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Intern
Joined: 10 Aug 2007
Posts: 30

Re: 11 questions
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20 Aug 2007, 22:32
Sergey_is_cool wrote: 9. 23. (0.8)^5 / (0.4)^4=
(A) 3/32 (B) 5/64 (C) 1/2 (D) 1 (E) 2
I got 5/64.... I found this one to be easier if I converted to fractions, so start with:
(4/5)^5

(2/5)^4
apply properties of negative exponents:
(5/4)^5

(5/2)^4
perform the division:
(5/4)^5 * (2/5)^4
bring out a 5/4 to get equal exponents:
(5/4) * (5/4)^4 * (2/5)^4
combine likeexponents:
(5/4) * (1/2)^4
and compute:
(5/4) * (1/16) = 5/64



Intern
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Posts: 30

Re: 11 questions
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20 Aug 2007, 22:38
[quote="Sergey_is_cool"]
10. There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?
(A) 20
(B) 25
(C) 40
(D) 60
(E) 125
[/quote]
I got 20 for this:
5!/(3! * 1! * 1!) = (5 * 4 * 3 * 2)/(3 * 2) = 20



Manager
Joined: 03 Sep 2006
Posts: 232

Re: 11 questions
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Updated on: 20 Aug 2007, 22:47
9. 23. (0.8)^5 / (0.4)^4=
I did it even easier:
(4/5)^5 = (5/4)^5 = 5^5 / 4^5
(2/5)^4 = (5/2)^4 = 5^4 / 2^4
5^5 * 2^4 ... 5 * 16
 = 
4^5 * 5^4 ... 16 * 16 * 4
Ans. 5/64, hence B
What about 11? I'm stuck ...
Originally posted by Whatever on 20 Aug 2007, 22:46.
Last edited by Whatever on 20 Aug 2007, 22:47, edited 1 time in total.



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Re: 11 questions
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20 Aug 2007, 22:46
Sergey_is_cool wrote: 11. A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales in 1997 were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?
(A) 1: 2 (B) 4: 5 (C) 1: 1 (D) 3: 2 (E) 5: 3
Gonna guess A) 1:2 on this.
say x = car sales and y = truck sales:
.89x + 1.07y = 1.01 (x+y)
.89x + 1.07y = 1.01x + 1.01y
.06y = .12x
.06/.12 = x/y
1/2 = x/y



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Re: 11 questions
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20 Aug 2007, 23:09
entranced wrote: To this this, we need to examine the function. Since the tens digit is multiplied by three, if we play around with this digit, we should be able to modify the final result by 9.
Could you please explain more about "if we play around with this digit" and how you get numbers 113 and 193  what was the logic?



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Re: 11 questions
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20 Aug 2007, 23:15
entranced wrote: Sergey_is_cool wrote: 6. The function f is defined for each positive threedigit integer n by f(n) = 2x3y5z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are threedigit positive integers such that f(m)=9f(v), them mv=? (A) 8 (B) 9 (C) 18 (D) 20 (E) 80
I think the answer to this is 80. You want to f(m) to equal 9 * f(v). To this this, we need to examine the function. Since the tens digit is multiplied by three, if we play around with this digit, we should be able to modify the final result by 9. let's say V = 113 > f(v) = 2 * 1 * 3 * 1 * 5 * 3 = 90 let's say M = 193 > f(m) = 2 * 1 * 3 * 9 * 5 * 3 = 810 810 = 90 * 9 so, 193  113 = 80
What if m=119 and v=111?
f(m) = 2 x 1 x 3 x 1 x 5 x 9 = 270
f(v) = 2 x 1 x 3 x 1 x 5 x 1 = 30
f(m)/9 = f(v)
30 = f(v)
m  v = 8 so isn't A also correct?



Manager
Joined: 14 May 2006
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Here are my answers:
C, C, B, C, B, ?, E, D, ?, A, A
Can you please post the OAs?



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Joined: 07 Jul 2004
Posts: 4804
Location: Singapore

Re: 11 questions
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20 Aug 2007, 23:58
entranced wrote: Sergey_is_cool wrote: 6. The function f is defined for each positive threedigit integer n by f(n) = 2x3y5z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are threedigit positive integers such that f(m)=9f(v), them mv=? (A) 8 (B) 9 (C) 18 (D) 20 (E) 80
I think the answer to this is 80. You want to f(m) to equal 9 * f(v). To this this, we need to examine the function. Since the tens digit is multiplied by three, if we play around with this digit, we should be able to modify the final result by 9. let's say V = 113 > f(v) = 2 * 1 * 3 * 1 * 5 * 3 = 90 let's say M = 193 > f(m) = 2 * 1 * 3 * 9 * 5 * 3 = 810 810 = 90 * 9 so, 193  113 = 80
hmm... not so sure about this one. I did it the say way, but end up with 8 as the answer. I'm sure I can get 80 if i manipulate another way.
If f(m) = 810, then m = (2*1)(3*3)(5*9) > m = 139
Then f(v) = 90, then m = (2*1)(3*3)(5*1) > v = 131
THen mv = 8



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At a certain food stand, the price of each apple is $0.4 and the price of each orange is $0.6. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is $0.56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is $0.52?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Total number of fruits brought = 10
Total number of apples brought = a
Total number of oranges brought = 10a
Total price of the fruits = 5.6 = 0.4a + (10a)0.6
5.6 = 0.4a + 6  0.6a
0.2a = 0.4
a = 2
We want average price to be 0.52. So assuming she returns x oranges.
# of apples = 2
# of oranges = 8x
Average price = [2(0.4) + (8x)(0.6)]/[2 + 8x] = 0.52
x = 5



GMAT Club Legend
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Posts: 4804
Location: Singapore

Working alone at its constant rate, machine K took 3 hours to produce ¼ of the units produced last Friday. Then machine M started working and the two machines, working simultaneously at their respective constant rates, took 6 hours to produce the rest of the unites produced last Friday. How many hours would it have taken machine
(A) 8
(B) 12
(C) 16
(D) 24
(E) 30
Let number of units be 24
K > 6 units in 3 hours > 2 units per hour
K+M > 18 units in 6 hours > 3 units per hour
M > m units per hour
So K+M = 2 + m units per hour = 3; m = 1 unit per hour. To produce 24 units, machine M would need 24 hours.



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(0.8)^5 / (0.4)^4=
(A) 3/32
(B) 5/64
(C) 1/2
(D) 1
(E) 2
(0.8)^5 / (0.4)^4
= 2^5(0.4)^5/(0.4)^4
= 1/32(0.4)^1
= 5/64



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 4804
Location: Singapore

There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?
(A) 20
(B) 25
(C) 40
(D) 60
(E) 125
# of arrangements = 5!/3! = 20 ways



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 4804
Location: Singapore

A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales in 1997 were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?
(A) 1: 2
(B) 4: 5
(C) 1: 1
(D) 3: 2
(E) 5: 3
Total revenue in 1996 = x
Total revenue for cars in 1996 = c
Total revenue for trucks in 1996 = xc
Total revenue in 1997 = 1.01x
Total revenue for cars in 1996 = 0.89c
Total revenue for trucks in 1997 = 1.07(xx)
We want to find c/xc
0.89c + 1.07x  1.07c c  x + c = 0.01x
0.18c + 0.06x = 0
x = 3c
So total revenue for trucks in 1996 = xc = 2c
and c/xc = 1:2



Senior Manager
Joined: 03 May 2007
Posts: 257

BCC145 wrote: Here are my answers:
C, C, B, C, B, ?, E, D, ?, A, A
Can you please post the OAs?
OA are
C, C, B, C, B, D, E, D, A, B, A



Manager
Joined: 01 Oct 2007
Posts: 83

Should question 6 read:
The function f is defined for each positive threedigit integer n by f(n) = 2^x*3^y*5^z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are threedigit positive integers such that f(m)=9f(v), them mv=?
Then it would have a unique solution mv = 20, which is reported as the OA.



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Re: A certain club has 20 members. What is the ratio of the
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30 Dec 2018, 12:10
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Re: A certain club has 20 members. What is the ratio of the &nbs
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30 Dec 2018, 12:10



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