PrashantPonde wrote:

Question: If 3x + 5 < x + 11, is x prime?

Simplify equation to x<3. Hence question becomesis x=2?

(1) the sum of x and y is even

INSUFFICIENT : We dont know y

(2) The product of x and y isodd.

INSUFFICIENT : We dont know whether y is integer

e.g. x=2, y=1.5 then xy=3 (ODD)

e.g. x=1, y=3 then xy=3(ODD)

x can be 1 or 2, hence its not sufficient.

Combining (1) &(2)

SUFFICIENT :

If x = 2, y must be 1.5 or -1.5 to make xy (ODD), however x + y cannot be EVEN.

If x

Simplify equation to x<3. Hence question becomesis x=2?

(1) the sum of x and y is even

INSUFFICIENT : We dont know y

(2) The product of x and y isodd.

INSUFFICIENT : We dont know whether y is integer

e.g. x=2, y=1.5 then xy=3 (ODD)

e.g. x=1, y=3 then xy=3(ODD)

x can be 1 or 2, hence its not sufficient.

Combining (1) &(2)

SUFFICIENT :

If x = 2, y must be 1.5 or -1.5 to make xy (ODD), however x + y cannot be EVEN.

If x

...

]]>

PrashantPonde wrote:

Question: If 3x + 5 < x + 11, is x prime?

Simplify equation to x<3. Hence question becomesis x=2?

(1) the sum of x and y is even

INSUFFICIENT : We dont know y

(2) The product of x and y isodd.

INSUFFICIENT : We dont know whether y is integer

e.g. x=2, y=1.5 then xy=3 (ODD)

e.g. x=1, y=3 then xy=3(ODD)

x can be 1 or 2, hence its not sufficient.

Combining (1) &(2)

SUFFICIENT :

If x = 2, y must be 1.5 or -1.5 to make xy (ODD), however x + y cannot be EVEN.

If x

Simplify equation to x<3. Hence question becomesis x=2?

(1) the sum of x and y is even

INSUFFICIENT : We dont know y

(2) The product of x and y isodd.

INSUFFICIENT : We dont know whether y is integer

e.g. x=2, y=1.5 then xy=3 (ODD)

e.g. x=1, y=3 then xy=3(ODD)

x can be 1 or 2, hence its not sufficient.

Combining (1) &(2)

SUFFICIENT :

If x = 2, y must be 1.5 or -1.5 to make xy (ODD), however x + y cannot be EVEN.

If x

...]]>

I don't see how C can be right. If I follow your reasoning, I come to E. (I chose B)

Can anyone explain the calculations?

If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

Given:AD=6 . Question:area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio1:\sqrt{3}:2 --> as AD=6 (larger leg opposite 60 degrees angle)

...

]]>

I don't see how C can be right. If I follow your reasoning, I come to E. (I chose B)

Can anyone explain the calculations?

If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

Given:AD=6 . Question:area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio1:\sqrt{3}:2 --> as AD=6 (larger leg opposite 60 degrees angle)

...]]>

Bunuel wrote:

PathFinder007 wrote:

Hi Bunnel,

Could you please provide your comments on this?

I am using following logic

|3x-7| = 2x +2. here L.H.S is positive so 2x+2 >= 0---> 2x>=-2--> x>=-1

so if if x>= -1 then |3x-7| = -3x+7 = 2x+2

-5x = -5

x = 1

only 1 value so sufficient. Ans A ( my question is why we are considering other scenario to get 9)

in st2

x^2 = 9x

x^2-9x = 0

x(x-9) = 0

x= 0 or 9

here we are getting 2 values so not sufficient.

...

]]>

Bunuel wrote:

PathFinder007 wrote:

Hi Bunnel,

Could you please provide your comments on this?

I am using following logic

|3x-7| = 2x +2. here L.H.S is positive so 2x+2 >= 0---> 2x>=-2--> x>=-1

so if if x>= -1 then |3x-7| = -3x+7 = 2x+2

-5x = -5

x = 1

only 1 value so sufficient. Ans A ( my question is why we are considering other scenario to get 9)

in st2

x^2 = 9x

x^2-9x = 0

x(x-9) = 0

x= 0 or 9

here we are getting 2 values so not sufficient.

...]]>

Is |2a – b| < 7?

(1) 2a – b < 7

(2) a = b + 3

Guyz, I can't agree with official answer. What are your thoughts on this?

Hi Bunnel,

Could you please provide your comments on this. I am using following logic

is |2a-b| <7

is -7<2a-b<7

here in st1 2a-b <7 but we dont know lower limit of 2a-b so clearly not sufficient.

in St2 a= b+3 . ( nothing new information)

I want to know how to combine both the statement and proceed further?

Thanks

...

]]>

Is |2a – b| < 7?

(1) 2a – b < 7

(2) a = b + 3

Guyz, I can't agree with official answer. What are your thoughts on this?

Hi Bunnel,

Could you please provide your comments on this. I am using following logic

is |2a-b| <7

is -7<2a-b<7

here in st1 2a-b <7 but we dont know lower limit of 2a-b so clearly not sufficient.

in St2 a= b+3 . ( nothing new information)

I want to know how to combine both the statement and proceed further?

Thanks

...]]>

(2^{2x})^2 = 60^2

2^{2x} = 60 .............. (1)

(2^{1-x})^2 = \frac{2^2}{2^{2x}}

= \frac{4}{60}.............. (From 1)

= \frac{1}{15}

Answer = B

]]>

(2^{2x})^2 = 60^2

2^{2x} = 60 .............. (1)

(2^{1-x})^2 = \frac{2^2}{2^{2x}}

= \frac{4}{60}.............. (From 1)

= \frac{1}{15}

Answer = B]]>

Service charge = 25

Fraction = \frac{25}{10000-y}

Answer = A

Bunuel, kindly write the OA in mathematical form

]]>

Service charge = 25

Fraction = \frac{25}{10000-y}

Answer = A

Bunuel, kindly write the OA in mathematical form ]]>

Bunuel wrote:

SOLUTIONs:

But x cannot be negative as it equals to the even (4th) root of some expression (\sqrt{expression}\geq{0}), thus only two solution are valid x=0 and x=3.

But x cannot be negative as it equals to the even (4th) root of some expression (\sqrt{expression}\geq{0}), thus only two solution are valid x=0 and x=3.

Can sumone plz explain the part in the quote? I dont quite get it I read WoundedTiger's post (one above) as well i still do not get it a lil help in layman terms plz

If the entire term under

...

]]>

Bunuel wrote:

SOLUTIONs:

But x cannot be negative as it equals to the even (4th) root of some expression (\sqrt{expression}\geq{0}), thus only two solution are valid x=0 and x=3.

But x cannot be negative as it equals to the even (4th) root of some expression (\sqrt{expression}\geq{0}), thus only two solution are valid x=0 and x=3.

Can sumone plz explain the part in the quote? I dont quite get it I read WoundedTiger's post (one above) as well i still do not get it a lil help in layman terms plz

If the entire term under

...]]>

\frac{6}{3} = 2 & \frac{-6}{-3} = 2

Checking all options:

A: a>0 ............ Fail for a = -6; b = -3

B: b>0 ............... Fail for a = -6; b =-3

C: ab>0 ................ OK for both (6*3 = 18 & -6 * -3 = 18)

D: a-b > 0 ........... Fail for a = -6; b = -3

E: a+b > 0 ........... Fail for a = -6; b = -3

Answer = C

...

]]>

\frac{6}{3} = 2 & \frac{-6}{-3} = 2

Checking all options:

A: a>0 ............ Fail for a = -6; b = -3

B: b>0 ............... Fail for a = -6; b =-3

C: ab>0 ................ OK for both (6*3 = 18 & -6 * -3 = 18)

D: a-b > 0 ........... Fail for a = -6; b = -3

E: a+b > 0 ........... Fail for a = -6; b = -3

Answer = 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).

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.]]>

After adjusting powers of 10, Answer = B

5.9 * 10^9

]]>

After adjusting powers of 10, Answer = B

5.9 * 10^9]]>

Narenn wrote:

Refer the attached Image

Hi Narenn,

How did you get the 2 there in Qty X? Kindly explain.

He's using allegation method to solve the problem

Mixture X = 80 gallon & Mixture Y = 120

There ratio = 80:120 = 2:3

From here, the values 2 & 3 are put in the allegation expression constructed

]]>

Narenn wrote:

Refer the attached Image

Hi Narenn,

How did you get the 2 there in Qty X? Kindly explain.

He's using allegation method to solve the problem

Mixture X = 80 gallon & Mixture Y = 120

There ratio = 80:120 = 2:3

From here, the values 2 & 3 are put in the allegation expression constructed]]>

Refer diagram below

Answer = 1/2

Bunuel, kindly update the OA.

Thanks in advance

Attachments

all.png [ 3.61 KiB | Viewed 15 times ]

]]>

Refer diagram below

Answer = 1/2

Bunuel, kindly update the OA.

Thanks in advance

Attachments

all.png [ 3.61 KiB | Viewed 15 times ]

]]>

(1) The product rs is divisible by 7. Since r and s are integers and 7 is prime, then either r or s must be divisible by 7. We don't know which Bot sufficient.

(2) s is not divisible by 7. Not sufficient.

(1)+(2) From (1) we know that either r or s must be divisible by 7 and from (2) we kn ow that it's not s, so it must be r. Sufficient.

Answer: C.

P.S. Please readcarefully and follow: rules-for-posting-please-read-this-before-posting-133935.htmlrules-for-posting-please-read-this-before-posting-133935.htmlif-n-is-the-product-of-all-positive-integers-less-than-103218.htmlif-n-is-the-product-of-all-positive-integers-less-than-103218.html