Bunuel wrote:

1. If 357^x*117^y=a, where x and y are positive integers, what is the units digit of a?

(1)100<y^2<x^2<169

(2)x^2-y^2=23

(1)100<y^2<x^2<169 --> since bothx andy are positive integers thenx^2 andy^2 are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 -->y=11 andx=12 . Sufficient.(As cyclicity of units digit of7 in integer power is4 , therefore the units digit of7^{23} is the same as the units digit

(1)100<y^2<x^2<169

(2)x^2-y^2=23

(1)100<y^2<x^2<169 --> since bothx andy are positive integers thenx^2 andy^2 are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 -->y=11 andx=12 . Sufficient.(As cyclicity of units digit of7 in integer power is4 , therefore the units digit of7^{23} is the same as the units digit

...

]]>

Bunuel wrote:

1. If 357^x*117^y=a, where x and y are positive integers, what is the units digit of a?

(1)100<y^2<x^2<169

(2)x^2-y^2=23

(1)100<y^2<x^2<169 --> since bothx andy are positive integers thenx^2 andy^2 are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 -->y=11 andx=12 . Sufficient.(As cyclicity of units digit of7 in integer power is4 , therefore the units digit of7^{23} is the same as the units digit

(1)100<y^2<x^2<169

(2)x^2-y^2=23

(1)100<y^2<x^2<169 --> since bothx andy are positive integers thenx^2 andy^2 are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 -->y=11 andx=12 . Sufficient.(As cyclicity of units digit of7 in integer power is4 , therefore the units digit of7^{23} is the same as the units digit

...]]>

Enael wrote:

Bunuel, even though the answer doesn't change, I still don't grasp how WATER = SOLUTION evaporated.

I have 10,000 gallons of both: NaCl and Water.

5% of NaCl = 500 gallons so water HAS to be 9,500 gallons.

It explicitly states that 2,500 gallons of water are evaporated, thus our base reduced should be 9,500 - 2,500 = 7,000 gallons of water.

So: NaCl / (NaCl+Water) = 500/(7000+500) = 1/15 = almost 7%.

As said, the answer doesn't change but the concept behind

know if theI have 10,000 gallons of both: NaCl and Water.

5% of NaCl = 500 gallons so water HAS to be 9,500 gallons.

It explicitly states that 2,500 gallons of water are evaporated, thus our base reduced should be 9,500 - 2,500 = 7,000 gallons of water.

So: NaCl / (NaCl+Water) = 500/(7000+500) = 1/15 = almost 7%.

As said, the answer doesn't change but the concept behind

...

]]>

Enael wrote:

Bunuel, even though the answer doesn't change, I still don't grasp how WATER = SOLUTION evaporated.

I have 10,000 gallons of both: NaCl and Water.

5% of NaCl = 500 gallons so water HAS to be 9,500 gallons.

It explicitly states that 2,500 gallons of water are evaporated, thus our base reduced should be 9,500 - 2,500 = 7,000 gallons of water.

So: NaCl / (NaCl+Water) = 500/(7000+500) = 1/15 = almost 7%.

As said, the answer doesn't change but the concept behind

know if theI have 10,000 gallons of both: NaCl and Water.

5% of NaCl = 500 gallons so water HAS to be 9,500 gallons.

It explicitly states that 2,500 gallons of water are evaporated, thus our base reduced should be 9,500 - 2,500 = 7,000 gallons of water.

So: NaCl / (NaCl+Water) = 500/(7000+500) = 1/15 = almost 7%.

As said, the answer doesn't change but the concept behind

...]]>

Sir,

How can we be sure that x+y be < 1 in the above question. I mean some constraints must make them <1 ?

Not sure I understand what you mean. Can you please elaborate?

]]>

Sir,

How can we be sure that x+y be < 1 in the above question. I mean some constraints must make them <1 ?

Not sure I understand what you mean. Can you please elaborate?]]>

why there are no answers to some questions??

That's not true. Links to solutions are here: devil-s-dozen-129312.html#p1060761

]]>

why there are no answers to some questions??

That's not true. Links to solutions are here: devil-s-dozen-129312.html#p1060761]]>

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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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suppose the distance is d.

and the initial constant speed be x and bob's later constant speed be y.

wendy travels d distance in d/x time.

and bob travels 2d distance in d/2x + 1.5d/y time ( travels half a distance by x speed and 1.5 times d by y speed )

Hence the time for which wendy waits is (d/2x + 1.5d/y) - d/x

1) doesnt say anything about time taken by Wendy ( d/x) . Hence insufficient.

2 ) it says that bob takes 32 minutes more alone than when he

...

]]>

suppose the distance is d.

and the initial constant speed be x and bob's later constant speed be y.

wendy travels d distance in d/x time.

and bob travels 2d distance in d/2x + 1.5d/y time ( travels half a distance by x speed and 1.5 times d by y speed )

Hence the time for which wendy waits is (d/2x + 1.5d/y) - d/x

1) doesnt say anything about time taken by Wendy ( d/x) . Hence insufficient.

2 ) it says that bob takes 32 minutes more alone than when he

...]]>

the perfect square of whatever a integer number has the following possible digit 1 ; 4 ; 5 ; 6; 9

=> eliminate: C and E

A and D is divisible by 2, but not by 4 => A and D is not a perfect square

B is correct

]]>

the perfect square of whatever a integer number has the following possible digit 1 ; 4 ; 5 ; 6; 9

=> eliminate: C and E

A and D is divisible by 2, but not by 4 => A and D is not a perfect square

B is correct]]>

=> 20*pie is correct

]]>

=> 20*pie is correct]]>

sagnik2422 wrote:

12 + 13 + 14 + ... 51 + 52 + 53 = ?

A. 1361

B. 1362

C. 1363

D. 1364

E. 1365

explanation showed adding in pairs which added to 65 and then y - x + 1 = 53 - 12 + 1 = 42

and then 21 pairs add to 65 and 65 x 21 and looked at the last digit which was 5

MY QUESTION : WHY IS IT 21 PAIRS? How would we know this?

A. 1361

B. 1362

C. 1363

D. 1364

E. 1365

explanation showed adding in pairs which added to 65 and then y - x + 1 = 53 - 12 + 1 = 42

and then 21 pairs add to 65 and 65 x 21 and looked at the last digit which was 5

MY QUESTION : WHY IS IT 21 PAIRS? How would we know this?

You can do this question in different ways:

Sum = 12 + 13 + 14 + ... 51 + 52 + 53

Method 1:

Sum of n consecutive positive integers starting from 1 is given as n(n+1)/2

Sum

...

]]>

sagnik2422 wrote:

12 + 13 + 14 + ... 51 + 52 + 53 = ?

A. 1361

B. 1362

C. 1363

D. 1364

E. 1365

explanation showed adding in pairs which added to 65 and then y - x + 1 = 53 - 12 + 1 = 42

and then 21 pairs add to 65 and 65 x 21 and looked at the last digit which was 5

MY QUESTION : WHY IS IT 21 PAIRS? How would we know this?

A. 1361

B. 1362

C. 1363

D. 1364

E. 1365

explanation showed adding in pairs which added to 65 and then y - x + 1 = 53 - 12 + 1 = 42

and then 21 pairs add to 65 and 65 x 21 and looked at the last digit which was 5

MY QUESTION : WHY IS IT 21 PAIRS? How would we know this?

You can do this question in different ways:

Sum = 12 + 13 + 14 + ... 51 + 52 + 53

Method 1:

Sum of n consecutive positive integers starting from 1 is given as n(n+1)/2

Sum

...]]>

A man chooses an outfit from 3 different shirts, 2 different pairs of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated?

A.(\frac{1}{3})^6*(\frac{1}{2})^3

B.(\frac{1}{3})^6*(\frac{1}{2})

C.(\frac{1}{3})^4

D.(\frac{1}{3})^2*(\frac{1}{2})

E.5*(\frac{1}{3})^2

Happy solving!

...

]]>

A man chooses an outfit from 3 different shirts, 2 different pairs of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated?

A.(\frac{1}{3})^6*(\frac{1}{2})^3

B.(\frac{1}{3})^6*(\frac{1}{2})

C.(\frac{1}{3})^4

D.(\frac{1}{3})^2*(\frac{1}{2})

E.5*(\frac{1}{3})^2

Happy solving!

...]]>

niyantg wrote:

Hello Bunuel

I got it till that - b is a multiple of 12.

But couldnt follow after that. How did you get that a is also a multiple of 12.

Please clarify.

Thankyou

I got it till that - b is a multiple of 12.

But couldnt follow after that. How did you get that a is also a multiple of 12.

Please clarify.

Thankyou

We are not whethera orb is divisible by 12, we are asked whether the product ofa andb (ab ) is divisible by 12. Knowing thatb is divisible by 12 (and taking into account thata is and integer) is sufficient to say thatab will indeed be divisible by 12, no matter what is the value ofa .

Does this make sense?

...

]]>

niyantg wrote:

Hello Bunuel

I got it till that - b is a multiple of 12.

But couldnt follow after that. How did you get that a is also a multiple of 12.

Please clarify.

Thankyou

I got it till that - b is a multiple of 12.

But couldnt follow after that. How did you get that a is also a multiple of 12.

Please clarify.

Thankyou

We are not whethera orb is divisible by 12, we are asked whether the product ofa andb (ab ) is divisible by 12. Knowing thatb is divisible by 12 (and taking into account thata is and integer) is sufficient to say thatab will indeed be divisible by 12, no matter what is the value ofa .

Does this make sense?

...]]>

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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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Bunuel wrote:

krit wrote:

If x^2 + y^2 = 29, what is the value of (x-y)^2

(1) xy = 10

(2) x = 5

Hi guys, I'd like some help here. The answer is A. I see why (1) is sufficient, but I don't get why (2) is not. We can just simply plug x = 5 into the equation, find y, and then solve for (x-y)^2

Thanks

(1) xy = 10

(2) x = 5

Hi guys, I'd like some help here. The answer is A. I see why (1) is sufficient, but I don't get why (2) is not. We can just simply plug x = 5 into the equation, find y, and then solve for (x-y)^2

Thanks

Given:x^2+y^2=29 .

Question:(x-y)^2=x^2-2xy+y^2=(x^2+y^2)-2xy=29-2xy=? So, basically we should find the value ofxy

(1) xy = 10. Directly gives us the value we needed. Sufficient.

(2) x = 5. Now even if

...

]]>

Bunuel wrote:

krit wrote:

If x^2 + y^2 = 29, what is the value of (x-y)^2

(1) xy = 10

(2) x = 5

Hi guys, I'd like some help here. The answer is A. I see why (1) is sufficient, but I don't get why (2) is not. We can just simply plug x = 5 into the equation, find y, and then solve for (x-y)^2

Thanks

(1) xy = 10

(2) x = 5

Hi guys, I'd like some help here. The answer is A. I see why (1) is sufficient, but I don't get why (2) is not. We can just simply plug x = 5 into the equation, find y, and then solve for (x-y)^2

Thanks

Given:x^2+y^2=29 .

Question:(x-y)^2=x^2-2xy+y^2=(x^2+y^2)-2xy=29-2xy=? So, basically we should find the value ofxy

(1) xy = 10. Directly gives us the value we needed. Sufficient.

(2) x = 5. Now even if

...]]>

Let me try finding that question and pasting my approach.

Thanks for the reply Bunuel.

]]>

Let me try finding that question and pasting my approach.

Thanks for the reply Bunuel.]]>

Prashant,

Why do we have to have to multiply 2/8x 1/7

According to me,

The probability that one tulip is selected from 8 = 1/8

So there are 7 remaining

The probability that 1 tulip is selected from 7 = 1/7

Therefore selectng 2 tulips

= 1/8*1/7

= 1/56

Probability that both flowers are not tulip = 1-1/56

= 55/56

The probability that one tulip is selected from 8 is not 1/8 but 2/8 as out of 8 no of tulips is 2 not 1.

...

]]>

Prashant,

Why do we have to have to multiply 2/8x 1/7

According to me,

The probability that one tulip is selected from 8 = 1/8

So there are 7 remaining

The probability that 1 tulip is selected from 7 = 1/7

Therefore selectng 2 tulips

= 1/8*1/7

= 1/56

Probability that both flowers are not tulip = 1-1/56

= 55/56

The probability that one tulip is selected from 8 is not 1/8 but 2/8 as out of 8 no of tulips is 2 not 1.

...]]>

What is the perimeter of the triangle in the figure above?

Notice that y = z and y= x, so, x = y=z:

Attachment:

Untitled.png [ 10.56 KiB | Viewed 48 times ]

(1)The side across from interior angle z measures 2. We the lengths of the legs of a 45-45-90 right triangle. We can get the third side and calculate the perimeter . Sufficient.

(2)2x > y. We knew this from the stem. Not

...

]]>

What is the perimeter of the triangle in the figure above?

Notice that y = z and y= x, so, x = y=z:

Attachment:

Untitled.png [ 10.56 KiB | Viewed 48 times ]

(1)The side across from interior angle z measures 2. We the lengths of the legs of a 45-45-90 right triangle. We can get the third side and calculate the perimeter . Sufficient.

(2)2x > y. We knew this from the stem. Not

...]]>

What is wrong with my methodology below:

8 spots in line: _ _ _ _ _ _ _ _

ABC must be before DEF so possibilities are:

1) _ _ _ D E F _ _ For this option, ABC must occupy first three spots, and 3!*3!*2! = 72 ways

2) _ _ _ _ D E F _ For this option, ABC must occupy 3 of the first 4 slots, and D and E can be wherever, so 4!*2!*3! = 288 ways

3) _ _ _ _ _ D E F For this option, the remaining 5 people can be arranged in 5! ways = 720 ways

For a total of 1080 ways. Obviously this is wrong but I

...

]]>

What is wrong with my methodology below:

8 spots in line: _ _ _ _ _ _ _ _

ABC must be before DEF so possibilities are:

1) _ _ _ D E F _ _ For this option, ABC must occupy first three spots, and 3!*3!*2! = 72 ways

2) _ _ _ _ D E F _ For this option, ABC must occupy 3 of the first 4 slots, and D and E can be wherever, so 4!*2!*3! = 288 ways

3) _ _ _ _ _ D E F For this option, the remaining 5 people can be arranged in 5! ways = 720 ways

For a total of 1080 ways. Obviously this is wrong but I

...]]>

Look at the diagrambelow:

Notice that "30% of the people like both strawberry and apple jam" doesn't mean that among these 30% (60) can not be some people who like raspberry as well. Both strawberry and apple jam is the intersection of these two groups, if we refer to the diagram it's the yellow segment in it.

Next, no formula is needed to solve this question: 112 like strawberry jam, 88 like apple jam, 60 people like both strawberry

Notice that "30% of the people like both strawberry and apple jam" doesn't mean that among these 30% (60) can not be some people who like raspberry as well. Both strawberry and apple jam is the intersection of these two groups, if we refer to the diagram it's the yellow segment in it.

Next, no formula is needed to solve this question: 112 like strawberry jam, 88 like apple jam, 60 people like both strawberry

...

]]>

Look at the diagrambelow:

Notice that "30% of the people like both strawberry and apple jam" doesn't mean that among these 30% (60) can not be some people who like raspberry as well. Both strawberry and apple jam is the intersection of these two groups, if we refer to the diagram it's the yellow segment in it.

Next, no formula is needed to solve this question: 112 like strawberry jam, 88 like apple jam, 60 people like both strawberry

Notice that "30% of the people like both strawberry and apple jam" doesn't mean that among these 30% (60) can not be some people who like raspberry as well. Both strawberry and apple jam is the intersection of these two groups, if we refer to the diagram it's the yellow segment in it.

Next, no formula is needed to solve this question: 112 like strawberry jam, 88 like apple jam, 60 people like both strawberry

...]]>

The figure above shows the dimensions of a semicircular cross section of a one-way tunnel. The single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel. If vehicles must clear the top of the tunnel by at least 1/2 foot when they are inside the traffic lane, what should be the limit on the height of vehicles that are allowed to use the tunnel?

A. 5½ ft

B. 7½ ft

C. 8 ½ ft

D. 9½ ft

E. 10 ft

...

]]>

The figure above shows the dimensions of a semicircular cross section of a one-way tunnel. The single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel. If vehicles must clear the top of the tunnel by at least 1/2 foot when they are inside the traffic lane, what should be the limit on the height of vehicles that are allowed to use the tunnel?

A. 5½ ft

B. 7½ ft

C. 8 ½ ft

D. 9½ ft

E. 10 ft

...]]>

Hi Bunuel,

I have a question here. I use the same method as you --> finding the prime factors and its power. But I didn't go through checking the answer choice in the factorial step. I only add the powers together 5+2+1 and I got 8. So, I'm not sure if this method correct or I just got it correct by luck...

Thanks for your help in advance.

Bunuel wrote:

...

]]>

Hi Bunuel,

I have a question here. I use the same method as you --> finding the prime factors and its power. But I didn't go through checking the answer choice in the factorial step. I only add the powers together 5+2+1 and I got 8. So, I'm not sure if this method correct or I just got it correct by luck...

Thanks for your help in advance.

Bunuel wrote:

...]]>

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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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I have been taking the MANHATTAN CHALLANGE SETS. How difficult are these? Is there any correlation between these and the actual gmat. I scored 50,49,49,49 in quant in 4 kaplan tests. I want to get to 51. I saw a post showing correlation between gmat club tests and the real gmat.

Thanks!!!!

]]>

I have been taking the MANHATTAN CHALLANGE SETS. How difficult are these? Is there any correlation between these and the actual gmat. I scored 50,49,49,49 in quant in 4 kaplan tests. I want to get to 51. I saw a post showing correlation between gmat club tests and the real gmat.

Thanks!!!!]]>

If the variables, X, Y, and Z take on only the values 10, 20, 30, 40, 50, 60, or 70 with frequencies indicated by the shaded regions above, for which of the frequency distributions is the mean equal to the median?

(A) X only

(B) Y only

(C) Z only

(D) X and Y

(E) X and Z

The frequency distributions for both X and Z are symmetric about 40, which means that both X and Z havemean = median = 40 .

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-the-variables-x-y-and-z-take-on-only-the-values-168633.html

...

]]>

If the variables, X, Y, and Z take on only the values 10, 20, 30, 40, 50, 60, or 70 with frequencies indicated by the shaded regions above, for which of the frequency distributions is the mean equal to the median?

(A) X only

(B) Y only

(C) Z only

(D) X and Y

(E) X and Z

The frequency distributions for both X and Z are symmetric about 40, which means that both X and Z havemean = median = 40 .

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-the-variables-x-y-and-z-take-on-only-the-values-168633.html

...]]>

The y intercept of a line L is 4. If the slope of L is negative, which of the fillowing could be the x intercept of L.

I -1

II 0

III 6

A) I only

B) II only

C) III only

D) I and II

E) I and III

With a Y-intercept of (0,4) and a negative slope, there is NO WAY for the x intercept to be less than or equal to 0

Answer is C

]]>

The y intercept of a line L is 4. If the slope of L is negative, which of the fillowing could be the x intercept of L.

I -1

II 0

III 6

A) I only

B) II only

C) III only

D) I and II

E) I and III

With a Y-intercept of (0,4) and a negative slope, there is NO WAY for the x intercept to be less than or equal to 0

Answer is C]]>

shrive555 wrote:

In 2003, the number of girls attending Jefferson High School was equal to the number of boys. In 2004, the population of girls and the population of boys both increased by 20 percent. Which of the following could be the total student population at Jefferson High School in 2004?

A. 4832

B. 5034

C. 5058

D. 5076

E. 5128

A. 4832

B. 5034

C. 5058

D. 5076

E. 5128

There is 1 key hint in this questions that will help you narrow down the answer.

"In 2004, the population of girls and the population of boys both increased by 20

...

]]>

shrive555 wrote:

In 2003, the number of girls attending Jefferson High School was equal to the number of boys. In 2004, the population of girls and the population of boys both increased by 20 percent. Which of the following could be the total student population at Jefferson High School in 2004?

A. 4832

B. 5034

C. 5058

D. 5076

E. 5128

A. 4832

B. 5034

C. 5058

D. 5076

E. 5128

There is 1 key hint in this questions that will help you narrow down the answer.

"In 2004, the population of girls and the population of boys both increased by 20

...]]>

Bunuel wrote:

suyash23n wrote:

A certain research group plans to create computer models of x% of a list of 10,000 bacterial species known to inhabit the human body. After a budget cut, the group finds it must reduce this selection by (x − 5)%. In terms of x, how many species of bacteria will the group be able to model?

a) x*x – 5x

b) (x)(105 – x)

c) (100)(105 – x)

d) (100)(95 – x)

e) (x-5)/100

a) x*x – 5x

b) (x)(105 – x)

c) (100)(105 – x)

d) (100)(95 – x)

e) (x-5)/100

The number of species that the group plans to model =10,000*\frac{x}{100} = 100x .

After a budget cut, the group finds it must reduce

...

]]>

Bunuel wrote:

suyash23n wrote:

A certain research group plans to create computer models of x% of a list of 10,000 bacterial species known to inhabit the human body. After a budget cut, the group finds it must reduce this selection by (x − 5)%. In terms of x, how many species of bacteria will the group be able to model?

a) x*x – 5x

b) (x)(105 – x)

c) (100)(105 – x)

d) (100)(95 – x)

e) (x-5)/100

a) x*x – 5x

b) (x)(105 – x)

c) (100)(105 – x)

d) (100)(95 – x)

e) (x-5)/100

The number of species that the group plans to model =10,000*\frac{x}{100} = 100x .

After a budget cut, the group finds it must reduce

...]]>

Cost Price = \frac{96*100}{120} = 80

Second Share sold at 96 which includes 20% loss

Cost Price= \frac{96 * 100}{80} = 120

Total Cost Price = 120 + 80 = 200

Total Selling Price = 96 * 2 = 192

Loss = 8

Answer = C

]]>

Cost Price = \frac{96*100}{120} = 80

Second Share sold at 96 which includes 20% loss

Cost Price= \frac{96 * 100}{80} = 120

Total Cost Price = 120 + 80 = 200

Total Selling Price = 96 * 2 = 192

Loss = 8

Answer = C]]>

Hello Everyone,

I need advice whether to re-take the GMAT.

My story, very briefly (hopefully there will be a detailed happy debrief someday :D):

I decided to get an MBA last summer (2013), I studied, and studied then took the test 3 times (600, 570, 570). Had a few personal issues that complicated things abit at the time, I stopped preparing April > June. Resumed studying, took the test today and got a 650 (Q42, V38). My last practice scores were 700 and 690 (Veritas prep), also I've been

...

]]>

Hello Everyone,

I need advice whether to re-take the GMAT.

My story, very briefly (hopefully there will be a detailed happy debrief someday :D):

I decided to get an MBA last summer (2013), I studied, and studied then took the test 3 times (600, 570, 570). Had a few personal issues that complicated things abit at the time, I stopped preparing April > June. Resumed studying, took the test today and got a 650 (Q42, V38). My last practice scores were 700 and 690 (Veritas prep), also I've been

...]]>

Thanks for the Reply Karishma. I follow your posts and your explanations are top notch.

If I went with the same logic, the answer would be 4 lesser (eliminating the number of games by the group winners lost in the Prelims) = 79 Games.

Can I generalize and use the above presented logic for say, a single elimination ( 1 Loss only) or triple elimination (3 Losses) format too?

Yes, you are correct. 4 fewer games is all you can

...

]]>

Thanks for the Reply Karishma. I follow your posts and your explanations are top notch.

If I went with the same logic, the answer would be 4 lesser (eliminating the number of games by the group winners lost in the Prelims) = 79 Games.

Can I generalize and use the above presented logic for say, a single elimination ( 1 Loss only) or triple elimination (3 Losses) format too?

Yes, you are correct. 4 fewer games is all you can

...]]>

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]]>

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x^2-2x+A>0

A>-(x^2-2x)---------1.

Basically looked at the range of x^2-2x

Max(x^2-2x)=Infinity

Min(x^2-2x)=-1

Therefore, from 1.

A>-(infinity)

and A>-(-1), i.e. A>1

which is sufficient!

]]>

x^2-2x+A>0

A>-(x^2-2x)---------1.

Basically looked at the range of x^2-2x

Max(x^2-2x)=Infinity

Min(x^2-2x)=-1

Therefore, from 1.

A>-(infinity)

and A>-(-1), i.e. A>1

which is sufficient!]]>

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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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15% Discount = 28.8 * \frac{85}{100}

= 24.48

Answer = D

]]>

15% Discount = 28.8 * \frac{85}{100}

= 24.48

Answer = D]]>

Time taken to travel distance p= \frac{p}{30}

Time taken to travel remaining distance= \frac{100-p}{50}

Total time consumed= \frac{p}{30} + \frac{100-p}{50}

= \frac{300+2p}{150}

Total Distance Travelled = 100

Average Velocity= \frac{100}{\frac{300+2p}{150}}

= \frac{15000}{300+2p}

Answer = E

...

Attachments

sw.png [ 2.31 KiB | Viewed 55 times ]

]]>

Time taken to travel distance p= \frac{p}{30}

Time taken to travel remaining distance= \frac{100-p}{50}

Total time consumed= \frac{p}{30} + \frac{100-p}{50}

= \frac{300+2p}{150}

Total Distance Travelled = 100

Average Velocity= \frac{100}{\frac{300+2p}{150}}

= \frac{15000}{300+2p}

Answer = E

...

Attachments

sw.png [ 2.31 KiB | Viewed 55 times ]

]]>

.

the answer is

.

2=2

6=3*2

9=3*3

12=3*2*2

15=3*5

30=2*3*5

The least possible value of X is X=2*2*3*3*5 (Not divisible by 8).

However, a greater value of X could be X=2*2*2*3*3*5 (divisible by 8)

]]>

.

the answer is

.

2=2

6=3*2

9=3*3

12=3*2*2

15=3*5

30=2*3*5

The least possible value of X is X=2*2*3*3*5 (Not divisible by 8).

However, a greater value of X could be X=2*2*2*3*3*5 (divisible by 8)]]>

The theatre made an arrangement with the acting company that the theatre owners get $5 for every ticket to the event sold to the event. How much the theatre make?

(1) The total revenue from ticket sales was $700. We don't know how many tickets were sold. Consider this: the number of tickets sold could be 100 ($7 for a ticket), which would make the owners revenue equal to 100*5 or the number of tickets sold could be 10 ($70 for a ticket), which would make the owners revenue equal to 10*5. Not

...

]]>

The theatre made an arrangement with the acting company that the theatre owners get $5 for every ticket to the event sold to the event. How much the theatre make?

(1) The total revenue from ticket sales was $700. We don't know how many tickets were sold. Consider this: the number of tickets sold could be 100 ($7 for a ticket), which would make the owners revenue equal to 100*5 or the number of tickets sold could be 10 ($70 for a ticket), which would make the owners revenue equal to 10*5. Not

...]]>

imhimanshu wrote:

In the xy-coordinate plane, Is point R equidistant from points (-3,-3) and (1,-3)

1. The x coordinate of point R is -1

2. Point R lies on line y = -3.

Expert, please provide your reasoning.

Thanks

1. The x coordinate of point R is -1

2. Point R lies on line y = -3.

Expert, please provide your reasoning.

Thanks

Hi,

As per belowdiagram,

All the points which lie on x=-1, will be equidistant from the given points,

Using (1),

x coordinate is -1. Sufficient.

Using (2),

point lies on y=-3, the point may lie anywhere on the line joining the two points, and

...

]]>

imhimanshu wrote:

In the xy-coordinate plane, Is point R equidistant from points (-3,-3) and (1,-3)

1. The x coordinate of point R is -1

2. Point R lies on line y = -3.

Expert, please provide your reasoning.

Thanks

1. The x coordinate of point R is -1

2. Point R lies on line y = -3.

Expert, please provide your reasoning.

Thanks

Hi,

As per belowdiagram,

All the points which lie on x=-1, will be equidistant from the given points,

Using (1),

x coordinate is -1. Sufficient.

Using (2),

point lies on y=-3, the point may lie anywhere on the line joining the two points, and

...]]>

Statement 1:

Given:

z>x+y+1 <---- Equ. 1

&

x+y+z>0

Now add z to both sides of Equ. 1

2z>x+y+z+1

Since x+y+z > 0

2z>GT0+1 (GT0 means a quantity greater than 0)

2z>GT1

z>GT(1/2)

So we cannot conclude that z>1

Therefore, insufficient!

Statement 2:

x+y+1<0

Adding z to both sides of the Equation

x+y+z+1<Z

GT0+1<z

or

GT1<z

Therefore, z>1

Sufficient!

Ans: 'B'

I hope the explanation is clear enough

...

]]>

Statement 1:

Given:

z>x+y+1 <---- Equ. 1

&

x+y+z>0

Now add z to both sides of Equ. 1

2z>x+y+z+1

Since x+y+z > 0

2z>GT0+1 (GT0 means a quantity greater than 0)

2z>GT1

z>GT(1/2)

So we cannot conclude that z>1

Therefore, insufficient!

Statement 2:

x+y+1<0

Adding z to both sides of the Equation

x+y+z+1<Z

GT0+1<z

or

GT1<z

Therefore, z>1

Sufficient!

Ans: 'B'

I hope the explanation is clear enough

...]]>

noboru wrote:

Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?

Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5.

Expected value of three dice is 3*3.5=10.5.

Mary scored 10 so the probability to get the sum more then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2.

P=1/2.

...

]]>

noboru wrote:

Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?

Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5.

Expected value of three dice is 3*3.5=10.5.

Mary scored 10 so the probability to get the sum more then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2.

P=1/2.

...]]>

The trick here is to know that remainder is always non-negative integer less than divisor0\leq{r}<d , so in our case0\leq{r}<7 .

So the remainder upon division of any integer by 7 can be: 0, 1, 2, 3, 4, 5, or 6 (7 values).

(1) The range of the seven remainders is 6 --> if we pick 6 different multiples of 7 (all remainders 0) and the 7th number 6 (remainder

...

]]>

The trick here is to know that remainder is always non-negative integer less than divisor0\leq{r}<d , so in our case0\leq{r}<7 .

So the remainder upon division of any integer by 7 can be: 0, 1, 2, 3, 4, 5, or 6 (7 values).

(1) The range of the seven remainders is 6 --> if we pick 6 different multiples of 7 (all remainders 0) and the 7th number 6 (remainder

...]]>

Q:---40------(x+6)/40-------x+6

The time should be equal, so we solve x/10 = (x+6)/40

x=2

so time = 2/10 min = 12 sec

]]>

Q:---40------(x+6)/40-------x+6

The time should be equal, so we solve x/10 = (x+6)/40

x=2

so time = 2/10 min = 12 sec]]>

]]>

Insufficient.

Stmt2: a<b

the coordinates may be (3,6) or (-5,-3).

b may be Positive or Negative.

Insufficient.

Combining 1 + 2

Line K must pass from either 2nd(-,+) or 4th(+,-) quadrants.

a<b .

Therefore B is Positive.

Hence C;

]]>

Insufficient.

Stmt2: a<b

the coordinates may be (3,6) or (-5,-3).

b may be Positive or Negative.

Insufficient.

Combining 1 + 2

Line K must pass from either 2nd(-,+) or 4th(+,-) quadrants.

a<b .

Therefore B is Positive.

Hence C;]]>

b=10 , d=n<5 , Select 2

The total possibilities to select 2 out of 10 bulbs: 10C2 = 45

.

Total possibilities to select 2 out of n defective bulbs: nC2 = 1/2 * n(n-1)

.

(1) => (1/2 * n(n-1))/45 =1/5 => n={3;-2} => (1) sufficient

.

Total possibilities to either select 2 out of 10-n bulbs or n defective bulbs: nC2 + (10-n)C2 = 1/2 * n(n-1) + 1/2 * (10-n)(9-n)

.

(2) => the probability to either select 2 out of 10-n bulbs or n defective bulbs is 8/15

we have the equation:

(1/2

...

]]>

b=10 , d=n<5 , Select 2

The total possibilities to select 2 out of 10 bulbs: 10C2 = 45

.

Total possibilities to select 2 out of n defective bulbs: nC2 = 1/2 * n(n-1)

.

(1) => (1/2 * n(n-1))/45 =1/5 => n={3;-2} => (1) sufficient

.

Total possibilities to either select 2 out of 10-n bulbs or n defective bulbs: nC2 + (10-n)C2 = 1/2 * n(n-1) + 1/2 * (10-n)(9-n)

.

(2) => the probability to either select 2 out of 10-n bulbs or n defective bulbs is 8/15

we have the equation:

(1/2

...]]>

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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(x+y)(x-y)=169

(x+y)=169/7..... as from 1. we know that x-y=7

hence perimeter is 2(x+y)= 2(169/7)....

Hence C

]]>

(x+y)(x-y)=169

(x+y)=169/7..... as from 1. we know that x-y=7

hence perimeter is 2(x+y)= 2(169/7)....

Hence C]]>

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).

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Hi Bunuel,

To keep it straight, just because it says its a rectangle does not mean we have to have two 30-60-90 triangles, but if we put together two 30-60-90 triangles we get a rectangle? Correct? I picked "A" because I thought that since it said we had a rectangle, we had to have two of these triangles. From the discussion above, it looks like this is not a mandatory condition of a rectangle.

Correct. But you can get a rectangle by putting together any two congruent right triangles,

...

]]>

Hi Bunuel,

To keep it straight, just because it says its a rectangle does not mean we have to have two 30-60-90 triangles, but if we put together two 30-60-90 triangles we get a rectangle? Correct? I picked "A" because I thought that since it said we had a rectangle, we had to have two of these triangles. From the discussion above, it looks like this is not a mandatory condition of a rectangle.

Correct. But you can get a rectangle by putting together any two congruent right triangles,

...]]>