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(1) definitely not sufficient alone, cause we need to know how many men and women (not how many total of men and women).

(2) is also not sufficient. Let's assume that x and y both larger than or equal to 3, so the number of committees that can be formed is (3Cx * 1Cy) + (1Cx * 3Cy)

Clearly if we swap x and y this sum is not changed, so (2) is always right for x, y >= 3

Combine (1) and (2) we see that (x, y) can be (3, 9), (4, 8), (5, 7) or (6, 6), hence we still can't calculate the result.

=>

...

]]>

(1) definitely not sufficient alone, cause we need to know how many men and women (not how many total of men and women).

(2) is also not sufficient. Let's assume that x and y both larger than or equal to 3, so the number of committees that can be formed is (3Cx * 1Cy) + (1Cx * 3Cy)

Clearly if we swap x and y this sum is not changed, so (2) is always right for x, y >= 3

Combine (1) and (2) we see that (x, y) can be (3, 9), (4, 8), (5, 7) or (6, 6), hence we still can't calculate the result.

=>

...]]>

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Still unclear how you got only 50 percent of the family arguing. Would be helpful if you took numbers to show that. For example, 6 men and 6 women. 6 men argue with each other and 6 women argue with each other.

That's 12 people (100%) arguing. 8 men and 4 women would not change the percentage.

Posted from my mobile device

Hi Rahulkashyap, Here is how I got 50 percent arguing with each other.

The question states :The greater the percentage of family members who argue with each other, the greater

...

]]>

Still unclear how you got only 50 percent of the family arguing. Would be helpful if you took numbers to show that. For example, 6 men and 6 women. 6 men argue with each other and 6 women argue with each other.

That's 12 people (100%) arguing. 8 men and 4 women would not change the percentage.

Posted from my mobile device

Hi Rahulkashyap, Here is how I got 50 percent arguing with each other.

The question states :The greater the percentage of family members who argue with each other, the greater

...]]>

2. Head gets 3 points and tails 1 point

Hence, 3H + T = 52 -----> 3H + (H-4) = 52. Sufficient.

Answer : D

]]>

2. Head gets 3 points and tails 1 point

Hence, 3H + T = 52 -----> 3H + (H-4) = 52. Sufficient.

Answer : D]]>

I think this is a high-quality question and I don't agree with the explanation. x > -1 means x could be 1 and 1 is not less that 1. Can anyone elaborate?

This is a tough question. I suggest you re-read the WHOLE thread carefully.

]]>

I think this is a high-quality question and I don't agree with the explanation. x > -1 means x could be 1 and 1 is not less that 1. Can anyone elaborate?

This is a tough question. I suggest you re-read the WHOLE thread carefully.]]>

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cledgard wrote:

I solved it in a way that I think is more direct:

I get the probability that each cup is not a specific type of tea (type A):

1st cup, Probability it is not type A = 3/9

2nd cup, Probability it is not type A = 3/8, since we only have 8 cups left, and 3 of them are type A

3rd cup, Probability it is not type A = 3/7

4th cup, Probability it is not type A = 3/6

We do the same for types B and C (multiply by 3)

So : 3/9 * 3/8 * 3/7 * 3/6 * 3 = 5/14

I get the probability that each cup is not a specific type of tea (type A):

1st cup, Probability it is not type A = 3/9

2nd cup, Probability it is not type A = 3/8, since we only have 8 cups left, and 3 of them are type A

3rd cup, Probability it is not type A = 3/7

4th cup, Probability it is not type A = 3/6

We do the same for types B and C (multiply by 3)

So : 3/9 * 3/8 * 3/7 * 3/6 * 3 = 5/14

1st cup : 3/9 is actually the probability

...

]]>

cledgard wrote:

I solved it in a way that I think is more direct:

I get the probability that each cup is not a specific type of tea (type A):

1st cup, Probability it is not type A = 3/9

2nd cup, Probability it is not type A = 3/8, since we only have 8 cups left, and 3 of them are type A

3rd cup, Probability it is not type A = 3/7

4th cup, Probability it is not type A = 3/6

We do the same for types B and C (multiply by 3)

So : 3/9 * 3/8 * 3/7 * 3/6 * 3 = 5/14

I get the probability that each cup is not a specific type of tea (type A):

1st cup, Probability it is not type A = 3/9

2nd cup, Probability it is not type A = 3/8, since we only have 8 cups left, and 3 of them are type A

3rd cup, Probability it is not type A = 3/7

4th cup, Probability it is not type A = 3/6

We do the same for types B and C (multiply by 3)

So : 3/9 * 3/8 * 3/7 * 3/6 * 3 = 5/14

1st cup : 3/9 is actually the probability

...]]>

(5/4)^(−n)<16^(−1)

(4/5)^n<1/2^4

so left side should be less than 1/2

we raise to cube

(4/5)^3=64/125 and multiply right side by 64 just to compare, makes 64/128, so the inequality is still not satisfied but just a bit, it's clear that if multiply the left side by 4/5 one more time the inequality will be respected

But as we have 4 power on the right 1/2^4

then

...

]]>

(5/4)^(−n)<16^(−1)

(4/5)^n<1/2^4

so left side should be less than 1/2

we raise to cube

(4/5)^3=64/125 and multiply right side by 64 just to compare, makes 64/128, so the inequality is still not satisfied but just a bit, it's clear that if multiply the left side by 4/5 one more time the inequality will be respected

But as we have 4 power on the right 1/2^4

then

...]]>

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9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function[] rounds DOWN a number to the nearest integer. For example[1.5] =1,[2] =2,[-1.5] =-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of[a] +[b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order[a] +[b] = 1 to hold true, must be true that[a] =0 and[b] =1 (or vise-versa). Which means that

...

]]>

9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function[] rounds DOWN a number to the nearest integer. For example[1.5] =1,[2] =2,[-1.5] =-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of[a] +[b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order[a] +[b] = 1 to hold true, must be true that[a] =0 and[b] =1 (or vise-versa). Which means that

...]]>

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How is statement 1 sufficient?

If we pick numbers for statement 1 we can get two different answers.

if x=2 and y=3, then 100-3=97, Answer NO

if x=2 and y=9, then 100-9=81, Answer YES

Am I missing something?

Even I make such mistakes, be careful. How should one avoid such mistakes?

]]>

How is statement 1 sufficient?

If we pick numbers for statement 1 we can get two different answers.

if x=2 and y=3, then 100-3=97, Answer NO

if x=2 and y=9, then 100-9=81, Answer YES

Am I missing something?

Even I make such mistakes, be careful. How should one avoid such mistakes?]]>

I have a small doubt the circumference and the rotation around the circle gives the same meaning right? Because when we say revolving around the circle we mean it is 1 rotation= going around the circumference. Is this not right??

No, they are not the same in that circumference gives you the actual distance and rotations dictate the number of times the actual distance is multiplied. Circumference is constant, rotations change.

For e.g., if the wheel's circumference is 2 mts (just for the sake

...

]]>

I have a small doubt the circumference and the rotation around the circle gives the same meaning right? Because when we say revolving around the circle we mean it is 1 rotation= going around the circumference. Is this not right??

No, they are not the same in that circumference gives you the actual distance and rotations dictate the number of times the actual distance is multiplied. Circumference is constant, rotations change.

For e.g., if the wheel's circumference is 2 mts (just for the sake

...]]>

If a jar of candies is divided among 3 children, how many candies did the child that received the fewest pieces receive?

(1) The two children that received the greatest number of pieces received a total of 13 pieces.

(2) The two children that received the fewest number of pieces received a total of 11 pieces.

Let the three children be\(A\) ,\(B\) &\(C\) where candy distribution is\(A>B>C\) . we need to find\(C\) the lowest

Statement 1 implies that\(A+B = 13\) . or\(A=13-B\) .

Hence\((A,B)\) can be: (12,1),

...

]]>

If a jar of candies is divided among 3 children, how many candies did the child that received the fewest pieces receive?

(1) The two children that received the greatest number of pieces received a total of 13 pieces.

(2) The two children that received the fewest number of pieces received a total of 11 pieces.

Let the three children be\(A\) ,\(B\) &\(C\) where candy distribution is\(A>B>C\) . we need to find\(C\) the lowest

Statement 1 implies that\(A+B = 13\) . or\(A=13-B\) .

Hence\((A,B)\) can be: (12,1),

...]]>

let back = B

front = 1/3B

done for front = 1/2*1/3B = 1/6B

done for back = 2/3B

combining workdone for front and back = 1/6 + 4/6 = 5/6B

what left to be done = (front + back) - (work done for front and back) = (1/3B + B) - 5/6B = 4/3B - 5/6B = 1/2B

ans. D?

]]>

let back = B

front = 1/3B

done for front = 1/2*1/3B = 1/6B

done for back = 2/3B

combining workdone for front and back = 1/6 + 4/6 = 5/6B

what left to be done = (front + back) - (work done for front and back) = (1/3B + B) - 5/6B = 4/3B - 5/6B = 1/2B

ans. D?]]>

The test-makers LOVE to test this concept.

For more on this, watch the following video:

[you-tube] https://www.youtube.com/watch?v=9SCrYMCA88A[/you-tube]

...

]]>

The test-makers LOVE to test this concept.

For more on this, watch the following video:

[you-tube] https://www.youtube.com/watch?v=9SCrYMCA88A[/you-tube]

...]]>

[GMAT math practice question]

The formula for the sum of squares of integer between 1 and n is n(n+1)(2n+1) / 6 = 1^2 + 2^2 + … + n^2. The is the value of 11^2 + 12^2 + … + 20^2?

A. 2470

B. 2475

C. 2480

D. 2485

E. 2490

Sum of squares from 1 to 20\(= \frac{20*21*41}{6} = 2870\) (use the formula given)

Sum of squares from 1 to 10\(= \frac{10*11*21}{6} = 385\)

Hence sum of squares from 11 to 20\(=\) Sum of squares from 1 to 20\(-\) sum of squares from 1 to 10\(= 2870-385 = 2485\)

OptionD

...

]]>

[GMAT math practice question]

The formula for the sum of squares of integer between 1 and n is n(n+1)(2n+1) / 6 = 1^2 + 2^2 + … + n^2. The is the value of 11^2 + 12^2 + … + 20^2?

A. 2470

B. 2475

C. 2480

D. 2485

E. 2490

Sum of squares from 1 to 20\(= \frac{20*21*41}{6} = 2870\) (use the formula given)

Sum of squares from 1 to 10\(= \frac{10*11*21}{6} = 385\)

Hence sum of squares from 11 to 20\(=\) Sum of squares from 1 to 20\(-\) sum of squares from 1 to 10\(= 2870-385 = 2485\)

OptionD

...]]>

If 3x^2 + 2x + 9 = 2x^2 + 9x + 3, what all the possible values of x ?

A.-6 and -1

B.-6 and 1

C.-1 and 6

D.1 and 6

E.it cannot be determined from the information given.

Could someone expand on this pls?

\(3x^2 + 2x + 9 = 2x^2 + 9x + 3\)

=>\(x^2 - 7x + 6 = 0\)

=>\(x^2 - 6x - x + 6 = 0\)

=>\(x(x- 6) - (x - 6) = 0\)

=>\((x- 6) (x + 1) = 0\)

So as ab =0 => either a =0 or b=0 or both a and b are = 0

So here we can say x-6 =0 or x-1 =0

=> x= 6 or x=1

=>

...

]]>

If 3x^2 + 2x + 9 = 2x^2 + 9x + 3, what all the possible values of x ?

A.-6 and -1

B.-6 and 1

C.-1 and 6

D.1 and 6

E.it cannot be determined from the information given.

Could someone expand on this pls?

\(3x^2 + 2x + 9 = 2x^2 + 9x + 3\)

=>\(x^2 - 7x + 6 = 0\)

=>\(x^2 - 6x - x + 6 = 0\)

=>\(x(x- 6) - (x - 6) = 0\)

=>\((x- 6) (x + 1) = 0\)

So as ab =0 => either a =0 or b=0 or both a and b are = 0

So here we can say x-6 =0 or x-1 =0

=> x= 6 or x=1

=>

...]]>

2. Current Rooms/Members must not be an integer

Evaluating the answer options for the conditions

1. Both Options B and D are possible

2. All options possible except D

The answer option which satisfies both the conditions is Option B

]]>

2. Current Rooms/Members must not be an integer

Evaluating the answer options for the conditions

1. Both Options B and D are possible

2. All options possible except D

The answer option which satisfies both the conditions is Option B]]>

Machines A and B each produce tablets at their respective constant rates. Machine A has produced 30 tablets when Machine B is turned on. Both machines continue to run until Machine B’s total production catches up to Machine A’s total production. How many tablets does Machine A produce in the time that it takes Machine B to catch up?

(1) Machine A’s rate is twice the difference between the rates of the two machines.

(2) The sum of Machine A’s rate and Machine B’s rate is five

...

]]>

Machines A and B each produce tablets at their respective constant rates. Machine A has produced 30 tablets when Machine B is turned on. Both machines continue to run until Machine B’s total production catches up to Machine A’s total production. How many tablets does Machine A produce in the time that it takes Machine B to catch up?

(1) Machine A’s rate is twice the difference between the rates of the two machines.

(2) The sum of Machine A’s rate and Machine B’s rate is five

...]]>

In how many ways can digits 0, 2, 4, 7 be arranged to make a 4 digit number without repetition

A. 12

B. 24

C. 18

D. 4

E. 28

Using FCP:

____ * ____ * ____ * ____

Conditions: 1) must be a 4-digit number; and 2) numbers from which to choose, no repetition: (0, 2, 4, 7)

First slot can be 2, 4 or 7 - there are threepossibilities

_3 _

Second slot also has only three possibilities; once 2, 4, or 7 is chosen, there are three numbers left

_3 * _3

Then, third slot, there are two numbers that

...

]]>

In how many ways can digits 0, 2, 4, 7 be arranged to make a 4 digit number without repetition

A. 12

B. 24

C. 18

D. 4

E. 28

Using FCP:

____ * ____ * ____ * ____

Conditions: 1) must be a 4-digit number; and 2) numbers from which to choose, no repetition: (0, 2, 4, 7)

First slot can be 2, 4 or 7 - there are threepossibilities

_3 _

Second slot also has only three possibilities; once 2, 4, or 7 is chosen, there are three numbers left

_3 * _3

Then, third slot, there are two numbers that

...]]>

Appreciate if you can explain a bit further why we assume m=2 in your explanation.

Thanks!

]]>

Appreciate if you can explain a bit further why we assume m=2 in your explanation.

Thanks!]]>

OPTION 1) 3 km/hr and 8 km/hr

2) 5 km/hr and 12 km/hr

3) 4 km/hr and 11 km/hr

4) 3 km/hr and 12 km/hr

5) 4 km/hr and 10 km/hr

]]>

OPTION 1) 3 km/hr and 8 km/hr

2) 5 km/hr and 12 km/hr

3) 4 km/hr and 11 km/hr

4) 3 km/hr and 12 km/hr

5) 4 km/hr and 10 km/hr]]>

What is the Greatest Common Factor (GCF) of \(18x^8y^{(20)}\) and \(24x^{(12)}y^{(15)}\)?

A. \(3x^4y^5\)

B. \(6x^4y^5\)

C. \(3x^8y^{(15)}\)

D. \(6x^8y^{(15)}\)

E. \(72x^{(12)}y^{(20)}\)

To find GCF, find the prime factorization of both expressions.

\(18x^8y^{(20)}\) =

\(2 * 3 * 3 * x^8 * y^{(20)}\)

\(24x^{(12)}y^{(15)}\) =

\(2 * 2 * 2 * 3 * x^{(12)} * y^{(15)}\)

Find all the factors they have in common, "to the lowest power."

...

]]>

What is the Greatest Common Factor (GCF) of \(18x^8y^{(20)}\) and \(24x^{(12)}y^{(15)}\)?

A. \(3x^4y^5\)

B. \(6x^4y^5\)

C. \(3x^8y^{(15)}\)

D. \(6x^8y^{(15)}\)

E. \(72x^{(12)}y^{(20)}\)

To find GCF, find the prime factorization of both expressions.

\(18x^8y^{(20)}\) =

\(2 * 3 * 3 * x^8 * y^{(20)}\)

\(24x^{(12)}y^{(15)}\) =

\(2 * 2 * 2 * 3 * x^{(12)} * y^{(15)}\)

Find all the factors they have in common, "to the lowest power."

...]]>

There are total of 8 (p,q) pairs possible so that the absolute difference between the two numbers to be 2: (1, 3), (2, 4), (3, 5), (3, 1), (4, 6), (4, 2), (5, 7), (5, 3) (first # is chosen from set P and second # is chosen from set Q). 4 pairs contain the number 3 in it, so P=4/8=1/2.

Bunuel - according to my inference from the question we are not specifically told to select first no. from set P and the second no. from set Q. Why are we specifically following the order ?

Could u please

...

]]>

There are total of 8 (p,q) pairs possible so that the absolute difference between the two numbers to be 2: (1, 3), (2, 4), (3, 5), (3, 1), (4, 6), (4, 2), (5, 7), (5, 3) (first # is chosen from set P and second # is chosen from set Q). 4 pairs contain the number 3 in it, so P=4/8=1/2.

Bunuel - according to my inference from the question we are not specifically told to select first no. from set P and the second no. from set Q. Why are we specifically following the order ?

Could u please

...]]>

Is xy < 10?

1) x < 2

2) y < 5

Target question: Is xy< 10?

Statement 1: x < 2

Since there's no information about y, this statement is not sufficient.

However, if we're not certain about this insufficiency, we can TEST some values.

There are several values of x and y that satisfy statement 1. Here are two:

Case a x = 1 and y = 1, which means xy = (1)(1) = 1. So,xy < 10

Case b x = 1 and y = 20, which means xy = (1)(20) = 1. So,xy > 10

Since we cannot answer the target

...

]]>

Is xy < 10?

1) x < 2

2) y < 5

Target question: Is xy< 10?

Statement 1: x < 2

Since there's no information about y, this statement is not sufficient.

However, if we're not certain about this insufficiency, we can TEST some values.

There are several values of x and y that satisfy statement 1. Here are two:

Case a x = 1 and y = 1, which means xy = (1)(1) = 1. So,xy < 10

Case b x = 1 and y = 20, which means xy = (1)(20) = 1. So,xy > 10

Since we cannot answer the target

...]]>

Bunuel wrote:

zisis wrote:

Machine A and machine B are each used to manufacture 660 sprockets. It takes machine A 10 hours longer to produce 660 sprockets than machine B. Machine B produces 10 percent more sprockets per hour than machine A. How many sprockets per hour does machine A produces?

6

6.6

60

100

110

book give a backsolving solution which I am not a big fan of...........please explain method...

6

6.6

60

100

110

book give a backsolving solution which I am not a big fan of...........please explain method...

Let time needed for machine A to produce 660 sprockets be\(a\) hours, then the rate of machine A would be\( rate_A=\frac{job \ done}{time}\)

...

]]>

Bunuel wrote:

zisis wrote:

Machine A and machine B are each used to manufacture 660 sprockets. It takes machine A 10 hours longer to produce 660 sprockets than machine B. Machine B produces 10 percent more sprockets per hour than machine A. How many sprockets per hour does machine A produces?

6

6.6

60

100

110

book give a backsolving solution which I am not a big fan of...........please explain method...

6

6.6

60

100

110

book give a backsolving solution which I am not a big fan of...........please explain method...

Let time needed for machine A to produce 660 sprockets be\(a\) hours, then the rate of machine A would be\( rate_A=\frac{job \ done}{time}\)

...]]>

generally 5-8 wrong answer can be accommodated to score Q50.

Try to solve each questions but if any question is taking more than 3 min then skip

most important thing is to read explanation of all the questions

]]>

generally 5-8 wrong answer can be accommodated to score Q50.

Try to solve each questions but if any question is taking more than 3 min then skip

most important thing is to read explanation of all the questions]]>

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Dear Experts Bunuel, VeritasPrepKarishma,

In case of inequalities, can we take square roots on both sides?

Thanks in advance.

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).

Check for more here:Manipulating Inequalities (adding, subtracting, squaring etc.).

Hope it helps.

...

]]>

Dear Experts Bunuel, VeritasPrepKarishma,

In case of inequalities, can we take square roots on both sides?

Thanks in advance.

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).

Check for more here:Manipulating Inequalities (adding, subtracting, squaring etc.).

Hope it helps.

...]]>

Dear Brent,

Based on what you stated, it should be case i. It is the only case that orange greater than $6 and be able to divide by $2 and hence gives 1 apple.

Good catch - thanks!

I've edited my response accordingly.

Cheers,

Brent

]]>

Dear Brent,

Based on what you stated, it should be case i. It is the only case that orange greater than $6 and be able to divide by $2 and hence gives 1 apple.

Good catch - thanks!

I've edited my response accordingly.

Cheers,

Brent]]>

If x is a positive integer, what is the units digit of 3x?

(1) x = 10k^2 + 1, where k is a positive integer.

(2) The units digit of x^2 is 1.

St 1

This statement basically tells us that the units digit of X will always be 1 because no matter what K is, and K must be a positive integer, 10k^2 will always be a multiple of 10 and thus have a units digit of 0. If this is true then x will always have a units digit of 1 and if x always has a units digit of 1 then the units digit of 3x will always

...

]]>

If x is a positive integer, what is the units digit of 3x?

(1) x = 10k^2 + 1, where k is a positive integer.

(2) The units digit of x^2 is 1.

St 1

This statement basically tells us that the units digit of X will always be 1 because no matter what K is, and K must be a positive integer, 10k^2 will always be a multiple of 10 and thus have a units digit of 0. If this is true then x will always have a units digit of 1 and if x always has a units digit of 1 then the units digit of 3x will always

...]]>

Remember + is used when there are more ways of doing the same task in different ways.

and * is used when the task at hand is not complete .

Hope this helps

]]>

Remember + is used when there are more ways of doing the same task in different ways.

and * is used when the task at hand is not complete .

Hope this helps]]>

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mau5 wrote:

Which of the following is the largest?

(A) \(2^{27.3}\)

(B) \(3^{18.2}\)

(C) \(5^{11.1}\)

(D)\(7^{9.1}\)

(E)\(11^{5.1}\)

Fresh from Manhattan Prep (Challenge of the week)

(A) \(2^{27.3}\)

(B) \(3^{18.2}\)

(C) \(5^{11.1}\)

(D)\(7^{9.1}\)

(E)\(11^{5.1}\)

Fresh from Manhattan Prep (Challenge of the week)

First look for some commonality - either power or base.

I see that 27.3 is 3*9.1 and 18.2 is 2*9.1.

So

(A)\(2^{27.3} = 8^{9.1}\)

(B)\(3^{18.2} = 9^{9.1}\)

(D)\(7^{9.1}\)

Out of (A), (B) and (D), (B) is the largest.

We can easily compare\((C) 5^{11}\) with\((E) 11^{5}\) . Out of a^b and b^a, greater is usually the one with

...

]]>

mau5 wrote:

Which of the following is the largest?

(A) \(2^{27.3}\)

(B) \(3^{18.2}\)

(C) \(5^{11.1}\)

(D)\(7^{9.1}\)

(E)\(11^{5.1}\)

Fresh from Manhattan Prep (Challenge of the week)

(A) \(2^{27.3}\)

(B) \(3^{18.2}\)

(C) \(5^{11.1}\)

(D)\(7^{9.1}\)

(E)\(11^{5.1}\)

Fresh from Manhattan Prep (Challenge of the week)

First look for some commonality - either power or base.

I see that 27.3 is 3*9.1 and 18.2 is 2*9.1.

So

(A)\(2^{27.3} = 8^{9.1}\)

(B)\(3^{18.2} = 9^{9.1}\)

(D)\(7^{9.1}\)

Out of (A), (B) and (D), (B) is the largest.

We can easily compare\((C) 5^{11}\) with\((E) 11^{5}\) . Out of a^b and b^a, greater is usually the one with

...]]>

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CROSS MULTIPLY

= - X × 0 > X + 1

= 0 > × + 1

]]>

CROSS MULTIPLY

= - X × 0 > X + 1

= 0 > × + 1]]>

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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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7 boys

total = 11

3 is the number of group to be formed

using this formula: n! (n-r!)r! therefore 11! - 7!

- (11-3!)3! (7-3!)3!

...

]]>

7 boys

total = 11

3 is the number of group to be formed

using this formula: n! (n-r!)r! therefore 11! - 7!

- (11-3!)3! (7-3!)3!

...]]>

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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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https://gmatclub.com/forum/how-many-odd ... 06082.htmlhttps://gmatclub.com/forum/how-many-odd ... 06082.html