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The average (arithmetic mean) of four numbers is equal to three times the largest number. If the largest number is equal to 3, what is the sum of the other three numbers?

A. 28.

B. 33.

C. 35.

D. 38.

E. 42.

( a + b + c + 3 )/4 = 9

So, ( a + b + c + 3 ) = 36

Or, ( a + b + c ) = 33

Hence correct answer will be (B)

]]>

The average (arithmetic mean) of four numbers is equal to three times the largest number. If the largest number is equal to 3, what is the sum of the other three numbers?

A. 28.

B. 33.

C. 35.

D. 38.

E. 42.

( a + b + c + 3 )/4 = 9

So, ( a + b + c + 3 ) = 36

Or, ( a + b + c ) = 33

Hence correct answer will be (B) ]]>

What is the reciprocal of\(\frac{(AB)}{(A + B)^2}\) ?

A.\(\frac{(AB)}{(A^2 + 2AB + B^2)}\) .

B.\(\frac{A}{B} + \frac{B}{A} + AB\) .

C.\(\frac{B}{A} + \frac{A}{B} + 2\) .

D.\(\frac{(A^2 + B^2)}{AB}\) .

E.\(\frac{A}{B} + 2AB\) .

...

]]>

What is the reciprocal of\(\frac{(AB)}{(A + B)^2}\) ?

A.\(\frac{(AB)}{(A^2 + 2AB + B^2)}\) .

B.\(\frac{A}{B} + \frac{B}{A} + AB\) .

C.\(\frac{B}{A} + \frac{A}{B} + 2\) .

D.\(\frac{(A^2 + B^2)}{AB}\) .

E.\(\frac{A}{B} + 2AB\) .

...]]>

The mall charges 50 cents for the first hour of parking and $3 for each additional hour until the customer reaches 4 hours, after that the parking fee is $1 per hour. If a certain customer parked his in the mall for 7 hours and 30 minutes, how much is he going to pay?

A. $11.5.

B. $12.

C. $13.

D. $14.5.

E. $15.

Charges for 7 Hours = ( First hour @ $ 0.50 ) + ( 3 hours @ $ 3 ) + ( 3.5 hours @ $ 1 )

Charges for 7 Hours = ( 1 @ $ 0.50 ) + ( 3 hours @ $ 3 ) + ( 3.5 hours @ $ 1 )

Charges for 7 Hours

...

]]>

The mall charges 50 cents for the first hour of parking and $3 for each additional hour until the customer reaches 4 hours, after that the parking fee is $1 per hour. If a certain customer parked his in the mall for 7 hours and 30 minutes, how much is he going to pay?

A. $11.5.

B. $12.

C. $13.

D. $14.5.

E. $15.

Charges for 7 Hours = ( First hour @ $ 0.50 ) + ( 3 hours @ $ 3 ) + ( 3.5 hours @ $ 1 )

Charges for 7 Hours = ( 1 @ $ 0.50 ) + ( 3 hours @ $ 3 ) + ( 3.5 hours @ $ 1 )

Charges for 7 Hours

...]]>

In the Biotechnology class of 2000, there were X graduates. 32 of the graduates found a job, 45 continued on to their second degree and 13 did both. If only 9 people didn't do both, What is X equal to?

A. 69.

B. 71.

C. 73.

D. 75.

E. 76.

Attachment:

Set.png [ 7.75 KiB | Viewed 18 times ]

Total no of Graduates = ( 32 + 47 - 15 ) + 9 => 73

Hence correct answer must be (C)

...

]]>

In the Biotechnology class of 2000, there were X graduates. 32 of the graduates found a job, 45 continued on to their second degree and 13 did both. If only 9 people didn't do both, What is X equal to?

A. 69.

B. 71.

C. 73.

D. 75.

E. 76.

Attachment:

Set.png [ 7.75 KiB | Viewed 18 times ]

Total no of Graduates = ( 32 + 47 - 15 ) + 9 => 73

Hence correct answer must be (C)

...]]>

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Definition of mod:

|x| = x when x >= 0 (x is either positive or 0)

|x| = -x when x < 0

How is this statement correct??

|x| = -x when x < 0

Modulus is always positive!![/quote]

You could easily test this by picking numbers. Say x = -5 then |-5| = -(-5) = 5 = positive.

]]>

Definition of mod:

|x| = x when x >= 0 (x is either positive or 0)

|x| = -x when x < 0

How is this statement correct??

|x| = -x when x < 0

Modulus is always positive!![/quote]

You could easily test this by picking numbers. Say x = -5 then |-5| = -(-5) = 5 = positive.]]>

There are three foam generators in the factory, each of the first two can generate 14 liters of foam in one hour and the third can generate 18 liters in an hour. The three generators start working together at the same time and after one hour and a half one of the first generators stops working and two hours after that the third generator stops working and only one generator is left. If 5 hours after they all started to work the last generator stops working, how many liters of foam were generated?

A.

...

]]>

There are three foam generators in the factory, each of the first two can generate 14 liters of foam in one hour and the third can generate 18 liters in an hour. The three generators start working together at the same time and after one hour and a half one of the first generators stops working and two hours after that the third generator stops working and only one generator is left. If 5 hours after they all started to work the last generator stops working, how many liters of foam were generated?

A.

...]]>

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It costs $4 for the first 1/4 hour to use the laundry machine at the Laundromat. After the first ¼ hour it costs $12 per hour. If a certain customer uses the laundry machine for 3 hours and 25 minutes, how much will it cost him?

A. $25.

B. $32.

C. $36.

D. $40.

E. $42.

Total Cost = $4 ( For first 15min ) + $ 12 ( For time exceeding first 15min )

3 hrs 25 min = 205 min

First 15 min ------> $4

Time left is 190 min...

Now, 60 min costs $ 12

1 min costs $ 12/60

190 min costs $ 12/60*190 =>

...

]]>

It costs $4 for the first 1/4 hour to use the laundry machine at the Laundromat. After the first ¼ hour it costs $12 per hour. If a certain customer uses the laundry machine for 3 hours and 25 minutes, how much will it cost him?

A. $25.

B. $32.

C. $36.

D. $40.

E. $42.

Total Cost = $4 ( For first 15min ) + $ 12 ( For time exceeding first 15min )

3 hrs 25 min = 205 min

First 15 min ------> $4

Time left is 190 min...

Now, 60 min costs $ 12

1 min costs $ 12/60

190 min costs $ 12/60*190 =>

...]]>

shasadou wrote:

On a fishing expedition, a group of 13 fishermen caught a total of 160 fish. \ Did any one fisherman catch more than 15 fish?

(1) The fisherman who caught the third-most fish caught 11 fish.

(2) The fisherman who caught the second-most fish caught 12 fish

(1) The fisherman who caught the third-most fish caught 11 fish.

(2) The fisherman who caught the second-most fish caught 12 fish

Hi,

for 160 fish, the average catch is 160/13 that is 11 and half approx..

lets see the statements..

1) The fisherman who caught the third-most fish caught 11 fish. for making the higher most catches the lowest posible, let thirdmost

...

]]>

shasadou wrote:

On a fishing expedition, a group of 13 fishermen caught a total of 160 fish. \ Did any one fisherman catch more than 15 fish?

(1) The fisherman who caught the third-most fish caught 11 fish.

(2) The fisherman who caught the second-most fish caught 12 fish

(1) The fisherman who caught the third-most fish caught 11 fish.

(2) The fisherman who caught the second-most fish caught 12 fish

Hi,

for 160 fish, the average catch is 160/13 that is 11 and half approx..

lets see the statements..

1) The fisherman who caught the third-most fish caught 11 fish. for making the higher most catches the lowest posible, let thirdmost

...]]>

]]>

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It takes Avery 3 hours to build a brick wall while Tom can do it in 2.5 hours. If the two start working together and after an hour Avery leaves, how much time will it take Tom to complete the wall on his own?

A. 25 minutes.

B. 30 minutes.

C. 40 minutes.

D. 55 minutes.

E. 1 hour and 20 minutes.

Avery takes 3 hours

Tom takes 2.5 hours

Let total work be 7.5 units..

Efficiency of Avery is 2.5 units/hr

Efficiency of Tom is 3 units/hr

Combined efficiency of Tom & Avery is 5.5 units/hr

Since

...

]]>

It takes Avery 3 hours to build a brick wall while Tom can do it in 2.5 hours. If the two start working together and after an hour Avery leaves, how much time will it take Tom to complete the wall on his own?

A. 25 minutes.

B. 30 minutes.

C. 40 minutes.

D. 55 minutes.

E. 1 hour and 20 minutes.

Avery takes 3 hours

Tom takes 2.5 hours

Let total work be 7.5 units..

Efficiency of Avery is 2.5 units/hr

Efficiency of Tom is 3 units/hr

Combined efficiency of Tom & Avery is 5.5 units/hr

Since

...]]>

Now 365 mod 7 = 1. That means, 365th day of the year 1990, i.e. 30/12/1990, was a Monday

(if[day of the year] mod 7 = 0 then its day will be same as 31st of Dec, 1989. ex. 7th Jan, 14th Jan and so on)

Similarly if[day of the year] mod 7 = 1 then its day will be same as 1st of Jan, 1990. ex 8th Jan, 15th Jan and so on)

As we now know 365th day of the year 1990, i.e. 30/12/1990, was a

...

]]>

Now 365 mod 7 = 1. That means, 365th day of the year 1990, i.e. 30/12/1990, was a Monday

(if[day of the year] mod 7 = 0 then its day will be same as 31st of Dec, 1989. ex. 7th Jan, 14th Jan and so on)

Similarly if[day of the year] mod 7 = 1 then its day will be same as 1st of Jan, 1990. ex 8th Jan, 15th Jan and so on)

As we now know 365th day of the year 1990, i.e. 30/12/1990, was a

...]]>

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Bunuel wrote:

Official Solution:

How many different prime factors does positive integer \(n\) have?

(1)\(44 < n^2 < 99\) . This implies that\(n\) can be 7, 8, or 9. Each of these numbers has 1 prime: 7, 2, and 3, respectively. Sufficient.

(2)\(8n^2\) has twelve factors. For\(8n^2=2^3n^2\) to have twelve factors\(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors\(= (3+1)(2+1)=12\) . Sufficient.

Answer: D

How many different prime factors does positive integer \(n\) have?

(1)\(44 < n^2 < 99\) . This implies that\(n\) can be 7, 8, or 9. Each of these numbers has 1 prime: 7, 2, and 3, respectively. Sufficient.

(2)\(8n^2\) has twelve factors. For\(8n^2=2^3n^2\) to have twelve factors\(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors\(= (3+1)(2+1)=12\) . Sufficient.

Answer: D

...

]]>

Bunuel wrote:

Official Solution:

How many different prime factors does positive integer \(n\) have?

(1)\(44 < n^2 < 99\) . This implies that\(n\) can be 7, 8, or 9. Each of these numbers has 1 prime: 7, 2, and 3, respectively. Sufficient.

(2)\(8n^2\) has twelve factors. For\(8n^2=2^3n^2\) to have twelve factors\(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors\(= (3+1)(2+1)=12\) . Sufficient.

Answer: D

How many different prime factors does positive integer \(n\) have?

(1)\(44 < n^2 < 99\) . This implies that\(n\) can be 7, 8, or 9. Each of these numbers has 1 prime: 7, 2, and 3, respectively. Sufficient.

(2)\(8n^2\) has twelve factors. For\(8n^2=2^3n^2\) to have twelve factors\(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors\(= (3+1)(2+1)=12\) . Sufficient.

Answer: D

...]]>

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0! is equal to 1? i know i might be asking a very basic question but please enlighten me

Yes, \(0! = 1\).

]]>

0! is equal to 1? i know i might be asking a very basic question but please enlighten me

Yes, \(0! = 1\).

Answer choice B.

]]>

Answer choice B.]]>

A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

A. 37.5

B. 75

C. 100

D. 150

E. 175

Can you show me how to solve this without backsolving, ie the real algabraic way?

Got the right answer but the wording is super bad!

...

]]>

A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

A. 37.5

B. 75

C. 100

D. 150

E. 175

Can you show me how to solve this without backsolving, ie the real algabraic way?

Got the right answer but the wording is super bad!

...]]>

from home to parlor

speed V

time 60 min

distance S

We get S=60*V-------(1)

from parlor to home

speed V/2

time t min

distance S

S=t*V/2 -------(2)

from 1 and 2 we get

60+t =S/V + 2S/V

t=3*(S/V) - 60=180 - 60=120

TOTAL TIME FOR TWO ROUND TRIPS :2*(120+ 60) =360 MIN----E

Attachments

TD.jpg [ 10.37 KiB | Viewed 55 times ]

]]>

from home to parlor

speed V

time 60 min

distance S

We get S=60*V-------(1)

from parlor to home

speed V/2

time t min

distance S

S=t*V/2 -------(2)

from 1 and 2 we get

60+t =S/V + 2S/V

t=3*(S/V) - 60=180 - 60=120

TOTAL TIME FOR TWO ROUND TRIPS :2*(120+ 60) =360 MIN----E

Attachments

TD.jpg [ 10.37 KiB | Viewed 55 times ]

]]>

Mike, Jim and Bob are all professional fisherman. Mike can catch 21 fish in one hour, Jim can catch twice as much and Bob can catch 50% more than Jim. If the three started to fish together and after 40 minutes Mike and Bob left, how many fish did the three fishermen catch in one hour?

A. 64.

B. 72.

C. 86.

D. 98.

E. 112.

Hi,

JUST another way...

All of them catch fishes in relation to number21....

and there is nothing in time that should cancel out 7, so our ANS should be a multiple of 7.. .

...

]]>

Mike, Jim and Bob are all professional fisherman. Mike can catch 21 fish in one hour, Jim can catch twice as much and Bob can catch 50% more than Jim. If the three started to fish together and after 40 minutes Mike and Bob left, how many fish did the three fishermen catch in one hour?

A. 64.

B. 72.

C. 86.

D. 98.

E. 112.

Hi,

JUST another way...

All of them catch fishes in relation to number21....

and there is nothing in time that should cancel out 7, so our ANS should be a multiple of 7.. .

...]]>

chetan2u wrote:

amitpaul527 wrote:

If a,b, and c are prime numbers, do \((a+b)\) and \(c\) have a common factor that is greater than 1?

(1) a,b, and c are all different prime numbers

(2) \(c\neq{2}\)

(1) a,b, and c are all different prime numbers

(2) \(c\neq{2}\)

Hi,

lets analyze the Q

1) If all three are same prime ans will be NO.. -- (3+3)=6 and 3-- factors 1 and 3

chetan2u, can you please explain why the answer is no when the primes are all the same. 3 is the common factor of 3 and 6, so the answer should be yes. What am I missing? Sorry if the question is confusing, my brain is fried right

...

]]>

chetan2u wrote:

amitpaul527 wrote:

If a,b, and c are prime numbers, do \((a+b)\) and \(c\) have a common factor that is greater than 1?

(1) a,b, and c are all different prime numbers

(2) \(c\neq{2}\)

(1) a,b, and c are all different prime numbers

(2) \(c\neq{2}\)

Hi,

lets analyze the Q

1) If all three are same prime ans will be NO.. -- (3+3)=6 and 3-- factors 1 and 3

chetan2u, can you please explain why the answer is no when the primes are all the same. 3 is the common factor of 3 and 6, so the answer should be yes. What am I missing? Sorry if the question is confusing, my brain is fried right

...]]>

Statement 2 tells us 494 < a < b < 497

As per this statement a can be only 495 and b can be only 496. so we get a definite value for a+b. So sufficient

Correct Answer - B

]]>

Statement 2 tells us 494 < a < b < 497

As per this statement a can be only 495 and b can be only 496. so we get a definite value for a+b. So sufficient

Correct Answer - B]]>

]]>

Thank you, Bunuel!

]]>

Thank you, Bunuel!]]>

The critical part here is knowing area of "Petal of flower"...

Assume that you draw four semicircles within square ...

so 8^2 = 4*area of half circle - 4*Area of that "petal of flower"

FROM ABOVE you can derive are of petal of flower as 8pi-16 ... and then put that value in (1) above....

Hoping to get first Kudo on the forum....

]]>

The critical part here is knowing area of "Petal of flower"...

Assume that you draw four semicircles within square ...

so 8^2 = 4*area of half circle - 4*Area of that "petal of flower"

FROM ABOVE you can derive are of petal of flower as 8pi-16 ... and then put that value in (1) above....

Hoping to get first Kudo on the forum....]]>

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]]>

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]]>

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speed = s1

d1=d2

d1=s1*(5/6)

d2=s1*(7/8)*t

s1*5/6=s1*7/8*t

t= 5*8/6*7=40/42= 20/21 = 95% = ~57 minutes

]]>

speed = s1

d1=d2

d1=s1*(5/6)

d2=s1*(7/8)*t

s1*5/6=s1*7/8*t

t= 5*8/6*7=40/42= 20/21 = 95% = ~57 minutes]]>

multiply eq 2 with 5 and subtract from eq 1

5x+15y=850

5x+5y=300

10y= 550

y=55 and x = 5

therefore $15 bills = 5

]]>

multiply eq 2 with 5 and subtract from eq 1

5x+15y=850

5x+5y=300

10y= 550

y=55 and x = 5

therefore $15 bills = 5]]>

A question.. Why not E?..

When checking A and B the the \(4^1\) can also be seen as \(2^2\)

So:

\(3^a*2^(2+b) = 2^2*3^2\) where 2^(2+b) is \(2^2*2^0\)

and we get that B can be 1 or 0.

Thanks!

]]>

A question.. Why not E?..

When checking A and B the the \(4^1\) can also be seen as \(2^2\)

So:

\(3^a*2^(2+b) = 2^2*3^2\) where 2^(2+b) is \(2^2*2^0\)

and we get that B can be 1 or 0.

Thanks!]]>

]]>

]]>

msk0657 wrote:

Hi Karishma,

Your answer is different from the OA, which is D.

Please find the same kind of question here m09-183831.html

By comparing these two questions, I am assuming that in the question it should have details regarding the number of first games.

Your answer is different from the OA, which is D.

Please find the same kind of question here m09-183831.html

By comparing these two questions, I am assuming that in the question it should have details regarding the number of first games.

That question and this question are different. That question gives you the number of games in the first lot.

"A chess player

...

]]>

msk0657 wrote:

Hi Karishma,

Your answer is different from the OA, which is D.

Please find the same kind of question here m09-183831.html

By comparing these two questions, I am assuming that in the question it should have details regarding the number of first games.

Your answer is different from the OA, which is D.

Please find the same kind of question here m09-183831.html

By comparing these two questions, I am assuming that in the question it should have details regarding the number of first games.

That question and this question are different. That question gives you the number of games in the first lot.

"A chess player

...]]>

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.]]>

Now as (1) says that n is prime, so it cannot be zero. Also (2) says that n is even so it cannot be zero. So forget zero.

Take n = 1,2,3…in each case, we will get 100+8 or 10000+8 or 1000000+3 ……

All of these numbers will be divisible by 18 because each of these sums are ending with 8 which is divisible by 2 AND

Sum of all digits is 9 ..(because there are lot of zeros and 1 and 8 in every sum).

As every number is

...

]]>

Now as (1) says that n is prime, so it cannot be zero. Also (2) says that n is even so it cannot be zero. So forget zero.

Take n = 1,2,3…in each case, we will get 100+8 or 10000+8 or 1000000+3 ……

All of these numbers will be divisible by 18 because each of these sums are ending with 8 which is divisible by 2 AND

Sum of all digits is 9 ..(because there are lot of zeros and 1 and 8 in every sum).

As every number is

...]]>

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

]]>

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.]]>