Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43917

The length of a certain rectangle is 4 inches longer than its width. [#permalink]
Show Tags
09 Sep 2015, 23:27
Question Stats:
72% (02:13) correct 28% (03:03) wrong based on 117 sessions
HideShow timer Statistics



Manager
Joined: 07 Apr 2015
Posts: 179

The length of a certain rectangle is 4 inches longer than its width. [#permalink]
Show Tags
10 Sep 2015, 00:38
1
This post received KUDOS
l = w+4
\(l*w = 221\) \(w+4*w = 221\) \(w^2+4w221=0\)
\((w+17)(w13)\)
w is 17 or 13, only 13 is possible. Hence L = 13+4 = 17
\(13^2+17^2 = x^2\) \(169+289 = x^2\) \(458 = x^2\)
20*20 is 400, therefore x has to be greater 20. 22*22 (just tried that one) is 484, therefore it has to be below that. Hence C
(P.S.: No Idea how to do that task in <2 minutes, to much calculation involved)



Manager
Joined: 10 Aug 2015
Posts: 103

The length of a certain rectangle is 4 inches longer than its width. [#permalink]
Show Tags
10 Sep 2015, 03:51
2
This post received KUDOS
Solution: a(a+4) = 221. 221 is near to 15^2. So, look for two no. around 15 whose product will have 1 in its uits place.You dont have to look far, its 13 and 17. Length required = sqrt(13^2 + 17^2) = sqrt(169 + 289) = sqrt(458) 21^2 = 441 and 22^2 = 484.
Option C



Manager
Joined: 29 Jul 2015
Posts: 159

Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
Show Tags
10 Sep 2015, 06:10
Bunuel wrote: 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
Kudos for a correct solution. Let the width of the rectangle be x,then the length will be x+4. The area is 221. so, x(x+4)=221 or \(x^2+4x221=0\) or \((x13)(x+17)=0\) or x=13 width is 13 so the lenght will 13+4= 17 So, The diagonal of the rectangle will be \(\sqrt{13^2 + 17^2}\) = \(\sqrt{458}\) Clearly, the sum lies between \(21^2\) i.e 441 and \(22^2\) i.e 484 Answer: C



Intern
Joined: 20 Apr 2015
Posts: 10

Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
Show Tags
10 Sep 2015, 23:34
1
This post received KUDOS
L = 17, B = 13 169+289 = 458 is between 441 and 484 hence between 21 and 22. answer C



Director
Joined: 21 May 2013
Posts: 581

Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
Show Tags
11 Sep 2015, 02:01
Bunuel wrote: 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
Kudos for a correct solution. Width=x, Length=x+4 x*x+4=221 Solving for x, x=13,x+4=17 Therefore using pythagoras theorem, Diagonal^2=169+289=458 Diagonal=More than 21 but less than 22 Answer C



Math Expert
Joined: 02 Sep 2009
Posts: 43917

Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
Show Tags
13 Sep 2015, 08:00
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
Bunuel wrote: 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
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION: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. x^2 + 4x – 221 = 0 Beyond pure trial and error, we can look for nearby squares. 221 is nearly 225, which equals 15^2. 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 – 2^2 = (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: d^2 = 13^2 + 17^2 = 169 + 289 = 458. The square root of 458 is definitely larger than 20, since 20^2 = 400. Going up, we can compute 21^2 = 441 < 458, whereas 22^2 = 484 > 458. So the length of the diagonal must be between 21 and 22. The correct answer is C.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



NonHuman User
Joined: 09 Sep 2013
Posts: 13746

Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
Show Tags
16 May 2017, 19:19
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Intern
Joined: 05 Dec 2016
Posts: 7

Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
Show Tags
16 May 2017, 20:11
SHORTER METHOD
Let the dimensions be L & B Given: L  B= 4 & L.B= 221 Length of diagonal = (L^2 + B^2)^(1/2)
L^2 + B^2= √{(LB)^2 + 2L.B}=√{(4)^2 + 2.(221)}= √{16 + 442}= √(458)
Now 21^2=441 & 22^2= 484.
Hence 21< Diagonal length< 22. Hence C



Manager
Joined: 09 Jan 2016
Posts: 133
GPA: 3.4
WE: General Management (Human Resources)

Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]
Show Tags
17 May 2017, 10:45
Bunuel wrote: 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
Kudos for a correct solution. W* (w+4)= 221 221 has four factors 1, 13, 17,and liself W= 13 length= 13 +4 = 17 According to , Pythagorean theoram C^2= a^2 + b^2 C^2= 13^2+ 17^2 thus Diagonal could be \sqrt{458} and could be Between 21 and 22.




Re: The length of a certain rectangle is 4 inches longer than its width.
[#permalink]
17 May 2017, 10:45






