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Tricky problem sets

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Re: Tricky problem sets [#permalink] New post 06 Oct 2013, 00:48
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Sarah pours four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?

A) 1/4
B) 1/3
C) 3/8
D) 2/5
E) 1/2

[Reveal] Spoiler:
D

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Re: Tricky problem sets [#permalink] New post 06 Oct 2013, 02:41
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call a number "prime-looking" if it is composite but not divisible by 2,3, or 5. The three smallest prime-looking numbers are 49,77 and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000?

A) 100
B) 102
C) 104
D) 106
E) 108

[Reveal] Spoiler:
A


First Find the numbers divisible by 2 , 3 and 5 as follows

1000/2 + 1000/3 + 1000/5 - 1000/6 - 1000/15 - 1000/10 + 1000/30 . Then the numbers not divisible by 2 3 5 will be 1000 - Sum = 266 .

Some of these numbers will be prime. So remove the 168 prime numbers. But in the 266 numbers 2 3 5 are not included as they are also multiples of 2 3 5 (duh). so the actual total is 269 - 165 ( 2 3 5 are not in the previous set so they can be excluded from the number of prime numbers ) = 101. But we dont include 1 as it is neither prime nor composite. 101-1 = 100 prime looking numbers.
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Re: Tricky problem sets [#permalink] New post 06 Oct 2013, 02:47
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Sarah pours four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?

A) 1/4
B) 1/3
C) 3/8
D) 2/5
E) 1/2

[Reveal] Spoiler:
D


The container size doesn't matter because we are finding fractions of the liquid

initially - 4c ------ 4r
first transfer 2c ------ 4r + 2c
2nd transfer 2c + 2r + c ------- 2r + c


We are interested in the contents of only the first cup --- r's represent the cream so its 2:5(2 + 2 + 1)
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Re: Tricky problem sets [#permalink] New post 06 Oct 2013, 03:21
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If 60^a = 3 and 60^b = 5, then 12^ {\frac{{ 1-a-b}}{2(1-b)}}

A) \sqrt{3}
B) 2
C) \sqrt{5}
D) 3
E) \sqrt{12}

[Reveal] Spoiler:
B


This took me a while, maybe i went about it the wrong way but

60^b = 5 gives 12 = 5^(1/b - 1) = 5^((1-b)/b)

12 ^(1/1-b) = 5^(1/b) ----- (1)


the given question can be written as ----- \sqrt{12^(1/1-b) / ( 12^(a/1-b) * 12^(b/1-b) )}

the first term is 5^(1/b) = 60

the second term is (5^(1/b))^a = 60^a = 3

Third term is 5 , can be deduced from statement (1)


so \sqrt{(60/15)} = 2 !
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Re: Tricky problem sets [#permalink] New post 06 Oct 2013, 04:36
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fozzzy wrote:
New problem added! Kudos for correct solutions

If 60^a = 3 and 60^b = 5, then 12^ {\frac{{ 1-a-b}}{2(1-b)}}

A) \sqrt{3}
B) 2
C) \sqrt{5}
D) 3
E) \sqrt{12}

[Reveal] Spoiler:
B


I can't type the solution. This one is calculation intensive. I wonder whether they will ask such questions in actual GMAT.
But I have attached my calculation.

The answer is 2 (Option B)

Hope it helped.
Cheers
Qoofi
Attachments

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Re: Tricky problem sets [#permalink] New post 06 Oct 2013, 23:46
New problem added kudos for right solution OA later...

one morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

A) 3
B) 4
C) 5
D) 6
E) 7

[Reveal] Spoiler:
C


Solution

[Reveal] Spoiler:
Let c be the total amount of coffee, m of milk, and p the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), so

(\frac{c}{6}+ \frac{m}{4})p = c+m

Regrouping, we get 2c(6-p)=3m(p-4). Since both c,m are positive, it follows that 6-p and p-4 are also positive, which is only possible when p = 5.

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Last edited by fozzzy on 07 Oct 2013, 07:58, edited 3 times in total.
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Re: Tricky problem sets [#permalink] New post 06 Oct 2013, 23:57
New problem added OA later kudos for right solution...

The set {3,6,9,10} is augmented by a fifth element n, not equal to any of the other four. The median of the resulting set is equal to its mean. What is the sum of all possible values of n?

A) 7
B) 9
C) 19
D) 24
E) 26

[Reveal] Spoiler:
E

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Last edited by fozzzy on 07 Oct 2013, 07:53, edited 1 time in total.
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Re: Tricky problem sets [#permalink] New post 07 Oct 2013, 07:09
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New problem added kudos for right solution OA later... A positive integer n has 60 divisors and 7n has 80 divisors. what is the greatest integer k such that 7^k divides n? A) 0 B) 1 C) 2 D) 3 E) 4


I got this in my email , i think the post was deleted, but it seems like a good question...

So N has 60 divisors
7N has 80 divisors,this means that 7 is already a divisor of N, if 7 were a new factor, the number of factors would have been doubled. (See prime factorization)

Now 80 60 , HCF is 20. The other factors are 4 and 3. The change from 3 to 4 took place because of multiplying by 7. So the original power of 7 contained changed from 2 to 3.

N can thus be divided by 7^2 . c) 2
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Re: Tricky problem sets [#permalink] New post 07 Oct 2013, 07:23
New problem added kudos for right solution OA later...

A positive integer n has 60 divisors and 7n has 80 divisors. what is the greatest integer k such that 7^k divides n?

A) 0
B) 1
C) 2
D) 3
E) 4

[Reveal] Spoiler:
C

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Last edited by fozzzy on 07 Oct 2013, 07:52, edited 1 time in total.
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Re: Tricky problem sets [#permalink] New post 07 Oct 2013, 07:26
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New problem added! Kudos for correct solutions

If 60^a = 3 and 60^b = 5, then 12^ {\frac{{ 1-a-b}}{2(1-b)}}

A) \sqrt{3}
B) 2
C) \sqrt{5}
D) 3
E) \sqrt{12}

[Reveal] Spoiler:
B



Take LOG both side
aLog60=Log3
=> a=LOG 3 (with base 60)
simillarly, => b=log 5 (with base 60)

now substitute values of a and b in the equation and the required solution can be easily reached then
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Re: Tricky problem sets [#permalink] New post 07 Oct 2013, 07:31
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goforsriram wrote:
Quote:
Here is the Text of the New Reply (by fozzzy)

New problem added kudos for right solution OA later... A positive integer n has 60 divisors and 7n has 80 divisors. what is the greatest integer k such that 7^k divides n? A) 0 B) 1 C) 2 D) 3 E) 4


I got this in my email , i think the post was deleted, but it seems like a good question...

So N has 60 divisors
7N has 80 divisors,this means that 7 is already a divisor of N, if 7 were a new factor, the number of factors would have been doubled. (See prime factorization)

Now 80 60 , HCF is 20. The other factors are 4 and 3. The change from 3 to 4 took place because of multiplying by 7. So the original power of 7 contained changed from 2 to 3.

N can thus be divided by 7^2 . c) 2



Factors & divisors are same. So

N = 60 divisors = 20 * 3
7N = 80 divisors = 20 *4
we know, the number of factors of N = (a^p)*(b^q), is (p+1)*(q+1)
Hence
N = a^19 * 7^2.
7N = a^19 * 7^3

7^2 is divisible by N. So k = 2

Cheers
Qoofi
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Re: Tricky problem sets [#permalink] New post 07 Oct 2013, 07:35
fozzzy wrote:
New problem added OA later kudos for right solution...

The set {3,6,9,10} is augmented by a fifth element n, not equal to any of the other four. The median of the resulting set is equal to its mean. What is the sum of all possible values of n?

A) 7
B) 9
C) 19
D) 24
E) 26


This is plug in numbers type.

After inserting the number, the median and mean should be equal.
Lets take Option A : 7

(3,6,7,9,10). Here median is 7

Mean = (3+6+7+9+10) / 5 = 7

Hence answer is Option.

Cheers
Qoofi
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Re: Tricky problem sets [#permalink] New post 07 Oct 2013, 07:47
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fozzzy wrote:
New problem added OA later kudos for right solution...

The set {3,6,9,10} is augmented by a fifth element n, not equal to any of the other four. The median of the resulting set is equal to its mean. What is the sum of all possible values of n?

A) 7
B) 9
C) 19
D) 24
E) 26


There are three cases.

Case i) 6 is the median. Sum should be 5x6 (6 also mean) = 30. So the missing element is 2
Case ii)9 is the median. Sum should be 5x9 = 45 . Missing element is 17
Case iii) x is the median whose value is either 7 or 8 ( lies between 6 and 9 ). Missing element that satisfies condition is 7

Their sum is 26
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Re: Tricky problem sets [#permalink] New post 07 Oct 2013, 07:55
Qoofi wrote:
fozzzy wrote:
New problem added OA later kudos for right solution...

The set {3,6,9,10} is augmented by a fifth element n, not equal to any of the other four. The median of the resulting set is equal to its mean. What is the sum of all possible values of n?

A) 7
B) 9
C) 19
D) 24
E) 26


This is plug in numbers type.

After inserting the number, the median and mean should be equal.
Lets take Option A : 7

(3,6,7,9,10). Here median is 7

Mean = (3+6+7+9+10) / 5 = 7

Hence answer is Option.

Cheers
Qoofi


Oops I made a huge mistake. Apologies.
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Re: Tricky problem sets [#permalink] New post 07 Oct 2013, 08:00
I have added the OA's and given kudos for correct solutions! you guys missed a problem... I added the solution to that one Angela's family...

@qoofi you fell for the trap answer choice, I highlighted that portion in bold.
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Re: Tricky problem sets [#permalink] New post 08 Oct 2013, 06:11
New problem added.. OA later.. Kudos for right solution

John ordered 4 pairs of black socks and some additional pair of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by 50%. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:

A) 4:1
B) 2:1
C) 1:4
D) 1:2
E) 1:8

[Reveal] Spoiler:
C

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Last edited by fozzzy on 09 Oct 2013, 04:03, edited 1 time in total.
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Re: Tricky problem sets [#permalink] New post 08 Oct 2013, 06:18
New problem added.. OA later.. Kudos for right solution

A drawer in a darkened room contains 100 red socks, 80 green socks, 60 blue socks and 40 black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least 10 pairs? ( A pair of socks is two socks of the same color. No sock may be counted in more than one pair).

a) 21
b) 23
c) 24
d) 30
e) 50

[Reveal] Spoiler:
B

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Last edited by fozzzy on 09 Oct 2013, 04:37, edited 1 time in total.
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Re: Tricky problem sets [#permalink] New post 08 Oct 2013, 06:43
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John ordered 4 pairs of black socks and some additional pair of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by 50%. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:

A) 4:1
B) 2:1
C) 1:4
D) 1:2
E) 1:8


Hi.

Lets say Cost of pair of Black Socks = A & Cost of pair of Blue Socks = B
Lets x be the number of pairs of Blue socks
A = 2 B (Given)

New Incorrect Bill is greater than Original Bill by 50%, Which is New Bill = 1.5(Original Bill)

Original Bill = 4A + xB = 8B + xB = B(8+x)
New Bill = xA + 4B = 2xB + 4B = B(2x+4)
B(8+x) = (1.5)B(2x+4)

Cancelling B on both sides and solving, we get x = 16

We need the ratio of socks in the original buy = 4:x = 4: 16 = 1:4

Option C

Cheers
Qoofi
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Re: Tricky problem sets [#permalink] New post 08 Oct 2013, 06:49
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New problem added.. OA later.. Kudos for right solution

A drawer in a darkened room contains 100 red socks, 80 green socks, 60 blue socks and 40 black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least 10 pairs? ( A pair of socks is two socks of the same color. No sock may be counted in more than one pair).

a) 21
b) 23
c) 24
d) 30
e) 50


Hi

Kind of simple one. We need 10 pairs of socks.

To find the smallest number of socks that must be selected :

We can get 9 pairs of socks from selecting 18 socks ( 9*2)
To get the 10th pair, let choose 4 more socks. In this 4 socks, we would get one of the 4 colors of socks. But one sock of a color doesn't make it a pair

So choose 1 more socks to make it to a pair of sock, which is the 10th Pair of sock.

So total number of socks select = 18+4+1 = 23

Option B

Hope it is clear

Cheers
Qoofi
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Re: Tricky problem sets [#permalink] New post 08 Oct 2013, 06:52
fozzzy wrote:
New problem added.. OA later.. Kudos for right solution

A drawer in a darkened room contains 100 red socks, 80 green socks, 60 blue socks and 40 black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least 10 pairs? ( A pair of socks is two socks of the same color. No sock may be counted in more than one pair).

a) 21
b) 23
c) 24
d) 30
e) 50


If u need 1 pair of socks , you need to pick 5 socks. So the least number u will get is 1 pair. Similarly if i pick 7 socks, then 2 pairs are guaranteed. 11 socks will give me 3.

so 2s + 3 socks will give me S pairs. Thus for 10 pairs u need 23 socks
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Re: Tricky problem sets   [#permalink] 08 Oct 2013, 06:52
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