]]>

So first number would be X/2-5

We know that Sum = Average * No of Terms

So Sum = 20* 7 = 140

To maximise the Last Digit we have to Minimise all six digits

Arrange the Digits in acceding order

X/2-5 ____ ____ 20 ____ ____ X

To minimise we have to choose 2nd and 3rd digit as x/2-5 and 5th and 6th as 20.

Sum it up and equate to 140

we get X=38

So B is the Answer

]]>

So first number would be X/2-5

We know that Sum = Average * No of Terms

So Sum = 20* 7 = 140

To maximise the Last Digit we have to Minimise all six digits

Arrange the Digits in acceding order

X/2-5 ____ ____ 20 ____ ____ X

To minimise we have to choose 2nd and 3rd digit as x/2-5 and 5th and 6th as 20.

Sum it up and equate to 140

we get X=38

So B is the Answer]]>

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.]]>

Q: Average>6? Y/N

St.1: 2,4,12,12,12 => M=42/5>6 OR 2,4,2,2,12 => M=22/5<6. INSUFF

St.2: 2,2,10,10,12 => M=36/5>6 as well as other scenarios. SUFF

B

]]>

Q: Average>6? Y/N

St.1: 2,4,12,12,12 => M=42/5>6 OR 2,4,2,2,12 => M=22/5<6. INSUFF

St.2: 2,2,10,10,12 => M=36/5>6 as well as other scenarios. SUFF

B]]>

1) Study Material's - I assumed the GMAT website would provide enough, boy was I wrong. It almost feels like they stole my money. I prefer to study old school text book style - Concept - > Questions - Intermediate Concept - > Intermediate Questions - >Advanced Concept -> Advanced Questions - where do I go? Is there a one book meets all that covers the five modules of the Quantitative

...

]]>

1) Study Material's - I assumed the GMAT website would provide enough, boy was I wrong. It almost feels like they stole my money. I prefer to study old school text book style - Concept - > Questions - Intermediate Concept - > Intermediate Questions - >Advanced Concept -> Advanced Questions - where do I go? Is there a one book meets all that covers the five modules of the Quantitative

...]]>

KillerSquirrel wrote:

Mixture problems made Easy

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see attachment

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i did not understand even a single strategy .. would you please simplify any one of them with the image structure??

My personal opinion is that a time line is easier than 4 arrows,

...

]]>

KillerSquirrel wrote:

Mixture problems made Easy

solving mixture problems fast ! - (30 day money back guarantee )

see attachment

Thanks

solving mixture problems fast ! - (30 day money back guarantee )

see attachment

Thanks

i did not understand even a single strategy .. would you please simplify any one of them with the image structure??

My personal opinion is that a time line is easier than 4 arrows,

...]]>

Bunuel wrote:

SOLUTION

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1

(B) 4

(C) 2 − 2x

(D) 4x − 2

(E) 6x −1

\(\sqrt{3-2x} = \sqrt{2x} +1\) --> square both sides:\((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) -->\(3-2x=2x+2*\sqrt{2x}+1\) --> rearrange so that to have root at one side:\(2-4x=2*\sqrt{2x}\) --> reduce by 2:\(1-2x=\sqrt{2x}\) --> square again:\((1-2x)^2=(\sqrt{2x})^2\) -->\(1-4x+4x^2=2x\) --> rearrange again:\(4x^2=6x-1\) .

Answer: E.

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1

(B) 4

(C) 2 − 2x

(D) 4x − 2

(E) 6x −1

\(\sqrt{3-2x} = \sqrt{2x} +1\) --> square both sides:\((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) -->\(3-2x=2x+2*\sqrt{2x}+1\) --> rearrange so that to have root at one side:\(2-4x=2*\sqrt{2x}\) --> reduce by 2:\(1-2x=\sqrt{2x}\) --> square again:\((1-2x)^2=(\sqrt{2x})^2\) -->\(1-4x+4x^2=2x\) --> rearrange again:\(4x^2=6x-1\) .

Answer: E.

...

]]>

Bunuel wrote:

SOLUTION

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1

(B) 4

(C) 2 − 2x

(D) 4x − 2

(E) 6x −1

\(\sqrt{3-2x} = \sqrt{2x} +1\) --> square both sides:\((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) -->\(3-2x=2x+2*\sqrt{2x}+1\) --> rearrange so that to have root at one side:\(2-4x=2*\sqrt{2x}\) --> reduce by 2:\(1-2x=\sqrt{2x}\) --> square again:\((1-2x)^2=(\sqrt{2x})^2\) -->\(1-4x+4x^2=2x\) --> rearrange again:\(4x^2=6x-1\) .

Answer: E.

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1

(B) 4

(C) 2 − 2x

(D) 4x − 2

(E) 6x −1

\(\sqrt{3-2x} = \sqrt{2x} +1\) --> square both sides:\((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) -->\(3-2x=2x+2*\sqrt{2x}+1\) --> rearrange so that to have root at one side:\(2-4x=2*\sqrt{2x}\) --> reduce by 2:\(1-2x=\sqrt{2x}\) --> square again:\((1-2x)^2=(\sqrt{2x})^2\) -->\(1-4x+4x^2=2x\) --> rearrange again:\(4x^2=6x-1\) .

Answer: E.

...]]>

3.500 * (7/200) = 122,5. As there can not be half a adviser, round up to 123, which is the minimum amount of advisers. Answer D.

]]>

3.500 * (7/200) = 122,5. As there can not be half a adviser, round up to 123, which is the minimum amount of advisers. Answer D.]]>

81 = q * 2^3 + 17

64 = q+z^3

Then I tried answer possibilities of 2, 4 and 8.

Only 2^3 = 64 fulfills the equation and yields a remainder of 17. So B.

]]>

81 = q * 2^3 + 17

64 = q+z^3

Then I tried answer possibilities of 2, 4 and 8.

Only 2^3 = 64 fulfills the equation and yields a remainder of 17. So B.]]>

required value of b ?

Got tricked initially (did the same mistake ), good question ( need to take note thata,b are different non-zero integers)

stmt 1: a^b =ab

- (a,b) =

3^1 = 3*1

4^1 = 4*1...

b is always1---> b =1 got one solution , so Sufficient

stmt 2: a^b - a^b-1=2

- Simplify : a^b (1-\(\frac{1}{a}\)) =2

this equation works for values as above (a,b) =

2^2 (1-\(\frac{1}{2}\)) => 4 (\(\frac{1}{2}\)) = 2

3^1 (1-\(\frac{1}{3}\)) => 3 (\(\frac{2}{3}\)) = 2

(a,b) = (3,1) works ---> b got one solution , so Sufficient

...

]]>

required value of b ?

Got tricked initially (did the same mistake ), good question ( need to take note thata,b are different non-zero integers)

stmt 1: a^b =ab

- (a,b) =

3^1 = 3*1

4^1 = 4*1...

b is always1---> b =1 got one solution , so Sufficient

stmt 2: a^b - a^b-1=2

- Simplify : a^b (1-\(\frac{1}{a}\)) =2

this equation works for values as above (a,b) =

2^2 (1-\(\frac{1}{2}\)) => 4 (\(\frac{1}{2}\)) = 2

3^1 (1-\(\frac{1}{3}\)) => 3 (\(\frac{2}{3}\)) = 2

(a,b) = (3,1) works ---> b got one solution , so Sufficient

...]]>

http://gmatclub.com/forum/when-81-is-divided-by-the-cube-of-positive-integer-z-the-remainder-is-197386.html#p1522792

Wish you all the best!

]]>

http://gmatclub.com/forum/when-81-is-divided-by-the-cube-of-positive-integer-z-the-remainder-is-197386.html#p1522792

Wish you all the best! ]]>

Hi Japinder,

Here goes my approach:

Is integer x negative?

(A) x is not equal to |x|

(B) x = -|y - 2|, where y is an integer

Statement A - if X <> |X| then this implies that x is positive number - Sufficient

Statement B - this introduces another variable 'y' we know nothing about - Insufficient

Hence A is the correct option.

Dearaimtoteach

By going through the solution I've posted just now, you'll see that even though y is just a variable, wecan deduce from Statement 2 that x is

...

]]>

Hi Japinder,

Here goes my approach:

Is integer x negative?

(A) x is not equal to |x|

(B) x = -|y - 2|, where y is an integer

Statement A - if X <> |X| then this implies that x is positive number - Sufficient

Statement B - this introduces another variable 'y' we know nothing about - Insufficient

Hence A is the correct option.

Dearaimtoteach

By going through the solution I've posted just now, you'll see that even though y is just a variable, wecan deduce from Statement 2 that x is

...]]>

Hi,

I would like to follow up on the post of linglinrtw, I initially answered the good answer however when re-reading the text, in order to be able to answer the question, we need to make an assumption on the population (it is not either college degree or master degree, it could be something else).

Basically, in many overlapping problems, it is not clearly stated black or white; 1 or 0; and never in the answer this assumption is taken into account.

However, this kind of assuption would be exactly

...

]]>

Hi,

I would like to follow up on the post of linglinrtw, I initially answered the good answer however when re-reading the text, in order to be able to answer the question, we need to make an assumption on the population (it is not either college degree or master degree, it could be something else).

Basically, in many overlapping problems, it is not clearly stated black or white; 1 or 0; and never in the answer this assumption is taken into account.

However, this kind of assuption would be exactly

...]]>

Completely missed this info while solving the question!

]]>

Completely missed this info while solving the question!

]]>

]]>

x+1.5x= 3.8

x=1.5

1.5x=2.3

the periond 2000-2005 is doble of 2.3 so, it is 4.6

the correct is

1.5 and 4.6

]]>

x+1.5x= 3.8

x=1.5

1.5x=2.3

the periond 2000-2005 is doble of 2.3 so, it is 4.6

the correct is

1.5 and 4.6]]>

Why is the following logic wrong?

1) P(W and E) = P(W)*P(E) = 0 --> P(W) or P(E) or both are zero. Insufficient.

Think about it like this. (Represent the given info in a tree structure as shown below). We are given that every ball is either Red or Blue or White. We are also given that ball of each color can have either an even number or an odd number written on them.

Now from statement 1, we can see that

...

]]>

Why is the following logic wrong?

1) P(W and E) = P(W)*P(E) = 0 --> P(W) or P(E) or both are zero. Insufficient.

Think about it like this. (Represent the given info in a tree structure as shown below). We are given that every ball is either Red or Blue or White. We are also given that ball of each color can have either an even number or an odd number written on them.

Now from statement 1, we can see that

...]]>

Moving to st2 it doesn't say much apart from the difference which can be a combination of any possible numbers having a difference of 3. Not sufficient.

Combining both statements f-c=3 we can get 6-c= 3 the value of z by adding 6=f

...

]]>

Moving to st2 it doesn't say much apart from the difference which can be a combination of any possible numbers having a difference of 3. Not sufficient.

Combining both statements f-c=3 we can get 6-c= 3 the value of z by adding 6=f

...]]>

The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between

A. 0 and 0.5

B. 0.5 and 1

C. 1 and 1.5

D. 1.5 and 2

E. 2 and 2.5

Assume any variable X = LHS - RHS

Calculating values of X at 0, 1 and 2. The answer lies between the two value where the polarity of X changes.

X at 0 = 0

X at 1 = -13

X at 2 = 724

Value lies between 1 and 2

X at 1.5 =109 (approx.) --- polarity changes after value at 1 but remains same for value at 2

=> X lies between 1 and 1.5

Hence answer

...

]]>

The positive value of x that satisfies the equation (1 + 2x)^5 = (1 + 3x)^4 is between

A. 0 and 0.5

B. 0.5 and 1

C. 1 and 1.5

D. 1.5 and 2

E. 2 and 2.5

Assume any variable X = LHS - RHS

Calculating values of X at 0, 1 and 2. The answer lies between the two value where the polarity of X changes.

X at 0 = 0

X at 1 = -13

X at 2 = 724

Value lies between 1 and 2

X at 1.5 =109 (approx.) --- polarity changes after value at 1 but remains same for value at 2

=> X lies between 1 and 1.5

Hence answer

...]]>

Bunuel wrote:

Example:

5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.

5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.

5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.

5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.

5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

Bunuel,

I'm slightly confused by your explanation and the GC Math Book. IMO, in example 1 regardless of the fact that you should round to the nearest tenth, you should always start backwards, basically that the ten-thousandth

...

]]>

Bunuel wrote:

Example:

5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.

5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.

5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.

5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.

5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

Bunuel,

I'm slightly confused by your explanation and the GC Math Book. IMO, in example 1 regardless of the fact that you should round to the nearest tenth, you should always start backwards, basically that the ten-thousandth

...]]>

200 people responded to a survey that asked them to rank three different brands of soap. The percentage of respondents that ranked each brand 1st, 2nd, and 3rd are listed above. If no respondents rated the soaps in the order Y, Z, X, how many respondents rated the soap in the following order: X, Y, Z?

A. 100

B. 60

C. 50

D. 30

E. 25

Kudos for a correct solution.

[Reveal] Spoiler:

Attachment:

Brand_Rankings.png [ 34.63 KiB | Viewed 74 times ]

...

]]>

200 people responded to a survey that asked them to rank three different brands of soap. The percentage of respondents that ranked each brand 1st, 2nd, and 3rd are listed above. If no respondents rated the soaps in the order Y, Z, X, how many respondents rated the soap in the following order: X, Y, Z?

A. 100

B. 60

C. 50

D. 30

E. 25

Kudos for a correct solution.

[Reveal] Spoiler:

Attachment:

Brand_Rankings.png [ 34.63 KiB | Viewed 74 times ]

...]]>

Is x > 0

(1) x^2 > x

(2) x^3 > x

(1) x^2 > x

Or x(x-1) >0

x=4

x=-5

both satisfies the equation. Not Sufficient.

(2) x^3 > x

\(x(x^2-1)>0\)

\(X=2\)

\(X=\frac{-1}{2}\)

Both satisfies the equation so Not Sufficient.

Together:

x(x-1)(x+1) + x(x-1) > 0

x(x-1) (x+2)>0

clearly -2<x<0 and x>1 both satisfies this inequality .

Answer E

see attached image .

...

Attachments

gmatclub.jpg [ 12.62 KiB | Viewed 38 times ]

]]>

Is x > 0

(1) x^2 > x

(2) x^3 > x

(1) x^2 > x

Or x(x-1) >0

x=4

x=-5

both satisfies the equation. Not Sufficient.

(2) x^3 > x

\(x(x^2-1)>0\)

\(X=2\)

\(X=\frac{-1}{2}\)

Both satisfies the equation so Not Sufficient.

Together:

x(x-1)(x+1) + x(x-1) > 0

x(x-1) (x+2)>0

clearly -2<x<0 and x>1 both satisfies this inequality .

Answer E

see attached image .

...

Attachments

gmatclub.jpg [ 12.62 KiB | Viewed 38 times ]

]]>

wait time = T

Then, tw + T = tb

or, (d/3 + T) = (d/6 + 3d/2x)

or, T = (3d/2x - d/6)

Stmt 1) This helps us solve for x but not for d... so T can't be found .... NOT SUFFICIENT

Stmt 2) Bob spent 32 mins. more being alone than with Wendy... or to say as follows:

time spent alone = (3d/2)*(1/x) = 3d/2x

time spent with wendy = (d/2)*(1/3)

...

]]>

wait time = T

Then, tw + T = tb

or, (d/3 + T) = (d/6 + 3d/2x)

or, T = (3d/2x - d/6)

Stmt 1) This helps us solve for x but not for d... so T can't be found .... NOT SUFFICIENT

Stmt 2) Bob spent 32 mins. more being alone than with Wendy... or to say as follows:

time spent alone = (3d/2)*(1/x) = 3d/2x

time spent with wendy = (d/2)*(1/3)

...]]>

Can anyone help ?

This seems like a very easy question, but somehow I am having trouble understanding it:

(Source: MGMAT - Guide 1 - page. 19 Chapter 1)

1. In the decimal, 2Ad7, d represents a digit from 0 to 9. If the value of the decimal

rounded to the nearest tenth is less than 2.5, what are the possible values of d?

Official Answer: {0, 1, 2, 3, 4}: If d is 5 or greater, the decimal rounded to the nearest tenth will be 2.5

My questions is:

if d = 0 , then 2.407 > 2.41 > 2.4

if d

...

]]>

Can anyone help ?

This seems like a very easy question, but somehow I am having trouble understanding it:

(Source: MGMAT - Guide 1 - page. 19 Chapter 1)

1. In the decimal, 2Ad7, d represents a digit from 0 to 9. If the value of the decimal

rounded to the nearest tenth is less than 2.5, what are the possible values of d?

Official Answer: {0, 1, 2, 3, 4}: If d is 5 or greater, the decimal rounded to the nearest tenth will be 2.5

My questions is:

if d = 0 , then 2.407 > 2.41 > 2.4

if d

...]]>

Honestly, I done understand why B is not the answer..

If he spends at least $20 on the ticket then $20 is the least possible amount that he spent on the ticket??

Explanations would be welcomed..

When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can getthe single numerical value of this variable.

From (2) it's possible that the prices are20 , 25, 25, 30 OR25 , 25, 25, 25, OR21 , 22, 23, 34, ...

...

]]>

Honestly, I done understand why B is not the answer..

If he spends at least $20 on the ticket then $20 is the least possible amount that he spent on the ticket??

Explanations would be welcomed..

When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can getthe single numerical value of this variable.

From (2) it's possible that the prices are20 , 25, 25, 30 OR25 , 25, 25, 25, OR21 , 22, 23, 34, ...

...]]>

At 8am on Thursday, two workers, A and B, each start working independently to build identical decorative lamps. Worker A completes her lamp at 5pm on Friday, while Worker B completes her lamp sometime during the morning on Friday. If both workers adhere to working hours of 8am to 12pm and 1pm to 5pm each day, at which of the following times might the two workers have completed a single lamp had they worked together at their respective constant rates?

A. Thursday, 1:30pm

B. Thursday, 2:15pm

C.

...

]]>

At 8am on Thursday, two workers, A and B, each start working independently to build identical decorative lamps. Worker A completes her lamp at 5pm on Friday, while Worker B completes her lamp sometime during the morning on Friday. If both workers adhere to working hours of 8am to 12pm and 1pm to 5pm each day, at which of the following times might the two workers have completed a single lamp had they worked together at their respective constant rates?

A. Thursday, 1:30pm

B. Thursday, 2:15pm

C.

...]]>

(1) 2xy - y^2 = x^2 --> rearrange: x^2 -2xy + y^2 = 0 --> (x - y)^2 = 0 --> x - y =0. Sufficient.

(2) y^2 = -x^2 --> rearrange: y^2 + x^2 = 0 --> the sum of two non-negative values can be zero only if both are 0. Thus x = y = 0 --> x - y = 0. Sufficient.

Answer: D.

Similar questions to practice:

what-is-the-difference-of-the-integers-m-and-n-154567.htmlwhat-is-the-difference-of-the-integers-m-and-n-154567.html