There is one important detail worth noting about this DS prompt: The prompt does NOT state that N is an integer.

Hi usre123,

Your deduction that N is less than or equal to 18 is incomplete (for example, N CANNOT be 10). There is a "lower boundary" for what N can equal.

GMAT assassins aren't born, they're made,

Rich

]]>

There is one important detail worth noting about this DS prompt: The prompt does NOT state that N is an integer.

Hi usre123,

Your deduction that N is less than or equal to 18 is incomplete (for example, N CANNOT be 10). There is a "lower boundary" for what N can equal.

GMAT assassins aren't born, they're made,

Rich]]>

Cheers !!!

]]>

Cheers !!! ]]>

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Total = Group1 + Group2 - Both + Neither

Group1 = Books that are hardcover

Group2 = Books that are fiction

40 = 30 + 20 - Both + Neither

Both + 40 = 50 + Neither

Both = 10 + Neither

We know that the highest value for "Neither" is 10, since it is given that 30 of the 40 books are hardcover.

Both = 10 + 10 = 20.

C

]]>

Total = Group1 + Group2 - Both + Neither

Group1 = Books that are hardcover

Group2 = Books that are fiction

40 = 30 + 20 - Both + Neither

Both + 40 = 50 + Neither

Both = 10 + Neither

We know that the highest value for "Neither" is 10, since it is given that 30 of the 40 books are hardcover.

Both = 10 + 10 = 20.

C]]>

gmacforjyoab wrote:

A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could

order working alone in(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could

...

]]>

gmacforjyoab wrote:

A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could

order working alone in(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could

...]]>

no height in statement A, therefore it is useless.

looking at 2, imagine the cylinder having a right angled triangle with in it. we have the hypotenuse, 13, and only one pair of 5, 12 satisfies this. (we know height is greater than base)

]]>

no height in statement A, therefore it is useless.

looking at 2, imagine the cylinder having a right angled triangle with in it. we have the hypotenuse, 13, and only one pair of 5, 12 satisfies this. (we know height is greater than base)]]>

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As problem says h1+h2=x & h1=h2+37,

We replace these two relations in the equation and we have h2+h2+37=x OR : 2h2+37=x and after adding z to the x , we have 2h2+37=x

Or : 2h2= x-37 and finally we get :h2= (x-37)/2+z

so the shorter h2 is (x-37)/2 +z answer c

...

]]>

As problem says h1+h2=x & h1=h2+37,

We replace these two relations in the equation and we have h2+h2+37=x OR : 2h2+37=x and after adding z to the x , we have 2h2+37=x

Or : 2h2= x-37 and finally we get :h2= (x-37)/2+z

so the shorter h2 is (x-37)/2 +z answer c

...]]>

What is the sum a + b?

A. 260

B. 270

C. 280

D. 290

E. 300

Kudos for a correct solution.

The OA will be revealed on Sunday

ans is straight forward ..we have to find a+b.. now a = angle kyz+angle kzy

so a+b=angle kyz+angle kzy+b=angle kyz+180=90 + 180= 270

]]>

What is the sum a + b?

A. 260

B. 270

C. 280

D. 290

E. 300

Kudos for a correct solution.

The OA will be revealed on Sunday

ans is straight forward ..we have to find a+b.. now a = angle kyz+angle kzy

so a+b=angle kyz+angle kzy+b=angle kyz+180=90 + 180= 270]]>

CORRECT ANSWER D

(1) is sufficient as if we know a2 and a1 we can find a3 (a3=(a1+a2)/s), so on and so forth up to a7 = 10 as well

(2) is sufficient as we know a3 and a1 and therefore we can find a2 and aftewards all answers up to a7 and a7 itself, a7 =10

Therefore both affirmations are sufficient, CORRECT ANSWER D

Bunuel wrote:

In the sequence a1, a2, a3, …, an, an is determined for all values of n>2 by taking the average of all termsa_1 througha_{n-1} . Ifa_1=1 , what

...

]]>

CORRECT ANSWER D

(1) is sufficient as if we know a2 and a1 we can find a3 (a3=(a1+a2)/s), so on and so forth up to a7 = 10 as well

(2) is sufficient as we know a3 and a1 and therefore we can find a2 and aftewards all answers up to a7 and a7 itself, a7 =10

Therefore both affirmations are sufficient, CORRECT ANSWER D

Bunuel wrote:

In the sequence a1, a2, a3, …, an, an is determined for all values of n>2 by taking the average of all termsa_1 througha_{n-1} . Ifa_1=1 , what

...]]>

length of new land =3*2*underrroot2..

a=underrroot2*3*2*underrroot2=24

]]>

length of new land =3*2*underrroot2..

a=underrroot2*3*2*underrroot2=24]]>

to be div by 4 , last two digits have to div by 4 so it becomes 44..

finally to be div by 3, sum shud be div by 3.. so it becomes 444...

]]>

to be div by 4 , last two digits have to div by 4 so it becomes 44..

finally to be div by 3, sum shud be div by 3.. so it becomes 444...]]>

CORRECT ANSWER D. I and II only

Since we have only a few possibilites, then we don't need to list all prime numbers and do any division but only plug in the numbers

The contestant won 31 prizes, which is fixed.

I. is of course true as 31 is prime and for a prime number he would won the same number of prizes

II. is also true as for a non prime number such as 24 he would win 7 prizes more: 24+7 = 31

III. is not true as 38+7 =45 which is not the number of prizes

...

]]>

CORRECT ANSWER D. I and II only

Since we have only a few possibilites, then we don't need to list all prime numbers and do any division but only plug in the numbers

The contestant won 31 prizes, which is fixed.

I. is of course true as 31 is prime and for a prime number he would won the same number of prizes

II. is also true as for a non prime number such as 24 he would win 7 prizes more: 24+7 = 31

III. is not true as 38+7 =45 which is not the number of prizes

...]]>

What is the remainder when 10^49 + 2 is divided by 11?

A. 1

B. 2

C. 3

D. 5

E. 7

Kudos for a correct solution.

The OA will be revealed on Sunday

the property of 11 as divisor is (sum of odd digits -sum of even digits)=0 starting from rightmost digit..

this means for 10^even number, (sum of odd digits -sum of even digits) =1-0=1 so 1 is the remainder...

and for 10^odd number, (sum of odd digits -sum of even digits) =0-1=-1 so 10 is the remainder...

therefore 10^49 will have a

...

]]>

What is the remainder when 10^49 + 2 is divided by 11?

A. 1

B. 2

C. 3

D. 5

E. 7

Kudos for a correct solution.

The OA will be revealed on Sunday

the property of 11 as divisor is (sum of odd digits -sum of even digits)=0 starting from rightmost digit..

this means for 10^even number, (sum of odd digits -sum of even digits) =1-0=1 so 1 is the remainder...

and for 10^odd number, (sum of odd digits -sum of even digits) =0-1=-1 so 10 is the remainder...

therefore 10^49 will have a

...]]>

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CORRECT ANSWER B. 10 and 1/10 pounds, explanation:

It's a weighted averageproblem:

[3 x (11 and 2/3) + 7 x (9 and 3/7) ] 10 (to find the average package weight)

= (3 x 35/3 + 7 x 66/7) / 10 = (35 + 66) / 10 = 10 and 1/10 pounds

Bunuel wrote:

An ice cream store received shipments of ingredients on Tuesday and Wednesday. Tuesday's shipment had 3 different packages, with an average (arithmetic mean) of 11 and 2/3 pounds. Wednesday's shipment included 7 packages, weighing an average

...

]]>

CORRECT ANSWER B. 10 and 1/10 pounds, explanation:

It's a weighted averageproblem:

[3 x (11 and 2/3) + 7 x (9 and 3/7) ] 10 (to find the average package weight)

= (3 x 35/3 + 7 x 66/7) / 10 = (35 + 66) / 10 = 10 and 1/10 pounds

Bunuel wrote:

An ice cream store received shipments of ingredients on Tuesday and Wednesday. Tuesday's shipment had 3 different packages, with an average (arithmetic mean) of 11 and 2/3 pounds. Wednesday's shipment included 7 packages, weighing an average

...]]>

PS441: The 441 problem solving questions are from the Official Guide GMAT 10th Edition.

GMAT_TEST__1__Full Length with Solutions.pdf: Quant sections copied from GMAT Paper Tests.

GMAT 800plus Questions set for 700plus score.pdf: Many of the quant questions are duplicate of the questions from the GMATPrep Practice Test database. Please don't do these prior to taking the GMATPrep CATs, otherwise your score would not be representative of your true performance.

...

]]>

PS441: The 441 problem solving questions are from the Official Guide GMAT 10th Edition.

GMAT_TEST__1__Full Length with Solutions.pdf: Quant sections copied from GMAT Paper Tests.

GMAT 800plus Questions set for 700plus score.pdf: Many of the quant questions are duplicate of the questions from the GMATPrep Practice Test database. Please don't do these prior to taking the GMATPrep CATs, otherwise your score would not be representative of your true performance.

...]]>

Wont 2(a+b-c) result into an even integer irrespective of the statements below? Thus, making either statement sufficient.

No. For example, if a+b-c=1/2, then 2(a+b-c)=1, or if a+b-c=1/3, then 2(a+b-c)=2/3, not an integer at all.

]]>

Wont 2(a+b-c) result into an even integer irrespective of the statements below? Thus, making either statement sufficient.

No. For example, if a+b-c=1/2, then 2(a+b-c)=1, or if a+b-c=1/3, then 2(a+b-c)=2/3, not an integer at all.]]>

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Is x^2*y^4 an integer divisible by 9 ?

(1) x is an integer divisible by 3

(2) xy is an integer divisible by 9

[Reveal] Spoiler:

Statements (1) and (2) TOGETHER are NOT sufficient

Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 andx^2 = 9 which is divisible by 9?

Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 andx^2 = 9 which is divisible by 9?

Dear Bunuel, for statement 2 I chose x=1/2 and y = 18 and the result

...

]]>

Is x^2*y^4 an integer divisible by 9 ?

(1) x is an integer divisible by 3

(2) xy is an integer divisible by 9

[Reveal] Spoiler:

Statements (1) and (2) TOGETHER are NOT sufficient

Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 andx^2 = 9 which is divisible by 9?

Statements (1) and (2) combined are insufficient. If and then both S1 and S2 hold, but is not divisible by 9.

Why is s1 not sufficient. If x is divisible by 3 then the minimum value of x will be 3 andx^2 = 9 which is divisible by 9?

Dear Bunuel, for statement 2 I chose x=1/2 and y = 18 and the result

...]]>

1) No details are given about b; hence it is insufficient

2) b=3a -> b^4=81a^4.

The question a^2<b^4? will become ->a^2 < 81a^4 -> as a^2 is positive -> 1<81a^2 ? since we do not know anything more about a, specifically, whether its absolute value is greater than 1, statement 2 alone is insufficient

1+2) As a<-1, its absolute value is greater than 1. Along with the statement 2, with a^2>1 , the question can be answered.

Answer C

...

]]>

1) No details are given about b; hence it is insufficient

2) b=3a -> b^4=81a^4.

The question a^2<b^4? will become ->a^2 < 81a^4 -> as a^2 is positive -> 1<81a^2 ? since we do not know anything more about a, specifically, whether its absolute value is greater than 1, statement 2 alone is insufficient

1+2) As a<-1, its absolute value is greater than 1. Along with the statement 2, with a^2>1 , the question can be answered.

Answer C

...]]>

One question,

I thought that when 2 objects are moving towards each other we subtract the rates, as they cover this distance. When they more away from each other we add the rates because they create the distance. In this case, why did we add the rates?

Oh, don't worry, I just understood what I said and realised where I am wrong.

So, when 2 objects move towards thesame direction we subtract the rates.

However, when they movetowards each other oraway from each other we add the rates.

...

]]>

One question,

I thought that when 2 objects are moving towards each other we subtract the rates, as they cover this distance. When they more away from each other we add the rates because they create the distance. In this case, why did we add the rates?

Oh, don't worry, I just understood what I said and realised where I am wrong.

So, when 2 objects move towards thesame direction we subtract the rates.

However, when they movetowards each other oraway from each other we add the rates.

...]]>

_______R_____T______D

A.........90........3..........270

B.........54.......15.........810

All........60.......18........1080

Let me explain how we fill in the chart:

1) Starting with what we know, we add 90 and 60

2) Picking an easy number for the total distance (6*9=54), so I chose 540. Multiply by 2 to get the whole trip, and we get 1080. Add 1080 for all-distance.

3) 1/4 of the

...

]]>

_______R_____T______D

A.........90........3..........270

B.........54.......15.........810

All........60.......18........1080

Let me explain how we fill in the chart:

1) Starting with what we know, we add 90 and 60

2) Picking an easy number for the total distance (6*9=54), so I chose 540. Multiply by 2 to get the whole trip, and we get 1080. Add 1080 for all-distance.

3) 1/4 of the

...]]>

Hi Bunuel - can you explain why ab^(-1)=6^(-1) yields b/a = 6?

a*b^{(-1)}=6^{(-1)};

a*\frac{1}{b}=\frac{1}{6};

\frac{a}{b}=\frac{1}{6};

\frac{b}{a}=6.

Hope it's clear.

]]>

Hi Bunuel - can you explain why ab^(-1)=6^(-1) yields b/a = 6?

a*b^{(-1)}=6^{(-1)};

a*\frac{1}{b}=\frac{1}{6};

\frac{a}{b}=\frac{1}{6};

\frac{b}{a}=6.

Hope it's clear.]]>

1)case 1 when all five colours are different-7c5=21..

2)case 2 when three different colours and 2 of same type=6c3*2=30..

3)case 3 when two different colours and 3 of same type=6c2=15..

4)case 4 when one different colours and 2 of two different colors=5c1*4!/2!2!=15...

5)case 4 when 3 of one type and 2 of other colour=1...

total-82 ans E

]]>

1)case 1 when all five colours are different-7c5=21..

2)case 2 when three different colours and 2 of same type=6c3*2=30..

3)case 3 when two different colours and 3 of same type=6c2=15..

4)case 4 when one different colours and 2 of two different colors=5c1*4!/2!2!=15...

5)case 4 when 3 of one type and 2 of other colour=1...

total-82 ans E]]>

Distance = Speed * Time

= xt + xt = 2xt

Answer = C

Hi Paresh,

I also chose this answer. However, I do have one question:

Why is the rate x and not xt?

Because I thought that since the boat is doing x miles in one hour, we need to find how many miles it will do in t hours. So, it would do xt miles in t hours, making the rate xt. This ends in an answer 2xt^2. This doesn't exist in the answer choices and made me think that my rate must have an extra x which is caused the square, so I chose the

...

]]>

Distance = Speed * Time

= xt + xt = 2xt

Answer = C

Hi Paresh,

I also chose this answer. However, I do have one question:

Why is the rate x and not xt?

Because I thought that since the boat is doing x miles in one hour, we need to find how many miles it will do in t hours. So, it would do xt miles in t hours, making the rate xt. This ends in an answer 2xt^2. This doesn't exist in the answer choices and made me think that my rate must have an extra x which is caused the square, so I chose the

...]]>

Bunuel wrote:

KalEl wrote:

Ben and Sam set out together on a bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, Ben stops to fix a flat tire. If it takes Ben one hour to fix the flat tire and Sam continues to ride during this time, how many hours will it take Ben to catch up Sam and assuming he resumes his ride at 15 miles per hour?

A. 3

B. 3.33

C. 3.5

D. 4

E. 4.5

Don't forget Kudos if you like the question

A. 3

B. 3.33

C. 3.5

D. 4

E. 4.5

Don't forget Kudos if you like the question

...

]]>

Bunuel wrote:

KalEl wrote:

Ben and Sam set out together on a bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, Ben stops to fix a flat tire. If it takes Ben one hour to fix the flat tire and Sam continues to ride during this time, how many hours will it take Ben to catch up Sam and assuming he resumes his ride at 15 miles per hour?

A. 3

B. 3.33

C. 3.5

D. 4

E. 4.5

Don't forget Kudos if you like the question

A. 3

B. 3.33

C. 3.5

D. 4

E. 4.5

Don't forget Kudos if you like the question

...]]>

=26*28... clearly 13 is the ans

]]>

=26*28... clearly 13 is the ans]]>

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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]]>

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max value of xy - 19602

only E is greater

]]>

max value of xy - 19602

only E is greater]]>

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the solutions posted thus far are good but lots of computation, an area where i tend to make mistakes. here is how i broke it down.

6 checks are at 750. 20 checks are at 780.

difference between these checks is 30.

20 checks * 30 = 600 (easy math)

when removing 600 from the latter checks, all checks are now 'evenly filled' at 750, as if they were glasses filled with 750ml of water, for example.

what's an even way to distribute the 600 among all of the 26 checks? ==> 600/26, but this

...

]]>

the solutions posted thus far are good but lots of computation, an area where i tend to make mistakes. here is how i broke it down.

6 checks are at 750. 20 checks are at 780.

difference between these checks is 30.

20 checks * 30 = 600 (easy math)

when removing 600 from the latter checks, all checks are now 'evenly filled' at 750, as if they were glasses filled with 750ml of water, for example.

what's an even way to distribute the 600 among all of the 26 checks? ==> 600/26, but this

...]]>

samichange wrote:

I think the options to the above questions fall short of the correct answer

Here is my solution-

A- 2-------10^0 series-----------------1 (0+1)

B- 11,20------------10^1 series----------------------2 (1+1)

C-101,110,200---------10^2 series----------------------------3 (2+1)

D-1001,1010,1100,2000-------10^3 series--------------------------4 (3+1)

E-10001,10010,10100,11000,20000--------10^4 series-------------------------5 (4+1)

and so on......................................................10^21

Here is my solution-

A- 2-------10^0 series-----------------1 (0+1)

B- 11,20------------10^1 series----------------------2 (1+1)

C-101,110,200---------10^2 series----------------------------3 (2+1)

D-1001,1010,1100,2000-------10^3 series--------------------------4 (3+1)

E-10001,10010,10100,11000,20000--------10^4 series-------------------------5 (4+1)

and so on......................................................10^21

...

]]>

samichange wrote:

I think the options to the above questions fall short of the correct answer

Here is my solution-

A- 2-------10^0 series-----------------1 (0+1)

B- 11,20------------10^1 series----------------------2 (1+1)

C-101,110,200---------10^2 series----------------------------3 (2+1)

D-1001,1010,1100,2000-------10^3 series--------------------------4 (3+1)

E-10001,10010,10100,11000,20000--------10^4 series-------------------------5 (4+1)

and so on......................................................10^21

Here is my solution-

A- 2-------10^0 series-----------------1 (0+1)

B- 11,20------------10^1 series----------------------2 (1+1)

C-101,110,200---------10^2 series----------------------------3 (2+1)

D-1001,1010,1100,2000-------10^3 series--------------------------4 (3+1)

E-10001,10010,10100,11000,20000--------10^4 series-------------------------5 (4+1)

and so on......................................................10^21

...]]>

Given that both roots are prime number, it means roots cannot be even EXCEPT for 2, which is a prime number as well as a even number

In this case, only possible value of x = 2*61 = 122

]]>

Given that both roots are prime number, it means roots cannot be even EXCEPT for 2, which is a prime number as well as a even number

In this case, only possible value of x = 2*61 = 122]]>

If a=1/2, then 1/8 < 1/4+1 --->No

(2) Insufficient. If a=-5, then 25>-125 --->Yes

If a=1/2, then 1/4>1/8 --->No

(1)+(2) Insufficient. We can use the same numbers and we get two different answers

]]>

If a=1/2, then 1/8 < 1/4+1 --->No

(2) Insufficient. If a=-5, then 25>-125 --->Yes

If a=1/2, then 1/4>1/8 --->No

(1)+(2) Insufficient. We can use the same numbers and we get two different answers]]>

a is a positive integer (given)

St1: a and (a+1) each have exactly 6 factors.

No limit defined on the value of a as a can have any value between 1 and infinity.

Not sufficient

St2: a < 76

a can have any value between 1 and 76

Not sufficient

Combining :

a and (a+1) each have exactly 6 factors and a < 76.

key here is : a and (a+1) are co-prime integer.

Consider p, q, r, and s be 4 different prime numbers.

for number of factors to be 6

a = p^2 * q

(a+1) = r^2 * s

as

...

]]>

a is a positive integer (given)

St1: a and (a+1) each have exactly 6 factors.

No limit defined on the value of a as a can have any value between 1 and infinity.

Not sufficient

St2: a < 76

a can have any value between 1 and 76

Not sufficient

Combining :

a and (a+1) each have exactly 6 factors and a < 76.

key here is : a and (a+1) are co-prime integer.

Consider p, q, r, and s be 4 different prime numbers.

for number of factors to be 6

a = p^2 * q

(a+1) = r^2 * s

as

...]]>

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]]>

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On Monday morning, Chris receives tickets to a baseball game that will be played at 7pm on the next evening that it does not rain. However, Chris is only in town until Wednesday morning, at which point he must fly to another city. If there is a 40% chance of rain each of the next two evenings, what is the probability that Chris will be able to attend the game?

(A) 36%

(B) 60%

(C) 66%

(D) 80%

(E) 84%

This is more of an English question than math :D

...

]]>

On Monday morning, Chris receives tickets to a baseball game that will be played at 7pm on the next evening that it does not rain. However, Chris is only in town until Wednesday morning, at which point he must fly to another city. If there is a 40% chance of rain each of the next two evenings, what is the probability that Chris will be able to attend the game?

(A) 36%

(B) 60%

(C) 66%

(D) 80%

(E) 84%

This is more of an English question than math :D

...]]>

The area of a square garden is A square feet and the perimeter is p feet. If a=2p+9, what is the perimeter of the garden, in feet?

A. 28

B. 36

C. 40

D. 56

E. 64

Please edit question to read A= 2p+9. a is different from A and I spent 10 minutes scratching my head over this typo.

]]>

The area of a square garden is A square feet and the perimeter is p feet. If a=2p+9, what is the perimeter of the garden, in feet?

A. 28

B. 36

C. 40

D. 56

E. 64

Please edit question to read A= 2p+9. a is different from A and I spent 10 minutes scratching my head over this typo.]]>

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]]>

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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These types of Average Speed questions require a certain amount of organization. While you have the freedom to approach this Test in any way you choose, having a "default" way to create tables will help you to set up questions quicker over the course of the entire Exam.

Here's one way to organize the work in thisquestion:

Attachment:

GCDrtBox.png [ 462.05 KiB | Viewed 44 times ]

Note that the question asks for the Average Speed on DAY 2, so we have one more step:

R = 30

R - 10 = 20

Final Answer:

...

]]>

These types of Average Speed questions require a certain amount of organization. While you have the freedom to approach this Test in any way you choose, having a "default" way to create tables will help you to set up questions quicker over the course of the entire Exam.

Here's one way to organize the work in thisquestion:

Attachment:

GCDrtBox.png [ 462.05 KiB | Viewed 44 times ]

Note that the question asks for the Average Speed on DAY 2, so we have one more step:

R = 30

R - 10 = 20

Final Answer:

...]]>

]]>

For me, the easiest way to solve this question is as under:

7!*8*9*10 - 2(5!)^2

7! = 5040 & 5! = 120; Square of 120 = 12^2 *100 = 144*100=14400

Thus 5040*72*10 - 14400*2

As 504*72 = 36288;

36288*100 - 28800

3628800 - 28800 = 3600000 = 36*100000 = 36*10^5

Therefore n=5

Good ole' fashion brute force

Mind as well simplify it bit:

10! - 2*(5!)^2

5!(6*7*8*9*10-2(5!))

5!(30240-240)

5!(30000) ----> With the 5 and 2 from

...

]]>

For me, the easiest way to solve this question is as under:

7!*8*9*10 - 2(5!)^2

7! = 5040 & 5! = 120; Square of 120 = 12^2 *100 = 144*100=14400

Thus 5040*72*10 - 14400*2

As 504*72 = 36288;

36288*100 - 28800

3628800 - 28800 = 3600000 = 36*100000 = 36*10^5

Therefore n=5

Good ole' fashion brute force

Mind as well simplify it bit:

10! - 2*(5!)^2

5!(6*7*8*9*10-2(5!))

5!(30240-240)

5!(30000) ----> With the 5 and 2 from

...]]>

Look at each column starting from last.

The difference between each pair of numbers (beyond the first number) results same.

For example, in the last column:

-14+21 = 7. -28+35 = 7. -42+49 = 7. The first entry of that column is 7.

So we have 4*7 = 28 as sum of the last column.

Similarly, sum of preceding column = 4*6 = 24

Ones before that results in:

4*5

4*4

4*3

4*2

4*1

So result is 4* (1+2+3+4+5+6+7) = 112.

]]>

Look at each column starting from last.

The difference between each pair of numbers (beyond the first number) results same.

For example, in the last column:

-14+21 = 7. -28+35 = 7. -42+49 = 7. The first entry of that column is 7.

So we have 4*7 = 28 as sum of the last column.

Similarly, sum of preceding column = 4*6 = 24

Ones before that results in:

4*5

4*4

4*3

4*2

4*1

So result is 4* (1+2+3+4+5+6+7) = 112.]]>

Could you explain why you chose to take

The captain and the cook together have 2 heads and 3 legs

]]>

Could you explain why you chose to take

The captain and the cook together have 2 heads and 3 legs

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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]]>

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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This question can be solved by TESTing VALUES.

This prompt tells us some specific changes that occur to the price of a show ticket if you buy the ticket on the day of the show:

1) If the regular price is LESS than $10, then the ticket sells for HALF the price + $0.50

2) If the regular price is GREATER than/equal to $10, then the ticket sells for HALF the price + $1.00

We're told that someone purchases Y tickets on the day of the show; X of the tickets normally sell for $9, while the

...

]]>

This question can be solved by TESTing VALUES.

This prompt tells us some specific changes that occur to the price of a show ticket if you buy the ticket on the day of the show:

1) If the regular price is LESS than $10, then the ticket sells for HALF the price + $0.50

2) If the regular price is GREATER than/equal to $10, then the ticket sells for HALF the price + $1.00

We're told that someone purchases Y tickets on the day of the show; X of the tickets normally sell for $9, while the

...]]>