Let m and n be positive integers. Is mn > 100?

(1) m < 100n

(2) m – n = 101

Target question: Is mn > 100?

Given: m and n are positive integers.

Statement 1: m < 100n

This statement doesn't FEEL sufficient, so let's TEST some values.

There are several values of m and n that satisfy statement 1. Here are two:

Case a: m = 1 and n = 1. This satisfies statement 1. In this case,mn = (1)(1) = 1. So, mn < 100

Case b: m = 1 and n = 200. This satisfies statement 1. In

...

]]>

Let m and n be positive integers. Is mn > 100?

(1) m < 100n

(2) m – n = 101

Target question: Is mn > 100?

Given: m and n are positive integers.

Statement 1: m < 100n

This statement doesn't FEEL sufficient, so let's TEST some values.

There are several values of m and n that satisfy statement 1. Here are two:

Case a: m = 1 and n = 1. This satisfies statement 1. In this case,mn = (1)(1) = 1. So, mn < 100

Case b: m = 1 and n = 200. This satisfies statement 1. In

...]]>

]]>

By decrypting the second condition I will get (7^x-100) >0 and/or (7^x +100) >0

that means 7^x>100 which is fine, but don't we also need to take another equation into the account i.e 7^x > -100. So, from the second equation we are not sure about 7^x, since 7^x can be anywhere starting from -99.999.

...

]]>

By decrypting the second condition I will get (7^x-100) >0 and/or (7^x +100) >0

that means 7^x>100 which is fine, but don't we also need to take another equation into the account i.e 7^x > -100. So, from the second equation we are not sure about 7^x, since 7^x can be anywhere starting from -99.999.

...]]>

Hope the representation helps.

Attachments

B1.PNG [ 5.23 KiB | Viewed 9 times ]

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Hope the representation helps.

Attachments

B1.PNG [ 5.23 KiB | Viewed 9 times ]

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If it were not given that n and k are positive integers, then taking the underroot of this expression (Underoort n+k > 2 underroot n) would have resulted in absolute value form as l n + k l > 4 l n l. Am I right?

Just wanted to make sure that I get the concept right.

Thanks for your help.

Regards,

H

]]>

If it were not given that n and k are positive integers, then taking the underroot of this expression (Underoort n+k > 2 underroot n) would have resulted in absolute value form as l n + k l > 4 l n l. Am I right?

Just wanted to make sure that I get the concept right.

Thanks for your help.

Regards,

H]]>

Attachments

E1.PNG [ 5.02 KiB | Viewed 13 times ]

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Attachments

E1.PNG [ 5.02 KiB | Viewed 13 times ]

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m and n are positive integers. If an equation \(x^2+(m-2)x-(n-2)=0\) has only one root, what is the value of mn?

A. 1

B. 2

C. 4

D. 6

E. 8

\(x^2+(m-2)x-(n-2)=0\)

Or,\(x^2+mx-2x -n + 2=0\)

If x = 2

Or,\(x^2 -2x +mx-n + 2=0\)

Or,\(2^2 - 4 +2m-n + 2=0\)

Or,\(2m - n + 2=0\)

Here if m = 1 , n must be 4

Or,\(2*1 - 4 + 2=0\)

Hence, mn = 4

Answer will be (C) 4

...

]]>

m and n are positive integers. If an equation \(x^2+(m-2)x-(n-2)=0\) has only one root, what is the value of mn?

A. 1

B. 2

C. 4

D. 6

E. 8

\(x^2+(m-2)x-(n-2)=0\)

Or,\(x^2+mx-2x -n + 2=0\)

If x = 2

Or,\(x^2 -2x +mx-n + 2=0\)

Or,\(2^2 - 4 +2m-n + 2=0\)

Or,\(2m - n + 2=0\)

Here if m = 1 , n must be 4

Or,\(2*1 - 4 + 2=0\)

Hence, mn = 4

Answer will be (C) 4

...]]>

rohit8865 wrote:

stonecold wrote:

Is x+y even?

(1)A^x * A^y = 1

(2) A≠-1

(1)A^x * A^y = 1

(2) A≠-1

(1) if A=2 and x=y=0 then YES

if A=1 and x=2 & y=3 then NO-----not suff..

(2) similar cases as (1)

combining

check as above in (1)-----not suff

AnsE

stonecold

could u please confirm the OA

Hi rohit8865

I have edited the Question.

The Answer to the previous version was indeed E.

It should have been A≠1 instead of A≠-1

Regards

Stone Cold

...

]]>

rohit8865 wrote:

stonecold wrote:

Is x+y even?

(1)A^x * A^y = 1

(2) A≠-1

(1)A^x * A^y = 1

(2) A≠-1

(1) if A=2 and x=y=0 then YES

if A=1 and x=2 & y=3 then NO-----not suff..

(2) similar cases as (1)

combining

check as above in (1)-----not suff

AnsE

stonecold

could u please confirm the OA

Hi rohit8865

I have edited the Question.

The Answer to the previous version was indeed E.

It should have been A≠1 instead of A≠-1

Regards

Stone Cold

...]]>

Bunuel wrote:

If x and y are numbers such that (x + 11) (y – 11) = 0, what is the smallest possible value of x^2 + y^2 ?

A) 0

B) 11

C) 22

D) 121

E) 242

Kudos for a correct solution.

A) 0

B) 11

C) 22

D) 121

E) 242

Kudos for a correct solution.

From (x + 11)(y - 11)=0 it follows thateither x = -11or y = 11. Thus either x^2 = 121 or y^2 = 121.

Now, if x^2 = 121, then the least value of y^2 is 0, so the least value of x^2 + y^2 = 121 + 0 = 121.

Similarly if y^2 = 121, then the least value of x^2 is 0, so the least value of x^2 + y^2 = 0 + 121 = 121.

Answer:

...

]]>

Bunuel wrote:

If x and y are numbers such that (x + 11) (y – 11) = 0, what is the smallest possible value of x^2 + y^2 ?

A) 0

B) 11

C) 22

D) 121

E) 242

Kudos for a correct solution.

A) 0

B) 11

C) 22

D) 121

E) 242

Kudos for a correct solution.

From (x + 11)(y - 11)=0 it follows thateither x = -11or y = 11. Thus either x^2 = 121 or y^2 = 121.

Now, if x^2 = 121, then the least value of y^2 is 0, so the least value of x^2 + y^2 = 121 + 0 = 121.

Similarly if y^2 = 121, then the least value of x^2 is 0, so the least value of x^2 + y^2 = 0 + 121 = 121.

Answer:

...]]>

The formula for such problems is like

9 /4= 2

9/4^2=0

Total = 2

However answer should be 3 if we actually count it. Where am I going wrong?

Posted from my mobile device

rakaisraka

when u r finding 4^y means u have to count every 2's ..

suppose if it was 10! then it must have 1*2*...*6...*10

then it has 6=2*3 && 10=2*5

where one no. 2 from 6 and one no. 2 from 10 also counted as a 4 in 10!

let me make more clear if u have to find 6^y

...

]]>

The formula for such problems is like

9 /4= 2

9/4^2=0

Total = 2

However answer should be 3 if we actually count it. Where am I going wrong?

Posted from my mobile device

rakaisraka

when u r finding 4^y means u have to count every 2's ..

suppose if it was 10! then it must have 1*2*...*6...*10

then it has 6=2*3 && 10=2*5

where one no. 2 from 6 and one no. 2 from 10 also counted as a 4 in 10!

let me make more clear if u have to find 6^y

...]]>

The Average age of a class of 10 students is 15.An year later,one more student joins in.The average age of the class becomes 16 years. What is the age of the newest member of the class ?

A. 26 years

B. 22 years

C. 20 years

D. 18 years

E. 16 years

Total age of 10 students now is 150

Total age of 10 students after 1 year is 160

Now,\(\frac{160 + x}{11} = 16\)

Or,\(160 + x = 176\)

Or,\(x = 16\)

Hence, answer will be (E) 16

...

]]>

The Average age of a class of 10 students is 15.An year later,one more student joins in.The average age of the class becomes 16 years. What is the age of the newest member of the class ?

A. 26 years

B. 22 years

C. 20 years

D. 18 years

E. 16 years

Total age of 10 students now is 150

Total age of 10 students after 1 year is 160

Now,\(\frac{160 + x}{11} = 16\)

Or,\(160 + x = 176\)

Or,\(x = 16\)

Hence, answer will be (E) 16

...]]>

The number m is the average (arithmetic mean) of the positive numbers a and b. If m is 75% more than a, then m must be

A. 30% less than b

B. \(42 \frac{6}{7}\)% less than b

C. 50% less than b

D. \(66 \frac{2}{3}\)% less than b

E. 75% less than b

If m is 75% more than a

Let a = 4 so m = 7

The number m is the average (arithmetic mean) of the positive numbers a and b

So,\(\frac{a + b}{2} = m\)

Or,\(\frac{4 + b}{2} = 7\)

Or,\(4 + b = 14\)

Or,\(b = 10\)

Thus, m is 30% less than b, answer will definitely be (A)

...

]]>

The number m is the average (arithmetic mean) of the positive numbers a and b. If m is 75% more than a, then m must be

A. 30% less than b

B. \(42 \frac{6}{7}\)% less than b

C. 50% less than b

D. \(66 \frac{2}{3}\)% less than b

E. 75% less than b

If m is 75% more than a

Let a = 4 so m = 7

The number m is the average (arithmetic mean) of the positive numbers a and b

So,\(\frac{a + b}{2} = m\)

Or,\(\frac{4 + b}{2} = 7\)

Or,\(4 + b = 14\)

Or,\(b = 10\)

Thus, m is 30% less than b, answer will definitely be (A)

...]]>

If it takes a train 20 minutes to go from station A to station B, how long does it take the train to return from station B to station A?

(1) The average speed of the train from station A to station B is half its speed from station B to station A.

(2) The distance between the stations is 10 miles.

FROM STATEMENT - I ( SUFFICIENT )

There is an inverse relationship between Speed and Time..

The more the speed the lesser the time taken to cover the same distance..

Since ,\( \frac{Speed \ ( A->B)}{2}\)

...

]]>

If it takes a train 20 minutes to go from station A to station B, how long does it take the train to return from station B to station A?

(1) The average speed of the train from station A to station B is half its speed from station B to station A.

(2) The distance between the stations is 10 miles.

FROM STATEMENT - I ( SUFFICIENT )

There is an inverse relationship between Speed and Time..

The more the speed the lesser the time taken to cover the same distance..

Since ,\( \frac{Speed \ ( A->B)}{2}\)

...]]>

W2

640

Aavg

820

(W1-Aavg) / (Aavg-W2) = Ratio

(980-820) / (820 - 640)

160 / 180

8 / 9 Reverse it. 9:8

]]>

W2

640

Aavg

820

(W1-Aavg) / (Aavg-W2) = Ratio

(980-820) / (820 - 640)

160 / 180

8 / 9 Reverse it. 9:8]]>

z = abc.d36

If a, b, c and d denote the hundreds, tens, units and tenths digits respectively in the decimal representation of z above, is the number that results when z is rounded to the nearest integer divisible by 3?

(1) When z/100 is rounded to the nearest hundredths, the result is 1.85.

(2) When z+2 is rounded to the nearest tens, the result is 190.

This question has some tricks

(1) \(\frac{z}{100}=\overline{a.bcd36}

...

]]>

z = abc.d36

If a, b, c and d denote the hundreds, tens, units and tenths digits respectively in the decimal representation of z above, is the number that results when z is rounded to the nearest integer divisible by 3?

(1) When z/100 is rounded to the nearest hundredths, the result is 1.85.

(2) When z+2 is rounded to the nearest tens, the result is 190.

This question has some tricks

(1) \(\frac{z}{100}=\overline{a.bcd36}

...]]>

height is doubled so the new height is 2h

new volume = \(\pi\) * \((2r)^2\)*2h = 8 \(\pi\)\(r^2\)h = 8 * original volume of the cylinder

Correct Answer - A

]]>

height is doubled so the new height is 2h

new volume = \(\pi\) * \((2r)^2\)*2h = 8 \(\pi\)\(r^2\)h = 8 * original volume of the cylinder

Correct Answer - A]]>

h=r

Correct answer - A

]]>

h=r

Correct answer - A]]>

Drop a perpendicular from a vertex to opposite side..

Check the attachment

Attachments

IMG_20161209_194015.jpg [ 1.37 MiB | Viewed 38 times ]

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Drop a perpendicular from a vertex to opposite side..

Check the attachment

Attachments

IMG_20161209_194015.jpg [ 1.37 MiB | Viewed 38 times ]

]]>

T = 2.z hours

If z denotes the tenths digit in the decimal representation of the time T that a car takes to cover 258 miles, what is the value of z?

(1) The speed of the car is 1 mile per minute, rounded to the nearest integer

(2) When T is rounded to the nearest 100 minutes, the result is 200.

\(T=2.z \;\) hours

\(s=258\; \)miles

\(v=\frac{s}{T} \; \)miles/hour

(1) The speed of car is 1 mile/minute, and this number is already rounded to the nearest integer. We have

\(0.5 \;\) miles/minute

...

]]>

T = 2.z hours

If z denotes the tenths digit in the decimal representation of the time T that a car takes to cover 258 miles, what is the value of z?

(1) The speed of the car is 1 mile per minute, rounded to the nearest integer

(2) When T is rounded to the nearest 100 minutes, the result is 200.

\(T=2.z \;\) hours

\(s=258\; \)miles

\(v=\frac{s}{T} \; \)miles/hour

(1) The speed of car is 1 mile/minute, and this number is already rounded to the nearest integer. We have

\(0.5 \;\) miles/minute

...]]>

When you combine both statement u can confirm that d=2 because D cant be 0 as m,n,p are different..

So,option C

Sent from my C6902 using GMAT Club Forum mobile app

]]>

When you combine both statement u can confirm that d=2 because D cant be 0 as m,n,p are different..

So,option C

Sent from my C6902 using GMAT Club Forum mobile app]]>

]]>

If M is the mean of x,y,z ; then what is the mean of x+10,y+10 and z+10

A. M

B. 3M

C 10M

D. M+10

E. M-10

\(M = \frac{( x + y + z )}{3}\)

Or,\(x + y + z = 3M\)

Now,\(( x+10 ) + ( y+10 ) +( z+10 ) = ( x + y + z )+ 30 => 3M + 30\)

Thus, Average of ( x+10 ) + ( y+10 ) +( z+10 ) will be M + 10

Hence, correct answer will be (D)

...

]]>

If M is the mean of x,y,z ; then what is the mean of x+10,y+10 and z+10

A. M

B. 3M

C 10M

D. M+10

E. M-10

\(M = \frac{( x + y + z )}{3}\)

Or,\(x + y + z = 3M\)

Now,\(( x+10 ) + ( y+10 ) +( z+10 ) = ( x + y + z )+ 30 => 3M + 30\)

Thus, Average of ( x+10 ) + ( y+10 ) +( z+10 ) will be M + 10

Hence, correct answer will be (D)

...]]>

man, that's just out of this world!

]]>

man, that's just out of this world!]]>

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.]]>

If z is an odd integer, is 300z > 1500?

(1) \(0<\sqrt{z}<8\)

(2) \(1<z^2<45\)

Please correct if wrong!!

z > 5?

1) 0 < z < 64 --> not clear weather z is bigger than 5. #insufficient

2) 1 < z < 6.7 --> possible number 5 and 3, which means that z is not > 5! #sufficient

B)

]]>

If z is an odd integer, is 300z > 1500?

(1) \(0<\sqrt{z}<8\)

(2) \(1<z^2<45\)

Please correct if wrong!!

z > 5?

1) 0 < z < 64 --> not clear weather z is bigger than 5. #insufficient

2) 1 < z < 6.7 --> possible number 5 and 3, which means that z is not > 5! #sufficient

B)]]>

On the other hand, if the expression would be (a+b)^2 = 16 , would this only yield one possible answer, 4?

thank you

]]>

On the other hand, if the expression would be (a+b)^2 = 16 , would this only yield one possible answer, 4?

thank you]]>

Bunuel would you mind explaining the question: it says 5 times the quantity ie 5*( 7^x-1 - 5^x) why are you just multiplying 7^x-1 * 5 and not 5*?

]]>

Bunuel would you mind explaining the question: it says 5 times the quantity ie 5*( 7^x-1 - 5^x) why are you just multiplying 7^x-1 * 5 and not 5*?]]>

Sum of the first 100-even integers: 100(101) = 10100

Sum of the first 200-even integers: 200(201) = 40200

Sum of the the even integers between the first 100 and the first 200: 40200-10100 = 30100

--> 10100/30100

]]>

Sum of the first 100-even integers: 100(101) = 10100

Sum of the first 200-even integers: 200(201) = 40200

Sum of the the even integers between the first 100 and the first 200: 40200-10100 = 30100

--> 10100/30100]]>

kanav06 wrote:

I'm stumped!

what is wrong in the approach below?

y= 2x+c (we know that c=0). Thus,

y=2x........equation 1

next,

(4-y)/(x-3)= 2

Thus, 10= 2x+y....equation 2

From equation 1 & equation 2, I found x= 5/2 & y= 5.

Thus x+y= 7.5

I'm not understanding what is wrong in this process!

what is wrong in the approach below?

y= 2x+c (we know that c=0). Thus,

y=2x........equation 1

next,

(4-y)/(x-3)= 2

Thus, 10= 2x+y....equation 2

From equation 1 & equation 2, I found x= 5/2 & y= 5.

Thus x+y= 7.5

I'm not understanding what is wrong in this process!

y = 2x is the equation of the line - fine.

But these two - (3,y) and (x,4) are points. Here y and x are specific co-ordinate points.

Think of them instead as points (3, a) and (b, 4). You

...

]]>

kanav06 wrote:

I'm stumped!

what is wrong in the approach below?

y= 2x+c (we know that c=0). Thus,

y=2x........equation 1

next,

(4-y)/(x-3)= 2

Thus, 10= 2x+y....equation 2

From equation 1 & equation 2, I found x= 5/2 & y= 5.

Thus x+y= 7.5

I'm not understanding what is wrong in this process!

what is wrong in the approach below?

y= 2x+c (we know that c=0). Thus,

y=2x........equation 1

next,

(4-y)/(x-3)= 2

Thus, 10= 2x+y....equation 2

From equation 1 & equation 2, I found x= 5/2 & y= 5.

Thus x+y= 7.5

I'm not understanding what is wrong in this process!

y = 2x is the equation of the line - fine.

But these two - (3,y) and (x,4) are points. Here y and x are specific co-ordinate points.

Think of them instead as points (3, a) and (b, 4). You

...]]>

I have been preparing for my GMAT from past 2-3 months.I started with GC tests last month and have completed 10 quant tests.

In each of these tests I have consistently got 19-21 correct. I want to increase my accuracy in quant.

I have around 3 weeks left for my exam.Please help me with the strategy that I should follow in getting the maximum out of these tests and how should I increase my accuracy.

Thanks in advance!!

]]>

I have been preparing for my GMAT from past 2-3 months.I started with GC tests last month and have completed 10 quant tests.

In each of these tests I have consistently got 19-21 correct. I want to increase my accuracy in quant.

I have around 3 weeks left for my exam.Please help me with the strategy that I should follow in getting the maximum out of these tests and how should I increase my accuracy.

Thanks in advance!!]]>

Buttermilk=2x Orange Juice

Orange juice= 3 x Brandy, meaning Brandy=Orange.j/3

Equation will be 7.5=Buttermilk+orange juice +brandy which will be 7.5=2*orange.j+Orange.j+Orange.j/3

to get ridd off fraction, multiply both sides with3

22.5(gallons)=6*Orange.j+3*Orange.j+Orange.j , which is22.5(Gallons)=10*orange.j

Answer is given in pints so we multiply22.5*8=180 pints.

180=10*Orange.j

Number of Orange juice pints =18

Asnwer B

+1 if you liked post

...

]]>

Buttermilk=2x Orange Juice

Orange juice= 3 x Brandy, meaning Brandy=Orange.j/3

Equation will be 7.5=Buttermilk+orange juice +brandy which will be 7.5=2*orange.j+Orange.j+Orange.j/3

to get ridd off fraction, multiply both sides with3

22.5(gallons)=6*Orange.j+3*Orange.j+Orange.j , which is22.5(Gallons)=10*orange.j

Answer is given in pints so we multiply22.5*8=180 pints.

180=10*Orange.j

Number of Orange juice pints =18

Asnwer B

+1 if you liked post

...]]>

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.]]>

Given data => mean =50

And number of terms =5

Hence sum=250.

Now let the smallest term be a

The largest term = 3a+5

Mean =50=median

Third term will be 50 (if the set is arranged in increasing or Decreasing Order)

To maximise the largest term we must minimise each term

First Term =a

Second term =a

Third=50

Fourth =50

Fifth = 3a+5

Hence 5a+105=250

Thus,a=29

The largest term = 3*29+5 = 92

Hence E

]]>

Given data => mean =50

And number of terms =5

Hence sum=250.

Now let the smallest term be a

The largest term = 3a+5

Mean =50=median

Third term will be 50 (if the set is arranged in increasing or Decreasing Order)

To maximise the largest term we must minimise each term

First Term =a

Second term =a

Third=50

Fourth =50

Fifth = 3a+5

Hence 5a+105=250

Thus,a=29

The largest term = 3*29+5 = 92

Hence E]]>

On five chairs around a big round table in a class, each of the 5 students A, B, C, D, and E sit on a chair. A must sit opposite to the white board and B must sit beside A. How many seating arrangements are there?

A. 12

B. 24

C. 44

D. 48

E. 60

help me solve this question

Hi

On five chairs, the one opposite white board is A..

Say B sits on left side of A, the remaining 3 can be seated in 3! ways..

Similarly when B sits to right of A, another 6 ways..

Total 12 ways

...

]]>

On five chairs around a big round table in a class, each of the 5 students A, B, C, D, and E sit on a chair. A must sit opposite to the white board and B must sit beside A. How many seating arrangements are there?

A. 12

B. 24

C. 44

D. 48

E. 60

help me solve this question

Hi

On five chairs, the one opposite white board is A..

Say B sits on left side of A, the remaining 3 can be seated in 3! ways..

Similarly when B sits to right of A, another 6 ways..

Total 12 ways

...]]>

If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?

(1) There are 15 terms in S.

In every set of 3 consecutive multiples of 3, there will be one multiple of9:

{-9 , -6, -3}

{-6, -3,0 }

{-3,0 ,3}

{0 , 3, 6}

{3, 6,9 }

...

15 consecutive multiples of 3 could be break down into 5 different such sets, each of which will contain 1 multiple of 9. Therefore, there are 5 multiples of 9 in the set. Sufficient.

(2) The greatest term of S is 126. Clearly insufficient.

Answer:

...

]]>

If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?

(1) There are 15 terms in S.

In every set of 3 consecutive multiples of 3, there will be one multiple of9:

{-9 , -6, -3}

{-6, -3,0 }

{-3,0 ,3}

{0 , 3, 6}

{3, 6,9 }

...

15 consecutive multiples of 3 could be break down into 5 different such sets, each of which will contain 1 multiple of 9. Therefore, there are 5 multiples of 9 in the set. Sufficient.

(2) The greatest term of S is 126. Clearly insufficient.

Answer:

...]]>

Given data => Mean =50

Let the smallest number be x

so the largest number will be 3x+4

AS they are different =>

x<50

So the maximum value of 3a+4=> 154

Hence All the values must be less than 154

Also 3a+4>50

Hence a must be greater than 15.3

Hence the range of values of data set will be => (15.3,154)

Clearly E is out of Bound here.

Hence E

]]>

Given data => Mean =50

Let the smallest number be x

so the largest number will be 3x+4

AS they are different =>

x<50

So the maximum value of 3a+4=> 154

Hence All the values must be less than 154

Also 3a+4>50

Hence a must be greater than 15.3

Hence the range of values of data set will be => (15.3,154)

Clearly E is out of Bound here.

Hence E]]>

Let Mean=p

APQ=> 2(n+1)=39-p

Hence p+2n=37

Also -1(n+1)=15-p

Hence p=16+n

Clearly Both of these Equations are as the above method.

To know Everything about Mean or Statistics in general, refer to this Post -->

statistics-made-easy-all-in-one-topic-203966.html#p1563151

]]>

Let Mean=p

APQ=> 2(n+1)=39-p

Hence p+2n=37

Also -1(n+1)=15-p

Hence p=16+n

Clearly Both of these Equations are as the above method.

To know Everything about Mean or Statistics in general, refer to this Post -->

statistics-made-easy-all-in-one-topic-203966.html#p1563151]]>

It can be easily solved using smart numbers

Doris = D, Fred = F

Let's say D = 24 than F is 12 less = 12 --> 24 + 12 = Y = 36; So in Y or 36 Years Fred will be 12 + 36 = 48

Plugin 36 for Y, in a right answer you schould get 48 --> D (3*36)/2-6 = 48

I tried this method and I am completely stumped.

3*36=108

2-6=-4

108/4=27 not 48.

Can

...

]]>

It can be easily solved using smart numbers

Doris = D, Fred = F

Let's say D = 24 than F is 12 less = 12 --> 24 + 12 = Y = 36; So in Y or 36 Years Fred will be 12 + 36 = 48

Plugin 36 for Y, in a right answer you schould get 48 --> D (3*36)/2-6 = 48

I tried this method and I am completely stumped.

3*36=108

2-6=-4

108/4=27 not 48.

Can

...]]>

Hi Bunnel,

I tried solving this problem by using the 45-45-90 triangle approach for triangle QSR !! Why is this approach incorrect ?

First of all, you don't show your work at all and asks what's wrong with it. Next, why do you assume that QSR is 45-45-90 triangle?

]]>

Hi Bunnel,

I tried solving this problem by using the 45-45-90 triangle approach for triangle QSR !! Why is this approach incorrect ?

First of all, you don't show your work at all and asks what's wrong with it. Next, why do you assume that QSR is 45-45-90 triangle?]]>

Bunuel wrote:

[b] Now, \(\frac{m}{n}\) can be neither an irrational number (since it's the ratio of two integers)

I am unable to understand this part can you please explain or perhaps provide me a link where I can understand my knowledge gap.Thanks!

Check here: if-n-p-q-p-and-q-are-nonzero-integers-is-an-integer-101475.html and here: ...

]]>

Bunuel wrote:

[b] Now, \(\frac{m}{n}\) can be neither an irrational number (since it's the ratio of two integers)

I am unable to understand this part can you please explain or perhaps provide me a link where I can understand my knowledge gap.Thanks!

Check here:

\(\frac{a}{b}\) = 35.24 = 35 +0.24 = 35 + \(\frac{24}{100}\)

:Remainder would come from \(\frac{24}{100}\) = \(\frac{6}{25}\)

:Remainder\((\frac{a}{b})\) = 6

]]>

\(\frac{a}{b}\) = 35.24 = 35 +0.24 = 35 + \(\frac{24}{100}\)

:Remainder would come from \(\frac{24}{100}\) = \(\frac{6}{25}\)

:Remainder\((\frac{a}{b})\) = 6]]>

Sum(8)=8*z

Replacing 14 by 28

New sum=8z-14+28=8z+14

Hence New Mean = \(\frac{8z+14}{8} = z+\frac{7}{4}\)

Hence B

]]>

Sum(8)=8*z

Replacing 14 by 28

New sum=8z-14+28=8z+14

Hence New Mean = \(\frac{8z+14}{8} = z+\frac{7}{4}\)

Hence B]]>

What if all the elements of the Set are 0? Then the answer would be E. 0 is a positive number.

0 is neither positive nor negative even integer.

]]>

What if all the elements of the Set are 0? Then the answer would be E. 0 is a positive number.

0 is neither positive nor negative even integer.]]>

Sum(1)=6*8.5=>51

Sum(2)=5*7.2=>36

Hence the discarded number must be 51-36=15

Hence E

]]>

Sum(1)=6*8.5=>51

Sum(2)=5*7.2=>36

Hence the discarded number must be 51-36=15

Hence E]]>

I had this question, but I'm not sure of the answer. Please provide an explanation.

If a and b are positive integers, and \((2^{3})(3^{4})(5^{7}) = a^{3b}\), how many different possible values of b are there?

A. 2

B. 4

C. 6

D. 9

E. 12

Check here:

...

]]>

I had this question, but I'm not sure of the answer. Please provide an explanation.

If a and b are positive integers, and \((2^{3})(3^{4})(5^{7}) = a^{3b}\), how many different possible values of b are there?

A. 2

B. 4

C. 6

D. 9

E. 12

Check here: if-a-and-b-are-positive-integers-and-147953.html

...]]>

What is the minimum percentage increase in the mean of set X {-4, -1, 0, 6, 9} if its two smallest elements are replaced with two different primes?

(A) 25% (B) 50% (C) 75% (D) 100% (E) 200%

Stuck with above problem looking for some guidance and assistance.

Merging topics. Please refer to the discussion above.

]]>

What is the minimum percentage increase in the mean of set X {-4, -1, 0, 6, 9} if its two smallest elements are replaced with two different primes?

(A) 25% (B) 50% (C) 75% (D) 100% (E) 200%

Stuck with above problem looking for some guidance and assistance.

Merging topics. Please refer to the discussion above.]]>