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The number of stamps that Kaye and Alberto had were in the
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27 Jul 2010, 13:25
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The number of stamps that Kaye and Alberto had were in the ration of 5:3 respecctively. 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
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Re: Pls explain how to come up to the slolution
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27 Jul 2010, 13:54
kilukilam wrote: The number of stamps that Kaye and Alberto had were in the ration of 5:3 respecctively. 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 \(\frac{5x10}{3x+10}=\frac{7}{5}\) > \(x=30\) > \(K=150\) and \(A=90\). Question: \((K10)(A+10)=?\) > \((K10)(A+10)=40\). Answer: C.
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Re: Pls explain how to come up to the slolution
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27 Jul 2010, 13:47
so before exchange ratio is 5x:3x (x basically cancels out but they might have 10:6, 15:9 etc. which reduces to 5:3) so after exchange we have 5x10:3x+10 = 7:5 You can represent this in fractions like: (5x10)/(3x+10) = 7/5 and cross multiply to get 25x50 = 21x + 70 > 4x = 120 x = 30 NOW 30 IS NOT THE ANSWER  that is the multiplier so you need to find the number of stamps in possession so...
Kaye 7*30 =210 Alberto 5*30 = 150 210150 = 60 = D




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Re: Pls explain how to come up to the slolution
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28 Jul 2010, 06:51
shaselai wrote: i think the problem here is yes you get X=30 BUT the new result is the ratio 7x:5x where the +/ 10 has already been taken into account or maybe i am wrong...? Answer to this question is C (40), provided OA is wrong. When we write \(\frac{K}{A}=\frac{5x}{3x}\) it means that \(x\) is the multiple of the initial ratio of 5/3 and not 7/5. We found that \(x=30\): initially K had \(5*30=150\) stamps and now K has \(15010=140\) stamps. Initially A had \(3*30=90\) stamps and now A has \(90+10=100\) stamps. Current difference is \(140100=40\). Hope it's clear.
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The number of stamps that Kaye and Alberto had were in the ratio
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01 Jun 2013, 10:19
Throw x=30 back into the second equation you set up. That's the ratio after the transaction has taken place and the number of stamps is defined from that equation, not the first. Make sense?
x=30 5(30)10=140 3(50)+10=100
140100=40 stamps



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Re: The number of stamps that Kaye and Alberto had were in the
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01 Jun 2013, 21:13
kilukilam wrote: The number of stamps that Kaye and Alberto had were in the ration of 5:3 respecctively. 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 1. We know the initial ratio of the number of stamps. We can represent the initial number of stamps that Kaye and Alberto had, as 5x and 3x resp 2. We can represent the final number of stamps that Kaye and Alberto had as 5x10 and 3x + 10 resp. The final ratio is 5x10/3x+10 3. The final ratio is also given as 7:5 4. As (2) and (3) both represent the final ratio of number of stamps and so as both are equivalent we can equate the two. 5x10/ 3x+10 = 7/5 5. x=30 6. We now need to find out how many stamps Kaye and Alberto had after the gift. 7. Kaye had 5x10 = 140 stamps. Alberto had 3x10 = 100 stamps 8. So Kaye has 40 more stamps than Alberto after the gift. Correct choice is c.
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Re: The number of stamps that Kaye and Alberto had were in the
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12 Sep 2013, 13:16
kilukilam wrote: The number of stamps that Kaye and Alberto had were in the ration of 5:3 respecctively. 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 5+3 = 8 7+5= 12 LCM of 12 and 8 = 24 8*3 = 24. Thus (5*3) / (3*3) = 15/9 12*2 = 24. Thus (7*2) / (5*2) = 14/10 10 – 9 = 1unit What we need is 1410= 4 Therefore 10*4 = 40



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Re: The number of stamps that Kaye and Alberto had were in the
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10 Mar 2014, 11:47
Initially their counts is 5x:3x
After K gave 10 stamps to A their new ratio is 7y:5y
i.e. 5x10=7yEq 1 3x+10=5y  eq 2
Solving eq 1 qnd 2 we get y=20
so from 2nd ratio
7*20  5 *20 = 40



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Re: Pls explain how to come up to the slolution
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10 Mar 2014, 22:06
Bunuel wrote: kilukilam wrote: The number of stamps that Kaye and Alberto had were in the ration of 5:3 respecctively. 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 \(\frac{5x10}{3x+10}=\frac{7}{5}\) > \(x=30\) > \(K=150\) and \(A=90\). Question: \((K10)(A+10)=?\) > \((K10)(A+10)=40\). Answer: C. Once x = 30 is calculated, we require to calculate (5x10)  (3x+10) = 2x20 = 6020 = 40 = Answer = C



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Re: The number of stamps that Kaye and Alberto had were in the
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04 Feb 2015, 04:05
so you can think about it this way. For some number (x) Kaye had 5x stamps and Alberto had 3x then kaye gave 10 stamps and the ration became 7:5, we can turn this into an equation like so: (5x10)/(3x+10)=7/5 cross multiplying gives us: 21x+70=25x50 or 120=4x x=30 but remember, we are trying to figure out how many more stamps kaye had than alberto after she gave him ten. originally, kaye had 150 stamps and alberto had 90 then she gave alberto 10, so she had 140 and he had 100 the difference is 40 stamps (C)
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The number of stamps that Kaye and Alberto had were in the
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27 Mar 2015, 19:37
Bunuel wrote: Given: \(\frac{K}{A}=\frac{5x}{3x}\) (where \(x\) is a positive integer) and \(\frac{5x10}{3x+10}=\frac{7}{5}\) > \(x=30\) > \(K=150\) and \(A=90\).
Hi Bunuel, I find your method above a very elegant way to convert a 2 variable (K, A) problem into 1 variable (X). It is clearly the same as saying: \(\frac{K}{A}=\frac{5}{3}\) and \(\frac{K10}{A+10}=\frac{7}{5}\) but I find your way much clearer and straighforward (no factions, one eq, etc). Would you recommend to apply this in ALL (or most, at least) ratio problems?My doubt is that, under time pressure, adding an extra variable (X) may not seem the best idea at first glance, as it actually is. Thank you in advance.



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The number of stamps that Kaye and Alberto had were in the
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03 Feb 2016, 11:54
Hi
all the above methods would take close to 2.5 min as the timer above suggests. The problem is easy but time consuming. Kudos to the one who can suggest a shorter solution! thanks!
If i had to triage that in 30 secs I would kick out B, D and E because they are all the multiples of 15 = 5*3. Choosing between 20 and 40 I would pick C and hit next because the odds for correct answers are on C answers



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Re: The number of stamps that Kaye and Alberto had were in the
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04 Feb 2016, 00:10
kilukilam wrote: The number of stamps that Kaye and Alberto had were in the ration of 5:3 respecctively. 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 Another method is to use ratios: Note that the total number of stamps is the same in both the cases. On ratio scale, 5:3 gives 8 total stamps while 7:5 gives 12 total stamps. So total stamps must be at least LCM of 8 and 12 i.e. 24. So in first case, Kaye had 15 stamps while Alberto had 9. After the exchange, Kaye had 14 stamps while Alberto had 10. So Kaye gave 1 stamp to Alberto. But actually, Kaye gave 10 stamps to Alberto. So actual total number of stamps is 240. So the multiplier of 7:5 is 20 (because 12*20 = 240). Kaye had (75)*20 = 40 more stamps than Alberto. Answer (C)
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Re: The number of stamps that Kaye and Alberto had were in the
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15 Mar 2016, 06:58
Bunuel wrote: shaselai wrote: i think the problem here is yes you get X=30 BUT the new result is the ratio 7x:5x where the +/ 10 has already been taken into account or maybe i am wrong...? Answer to this question is C (40), provided OA is wrong. When we write \(\frac{K}{A}=\frac{5x}{3x}\) it means that \(x\) is the multiple of the initial ratio of 5/3 and not 7/5. We found that \(x=30\): initially K had \(5*30=150\) stamps and now K has \(15010=140\) stamps. Initially A had \(3*30=90\) stamps and now A has \(90+10=100\) stamps. Current difference is \(140100=40\). Hope it's clear. Sometimes you wonder what quant section would be like without Bunuel. The first solution to the post was ultimately wrong, but inexplicably, hapless folks like me were falling over themselves giving it kudos. Tis terrific how GMAT tests the neatness of ones reasoning.



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Re: The number of stamps that Kaye and Alberto had were in the
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09 Apr 2016, 14:04
kilukilam wrote: The number of stamps that Kaye and Alberto had were in the ration of 5:3 respecctively. 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 K/A = 5/3 (K10)/(A+10) = 7/5 K=5A/3 K10=(7A+70)/5 5K=7A+120 25A/3 = 7A+120 multiply by 3 25A=21A+360 4A=360 A=90 since A = 90, then K = 150 A+10=100; K10=140 140100=40 C=40.



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16 May 2016, 14:11
Similar solution to what has already been posted but I find it slightly easier to understand (maybe that's because it's what I did Given5A = 3K 7*(A+10) = 5*(K10) Expand and solve simultaneously 7*(A+10) = 5*(K10).........7A + 70 = 5K  50.........7A = 5K  120..........35A = 25K  600 5A = 3K .............................................................................35A = 21K Subtract to find that 4K = 600......K = 150 Apply the ratio ..........K ......... A ..........5 ..........3 ..........150 .......90 Gift ....10 .......+10 ..........140 ......100 Difference ....40.... (C) (Test: new ratio is 14:10 i.e 7:5)



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Re: The number of stamps that Kaye and Alberto had were in the
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05 Aug 2016, 10:32
40 is another game where you don't read the question carefuly. The question is asking after the exchange



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The number of stamps that Kaye and Alberto had were in the
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11 Oct 2016, 06:53
Bunuel Hi, In his answer to the question, thenumberofstampsthatkayeandalbertohadwereinthe98116.html#p755395, Bunuel writes (5x10)/(3x+10)=7/5 and not (5x10)/(3x+10)=7y/5y Why not? Why does he directly simplify the y's there?



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The number of stamps that Kaye and Alberto had were in the
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11 Oct 2016, 06:57
plalud wrote: Bunuel Hi, In his answer to the question, http://gmatclub.com/forum/thenumberof ... ml#p755395, Bunuel writes (5x10)/(3x+10)=7/5 and not (5x10)/(3x+10)=7y/5y Why not? Why does he directly simplify the y's there? \(\frac{7y}{5y} = \frac{7}{5}\), because y is reduced.
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Re: The number of stamps that Kaye and Alberto had were in the
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11 Oct 2016, 09:09
kilukilam wrote: The number of stamps that Kaye and Alberto had were in the ration of 5:3 respecctively. 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 One option is to solve the question using TWO VARIABLES. Let K = # of stamps K had after the exchangeLet A = # of stamps A had after the exchangeThis means that K+10 = # of stamps K had before the exchange This means that A10 = # of stamps A had before the exchange Note: Our goal is to find the value of KA The number of stamps that K and A (originally) had were in the ratio 5:3So, (K+10)/(A10) = 5/3 We want a prettier equation, so let's cross multiply to get 3(K+10) = 5(A10) Expand: 3K + 30 = 5A  50 Rearrange: 3K  5A = 80After K gave A 10 of her stamps, the ratio of the number K had to the number A had was 7:5So, K/A = 7/5 We want a prettier equation, so let's cross multiply to get 5K = 7A Rearrange to get: 5K  7A = 0At this point we have two equations: 5K  7A = 03K  5A = 80Our goal is to find the value of K  A. IMPORTANT: We need not solve for the individual values of K and A. This is great, because something nice happens when we subtract the blue equation from the red equation. We get: 2K  2A = 80 Now divide both sides by 2 to get: K  A = 40 Answer: Cheers, Brent
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