It is currently 23 Oct 2017, 14:26

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# The number of stamps that Kaye and Alberto had were in ratio

Author Message
TAGS:

### Hide Tags

Intern
Joined: 31 May 2006
Posts: 44

Kudos [?]: 11 [1], given: 0

The number of stamps that Kaye and Alberto had were in ratio [#permalink]

### Show Tags

05 Nov 2006, 16:07
1
KUDOS
4
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

63% (01:28) correct 37% (02:13) wrong based on 233 sessions

### HideShow timer Statistics

The number of stamps that Kaye and Alberto had were in the ration of 5:3 respectively. After Kaye gave Alberto 10 of her stamps, the ration of the number of Kaye had to the number of Alberto had was 7:5. As a result of the gift, Kaye had how many more stamps than Alberto

A. 20
B. 30
C. 40
D. 60
E. 90

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-number-of-stamps-that-kaye-and-alberto-had-were-in-the-98116.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Mar 2014, 02:05, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.

Kudos [?]: 11 [1], given: 0

Senior Manager
Joined: 24 Oct 2006
Posts: 339

Kudos [?]: 34 [0], given: 0

### Show Tags

05 Nov 2006, 21:34
5n-10=7x
3n+10=5x

7x-5x=2x=? Solving equations

15n-30=21x
15n+50=25x

80=4x
2x=40=ans

Kudos [?]: 34 [0], given: 0

Director
Joined: 11 Sep 2006
Posts: 515

Kudos [?]: 39 [0], given: 0

### Show Tags

13 Nov 2006, 20:25
The number of stamps that Kaye and Alberto had wer in the ration 5:3, respectively. After Kaye gave Alberto 10 of her stamps, the ratio of the number Kaye had to the number Alberto had was 7:5. As a result of this gift, Kaye had how many more stamps than Alberto?
20
30
40
60
90

I know how to do these, and yet I keep getting the wrong answer on this one...
_________________

...there ain't no such thing as a free lunch...

Kudos [?]: 39 [0], given: 0

Manager
Joined: 29 Aug 2006
Posts: 156

Kudos [?]: 6 [0], given: 0

### Show Tags

13 Nov 2006, 20:37
I get C

K=Kayle
A=Alberto

K/A =5/3--------(1)

(K-10)/(A+10)= 7/5------(2)

Substitute values from 1 in 2,

A=90
K=150

As a result of the gift, Alberto has A+10=100 and Kayle has K-10=140 and the diff between the two is 40.

Kudos [?]: 6 [0], given: 0

Director
Joined: 11 Sep 2006
Posts: 515

Kudos [?]: 39 [0], given: 0

### Show Tags

13 Nov 2006, 20:56
Priyah, I am terribly sorry to be so dense, but from whence did you derive your numerical values for K and A??
_________________

...there ain't no such thing as a free lunch...

Kudos [?]: 39 [0], given: 0

Manager
Joined: 29 Aug 2006
Posts: 156

Kudos [?]: 6 [0], given: 0

### Show Tags

13 Nov 2006, 21:16
K/A =5/3--------(1)

(K-10)/(A+10)= 7/5------(2)

eqn 2 can be simplified to
5(K-10)=7(A+10)

Now substitute K=5A/3 from eqn 1 in 2 and u will get the values.
25A-150=21A+210
4A=360
A=90

Kudos [?]: 6 [0], given: 0

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5032

Kudos [?]: 438 [0], given: 0

Location: Singapore

### Show Tags

14 Nov 2006, 01:01
Total number of stamps = x
# of stamps kaye owned = 5x/8
# of stamps Alberto owned =3x/8

New # of stamps kaye own = 5x/8 - 10
New # of stamps Alberto own = 3x/8 + 10

Ratio =

7/5 = (5x-80)/8 * 8/(3x+80)
7/5 = (5x-80)/(3x+80)
21x + 560 = 25x - 400
4x = 960
x = 240

Kaye now has 140
Alberto now has 100

So Kaye has 40 more stamps now.

Kudos [?]: 438 [0], given: 0

Senior Manager
Joined: 01 Oct 2006
Posts: 494

Kudos [?]: 39 [0], given: 0

### Show Tags

14 Nov 2006, 01:14
Let Kaye have x stamps alberto have x:y = 5:3 and after giving 10 stamps x-10:y+10 = 7:5
Put x =5y/3 in the eq above and solve to get value of y as 90 and then x=5y/3=150.
After giving 10 x=140 and y=100.
Thus kaye has 40 more stamps then alberto

Kudos [?]: 39 [0], given: 0

SVP
Joined: 05 Jul 2006
Posts: 1751

Kudos [?]: 432 [0], given: 49

### Show Tags

14 Nov 2006, 05:01
The number of stamps that Kaye and Alberto had wer in the ration 5:3, respectively. After Kaye gave Alberto 10 of her stamps, the ratio of the number Kaye had to the number Alberto had was 7:5. As a result of this gift, Kaye had how many more stamps than Alberto?
20
30
40
60
90

kaye stamps = 5x

Alberto stamps = 3x

after transaction

kay = 5x-10

Alberto = 3x+10

kaye/alberto = 5x-10/3x+10 = 7/5

25x-50 = 21x+70

4x = 120

x = 30

Kaye after = 140

alberto = 100

Kudos [?]: 432 [0], given: 49

Director
Joined: 11 Sep 2006
Posts: 515

Kudos [?]: 39 [0], given: 0

### Show Tags

14 Nov 2006, 08:10
OA is C - and I get it now! As always, thank you all very much...
_________________

...there ain't no such thing as a free lunch...

Kudos [?]: 39 [0], given: 0

Manager
Joined: 12 Oct 2006
Posts: 81

Kudos [?]: [0], given: 0

Location: NJ, US

### Show Tags

14 Nov 2006, 10:35
This is a great problem because while it is not real tough it is real time consuming. Does anyone know any shortcuts?

Kudos [?]: [0], given: 0

Senior Manager
Joined: 29 Jan 2011
Posts: 343

Kudos [?]: 244 [0], given: 87

### Show Tags

04 Jul 2011, 04:04
Can any one post the solution to solve this using random numbers? Since the ratio is 5: 3

And lets say I choose the total stamps to be 15... Then Kaye might have 15/5 = 3 stamps and Alberto might have 15/3 = 5 stamps ...but 3/5 is not 5:3? What am I doing wrong here?Can someone please help

Kudos [?]: 244 [0], given: 87

Current Student
Joined: 26 May 2005
Posts: 555

Kudos [?]: 239 [0], given: 13

### Show Tags

04 Jul 2011, 04:37
siddhans wrote:
Can any one post the solution to solve this using random numbers? Since the ratio is 5: 3

And lets say I choose the total stamps to be 15... Then Kaye might have 15/5 = 3 stamps and Alberto might have 15/3 = 5 stamps ...but 3/5 is not 5:3? What am I doing wrong here?Can someone please help

im sure u r following Manhattan gmat... they always ask u to pick smart numbers..

15 is not a smart number..U just cant pick it and reduce 10 from it and have a rati of 7
the current ratio is 5:3 and after giving away 10 .. the ratio becomes 7:5
picking numbers satisfying this is condition is tough unless ur very good at multiplication and see numbers very easily.
It took me 15 seconds to guess 5:3 shud be 150:90
so its better that u follow the std method;

5x-10/3x+10 = 7/5 , X = 30

Kudos [?]: 239 [0], given: 13

Senior Manager
Joined: 29 Jan 2011
Posts: 343

Kudos [?]: 244 [0], given: 87

### Show Tags

04 Jul 2011, 04:47
sudhir18n wrote:
siddhans wrote:
Can any one post the solution to solve this using random numbers? Since the ratio is 5: 3

And lets say I choose the total stamps to be 15... Then Kaye might have 15/5 = 3 stamps and Alberto might have 15/3 = 5 stamps ...but 3/5 is not 5:3? What am I doing wrong here?Can someone please help

im sure u r following Manhattan gmat... they always ask u to pick smart numbers..

15 is not a smart number..U just cant pick it and reduce 10 from it and have a rati of 7
the current ratio is 5:3 and after giving away 10 .. the ratio becomes 7:5
picking numbers satisfying this is condition is tough unless ur very good at multiplication and see numbers very easily.
It took me 15 seconds to guess 5:3 shud be 150:90
so its better that u follow the std method;

5x-10/3x+10 = 7/5 , X = 30

Yes, I was referring to smart numbers ..Looks like its difficult to solve this using smart numbers?

Kudos [?]: 244 [0], given: 87

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7677

Kudos [?]: 17405 [0], given: 232

Location: Pune, India

### Show Tags

05 Jul 2011, 03:57
siddhans wrote:

Yes, I was referring to smart numbers ..Looks like its difficult to solve this using smart numbers?

I would strongly suggest that you should stick to the method of solving which is quite straight forward as shown by yezz above. You may not be able to put your finger on the right numbers for a long time so its a gamble.
Nevertheless, if you do insist on using numbers to figure out, pick numbers smartly (I don't know what you mean by smart numbers except LCM/HCF is some cases and 100 in percentages.)

Let me explain what I mean by 'picking numbers smartly'
I see that the given ratio is 5:3
So actual numbers could be K-10 and A-6. If K gives 10 to A, ratio becomes 0:16. K has less than A but we need K to retain more than A so that the ratio is 7:5.
Actual numbers could be K-50 and A-30. If K gives 10 to A, ratio becomes 40:40 i.e. 1:1. The ratio has increased from before. Both have equal numbers. If the common multiplier chosen is less than 10, K will have less than A. If the common multiplier chosen is more than 10, K will retain more than A. So we need to go higher up.
Actual numbers could be K-100 and A-60. If K gives 10 to A, ratio becomes 9:7. Still less than 7:5. So we need to go further up
Actual numbers could be K-150 and A-90. If K gives 10 to A, ratio becomes 7:5.
There is a logic you are following to reach to the answer. Say, if I had jumped to K-200, A-120 directly, the new ratio would be 19:13 which is greater than 7:5 so I would try and find something in between K-100 and K-200. But using this approach, I am hoping that the multiplier will be an easy round number. Though it is true in most cases for GMAT, I would still not use this approach considering that the alternative standard method is quick and clean.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17405 [0], given: 232

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1853

Kudos [?]: 2633 [3], given: 193

Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: The number of stamps that Kaye and Alberto had were in ratio [#permalink]

### Show Tags

25 Mar 2014, 00:26
3
KUDOS
Stamps with Kale = 5x

Stamps with Alberto = 3x

So ratio $$= \frac{5x}{3x}$$

Kale gave 10 stamps to Alberto; so new ratio

= $$\frac{5x-10}{3x+10} = \frac{7}{5}$$............. (1)

We require to find the value of 5x-10 - (3x+10) = 2x-20 ......... (2)

Solving equation 1; we get x = 30

Placing value of x in equation (2)

= 60-20 = 40 = Answer C
_________________

Kindly press "+1 Kudos" to appreciate

Kudos [?]: 2633 [3], given: 193

Math Expert
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129505 [2], given: 12201

Re: The number of stamps that Kaye and Alberto had were in ratio [#permalink]

### Show Tags

25 Mar 2014, 02:05
2
KUDOS
Expert's post
The number of stamps that Kaye and Alberto had were in the ration of 5:3 respectively. After Kaye gave Alberto 10 of her stamps, the ration of the number of Kaye had to the number of Alberto had was 7:5. As a result of the gift, Kaye had how many more stamps than Alberto

A. 20
B. 30
C. 40
D. 60
E. 90

Given: $$\frac{K}{A}=\frac{5x}{3x}$$ (where $$x$$ is a positive integer) and $$\frac{5x-10}{3x+10}=\frac{7}{5}$$ --> $$x=30$$ --> $$K=150$$ and $$A=90$$.

Question: $$(K-10)-(A+10)=?$$ --> $$(K-10)-(A+10)=40$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-number-of-stamps-that-kaye-and-alberto-had-were-in-the-98116.html
_________________

Kudos [?]: 129505 [2], given: 12201

Re: The number of stamps that Kaye and Alberto had were in ratio   [#permalink] 25 Mar 2014, 02:05
Display posts from previous: Sort by