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# Four Problem Solving Questions from OG 12

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Manager
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Four Problem Solving Questions from OG 12 [#permalink]  08 Apr 2009, 14:43
1
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Difficulty:

(N/A)

Question Stats:

50% (00:00) correct 50% (00:00) wrong based on 2 sessions
If you know how to solve the following questions and you know how to explain your answers, please let me know. Thanks!

---------------------------------------------------------------------------------------------
1) When positive integer x is divided by positive y, the remainder is 9. If x / y = 96.12, what is the value of y?
a)96
b)75
c)48)
d)25
e)12
--------------------------------------------------------------------------------------------------------
2) If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the fist nonzero digit to the right of the decimal point?
a)Three
b)Four
c)Five
d)Six
e)Nine
--------------------------------------------------------------------------------------------------------
3) If x, y, and k are positive numbers such that (x / x+y)(10) + (y / x+y)(20) = k and if x<y, which of the following could be the value of k?

a)10
b)12
c)15
d)18
e)30
------------------------------------------------------------------------------------------------------
4)For any positive integer n, the sum of the fist n positive integers equals (n(n+1)) / 2. What is the sum of all even integers between 99 and 301?
a)10,100
b)20,200
c)22,650
d)40,200
e)45,150

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Re: Four Problem Solving Questions from OG 12 [#permalink]  08 Apr 2009, 18:09
2) t = (1) / (2^9 * 5^3)
t = 1/( 2^6 * 10 ^3)

1/2^6 = .01...

therefore t = .01 / 10^3 = .00001

Ans B. Four

3. (x / x+y)(10) + (y / x+y)(20) = k and x<y

10x+ 20y = K ( x+y)
x( 10 - k) = y( k-20)
y/x = 10- k / K -20 we know y> x so y/x > 0

therefore 15 < k < 20 only answer choice K = 18.

4. All even b/w 99 and 301

a = 100
L = 300
d = 2

Number of terms L = A + (n-1) d
300 = 100 + (n-1) 2
n = 101
Sum of the all even terms = S = n/2 ( a+ (n-1)d)
S = 20200
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Re: Four Problem Solving Questions from OG 12 [#permalink]  12 Apr 2009, 13:21
Can yo u please explain last part of question 3. How do you get the final inequality?
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Re: Four Problem Solving Questions from OG 12 [#permalink]  12 Apr 2009, 19:29
GMAT_700 wrote:
If you know how to solve the following questions and you know how to explain your answers, please let me know. Thanks!

---------------------------------------------------------------------------------------------
1) When positive integer x is divided by positive y, the remainder is 9. If x / y = 96.12, what is the value of y?
a)96
b)75
c)48)
d)25
e)12
--------------------------------------------------------------------------------------------------------

9/y=12/100 => 9/y=3/25 => 9*25/3=75
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Re: Four Problem Solving Questions from OG 12 [#permalink]  12 Apr 2009, 19:44
GMAT_700 wrote:
If you know how to solve the following questions and you know how to explain your answers, please let me know. Thanks!

--------------------------------------------------------------------------------------------------------
2) If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the fist nonzero digit to the right of the decimal point?
a)Three
b)Four
c)Five
d)Six
e)Nine

2^9*5^3=2^6*10^3

1/2^9=0.01...
1/10^3=0.001
0.01*0.001=0.00001
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Re: Four Problem Solving Questions from OG 12 [#permalink]  12 Apr 2009, 20:02
pmal04 wrote:
Can yo u please explain last part of question 3. How do you get the final inequality?

I would also like to know that.
Otherwise I think 15 also passes as an answer, following your logic: 10-15/15-20=-5/-5=1, and 1>0.
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Re: Four Problem Solving Questions from OG 12 [#permalink]  12 Apr 2009, 20:22
2
KUDOS
GMAT_700 wrote:
If you know how to solve the following questions and you know how to explain your answers, please let me know. Thanks!

------------------------------------------------------------------------------------------------------
4)For any positive integer n, the sum of the fist n positive integers equals (n(n+1)) / 2. What is the sum of all even integers between 99 and 301?
a)10,100
b)20,200
c)22,650
d)40,200
e)45,150

301-99=203 elements => 202/2=101 of them are even
(99+301)/2=200 - average of these 101 elements
the sum: 101*200=20200
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Re: Four Problem Solving Questions from OG 12 [#permalink]  14 Apr 2009, 18:59
the problem with K = 15 , the question says y > x in our case ( y =x) to yeild a result 1.

Last edited by tkarthi4u on 14 Apr 2009, 19:02, edited 1 time in total.
Senior Manager
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Re: Four Problem Solving Questions from OG 12 [#permalink]  14 Apr 2009, 19:01
peraspera wrote:
GMAT_700 wrote:
If you know how to solve the following questions and you know how to explain your answers, please let me know. Thanks!

---------------------------------------------------------------------------------------------
1) When positive integer x is divided by positive y, the remainder is 9. If x / y = 96.12, what is the value of y?
a)96
b)75
c)48)
d)25
e)12
--------------------------------------------------------------------------------------------------------

9/y=12/100 => 9/y=3/25 => 9*25/3=75

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Re: Four Problem Solving Questions from OG 12 [#permalink]  14 Apr 2009, 19:25
1
KUDOS
tkarthi4u wrote:
peraspera wrote:
GMAT_700 wrote:
If you know how to solve the following questions and you know how to explain your answers, please let me know. Thanks!

---------------------------------------------------------------------------------------------
1) When positive integer x is divided by positive y, the remainder is 9. If x / y = 96.12, what is the value of y?
a)96
b)75
c)48)
d)25
e)12
--------------------------------------------------------------------------------------------------------

9/y=12/100 => 9/y=3/25 => 9*25/3=75

from the question stem, x/y=qy+9
also from the question stem, x/y=96.12, therefore qy+9=96*100+12
from here we can assume the ratio of 9/y=12/100 or 9/y=3/25
From here y=(9*25)/3 = 75

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Schools: Northwestern, Booth, Duke, Berkley, Stanford, Harvard, IMD, IESE,
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Re: Four Problem Solving Questions from OG 12 [#permalink]  17 Aug 2010, 17:13
tkarthi4u wrote:
2) t = (1) / (2^9 * 5^3)
t = 1/( 2^6 * 10 ^3)

1/2^6 = .01...

therefore t = .01 / 10^3 = .00001

Ans B. Four

So I get how you get to t=(1/2^6)*10^-3 and then I'm lost in your description. Can you break it down a bit more for me
Manager
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Re: Four Problem Solving Questions from OG 12 [#permalink]  17 Aug 2010, 21:46
tkarthi4u wrote:
2)

3. (x / x+y)(10) + (y / x+y)(20) = k and x<y

10x+ 20y = K ( x+y)
x( 10 - k) = y( k-20)
y/x = 10- k / K -20 we know y> x so y/x > 0

therefore 15 < k < 20 only answer choice K = 18.

I guess when y>x, y/x > 1, and so k has to be greater than 15 but less than 20
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Re: Four Problem Solving Questions from OG 12 [#permalink]  19 Aug 2010, 06:35
---------------------------------------------------------------------------------------------
1) When positive integer x is divided by positive y, the remainder is 9. If x / y = 96.12, what is the value of y?
a)96
b)75
c)48)
d)25
e)12

x/y=q +9 and x/y=96.12

x=qy + 9, x= 96.12y, or x=96y + .12y

So, .12y=9, from here, y=9/.12=75

B.
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Re: Four Problem Solving Questions from OG 12 [#permalink]  27 Aug 2010, 06:06
2. 1/64=.01 and 1/10^3=.001

so, .01 x .001=0.00001
ans. B

Thanks for making clear.
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Re: Four Problem Solving Questions from OG 12   [#permalink] 27 Aug 2010, 06:06
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