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805+ Level|   Geometry|      
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Bunuel
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A rectangle with positive integer side lengths in cm has area A cm^2 a 25 Apr 2019, 02:27
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95% (hard)

Question Stats:

25% (02:25) correct 75% (02:44) wrong based on 75 sessions

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A rectangle with positive integer side lengths in cm has area A cm^2 and perimeter P cm. Which of the following numbers cannot equal A + P ?

(A) 100
(B) 102
(C) 104
(D) 106
(E) 108
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B
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A rectangle with positive integer side lengths in cm has area A cm^2 a 29 Apr 2019, 04:35
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To test numbers, we need to determine approx values.

Max A+P is for a square as square has max A for a given P.

Testing on squares:
For side 8, A+P= 64+32= 96
For side 9, A+P= 81+36= 117

All choices are within this range.

Let's test rectangles with sides:
9,8= 72+34= 106
10,7= 70+34= 104
11,6= 66+34= 100
Note that 102 has missed out from this sequence.
12,6= 72+36= 108

Ans B
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A rectangle with positive integer side lengths in cm has area A cm^2 a 06 Oct 2019, 21:27
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Bunuel Hello Sir,
Can you please provide OE or your explanation for this one? :)
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A rectangle with positive integer side lengths in cm has area A cm^2 a 09 Oct 2019, 20:15
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Bunuel
A rectangle with positive integer side lengths in cm has area A cm^2 and perimeter P cm. Which of the following numbers cannot equal A + P ?

(A) 100
(B) 102
(C) 104
(D) 106
(E) 108

Let L and W be the length and width of the rectangle, respectively. Since A = LW and P = 2L + 2W, we have:

A + P = LW + 2L + 2W = L(W + 2) + 2W

Let’s add and subtract 4 to the last expression:

A + P = L(W + 2) + 2W + 4 - 4 = L(W + 2) + 2(W + 2) - 4 = (W + 2)(L + 2) - 4

Using the last equality, we can test each answer choice:

A) A + P = 100

100 = (W + 2)(L + 2) - 4

(W + 2)(L + 2) = 104

We can let, for instance, L + 2 = 26 and W + 2 = 4. Then, L = 24 and W = 2. We see that 100 is a possible value for A + P.

B) A + P = 102

102 = (W + 2)(L + 2) - 4

(W + 2)(L + 2) = 106

The only way to factor 106 is 106 x 1 or 53 x 2. If W + 2 = 1, then W is negative and if W + 2 = 2, then W is zero. Neither are possible values for the width of a rectangle; therefore A + P = 102 is not possible.

Answer: B
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A rectangle with positive integer side lengths in cm has area A cm^2 a 09 Oct 2019, 21:25
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Bunuel
A rectangle with positive integer side lengths in cm has area A cm^2 and perimeter P cm. Which of the following numbers cannot equal A + P ?

(A) 100
(B) 102
(C) 104
(D) 106
(E) 108
Let one side of the rectangle is X
other side is X+k(where K >0 and Kmin=1 for X+k= integer)
A+P=X(X+k)+2[X+X+k]
A+P=X^2+X(k+4)+2k
According to the options
Plug
X=9
Just by looking at the equation we will get to know that this value will exceed the given values in the options.
so start with X=8
A+P= 96+10k For k=1- 106
X=6
A+P=60+8k For k=5-100, For K=6-108

X=4
A+P=32+6k-For K=12-104
B:)
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A rectangle with positive integer side lengths in cm has area A cm^2 a 11 Jul 2022, 07:44
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