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655-705 Level|   Geometry|         
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Bunuel
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HOT Competition 26 Aug/8PM: If the length and breadth of rectangle ABC Updated on: 26 Aug 2020, 20:31
Originally posted by Bunuel on 26 Aug 2020, 20:00.
Last edited by bb on 26 Aug 2020, 20:31, edited 3 times in total.
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If the length and breadth of rectangle ABCD are integers, is the perimeter of ABCD greater than 28?

(1) Area of rectangle is less than or equal to 48 cm2
(2) Diagonal of the rectangle is less than 10 cm.



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Red "are" has been added as it seems it was missing at 8:29 PM PST
-BB.

 

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B
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HOT Competition 26 Aug/8PM: If the length and breadth of rectangle ABC 26 Aug 2020, 20:24
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Q 2(l+b)>28
Q (l+b)>14

St.1
l x b<=48
l = 8 and b = 6 but not greater than 14 No
l = 48 and b = 1 Greater than 14 Yes
Insufficient

St 2
d<10
(l^2 + b^2)^1/2<10
Max value can be l = 8 and b = 6, but in these case diagonal will be equal to 10 not less than 10. Hence (l+b)<14
Sufficient
Hence B

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HOT Competition 26 Aug/8PM: If the length and breadth of rectangle ABC 26 Aug 2020, 20:24
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target is
2*(x+y)>28
or is (x+y)>14
where x & y are integer values of lengthe & breadth of rectangle
#1
Area of rectangle is less than or equal to 48 cm2
x*y>=48
x&y can be (48*1) & ( 6*8) ; we get yes & no to our target insufficient
#2
Diagonal of the rectangle is less than 10 cm.
x^2+y^2 <100
max value possible which satisfies this stmnt (9,4)
9^2+4^2 <100 ; 9+4 ; 13
8^2+5^2 <100 ; 8+5;13

hence sufficient for all other values
IMO B is correct

If the length and breadth of rectangle ABCD integers, is the perimeter of ABCD greater than 28?

(1) Area of rectangle is less than or equal to 48 cm2
(2) Diagonal of the rectangle is less than 10 cm.
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HOT Competition 26 Aug/8PM: If the length and breadth of rectangle ABC 26 Aug 2020, 20:29
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Given : If the length and breadth of rectangle ABCD integers,
To Find:is the perimeter of ABCD greater than 28?

(1) Area of rectangle is less than or equal to 48 cm^2
L*B = 48
Pair of L*B can be (24,2),(12,4),(6,8),(3,16)
Perimeter will be 2*(L+B)
=> 2*(24+2) => 52 > 28
=> 2*(12+4) => 32 > 28
=> 2*(6+8) => 28 = 28
=> 2*(3+16) => 38 > 28
(INSUFFICIENT)

(2) Diagonal of the rectangle is less than 10 cm.
10-8-6 is a right angle triplet
if the hypotenuse is less than 10,
the only integer Right triangle triplet less than 10-8-6 is 5-4-3
where 4 and 3 are vertices.

so calculating perimeter
2*(4+3) = 14
which is lower than < 28
(sufficient)

Answer is B
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HOT Competition 26 Aug/8PM: If the length and breadth of rectangle ABC 26 Aug 2020, 21:07
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Answer should be (B).
Area of rectangle = l*b <= 48 cm2
Trying to find if Perimeter = 2(l+b) > 28 => l+b > 14
Using all factors of 48 for area, (1,48), (2, 24), (3, 16).. and so on. all satisfy condition for perimeter > 28 except when L & b pair would be (6,8). That is l+b=14 and not > 14. So, A is out since condition is not enough.
Second scenario states, diagonal < 10. If that is the case, l2 + b2 < (10)^2, which means L & B are less than 8 & 6. In that case, perimeter is definitely < 28. So, this condition is enough.
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HOT Competition 26 Aug/8PM: If the length and breadth of rectangle ABC 26 Aug 2020, 21:26
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If the length and breadth of rectangle ABCD are integers, is the perimeter of ABCD greater than 28?

(1) Area of rectangle is less than or equal to 48 cm2
(2) Diagonal of the rectangle is less than 10 cm.

——
1) we can come up with 2 scenarios that provide a yes and no to answer. 12 and 4 and 6 and 8. Not sufficient.
2) sufficient because we are told integers and must be less than 10. Also — think about special triangles 6,8,10 and perimeter is less. Any diagonal less than 10 would be even less.

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HOT Competition 26 Aug/8PM: If the length and breadth of rectangle ABC 26 Aug 2020, 21:34
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Let length and width of rectangle ABCD be a and b respectively
The question is asking if 2(a+b) > 28 ?

St 1)
Area of rectangle is less than or equal to 48 cm2
ab < 48 or ab = 48

Plugging in a = 6, b = 8
2(6+8) > 28
28 > 28 -- NO

Plugging in a = 1, b = 20
2(1+20) > 28
42 > 28 -- YES

Insufficient

(2) Diagonal of the rectangle is less than 10 cm.

a^ + b^2 < 100

We know that at a = 6, b = 8
a^2 + b^2 = 100

Therefore
(a + b) < 6 + 8
(a + b) < 14
2(a + b) < 28

Sufficient

Correct Answer is B

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HOT Competition 26 Aug/8PM: If the length and breadth of rectangle ABC 26 Aug 2020, 22:15
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(1) Area of rectangle is less than or equal to 48 cm2
l*b<=48
l=1 b=48
perimeter=2(1+48)=98>28 yes
l=6 b=8
perimeter=2(6+8)=28>28 no
not sufficient

(2) Diagonal of the rectangle is less than 10 cm.
square root(l^2+b^2)<10
(l^2+b^2)<100 taking square in both of the sides
highest values it can take,l=9 b=4
perimeter=2(9+4)=26>28 no

sufficient
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