeter of rectangle: 2(l + w)

=> 2(l + w) = 10 OR 2(l + w) = 11 OR 2(l + w) = 12

=> l + w = 5 OR 5.5 OR 6

=> Possible combinations to get l + w = 5 => (l,w) = (1,4), (2,3), (3,2) and (4,1)

=> Possible combinations to get l + w = 6 => (l,w) = (2,4), (4,2), (3,3)

Area of rectangle: l * w

=> l * w = 4 OR l * w = 5 OR l * w = 6 OR l * w = 7 OR l * w = 8

=> l * w = 5 OR 7

=> Possible combinations to get l * w = 4 => (l,w) = (1,4), (2,2), and (4,1)

=> Possible combinations to get l * w = 6 => (l,w) = (2,3), (3,2)

=> Possible combinations to get l * w = 8 => (l,w) = (2,4), (4,2)

Desired outcome for both 'p' and 'a' are: (1,4), (2,3), (3,2), (4,1),(2,4), (4,2) = 6

Total outcomes: 16

Probability:  \frac{Desired }{ Total}

=>  \frac{6 }{ 16 }

=>  \frac{3}{8}

Answer C

lse19 · Answer

I took a slightly diff approach, experts pls verify

Instead of looking at rectangles that satisfy, I took those that do not so

the cases that do not satisfy would be
first i fix breadth at 1
1 1
1 2
1 3
fix breadth at 2
2 1
fix breadth at 3
3 1

this means 5 possibilities that do not satisfy when taking breadth first and moving length, if we reverse this we get same 5 from other side keeping length fixed. therefore total 10 possibilities that do not satisfy.

now, since total=16
do not satisfy= 10
therefore satisfy=16-10=6

therefore probability = 6/16 = 3/8
 Bunuel  please correct if wrong

EgmatQuantExpert · Answer

Given

• 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.

To Find

• The probability of choosing a rectangle has perimeter p and area a such that both 10 ≤ p ≤ 12 and 4 ≤ a ≤ 8.

Approach and Working Out

• As 16 rectangles are formed, we can say that the order of l and w matters.

• 10 ≤ p ≤ 12 
 o p = 2(l + w)
o l + w = 5, 6
o (l, w) = (1, 4); (2,3); (2,4); (3,3); (4, 1); (3, 2); (4,2) 
• 4 ≤ a ≤ 8.
 o a = lw
o 4 ≤ lw ≤ 8 
o Out of the possible sets to only (1, 4); (2,3); (2,4); (4, 1); (3, 2); (4,2)
  (3,3) will have an area of 9

• Answer =  \frac{6}{16}  =  \frac{3}{8}

Correct Answer: Option C

bumpbot · Answer

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