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# Collection of 12 DS questions

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Re: NEW SET of good DS(4) [#permalink]

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21 Oct 2010, 10:31
1
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Why is the answer to number 2 not A?

Let c be the price of computers and p be the price of printers. We know that c = 5p. All we want is the ratio of the revenue of computers and the revenue of printers. From A, we know that in the first half of the year, the ratio of c:p was 3:2, so the ratio was 3c/2p = 15p/2p = 15:2. In the second half of the year, the ratio of c:p was 2:1, so 2c/p = 10p/p = 10:1. Then the total ratio is (15/2)/2 + (10/1)/2 = 15/4 + 20/4 = 35:4. Why is this wrong?
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Re: NEW SET of good DS(4) [#permalink]

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21 Oct 2010, 15:22
1
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2. In 2003 Acme Computer priced its computers five times higher than its printers. What is the ratio of its gross revenue for computers and printers respectively in the year 2003?
(1) In the first half of 2003 it sold computers and printers in the ratio of 3:2, respectively, and in the second half in the ratio of 2:1.
(2) It sold each computer for $1000. TehJay wrote: Why is the answer to number 2 not A? Let c be the price of computers and p be the price of printers. We know that c = 5p. All we want is the ratio of the revenue of computers and the revenue of printers. From A, we know that in the first half of the year, the ratio of c:p was 3:2, so the ratio was 3c/2p = 15p/2p = 15:2. In the second half of the year, the ratio of c:p was 2:1, so 2c/p = 10p/p = 10:1. Then the total ratio is (15/2)/2 + (10/1)/2 = 15/4 + 20/4 = 35:4. Why is this wrong? Whats wrong is you cannot just add the two ratios. Consider the two extreme cases : Case 1 : First half they sell 3million computers and 2million printers. Second half they sell 2 computers and 1 printer. The final ratio will be pretty much 3:2 Case 2 : First half they sell 3 computers and 2 printers. Second half they sell 2million computers and 1million printers. The final ratio will be pretty much 2:1 To find the final ratio, you need to exactly how many were sold in each half. _________________ Math Expert Joined: 02 Sep 2009 Posts: 33103 Followers: 5783 Kudos [?]: 70980 [0], given: 9857 Re: NEW SET of good DS(4) [#permalink] ### Show Tags 29 Oct 2010, 04:18 Expert's post whichscore wrote: asterixmatrix wrote: scoregmat wrote: 6. How many perfect squares are less than the integer d? (1) 23 < d < 33 (2) 27 < d < 37 1) For this condition d = 25 is perfect square of 5. Can we say there are 9, 16... only 2 perfect squares are < d ...? so Suff.. 2) For this condition d = 36 is perfect square of 6. Can we say there are 9, 16 and 25.. only 3 perfect squares are < d ...? Suff.. D. Is this explanation correct ? I may be wrong but this is what I could make out from the question stem option1 - for this we have 23 < d < 33 where d can be 24 upto 32 so we can have 1,4,9,16 and 25 depending on value of d. If d = 24 then perfect squares should be 1,4,9,16. If d = 26 then we can have 1,4,9,16 and 25. Hence insuff option 2 -for this we have 27 < d < 37 where d can be 28 upto 36. So perfect squares less than d will be 1,4,9,16 and 25. even if d =36 we need to find squares less than 36 so it will have only 5 perfect squares. hence suff will go with B (I may be totally wrong with my understanding) 6. How many perfect squares are less than the integer d? (1) 23 < d < 33 (2) 27 < d < 37 First of all: a perfect square is a number which is the square of an integer. So, perfect squares are: 0=0^2, 1=1^2, 4=2^2, 9=3^2, ... (1) 23 < d < 33 --> if $$d>25$$ (for example 26, 27, ...) then there will be 6 perfect square less then d: 0, 1, 4, 9, 16, and 25 BUT if $$d\leq25$$ (for example 25 or 24) then there will be only 5 perfect square less then d: 0, 1, 4, 9, and 16. Not sufficient. (2) 27 < d < 37 --> no matter what the value of d is there will be 6 perfect square less then d: 0, 1, 4, 9, 16, and 25 (even for max and min values of d). Sufficient. Answer: B. _________________ Math Forum Moderator Joined: 20 Dec 2010 Posts: 2022 Followers: 154 Kudos [?]: 1441 [1] , given: 376 Re: NEW SET of good DS(4) [#permalink] ### Show Tags 03 Feb 2011, 13:22 1 This post received KUDOS These may be the lousy way to solve the questions, but this is how I attempted to solve the questions. The answers may be incorrect. I didn't match them with OA. ******************************************************************************************************** New questions: 1. When the positive integer x is divided by 4, is the remainder equal to 3? (1) When x/3 is divided by 2, the remainder is 1. (2) x is divisible by 5. Soln: Q: will "x/4" leave a remainder of 3? (1) Rephrase: (x/3)/2= x/6 leaves remainder 1. Values of x: 1,7,13,19 x%4 : 1,3 Remainder may be 3 or NOT 3. Not Sufficient. (2) x%5 = 0 Values of x: 5,10,15 x%4 : 1,2,3 Remainder may be 3 or NOT 3. Not Sufficient. Both: Values of x: 25,55,85,115 x%4 : 1,1,1,3 Remainder may be 3 or NOT 3. Not Sufficient. Ans: "E" ********************************************************************* 2. In 2003 Acme Computer priced its computers five times higher than its printers. What is the ratio of its gross revenue for computers and printers respectively in the year 2003? (1) In the first half of 2003 it sold computers and printers in the ratio of 3:2, respectively, and in the second half in the ratio of 2:1. (2) It sold each computer for$1000.

Q: What is ratio of: (total money received by selling computers)/(total money received by selling printers)

(1) It just talks about the ratio of number of computers sold. Doesn't talk about how much was it sold for. Not Sufficient.
(2) How many computer and how about the printer!! Not sufficient.

Both: Not sufficient

Ans: "E"

***************************************************************************

3. Last Tuesday a trucker paid $155.76, including 10 percent state and federal taxes, for diesel fuel. What was the price per gallon for the fuel if the taxes are excluded? (1) The trucker paid$0.118 per gallon in state and federal taxes on the fuel last Tuesday.
(2) The trucker purchased 120 gallons of the fuel last Tuesday.

1.1x=155.76
x=(155.76/1.1)
x- base price paid for the entire purchase of the fuel without any taxes- known value

(155.76/(1.1*10)) paid in the taxes

(1) (155.76/(1.1*10))/n=0.118
Number of gallons n can be found. Then x/n to find the price/gallon without taxes. Sufficient.

(2) (155.76/20)*0.9 will be the price per gallon without taxes. Sufficient.

Ans: "D"

**************************************************************************

4. What is the remainder when the positive integer x is divided by 8?
(1) When x is divided by 12, the remainder is 5.
(2) When x is divided by 18, the remainder is 11.

Same approach as question 1. I couldn't find any shortcuts for this.

1.
x -> 5,17,29,41,53,65
x%8 -> 5,1,5,1,5,1. Can be 1 or 5. Not sufficient.

2.
x -> 11,29,47,65
x%8 -> 3,5,7,1 Can be many remainders. Not sufficient.

Both:
x -> 29,65
x%8 -> 5,1. Can be many remainders. Not sufficient.

Ans: "E"

****************************************************************
5. Al, Pablo, and Marsha shared the driving on a 1500 mile trip, which of the three drove the greatest distance on the trip?
(1) Al drove 1 hour longer than Pablo but at an average of 5 miles per hour slower than Pablo.
(2) Marsha drove 9 hours and averaged 50 miles per hour

Total distance: 1500 miles

(1) Al: Drove some miles driving for t hours at k miles/hour of speed
Pablo: Drove some miles for t-1 hours (K+5) miles/hour of speed

Miles driven by Al: tk
Miles driven by Pablo: (t-1)(k+5)
Can't deduce anything beyond this.
Not sufficient.

(2) Marsha drove 9*50=450 miles
So Al+Pablo drove=1050

Definitely, Marsha didn't drive the maximum. Because, Al or Pablo could have driven around 1049 miles alone. Even if equally, they must have
driven 1050/2=525 each, which is > 450. Not sufficient.

Both:
Marsha drove 450 miles
Al + Pablo drove 1050 miles
tk + (t-1)(k+5) = 1050

We can't derive exactly who drove how much of distance. Not sufficient.

Ans: "E"

********************************************************************

6. How many perfect squares are less than the integer d?
(1) 23 < d < 33
(2) 27 < d < 37

(1) d could be 24 or 26
if d=24. Number of Perfect squares less than 24: 1^2=1, 2^2=4, 3^2=9, 4^2=16 = 4
if d=26. Number of Perfect squares less than 26: 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25 = 5
Not sufficient.

(2) 27 < d < 37
d can be any integer from 28 to 36, inclusive.
from 28 to 36; the number of perfect square less than the number will always be 5.

if d=28. Number of Perfect squares less than 26: 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25 = 5
if d=36. Number of Perfect squares less than 36: 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25 = 5

The question is trying to trick us by saying less than rather than "less than equal to".

We should not count 36 as the perfect square because 36 is not less than 36. It is equal to 36.

Sufficient

Ans: "B"

*****************************************************************

7. The integers m and p are such that 2 is less than m and m is less than p. Also, m is not a factor of p. If r is the remainder when p is divided by m, is r > 1.
(1) The greatest common factor of m and p is 2.
(2) The least common multiple of m and p is 30.

(1) m and p must both be even
since p is not a multiple of m, remainder can't be 0.
Can it be 1.
For the remainder to 1, p will have to be odd, which is not possible.
So, the remainder will always be more than 1.

Sufficient.

(2) Possible values for m and p are
2,15 -> Remainder: 1 Ans: No
3,10 -> Remainder: 1 Ans: No
5,6 -> Remainder: 1 Ans: No

It can't be 1,30 because 30 is a multiple of 1.
Sufficient.

Ans: "D"

******************************************************************************

8. A scientist is studying bacteria whose cell population doubles at constant intervals, at which times each cell in the population divides simultaneously. Four hours from now, immediately after the population doubles, the scientist will destroy the entire sample. How many cells will the population contain when the bacteria is destroyed?
(1) Since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells.
(2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample.

(1) We can infer that bacteria is doubling every 2 hours. So,
2x: two hours back
4x: now
4x-2x=3750
2x=3750
x=1875

r = 2
A = 1875
n = 4
A(5) = a*r^(n-1)
or
simply: 3750*2=now
in 2 hours: 3750*2*2
in 4 hours: 3750*2*2*2
Sufficient

2) I am not too sure what this statement is trying to say. However, here's an abortive effort.

Say, it is time=1:00PM now
We have to consider the time until 1+4=5:00PM
It is saying that from "t+4-1" or from 4:00PM to 5:00PM, the bacteria will grow from 20,000 to 40,000. This is kind of too straightforward to be
true.

In other words, it is(or is it) saying that bacteria's count will be 40,000 when scientist destroys the sample.
Sufficient

Ans: "D"

****************************************************************************************
9. Is x^2 equal to xy?
(1) x^2 - y^2 = (x+5)(y-5)
(2) x=y

(1) Following values satisfy the equation.
x=y=-5
x=y=5

Can't think of any other number. But, I feel this is a lame way to solve this even if this is true.

Sufficient.

(2) Sufficient.

Ans: "D"

***********************************************************************************************

10. A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?
(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.
(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

Possibilities
20<->1
1<->20
10<->2
2<->10
4<->5
5<->4

(1)
Consider
20<->1
Oranges doubled: 2

Consider
10<->2
5<->4

So answer could be 20 or 2. Not sufficient.

(2)
Means after doubling the basket count, it will be more than 20.

Possible only in 20<->1 scenario. 1 oranges.

Sufficient

Ans: "B"

*************************************************************************************************************
11. If p is a prime number greater than 2, what is the value of p?
(1) There are a total of 100 prime numbers between 1 and p+1
(2) There are a total of p prime numbers between 1 and 3912.

(1) p is the hundredth prime number. Can be found. Sufficient.
(2) The total number of prime numbers can be found between 1 and 3912. p will be that total. Sufficient.

Ans: "D"

*******************************************************************************
12. If x is a positive integer, what is the least common multiple of x, 6, and 9?
(1) The LCM of x and 6 is 30.
(2) The LCM of x and 9 is 45.

(1) x can be 5 or 30. LCM=90 for both 5,6,9 and 30,6,9. Sufficient.

(2) x can be 5 or 45. LCM=90 for both 5,6,9 and 45,6,9. Sufficient.

Ans: "D"

Thanks to the author.

****************************************************************************

Also you can check new set of PS problems: new-set-of-good-ps-85440.html
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Re: NEW SET of good DS(4) [#permalink]

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22 Feb 2011, 06:34
This one can only be E, as Pablo and Al could have driven for 50 hours at a very low speed or for 5 hours at a very high speed.. This would change the relations.

asterixmatrix wrote:
Bunuel wrote:
New questions:

5. Al, Pablo, and Marsha shared the driving on a 1500 mile trip, which of the three drove the greatest distance on the trip?
(1) Al drove 1 hour longer than Pablo but at an average of 5 miles per hour slower than Pablo.
(2) Marsha drove 9 hours and averaged 50 miles per hour

option 2 gives us that Marsha drives a total of 450 miles. No other info. hence insuff
option 1 gives us details about Al and Pablo. No info about Marsha. hence insuff

together we get that Al and Pablo driive 1050 miles
Let time be T for which Pablo drove then T+1 wil be the time that Al drove and
S be the speed with which Pablo drove then S-5 will be the speed with which Al drove
we get S*T + (T+1)* (S-5) =1050 this equation formed will give various answers
will go with E
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Re: NEW SET of good DS(4) [#permalink]

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08 Aug 2011, 03:09
Quote:
7. The integers m and p are such that 2 is less than m and m is less than p. Also, m is not a factor of p. If r is the remainder when p is divided by m, is r > 1.
(1) The greatest common factor of m and p is 2.
(2) The least common multiple of m and p is 30.

(1) m and p must both be even
since p is not a multiple of m, remainder can't be 0.
Can it be 1.
For the remainder to 1, p will have to be odd, which is not possible.
So, the remainder will always be more than 1.

Sufficient.

(2) Possible values for m and p are
2,15 -> Remainder: 1 Ans: No
3,10 -> Remainder: 1 Ans: No
5,6 -> Remainder: 1 Ans: No

It can't be 1,30 because 30 is a multiple of 1.
Sufficient.

Ans: "D"

(2) Possible values for m and p are
2,15 -> Remainder: 1 Ans: No
3,10 -> Remainder: 1 Ans: No
5,6 -> Remainder: 1 Ans: No
could be also use the combination: 6,10 -> remainder = 4 {thus, making option B insufficient}
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Re: NEW SET of good DS(4) [#permalink]

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02 Oct 2011, 05:08
asterixmatrix wrote:
7. The integers m and p are such that 2 is less than m and m is less than p. Also, m is not a factor of p. If r is the remainder when p is divided by m, is r > 1.
(1) The greatest common factor of m and p is 2.
(2) The least common multiple of m and p is 30.

we have 2<m<p

1. we know m and p are factors of 2. So will be even.
Lets take m = 8 and p=10 then remainder r>1
Let m = 6 and p = 12 then also remainder r>1.
Hence suff

2. given least common multiple is 30 then we can have values for m and p as (5,6) (10,15) (3,10)
for the above pairs r can be =1 or >1 hence insuff

will go with A

I am not sure A is the answer. Because m= 4 and p = 8 then the remainder r will be 0....

IMO it should be C.

See my logic below....

S1: GCF is 2 means at least 2 is present in m and p
Not suff for above mentioned reason.

S2. LCM is 30, means 2x3X5 are the common factors between these m and p
As you have said earlier not suff because m and p can be (5,6) (10,15) (3,10)

and remainder is not always > 1

Combining 2 statements

LCM X GCF = m X P
30 X 2 = m X p

so 60 can be ( 1 X 60, 2 X 30, 3 X 20, 4 X 15, 5 X 12, 6 X 10)

given the condition 2<m<p remainder is always greater than 1.
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Re: New questions: 1. When the positive integer x is divided by [#permalink]

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09 Mar 2012, 20:58
Quote:
These may be the lousy way to solve the questions, but this is how I attempted to solve the questions.

The answers may be incorrect. I didn't match them with OA.

********************************************************************************************************
New questions:

1. When the positive integer x is divided by 4, is the remainder equal to 3?
(1) When x/3 is divided by 2, the remainder is 1.
(2) x is divisible by 5.

Soln:

Q: will "x/4" leave a remainder of 3?

(1) Rephrase: (x/3)/2= x/6 leaves remainder 1.

Values of x: 1,7,13,19
x%4 : 1,3 Remainder may be 3 or NOT 3. Not Sufficient.

Hey Bunuel, Please let me know where I am going wrong.
I also made the equation -

(x/3)/2= x/6 leaves remainder 1
i.e x = 6a+ 6
However, as per your previous posts- it should be 6a+3.

Thanks
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Re: New questions: 1. When the positive integer x is divided by [#permalink]

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10 Mar 2012, 01:14
Expert's post
imhimanshu wrote:
Quote:
These may be the lousy way to solve the questions, but this is how I attempted to solve the questions.

The answers may be incorrect. I didn't match them with OA.

********************************************************************************************************
New questions:

1. When the positive integer x is divided by 4, is the remainder equal to 3?
(1) When x/3 is divided by 2, the remainder is 1.
(2) x is divisible by 5.

Soln:

Q: will "x/4" leave a remainder of 3?

(1) Rephrase: (x/3)/2= x/6 leaves remainder 1.

Values of x: 1,7,13,19
x%4 : 1,3 Remainder may be 3 or NOT 3. Not Sufficient.

Hey Bunuel, Please let me know where I am going wrong.
I also made the equation -

(x/3)/2= x/6 leaves remainder 1
i.e x = 6a+ 6
However, as per your previous posts- it should be 6a+3.

Thanks

Positive integer $$a$$ divided by positive integer $$d$$ yields a reminder of $$r$$ can always be expressed as $$a=qd+r$$, where $$q$$ is called a quotient and $$r$$ is called a remainder, note here that $$0\leq{r}<d$$ (remainder is non-negative integer and always less than divisor).

Hence, when x/3 is divided by 2, the remainder is 1 should be expressed as x/3=2q+1 --> x=6q+3. So, x can be 3, 9, 15, ... For all those values, x/3 (1, 3, 5, ...) divided by 2 yields the remainder of 1.

As for your formula x can be 6, 12, 18, ... For those values, x/3 (2, 4, 6, ...) divided by 2 yields the remainder of 0 not 1.

Hope it's clear.
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Re: New questions: 1. When the positive integer x is divided by [#permalink]

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10 Mar 2012, 05:02
Thanks Bunuel, I can see the mistake.
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Re: New questions: 1. When the positive integer x is divided by [#permalink]

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27 Mar 2012, 23:47
1. When the positive integer x is divided by 4, is the remainder equal to 3?
(1) When x/3 is divided by 2, the remainder is 1.
(2) x is divisible by 5.

its E

I. 7, 13, 19, 25...... (No./6 = remainder 1)
II. 5,10,15,25....( No. divisible by 5)

I & II . 25, 55,85,.... (LCM of 5 & 6 =30)

hence 1 3 1 is pattern of remainders for I & II together......

Hence E
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Re: Collection of 12 DS questions [#permalink]

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28 Mar 2012, 02:31
7. The integers m and p are such that 2 is less than m and m is less than p. Also, m is not a factor of p. If r is the remainder when p is divided by m, is r > 1.
(1) The greatest common factor of m and p is 2.
(2) The least common multiple of m and p is 30.

IMO A.

since m=2*x
p=2* y and y should be greater than x
hence p/m will always have remainder >1
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Re: Collection of 12 DS questions [#permalink]

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28 Mar 2012, 02:51
Expert's post
vdbhamare wrote:
7. The integers m and p are such that 2 is less than m and m is less than p. Also, m is not a factor of p. If r is the remainder when p is divided by m, is r > 1.
(1) The greatest common factor of m and p is 2.
(2) The least common multiple of m and p is 30.

IMO A.

since m=2*x
p=2* y and y should be greater than x
hence p/m will always have remainder >1

OA's (answers) are given in this post: collection-of-12-ds-questions-85441-20.html#p642315

Solution for #7:
The integers m and p are such that 2<m<p, and m is not a factor of p. If r is the remainder when p is divided by m, is r>1?

Given: $$2<m<p$$ and $$\frac{p}{m}\neq{integer}$$. $$p=xm+r$$. q: $$r=?$$

(1) The greatest common factor of m and p is 2 --> $$p$$ and $$m$$ are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since the remainder is not zero (as $$\frac{p}{m}\neq{integer}$$), then the remainder must be more than 1. Sufficient.

(2) The least common multiple of m and p is 30 --> if $$m=5$$ and $$p=6$$, then remainder=1=1 and thus the answer to the question will be NO. BUT if $$m=10$$ and $$p=15$$, then remainder=5>1 and thus the answer to the question will be YES. Two different answers. Not sufficient.

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Re: Collection of 12 DS questions [#permalink]

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08 Apr 2012, 12:21
Hey Bunuel,

I am a little confused by question 2.

2. In 2003 Acme Computer priced its computers five times higher than its printers. What is the ratio of its gross revenue for computers and printers respectively in the year 2003?
(1) In the first half of 2003 it sold computers and printers in the ratio of 3:2, respectively, and in the second half in the ratio of 2:1.
(2) It sold each computer for $1000. ======================================================== C-Price of a computer P- Price of a Printer x-number of computers sold y-number of printers sold C=5p were trying to find x*C/y*P. Well we know C/P=5. So x*C/y*P=(x/y)*5. All we need to find is the ratio of x/y. Statement 1) x/y=3/2 in the first half of the year, ie the company sold 3z computers and 2z printers in the 1st half. x/y=2/1 in the second half of the year, ie the company sold 2z computers and 1z printers in the 2nd half. Hence, x/y=5/3 for the entire year. Shouldn't that be sufficient? Is my error to assume that z is the same in both the first half and second half? The company may have sold computers 9 computers and 6 printers in the first (z=3), maintaing the 3/2 ratio, but in the second half they might have sold 4 computers and 2 printers (ie z=2). By changing the z for the first and second we get different # of computers and printers sold. Is this why I am wrong? Thank you so much Bunuel! Math Expert Joined: 02 Sep 2009 Posts: 33103 Followers: 5783 Kudos [?]: 70980 [0], given: 9857 Re: Collection of 12 DS questions [#permalink] ### Show Tags 08 Apr 2012, 12:38 Expert's post alphabeta1234 wrote: Hey Bunuel, I am a little confused by question 2. 2. In 2003 Acme Computer priced its computers five times higher than its printers. What is the ratio of its gross revenue for computers and printers respectively in the year 2003? (1) In the first half of 2003 it sold computers and printers in the ratio of 3:2, respectively, and in the second half in the ratio of 2:1. (2) It sold each computer for$1000.

========================================================

C-Price of a computer
P- Price of a Printer
x-number of computers sold
y-number of printers sold

C=5p

were trying to find x*C/y*P. Well we know C/P=5. So x*C/y*P=(x/y)*5. All we need to find is the ratio of x/y.

Statement 1) x/y=3/2 in the first half of the year, ie the company sold 3z computers and 2z printers in the 1st half. x/y=2/1 in the second half of the year, ie the company sold 2z computers and 1z printers in the 2nd half. Hence, x/y=5/3 for the entire year. Shouldn't that be sufficient?

Is my error to assume that z is the same in both the first half and second half? The company may have sold computers 9 computers and 6 printers in the first (z=3), maintaing the 3/2 ratio, but in the second half they might have sold 4 computers and 2 printers (ie z=2). By changing the z for the first and second we get different # of computers and printers sold. Is this why I am wrong?

Thank you so much Bunuel!

Exactly. Company could have sold ANY number of computers and printers in the ratio of 3/2 in the first half of the year (3 and 2, 6 and 4, ..., 300 and 200, ...). Similarly it could have sold ANY number of computers and printers in the ratio of 2/1 in the second half of the year (2 and 1, 4 and 2, ..., 2,000,000 and 1,000,000 ...). So, infinitely many combinations of ratios are possible for the whole year, and 5 to 3 is just one of them.
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Re: Collection of 12 DS questions [#permalink]

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08 Apr 2012, 14:30
9. Is x^2 equal to xy?
(1) x^2 - y^2 = (x+5)(y-5)
(2) x=y

I am confused. Statement one should be sufficient. We get x=y=5 or x=y=-5. If x=y the statement should be sufficient. We cannot mix and match these two pairs, ie x=5 and y=-5. If that were the case then statement 1 would be insufficient.

x^2=x*y
X=5=Y=5 ---> 25=(5)^2=(5)(5)=25
X=-5=-5-----> 25=(-5)^2=(-5)(-5)=25

Both pairs satisfy the above statements. What am I doing wrong?
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Re: Collection of 12 DS questions [#permalink]

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08 Apr 2012, 14:48
Expert's post
alphabeta1234 wrote:
9. Is x^2 equal to xy?
(1) x^2 - y^2 = (x+5)(y-5)
(2) x=y

I am confused. Statement one should be sufficient. We get x=y=5 or x=y=-5. If x=y the statement should be sufficient. We cannot mix and match these two pairs, ie x=5 and y=-5. If that were the case then statement 1 would be insufficient.

x^2=x*y
X=5=Y=5 ---> 25=(5)^2=(5)(5)=25
X=-5=-5-----> 25=(-5)^2=(-5)(-5)=25

Both pairs satisfy the above statements. What am I doing wrong?

Is x^2 equal to xy?

(1) x^2 – y^2 = (x + 5)(y - 5)
(2) x = y

Question: is $$x^2=xy$$? --> $$x(x-y)=0$$ --> Equation holds true if $$x=0$$ or/and $$x=y$$.

So, basically question asks: Is $$x=0$$ or/and $$x=y$$ true?

Obviously statement (2) is sufficient, as it gives directly that $$x=y$$.

(1) $$x^2-y^2 = (x + 5)(y - 5)$$ --> $$(x+y)(x-y)=(x + 5)(y-5)$$

If $$y=5$$ --> $$(x+5)(x-5)=(x+5)(5-5)$$ --> $$(x+5)(x-5)=0$$ --> Either $$x=5=y$$ and in this case answer to the question is YES OR $$x=-5$$, hence $$x$$ is not equal to $$y$$ (nor to zero) and in this case answer to the question is NO. So two different answers.

Not sufficient.

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Re: Collection of 12 DS questions [#permalink]

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26 Oct 2012, 05:23
Bunuel wrote:
alphabeta1234 wrote:
Hey Bunuel,

I am a little confused by question 2.

2. In 2003 Acme Computer priced its computers five times higher than its printers. What is the ratio of its gross revenue for computers and printers respectively in the year 2003?
(1) In the first half of 2003 it sold computers and printers in the ratio of 3:2, respectively, and in the second half in the ratio of 2:1.
(2) It sold each computer for $1000. ======================================================== C-Price of a computer P- Price of a Printer x-number of computers sold y-number of printers sold C=5p were trying to find x*C/y*P. Well we know C/P=5. So x*C/y*P=(x/y)*5. All we need to find is the ratio of x/y. Statement 1) x/y=3/2 in the first half of the year, ie the company sold 3z computers and 2z printers in the 1st half. x/y=2/1 in the second half of the year, ie the company sold 2z computers and 1z printers in the 2nd half. Hence, x/y=5/3 for the entire year. Shouldn't that be sufficient? Is my error to assume that z is the same in both the first half and second half? The company may have sold computers 9 computers and 6 printers in the first (z=3), maintaing the 3/2 ratio, but in the second half they might have sold 4 computers and 2 printers (ie z=2). By changing the z for the first and second we get different # of computers and printers sold. Is this why I am wrong? Thank you so much Bunuel! Exactly. Company could have sold ANY number of computers and printers in the ratio of 3/2 in the first half of the year (3 and 2, 6 and 4, ..., 300 and 200, ...). Similarly it could have sold ANY number of computers and printers in the ratio of 2/1 in the second half of the year (2 and 1, 4 and 2, ..., 2,000,000 and 1,000,000 ...). So, infinitely many combinations of ratios are possible for the whole year, and 5 to 3 is just one of them. In 2003 Acme Computer’s price for each of its computers was five times the price for each of its printers. What was the ratio of its gross revenue from computers to its gross revenue from printers in 2003? (1) In the first half of 2003, Acme sold computers and printers in a ratio of 3:2; in the second half of 2003, Acme sold computers and printers in the ratio of 2:1. (2) Acme’s 2003 price for each of its computers was$1,000.

Bunuel..i got that ans is E..but my question is.

Assume..in statement 1..if it wud have been given only one ratio for whole year..like 2 ratio 1 ?? will it b sufficient..i tried it..i got the ans sufficent ..

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Re: Collection of 12 DS questions [#permalink]

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25 Apr 2013, 11:19
3. Last Tuesday a trucker paid $155.76, including 10 percent state and federal taxes, for diesel fuel. What was the price per gallon for the fuel if the taxes are excluded? (1) The trucker paid$0.118 per gallon in state and federal taxes on the fuel last Tuesday.
(2) The trucker purchased 120 gallons of the fuel last Tuesday.

Ans: D

In the first statement, whether it was excluding or including tax is not given.

So this 1 statement is not sufficient.
Re: Collection of 12 DS questions   [#permalink] 25 Apr 2013, 11:19

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# Collection of 12 DS questions

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