Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Jun 2016, 02:40

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# 4 GMATPrep Qs

Author Message
Manager
Joined: 29 Jul 2009
Posts: 176
Followers: 4

Kudos [?]: 63 [0], given: 2

### Show Tags

27 Aug 2009, 18:32
00:00

Difficulty:

(N/A)

Question Stats:

100% (05:30) correct 0% (00:00) wrong based on 1 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi

for the 2nd question, I know that there should be no addition because the first 16 doesn't have a radical..so how was it solved?
Attachments

Untitled1.jpg [ 51.46 KiB | Viewed 1258 times ]

Untitled.jpg [ 22.99 KiB | Viewed 1258 times ]

Manager
Joined: 29 Jul 2009
Posts: 176
Followers: 4

Kudos [?]: 63 [0], given: 2

### Show Tags

27 Aug 2009, 18:34
for Q10,
10 M = 16 H
8 M = ?

8*16= 96/10 = 9.6
Attachments

Untitled3.jpg [ 31.78 KiB | Viewed 1247 times ]

Untitled2.jpg [ 35.92 KiB | Viewed 1248 times ]

Manager
Joined: 29 Jul 2009
Posts: 176
Followers: 4

Kudos [?]: 63 [0], given: 2

### Show Tags

27 Aug 2009, 18:36
and here's the last question...

Attachments

Untitled4.jpg [ 50.75 KiB | Viewed 1253 times ]

Manager
Joined: 08 Jul 2009
Posts: 174
Followers: 3

Kudos [?]: 58 [1] , given: 13

### Show Tags

27 Aug 2009, 19:01
1
KUDOS
Hi

for the 2nd question, I know that there should be no addition because the first 16 doesn't have a radical..so how was it solved?

Answer to the first should be D

St 1

If LCM of x and 6 is 30, then x is either 5 or 30. Either way the LCM of x,6,9 would be 90

If LCM of x and 9 is 45, the x is either 5 or 45. Either way the LCM of x,6,9 would be 90
Manager
Joined: 08 Jul 2009
Posts: 174
Followers: 3

Kudos [?]: 58 [0], given: 13

### Show Tags

27 Aug 2009, 19:04
Hi

for the 2nd question, I know that there should be no addition because the first 16 doesn't have a radical..so how was it solved?

The only sides that would work for the triangle are 8sqrt2 + 8Sqrt2 and 16 = 16sqrt2 + 16

This question can be solved by logically thinking about the isoscles right triangle. Its sides are x,x,xsqrt2.
Manager
Joined: 10 Aug 2009
Posts: 130
Followers: 3

Kudos [?]: 62 [1] , given: 10

### Show Tags

27 Aug 2009, 19:26
1
KUDOS
For the first question about LCM, each statement alone is sufficient.
6=3X2
9=3^2

To calculate LCM we take the highest count (power of a certain prime) of each prime factor found across all integers.
$$LCM(6,9) = 3^2\times 2$$

Statement 1:
$$LCM (6,x)=30=3\times 2\times 5$$
x must have only one "5" in its prime factorization.
x must have either no "3" or only one "3"
x must have either no "2" or only one "2"
It can be 5, 10,15 or 30. We do not need to know the exact value of x but the maximum powers of it's primes.
It is sufficient. Across three numbers (6,9,x), the highest power of prime "3" is 2, the highest power of prime "2" is 1 and the highest power of prime "5" is 1. Therefore, $$LCM (x,6,9)=3^2\time 2^1\times 5^1=90$$

Statement 2
$$LCM(x,9)=45=5\times 3\times 3$$
x has only one "5"
x has no or one or two "3".
x can be 5,15 or 45.
Again, it is sufficent. Across three numbers (6,9,x), the highest power of prime "3" is 2, the highest power of prime "2" is 1 and the highest power of prime "5" is 1. $$LCM (x,6,9)=3^2\time 2^1\times 5^1=90$$
Manager
Joined: 10 Aug 2009
Posts: 130
Followers: 3

Kudos [?]: 62 [0], given: 10

### Show Tags

27 Aug 2009, 19:49

I define hypotenuse as B and other side as A.

1) $$2A+B=16+16\times \sqrt{2}$$
2) $$2A^2=B^2$$

$$A=\frac{B}{\sqrt{2}}$$
$$\frac{2B}{\sqrt{2}}+B=16+16\times \sqrt{2}$$
$$B\times (1+\sqrt{2})=16\times (1+\sqrt{2})$$
$$B=16$$
Manager
Joined: 28 Jul 2009
Posts: 124
Location: India
Schools: NUS, NTU, SMU, AGSM, Melbourne School of Business
Followers: 6

Kudos [?]: 66 [1] , given: 12

### Show Tags

27 Aug 2009, 23:09
1
KUDOS
Q1 & Q2 Already posted by sher676 and LenaA.

For Q3) A certain company employs 6 senior....
Ans : 80
Soln : 6C3 * 4C1 = 20 * 4 = 80.
I see that you have added 20 and 4. However, the question mentions "and" . Hence, we need to multiply.

For Q4) 10 machines...
Ans : 20
Soln : I am getting 20 as well.
10 machines = 1/16th of job in each hour. Each machine working alone will do 1/160th of the job in one hour.
Therefore, for 8 machines, we would multiply 8, to the equation.
1/160 * 8 = 1 / 20 ie 20 hours. If I am wrong here somewhere, please point it out. It will help. What is the OA?

For Q5) What is the average of eleven consecutive...
Ans : D.
Lets take the first integer as n. Then, the question asks for [n+(n+1)+(n+2)...(n+10)]/11. Therefore, if the stmt helps in finding the value of n, the stmt is sufficient.
Stmt 1 : Average of first 9 integers is 7.
That is (9n+36)/9 = 7. Therefore, n = 3. Suff.
Stmt 2 : Average of last 9 integers is 9.
That is (9n+54)/9 = 9. Therefore, n = 3. Suff.

Hope this helps.
_________________

GMAT offended me. Now, its my turn!
Will do anything for Kudos! Please feel free to give one.

Manager
Joined: 13 Jan 2009
Posts: 170
Followers: 4

Kudos [?]: 22 [0], given: 9

### Show Tags

28 Aug 2009, 02:18
bhanushalinikhil wrote:
For Q4) 10 machines...
Ans : 20
Soln : I am getting 20 as well.
10 machines = 1/16th of job in each hour. Each machine working alone will do 1/160th of the job in one hour.
Therefore, for 8 machines, we would multiply 8, to the equation.
1/160 * 8 = 1 / 20 ie 20 hours. If I am wrong here somewhere, please point it out. It will help. What is the OA?

Another good approach without using equations.
10 machines can complete the job in 16 hours
then, 1 machine can complete the job in 160 hours
hence, 8 machines can complete the job 160/8=20hours.
Intern
Joined: 30 Jun 2009
Posts: 48
Followers: 1

Kudos [?]: 9 [0], given: 2

### Show Tags

28 Aug 2009, 06:07
LenaA wrote:
For the first question about LCM, each statement alone is sufficient.
6=3X2
9=3^2

To calculate LCM we take the highest count (power of a certain prime) of each prime factor found across all integers.
$$LCM(6,9) = 3^2\times 2$$

Statement 1:
$$LCM (6,x)=30=3\times 2\times 5$$
x must have only one "5" in its prime factorization.
x must have either no "3" or only one "3"
x must have either no "2" or only one "2"
It can be 5, 10,15 or 30. We do not need to know the exact value of x but the maximum powers of it's primes.
It is sufficient. Across three numbers (6,9,x), the highest power of prime "3" is 2, the highest power of prime "2" is 1 and the highest power of prime "5" is 1. Therefore, $$LCM (x,6,9)=3^2\time 2^1\times 5^1=90$$

Statement 2
$$LCM(x,9)=45=5\times 3\times 3$$
x has only one "5"
x has no or one or two "3".
x can be 5,15 or 45.
Again, it is sufficent. Across three numbers (6,9,x), the highest power of prime "3" is 2, the highest power of prime "2" is 1 and the highest power of prime "5" is 1. $$LCM (x,6,9)=3^2\time 2^1\times 5^1=90$$

HI LenaA,
I did not get your approach. How do you deduct the primes and the powers of prime?
It may sound a stupid question...
Thx
Manager
Joined: 10 Aug 2009
Posts: 130
Followers: 3

Kudos [?]: 62 [0], given: 10

### Show Tags

28 Aug 2009, 08:14
defoue wrote:
LenaA wrote:
For the first question about LCM, each statement alone is sufficient.
6=3X2
9=3^2

To calculate LCM we take the highest count (power of a certain prime) of each prime factor found across all integers.
$$LCM(6,9) = 3^2\times 2$$

Statement 1:
$$LCM (6,x)=30=3\times 2\times 5$$
x must have only one "5" in its prime factorization.
x must have either no "3" or only one "3"
x must have either no "2" or only one "2"
It can be 5, 10,15 or 30. We do not need to know the exact value of x but the maximum powers of it's primes.
It is sufficient. Across three numbers (6,9,x), the highest power of prime "3" is 2, the highest power of prime "2" is 1 and the highest power of prime "5" is 1. Therefore, $$LCM (x,6,9)=3^2\time 2^1\times 5^1=90$$

Statement 2
$$LCM(x,9)=45=5\times 3\times 3$$
x has only one "5"
x has no or one or two "3".
x can be 5,15 or 45.
Again, it is sufficent. Across three numbers (6,9,x), the highest power of prime "3" is 2, the highest power of prime "2" is 1 and the highest power of prime "5" is 1. $$LCM (x,6,9)=3^2\time 2^1\times 5^1=90$$

HI LenaA,
I did not get your approach. How do you deduct the primes and the powers of prime?
It may sound a stupid question...
Thx

Ok, I will try to explain by example.
If you have 3 numbers, ex. 6,9, and 15. What would be their LCM?
First, I write them as a product of primes:
6=3X2
9=3X3
15=5X3
Now look at each prime and count its occurance in each number:
Prime 3 - it happens two times in 9 and only one time in 6 and 15. Therefore, its highest count/power across the numbers is 2. $$3^2$$ will be included in LCM.
Prime 2 - it happens once in number 6 and it does not exist in 9 and 15. Hence its highest count/power is 1. $$2^1$$ will be included in LCM.
Prime 5 - it happens once in 15 and it does not exist in 9 or 6. So its highest count/power is 1. $$5^1$$ will be included in LCM
Based on the analysis above, LCM(6,9,15)=$$3^2\times 2\times 5$$. Again, when finding LCM take only the highest powers of primes (when you find GCF you take the lowest powers, in this example in would be 3$$^1\times 2^0\times 5^0$$)...When you think of the meaning of LCM and GCF, this is very logical.
With application to the specific question, it was enough to prove that x does not have the power of prime 3 greater than 2, x does not have the power of prime 2 greater than 1 and x has only one 5...You can get this info from each statement separately. Hence the answer D.

Hope it helps.
Manager
Joined: 29 Jul 2009
Posts: 176
Followers: 4

Kudos [?]: 63 [0], given: 2

### Show Tags

29 Aug 2009, 07:35
LenaA wrote:
For the first question about LCM, each statement alone is sufficient.
6=3X2
9=3^2

To calculate LCM we take the highest count (power of a certain prime) of each prime factor found across all integers.
$$LCM(6,9) = 3^2\times 2$$

Statement 1:
$$LCM (6,x)=30=3\times 2\times 5$$
x must have only one "5" in its prime factorization.
x must have either no "3" or only one "3"
x must have either no "2" or only one "2"
It can be 5, 10,15 or 30. We do not need to know the exact value of x but the maximum powers of it's primes.
It is sufficient. Across three numbers (6,9,x), the highest power of prime "3" is 2, the highest power of prime "2" is 1 and the highest power of prime "5" is 1. Therefore, $$LCM (x,6,9)=3^2\time 2^1\times 5^1=90$$

Statement 2
$$LCM(x,9)=45=5\times 3\times 3$$
x has only one "5"
x has no or one or two "3".
x can be 5,15 or 45.
Again, it is sufficent. Across three numbers (6,9,x), the highest power of prime "3" is 2, the highest power of prime "2" is 1 and the highest power of prime "5" is 1. $$LCM (x,6,9)=3^2\time 2^1\times 5^1=90$$

thank you....

Last edited by MontrealLady on 29 Aug 2009, 07:47, edited 1 time in total.
Manager
Joined: 29 Jul 2009
Posts: 176
Followers: 4

Kudos [?]: 63 [0], given: 2

### Show Tags

29 Aug 2009, 08:00
bhanushalinikhil wrote:
Q1 & Q2 Already posted by sher676 nd LenaA.

For Q3) A certain company employs 6 senior....
Ans : 80
Soln : 6C3 * 4C1 = 20 * 4 = 80.

can you please provide more explanation? why did you multiply 6*3 and then the total you multiplied by 20*4?
I see that you have added 20 and 4. However, the question mentions "and" . Hence, we need to multiply.

For Q4) 10 machines...
Ans : 20
Soln : I am getting 20 as well.
10 machines = 1/16th of job in each hour. Each machine working alone will do 1/160th of the job in one hour.
Therefore, for 8 machines, we would multiply 8, to the equation.
1/160 * 8 = 1 / 20 ie 20 hours. If I am wrong here somewhere, please point it out. It will help. What is the OA?

the correct answer is 20 thank you

For Q5) What is the average of eleven consecutive...
Ans : D.
Lets take the first integer as n. Then, the question asks for [n+(n+1)+(n+2)...(n+10)]/11. Therefore, if the stmt helps in finding the value of n, the stmt is sufficient.
Stmt 1 : Average of first 9 integers is 7.
That is (9n+36)/9 = 7. Therefore, n = 3. Suff.
Stmt 2 : Average of last 9 integers is 9.
That is (9n+54)/9 = 9. Therefore, n = 3. Suff.

Hope this helps.

OMG...that's easy!!!
Manager
Joined: 10 Aug 2009
Posts: 130
Followers: 3

Kudos [?]: 62 [0], given: 10

### Show Tags

29 Aug 2009, 08:30
LenaA wrote:

I define hypotenuse as B and other side as A.

1) $$2A+B=16+16\times \sqrt{2}$$
2) $$2A^2=B^2$$

$$A=\frac{B}{\sqrt{2}}$$
$$\frac{2B}{\sqrt{2}}+B=16+16\times \sqrt{2}$$
$$B\times (1+\sqrt{2})=16\times (1+\sqrt{2})$$

you took B as a common factor, so you should be left by 2/sqrt{2}+1

$$B=16$$

Hi,

what I did is correct for the reason that $$\frac{2}{\sqrt{2}}=\sqrt{2}$$
Manager
Joined: 10 Aug 2009
Posts: 130
Followers: 3

Kudos [?]: 62 [0], given: 10

### Show Tags

29 Aug 2009, 08:58

$$6C3\neq6 \times 3$$
$$6C3=\frac{6!}{(6-3)!\times 3!}=\frac{6!}{3!\times 3!}$$=20

The committee consists of 1 junior memeber and you have 4 junior memembers => so you have 4 options to choose a junior member or 4C1 options

The committee consists of 3 senior members and you have 6 senior member => you have to choose DIFFERENT sets of three people (ex. ABC and CBA will be considered the same set for this problem) since order does not important in this committee. So you have 6C3 options
In total you have $$4C1\times 6C3$$ options or 80 options in total.

Note if the senior officers would be arranged with some hierarchy (order does matter) then your options would be$$6\times 5\times 4$$ for choosing seniors and 4 options for choosing junior
and in total $$6\times 5\times4\times 4$$
Manager
Joined: 29 Jul 2009
Posts: 176
Followers: 4

Kudos [?]: 63 [0], given: 2

### Show Tags

02 Sep 2009, 18:46
LenaA wrote:
LenaA wrote:

I define hypotenuse as B and other side as A.

1) $$2A+B=16+16\times \sqrt{2}$$
2) $$2A^2=B^2$$

$$A=\frac{B}{\sqrt{2}}$$
$$\frac{2B}{\sqrt{2}}+B=16+16\times \sqrt{2}$$
$$B\times (1+\sqrt{2})=16\times (1+\sqrt{2})$$

you took B as a common factor, so you should be left by 2/sqrt{2}+1

$$B=16$$

Hi,

what I did is correct for the reason that $$\frac{2}{\sqrt{2}}=\sqrt{2}$$

Lena,

this is something new for me...can you please explain more?
what if it's 20/sqrt{5}....does it equal to 4?

I know that for example
2sqrt{2}*5sqrt{3} = 10sqrt{6}
3sqrt{2}+8sqrt{5}= impossible because the square roots are not the same
sqrt{10}/sqrt{5} = sqrt{2}

is what I said correct?
thanks,
Manager
Joined: 29 Jul 2009
Posts: 176
Followers: 4

Kudos [?]: 63 [0], given: 2

### Show Tags

02 Sep 2009, 18:58
LenaA wrote:

$$6C3\neq6 \times 3$$
$$6C3=\frac{6!}{(6-3)!\times 3!}=\frac{6!}{3!\times 3!}$$=20

The committee consists of 1 junior memeber and you have 4 junior memembers => so you have 4 options to choose a junior member or 4C1 options

The committee consists of 3 senior members and you have 6 senior member => you have to choose DIFFERENT sets of three people (ex. ABC and CBA will be considered the same set for this problem) since order does not important in this committee. So you have 6C3 options
In total you have $$4C1\times 6C3$$ options or 80 options in total.

Note if the senior officers would be arranged with some hierarchy (order does matter) then your options would be$$6\times 5\times 4$$ for choosing seniors and 4 options for choosing junior
and in total $$6\times 5\times4\times 4$$

is that similar to the following
for example, we have 4 meals and three meals together make a course meal. how many different course meals can we have?

4!/(4-3)!3! = 24/1!*3! = 4 different course meals
or do I simply do 4*3 = 12 course meals

which way is correct?

Manager
Joined: 10 Aug 2009
Posts: 130
Followers: 3

Kudos [?]: 62 [0], given: 10

### Show Tags

02 Sep 2009, 19:26
Think of 2 as $$\sqrt{2}\times \sqrt{2}$$
then you have $$\frac{\sqrt{2}\times \sqrt{2}}{\sqrt{2}}$$

or another way of thinking $$2^1\div 2^{\frac{1}{2}}=2^{1-\frac{1}{2}}$$

or just multiply numerator and denominator by $$\sqrt{2}$$:

$$\frac{2}{\sqrt{2}}\times \frac{\sqrt{2}}{sqrt{2}}$$
Manager
Joined: 10 Aug 2009
Posts: 130
Followers: 3

Kudos [?]: 62 [0], given: 10

### Show Tags

02 Sep 2009, 19:42
I did not get your example with meals.
But I will give you mine...also food related...

You are going to have a dessert and you have a selction of 4 chocolate cakes and 3 cheese cakes. You are going to eat 2 types of cake...how many desserts options can you have?
1) you are going to have one chocolate and one cheese cake then you have $$4\times 3=12$$ dessert options
2) there is no limitation you can have 2 chocolate or 2 cheese or one chocolate and one cheese...
then you have 7 types of cake and you can have $$7C2=\frac{7!}{5!2!}$$dessert options...

Last edited by LenaA on 03 Sep 2009, 04:36, edited 1 time in total.
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 65

Kudos [?]: 669 [0], given: 19

### Show Tags

02 Sep 2009, 20:02
For LCM Question:

1: If x and 6 has LCM of 30, then take only LCM between 30 and 9. Thats 90. Suff..
No need to ahve the value of x, however x could be 5, 15 or 30.

2: If x and 9 has LCM of 45, then take LCM only between 30 and 6. Thats 90. Suff..
No need to have the value of x, however x could be 5, 15 or 45.
Attachments

Q-LCM.jpg [ 15.66 KiB | Viewed 733 times ]

_________________
Re: 4 GMATPrep Qs   [#permalink] 02 Sep 2009, 20:02

Go to page    1   2    Next  [ 23 posts ]

Display posts from previous: Sort by