Problem Solving Questions

This topic has expert replies
Junior | Next Rank: 30 Posts
Posts: 22
Joined: Sat Jan 01, 2011 2:58 am
Thanked: 1 times

Problem Solving Questions

by Anamaiy » Mon Mar 14, 2011 10:16 am
Couple of questions I need help with : ( kindly post the solutions too)

The perimeter of square region S and rectangular region R are equal. If the sides of R are in the ratio 2:3, what is the ratio of the area of region R to the area of region S ?
(a). 25:16
(b). 24:25
(c). 5:6
(d). 4:5
(e). 4:9

Three boxes of supplies have an average(arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight in kilograms, of the lightest box ?
(a). 1
(b). 2
(c). 3
(d). 4
(e). 5

Thanks
Source: — Problem Solving |

Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pm
Thanked: 127 times
Followed by:10 members

by Night reader » Mon Mar 14, 2011 10:56 am
S=a^2, R=bc, b/c=2/3, 4a=2b+2c, find R/S-?
solution: a^2/bc-? a=(b+c)/2, ((b+c)^2)/4 : (bc)=(b^2+c^2+2bc)/4bc;
since 2c=3b, c=3b/2 we can rewrite (b^2+ 9b^2/4 + 3b^2)/ (6b^2) = (4b^2+9b^2+12b^2)/24b^2= 25/24. Since we need to find R/S reciprocal of 25/24 is 24/25

IOM b
Anamaiy wrote:Couple of questions I need help with : ( kindly post the solutions too)

The perimeter of square region S and rectangular region R are equal. If the sides of R are in the ratio 2:3, what is the ratio of the area of region R to the area of region S ?
(a). 25:16
(b). 24:25
(c). 5:6
(d). 4:5
(e). 4:9

Thanks
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com

Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pm
Thanked: 127 times
Followed by:10 members

by Night reader » Mon Mar 14, 2011 11:02 am
average=7, median=9, number of boxes=3, find max.weight of the lightest box-?
solution: a,b,c are three boxes, b is median and b=9, (a+b+c)=3*7=21, a+c=(21-9)=12, since c>=b (median) OR c>=9, a can be only (12-9)=3
IOM C
Anamaiy wrote: Three boxes of supplies have an average(arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight in kilograms, of the lightest box ?
(a). 1
(b). 2
(c). 3
(d). 4
(e). 5

Thanks
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com

User avatar
Legendary Member
Posts: 582
Joined: Tue Mar 08, 2011 12:48 am
Thanked: 61 times
Followed by:6 members
GMAT Score:740

by force5 » Mon Mar 14, 2011 2:50 pm
B, C

User avatar
GMAT Instructor
Posts: 613
Joined: Thu Mar 22, 2007 6:17 am
Location: madrid
Thanked: 171 times
Followed by:64 members
GMAT Score:790

by kevincanspain » Mon Mar 14, 2011 3:14 pm
Anamaiy wrote:Couple of questions I need help with : ( kindly post the solutions too)

The perimeter of square region S and rectangular region R are equal. If the sides of R are in the ratio 2:3, what is the ratio of the area of region R to the area of region S ?
(a). 25:16
(b). 24:25
(c). 5:6
(d). 4:5
(e). 4:9

Three boxes of supplies have an average(arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight in kilograms, of the lightest box ?
(a). 1
(b). 2
(c). 3
(d). 4
(e). 5

Thanks
If the sides of R are in a 2:3 ratio, we can call them 2x and 3x. The perimeter of R is 10x and the area 6x^2

Since the perimeter of S is equal to the perimeter of R (i.e. 10x) each side of S is 10x/4 =5x/2, and its area is 25x^2/4


The required ratio is 6x^2 : 25x^2/4 = 24/25

This problem becomes easier if you consider an example, x=1
Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid