BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Ratio d.s. problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Wed May 13, 2009 5:40 am
Location: MN
Thanked: 1 times
GMAT Score:580

Ratio d.s. problem

by LFalken » Mon Jun 15, 2009 2:35 pm
On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sight-seeing tour?

(1) On the sight-seeing tour, the ratio of the number of children to the number of men was 5 to 11
(2) The number of women on the sight-seeing tour was less than 30.

Answer = C


I chose "E" because I dont understand how simply stating the number of women there were less than (30) lets you find the answer? Could someone please clarify?

Thank you.
LFAL
Top
Source: — Data Sufficiency |
User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

Re: Ratio d.s. problem

by ssmiles08 » Mon Jun 15, 2009 2:51 pm
LFalken wrote:On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sight-seeing tour?

(1) On the sight-seeing tour, the ratio of the number of children to the number of men was 5 to 11
(2) The number of women on the sight-seeing tour was less than 30.

Answer = C


I chose "E" because I dont understand how simply stating the number of women there were less than (30) lets you find the answer? Could someone please clarify?

Thank you.
Let's work backwards:

Choice 2 is clearly insufficient: question gives the ratio of women to children. choice 2 give the number of women. To find the number of men is INSUFFICIENT.

Choice 1 tells us that men::children = 11:5

question stem gives the ratio of women::children = 5:2

the ratio of men to women to children is: 22:25:10

This is INSUFFICIENT as well b/c this only gives us the ratio and not the actual number.

together: the women < 30

22:25:10 (25 is multiplied by a factor of 1 to give us the number of women <30. If it is multiplied by a factor > 1 (ex:2) the # of women will be > 30)

so every ratio part is multiplied by a factor of 1
The actual number of men is 22.

(C)
Last edited by ssmiles08 on Mon Jun 15, 2009 4:15 pm, edited 1 time in total.
Top
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Tue Jan 27, 2009 8:35 pm
Location: Los Angeles, CA
Thanked: 1 times
GMAT Score:760

by walkingbanana » Mon Jun 15, 2009 3:01 pm
edit: dangit smiles! you beat me to the answer!

From the question stem:

W/C = 5/2

(1) Insufficient. This tells us that C/M = 5/11, which is in sufficient because we only have ratios and no info to determine the actual qty of men.

(2) Insufficient because this does not provide any information about the number of men.

(c) taking W/C = 5/2 and C/M = 5/11, we can find find lowest common multiple and equate all three terms together. W:C:M becomes 25:10:22. The qty of each group will have to be in multiples of each ratio. Since we know from statement 2 that # of women is less than 30, 25 if the only number that satisfies this requirement, which means that # of men = 22. Sufficient.
Top
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Wed May 13, 2009 5:40 am
Location: MN
Thanked: 1 times
GMAT Score:580

by LFalken » Mon Jun 15, 2009 3:36 pm
Thank you two! ssmiles08, you have been terrific answering all my questions, very much appreciated!
LFAL
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

by ssmiles08 » Mon Jun 15, 2009 3:44 pm
LFalken wrote:Thank you two! ssmiles08, you have been terrific answering all my questions, very much appreciated!
:)
Top
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Wed May 13, 2009 5:40 am
Location: MN
Thanked: 1 times
GMAT Score:580

by LFalken » Mon Jun 15, 2009 3:45 pm
I have a dumb question to ask, but can you show how to properly find the lowest common multiple of these ratios?
LFAL
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

by ssmiles08 » Mon Jun 15, 2009 4:03 pm
LFalken wrote:I have a dumb question to ask, but can you show how to properly find the lowest common multiple of these ratios?
I think you are asking how I got the ratio of 22:25:10.

set up the 2 ratios first

w::c
5::2

m::c
11::5

since both ratios have children in common, you need to find the LCM of the children: in this case the LCM of 5 and 2 is 10.

w::m::c
(5 :: - ::2) *5
( - ::11::5) *2

gives us

w::m::c
25:: - ::10
- ::22::10

since now both c's have 10 the ratio is now 25::22::10 (w::m::c)
Top
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Wed May 13, 2009 5:40 am
Location: MN
Thanked: 1 times
GMAT Score:580

by LFalken » Mon Jun 15, 2009 4:10 pm
ssmiles08 wrote:
LFalken wrote:I have a dumb question to ask, but can you show how to properly find the lowest common multiple of these ratios?
I think you are asking how I got the ratio of 22:25:10.

set up the 2 ratios first

w::c
5::2

m::c
11::5

since both ratios have children in common, you need to find the LCM of the children: in this case the LCM of 5 and 2 is 10.

w::m::c
(5 :: - ::2) *5
( - ::11::5) *2

gives us

w::m::c
25:: - ::10
- ::22::10

since now both c's have 10 the ratio is now 25::22::10 (w::m::c)

ssmiles08,

Perfect. Again, exactly what I needed. I have a long way to go it seems. thank you!
LFAL
Top
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Mon Jun 15, 2009 8:58 pm

by aks135 » Mon Jun 15, 2009 9:07 pm
ssmiles i agree with your explanation. But what i don't get is how the second option is being used to get the answer. Isn't the first option enough by itself?
Top
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Mon Jun 15, 2009 8:58 pm

Doubt

by aks135 » Mon Jun 15, 2009 9:16 pm
ssmiles i agree with your explanation. But what i don't understand is how you the second option is being used for deriving the answer. Shouldn't just A be enough?
Top
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pm
Thanked: 331 times
Followed by:11 members

by cramya » Mon Jun 15, 2009 9:17 pm
ssmiles i agree with your explanation. But what i don't get is how the second option is being used to get the answer. Isn't the first option enough by itself?
I hope smiles doesnt mind my explaining this.

A ratio between quantities could apply to many totals that satosfy this ratio

x:y:z

1:2:3 -> Total could be 6
1:2:3-> Total could be 12

x,y,z individual counts vary in both the scenarios above


With less than 30 women we know from the ratio 22:25:10 that there can possibley be only 22 men.

25*any other integer would take the number of women to more than 30


Hope this helps.

Regards,
CR
Top
Post new topic Post Reply
• Page 1 of 1