Hello
Could you please help me solve the following 2 problems:
1) The number of stamps that kay and Alberto had were in the ratio 5:3. After Kaye gave 10 stamps to Alberto, the ratio is 7:5. As a result, kaye had how many more stamps than Alberto?
2) y, n are positive integers and 450y=n^3. Which of the foll must be an integer?
(i) y/ 3*2^2*5
(ii) y/ 3^2*2*5
(iii) y/ 3*2*5^2
(i) only
(ii) only
(iii) only
(iv)None
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Rahul@gurome
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Solution to first question:
Let the number of stamps that Kay and Alberto have be 5x and 3x.
After Kay gave 10 stamps to Alberto, Kay has 5x-10 stamps and Alberto has 3x+10 stamps.
The ratio of number of stamps Kay has to the ratio of the number of stamps Alberto has is 5x-10:3x+10. This is given as 7:5.
Or (3x+10)*7 = (5x-10)*5.
Or 21x+70=25x-50.
Or x=30.
So Kay had (5x-10)-(3x+10) = (2x - 20) or 40 more stamps than Alberto.
Let the number of stamps that Kay and Alberto have be 5x and 3x.
After Kay gave 10 stamps to Alberto, Kay has 5x-10 stamps and Alberto has 3x+10 stamps.
The ratio of number of stamps Kay has to the ratio of the number of stamps Alberto has is 5x-10:3x+10. This is given as 7:5.
Or (3x+10)*7 = (5x-10)*5.
Or 21x+70=25x-50.
Or x=30.
So Kay had (5x-10)-(3x+10) = (2x - 20) or 40 more stamps than Alberto.
Rahul Lakhani
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Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Hi
The solution for 1st problem:
Number of stamps with Kay = K
Number of stamps with Alberto = A
Asked: K - A = ??
Given:
K:A = 5:3 i.e., k/A = 5/3 ==> 3K = 5A ==> 3K - 5A = 0
After Kay gives 10 stamps to Alberto => Kaye has (K - 10) stamps and Alberto has (A + 10) stamps
Therefore, (k - 10)/(A + 10) = 7/5 ==> 5(k - 10) = 7(A + 10) ==> 5K - 50 = 7A + 70 ==> 5K - 7A = 120
So now we have 2 equations and 2 variable
3K - 5A = 0
5K - 7A = 120
Solving these we get : K = 150 & A = 90
Therefore, (K - 10) - (A + 10) = 140 - 100 = 40
Answer: As a result Kaye has 40 more stamps than Alberto.
I am not very sure about the 2nd problem, but i think the answer is none.
I hope this helps.
The solution for 1st problem:
Number of stamps with Kay = K
Number of stamps with Alberto = A
Asked: K - A = ??
Given:
K:A = 5:3 i.e., k/A = 5/3 ==> 3K = 5A ==> 3K - 5A = 0
After Kay gives 10 stamps to Alberto => Kaye has (K - 10) stamps and Alberto has (A + 10) stamps
Therefore, (k - 10)/(A + 10) = 7/5 ==> 5(k - 10) = 7(A + 10) ==> 5K - 50 = 7A + 70 ==> 5K - 7A = 120
So now we have 2 equations and 2 variable
3K - 5A = 0
5K - 7A = 120
Solving these we get : K = 150 & A = 90
Therefore, (K - 10) - (A + 10) = 140 - 100 = 40
Answer: As a result Kaye has 40 more stamps than Alberto.
I am not very sure about the 2nd problem, but i think the answer is none.
I hope this helps.
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selango
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K/A=5/3--A
K-10/A+10=7/5 --B
5K-50=7A+70
15K-150=21A+210
From equation A,
3K=5A -->15K=25A
-->25A-150=21A+210-->A=90 &K=150
After Kay gave alberto 10 stamps,A=100,K=140
-->Kay had 40 more stamps than alberto
K-10/A+10=7/5 --B
5K-50=7A+70
15K-150=21A+210
From equation A,
3K=5A -->15K=25A
-->25A-150=21A+210-->A=90 &K=150
After Kay gave alberto 10 stamps,A=100,K=140
-->Kay had 40 more stamps than alberto
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amising6
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The number of stamps that kay and Alberto had were in the ratio 5:3. After Kaye gave 10 stamps to Alberto, the ratio is 7:5. As a result, kaye had how many more stamps than Alberto
let the total number of stamp be X
so kay had 5x stamps
alberto had 3x stamp
after kay gave 10 stamp to alberto
now kay has 5x-10 stamp
alberto has 3x+10 stamp
so we know 5x-10/3x+10=7/5
solving we get x as 30
case 1)kaye had how many more stamps than Alberto before the transaction of stamps took place
i.e kay had 5*30=150
albert had 3*30=90
kay had 60 stamps more
case 2)kaye had how many more stamps than Alberto after the stamp transaction took place
kaye had (5x-10) - (3x+10)=40 stamps more
problem 2)
450y=n^3. i.e
5*5*3*3*2*y= n^3 (now since y and n are positive integer )
to make L.H.S a cube y will have to take value of 5*3*2*2
so now we have
if y is 5*3*2*2 then n will be 30
now taking this value of Y and N
we can say
(i) y/ 3*2^2*5
will only be integer
remaining two option will give fraction value
AKS
"ideation without execution is delusion"
let the total number of stamp be X
so kay had 5x stamps
alberto had 3x stamp
after kay gave 10 stamp to alberto
now kay has 5x-10 stamp
alberto has 3x+10 stamp
so we know 5x-10/3x+10=7/5
solving we get x as 30
case 1)kaye had how many more stamps than Alberto before the transaction of stamps took place
i.e kay had 5*30=150
albert had 3*30=90
kay had 60 stamps more
case 2)kaye had how many more stamps than Alberto after the stamp transaction took place
kaye had (5x-10) - (3x+10)=40 stamps more
problem 2)
450y=n^3. i.e
5*5*3*3*2*y= n^3 (now since y and n are positive integer )
to make L.H.S a cube y will have to take value of 5*3*2*2
so now we have
if y is 5*3*2*2 then n will be 30
now taking this value of Y and N
we can say
(i) y/ 3*2^2*5
will only be integer
remaining two option will give fraction value
AKS
"ideation without execution is delusion"
