Manhattan GMAT Challenge Problem of the Week – 31 May 2011:

by on May 31st, 2011

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Question

The length of a certain rectangle is 4 inches longer than its width. If the area of the rectangle is 221 square inches, then the length of the rectangle’s diagonal, in inches, is

(A) between 19 and 20
(B) between 20 and 21
(C) between 21 and 22
(D) between 22 and 23
(E) between 23 and 24

Answer

First, set up an equation for the area of the rectangle. If x is the width, then we have

x(x + 4) = 221

Note that if we put this into standard quadratic form and then try to factor, we wind up back where we started, in some sense: we are looking for two numbers that multiply to 221 and that differ by 4.

+ 4x – 221 = 0

Beyond pure trial and error, we can look for nearby squares. 221 is nearly 225, which equals . So we might try numbers near 15. As it turns out, 221 = 13 × 17. We might even get there by noticing a difference of squares:

221 = 225 – 4 = = (15 – 2)(15 + 2) = 13 × 17.

As a last resort, we could always use the quadratic formula, which gets us the roots of the equation as well.

Now, the diagonal of the rectangle will be given by the Pythagorean Theorem:

= +
= 169 + 289
= 458.

The square root of 458 is definitely larger than 20, since = 400. Going up, we can compute = 441 < 458, whereas = 484 > 458. So the length of the diagonal must be between 21 and 22.

The correct answer is C.

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4 comments

• I did it like this :

Let length = l and breadth = b

Then l = b + 4

So b * (b+4) = 221

=>  b *(b+4)  = 17 * 13

So Diagonal = Root( 17^2 + 13^2)

=> Diagonal = root(458) > 21^2 and <  22^2

So answer is C.

• yeah the answer is C. it looked right in the eye and said its me its me. I heard it.

• hey where can i find more questions , please post the link

• x(x+4) = 221

diagonal^2 = x^2 + (x+4)^2
= x^2 + x^2 + 8x + 16
= 2x(x+4) + 16
= 2x221 + 16
= 442 + 16
= 458

21^2 = 441