Are x and y both positive?

(1) 2x - 2y = 1

(2) x/y > 1

OPEN DISCUSSION OF THIS QUESTION IS HERE: are-x-and-y-both-positive-1-2x-2x-1-2-x-y-63377.html

]]>

Are x and y both positive?

(1) 2x - 2y = 1

(2) x/y > 1

OPEN DISCUSSION OF THIS QUESTION IS HERE: are-x-and-y-both-positive-1-2x-2x-1-2-x-y-63377.html]]>

If x=(10^10)-47, what is the sum of all the digit of x?

A. 40

B. 45

C. 50

D. 55

E. 80

*An answer will be posted in 2 days.

\(10^{10} = 1 \text{ followed by 10 zeroes}\\

100-47 = 53 \text{ (}2\text{ digits)}\\

\text{There are 8 preceding digits, all of which will be }9. \\9 \times 8 = 72\\

72+5+3 = 80\\

\text{The initial 1 is ignored due to the subtraction.}\)

[Reveal] Spoiler:

E

...

]]>

If x=(10^10)-47, what is the sum of all the digit of x?

A. 40

B. 45

C. 50

D. 55

E. 80

*An answer will be posted in 2 days.

\(10^{10} = 1 \text{ followed by 10 zeroes}\\

100-47 = 53 \text{ (}2\text{ digits)}\\

\text{There are 8 preceding digits, all of which will be }9. \\9 \times 8 = 72\\

72+5+3 = 80\\

\text{The initial 1 is ignored due to the subtraction.}\)

[Reveal] Spoiler:

E

...]]>

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If a and b are positive integers, is ab an even?

1) (a+1)b=even

2) (a+1)^b=odd

*An answer will be posted in 2 days.

is\(ab\) an even ?

which mean only\(a\) or only\(b\) or both\(a\) and\(b\) are even.

1)\((a+1)b\) = even, which can mean,\(b\) alone is even or\(a\) alone is odd or both\(b\) is even and\(a\) is odd. Not sufficient. Eliminate A and D

2)\((a+1)^b\) = odd, which means no matter what\(b\) is ,\(a\) has to be even. Sufficient.

Answer is B.

...

]]>

If a and b are positive integers, is ab an even?

1) (a+1)b=even

2) (a+1)^b=odd

*An answer will be posted in 2 days.

is\(ab\) an even ?

which mean only\(a\) or only\(b\) or both\(a\) and\(b\) are even.

1)\((a+1)b\) = even, which can mean,\(b\) alone is even or\(a\) alone is odd or both\(b\) is even and\(a\) is odd. Not sufficient. Eliminate A and D

2)\((a+1)^b\) = odd, which means no matter what\(b\) is ,\(a\) has to be even. Sufficient.

Answer is B.

...]]>

At noon, both clocks are at noon.

At 1:00 -

correct clock is at 1:00

faster clock is at 1:01:40 (1 hour + gained 100 seconds)

At 2:00 -

correct clock is at 2:00

faster clock is at 2:03:20 (2 hours + gained 200 seconds)

.

.

.

.

In 9 hours, faster clock would gain 900 seconds which is equal to 15 minutes. It will show 9:15. At that point, the correct clock would show 9:00.

Answer (D).

]]>

At noon, both clocks are at noon.

At 1:00 -

correct clock is at 1:00

faster clock is at 1:01:40 (1 hour + gained 100 seconds)

At 2:00 -

correct clock is at 2:00

faster clock is at 2:03:20 (2 hours + gained 200 seconds)

.

.

.

.

In 9 hours, faster clock would gain 900 seconds which is equal to 15 minutes. It will show 9:15. At that point, the correct clock would show 9:00.

Answer (D).]]>

Thanks for the post. This helps!

I think there is a problem in Q37.

37. How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a)2520

b)3150

c)3360

d)6000

e)7500

Solution provided is:

The best answer is A.

The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has

...

]]>

Thanks for the post. This helps!

I think there is a problem in Q37.

37. How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a)2520

b)3150

c)3360

d)6000

e)7500

Solution provided is:

The best answer is A.

The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has

...]]>

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What is x?

(1) |x| < 2

(2) |x| = 3x – 2

Question : x = ?

Statement 1: |x| < 2

i.e. -2 < x < 2

NOTSUFFICIENT

Statement 1: |x| = 3x – 2

i.e. x = 3x – 2 and -x = 3x – 2

i.e. x = 1 and x = 1/2

But 1/2 doesn't satisfy the given statement hence that is not acceptable. Therefore,

x=2 is the only acceptable solution SUFFICIENT

Answer: Option B

...

]]>

What is x?

(1) |x| < 2

(2) |x| = 3x – 2

Question : x = ?

Statement 1: |x| < 2

i.e. -2 < x < 2

NOTSUFFICIENT

Statement 1: |x| = 3x – 2

i.e. x = 3x – 2 and -x = 3x – 2

i.e. x = 1 and x = 1/2

But 1/2 doesn't satisfy the given statement hence that is not acceptable. Therefore,

x=2 is the only acceptable solution SUFFICIENT

Answer: Option B

...]]>

Over the course of a year, a certain house appreciated in value by 10 percent while the house next door decreased in value by 10 percent as a result of foundation damage. At the end of the year, the reduced price of the second house was what percentage of the increased price of the first house?

(1) The amount by which the first house increased in value was half as much as the amount by which the second house decreased in value.

(2) At the end of the year, the second house was worth $70,000 more

...

]]>

Over the course of a year, a certain house appreciated in value by 10 percent while the house next door decreased in value by 10 percent as a result of foundation damage. At the end of the year, the reduced price of the second house was what percentage of the increased price of the first house?

(1) The amount by which the first house increased in value was half as much as the amount by which the second house decreased in value.

(2) At the end of the year, the second house was worth $70,000 more

...]]>

A truck driver wants to load as many identical cylindrical canisters of olive oil as can fit into the 3-meter × 4-meter × 9-meter storage space of his truck. How many canisters can he load into the truck?

(1) Each canister has a volume of 62, 500π cubic centimeters.

(2) The height of each canister is four times the radius.

Question : many canisters can he load into the truck?

Note: To find the number of cannisters we require

1) Dimensions of Cylinder i.e. Radius and Height of cylindrical

...

]]>

A truck driver wants to load as many identical cylindrical canisters of olive oil as can fit into the 3-meter × 4-meter × 9-meter storage space of his truck. How many canisters can he load into the truck?

(1) Each canister has a volume of 62, 500π cubic centimeters.

(2) The height of each canister is four times the radius.

Question : many canisters can he load into the truck?

Note: To find the number of cannisters we require

1) Dimensions of Cylinder i.e. Radius and Height of cylindrical

...]]>

]]>

]]>

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What is the value of y?

(1) y^2 + 8y + 16 = 0

(2) y < 0

Question : y = ?

Statement 1: y^2 + 8y + 16 = 0

y^2 + 8y + 16 = 0 ===> y^2 + 4y +4y + 16 = 0 ===> (y+4)^2 = =

i.e. y = -4

SUFFICIENT

Statement 1: y<0

NOT SUFFICIENT

Answer: Option A

]]>

What is the value of y?

(1) y^2 + 8y + 16 = 0

(2) y < 0

Question : y = ?

Statement 1: y^2 + 8y + 16 = 0

y^2 + 8y + 16 = 0 ===> y^2 + 4y +4y + 16 = 0 ===> (y+4)^2 = =

i.e. y = -4

SUFFICIENT

Statement 1: y<0

NOT SUFFICIENT

Answer: Option A]]>

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clearly there are more than 4 2's

Thus number of 18 => 4

SMASH that D

]]>

clearly there are more than 4 2's

Thus number of 18 => 4

SMASH that D]]>

This implies that x is a positive multiple of 10 and upon rounding to nearest tens, we will have the same number. Answer to main question is no.

Sufficient.

St2: If x is divided by 5, the remainder is odd.

If x = 13, rounding to nearest tens -> 10. Answer to main question is yes.

If x = 28, rounding to nearest tens -> 30. Answer to main question is no.

Insufficient.

Answer (A).

]]>

This implies that x is a positive multiple of 10 and upon rounding to nearest tens, we will have the same number. Answer to main question is no.

Sufficient.

St2: If x is divided by 5, the remainder is odd.

If x = 13, rounding to nearest tens -> 10. Answer to main question is yes.

If x = 28, rounding to nearest tens -> 30. Answer to main question is no.

Insufficient.

Answer (A).]]>

I have another possible approach but would need help from the experts. Notice that the first term is 10 so the entire expression will end in a 0. Now, let's use concept of units digits and check each options.

a. Unit digit = 1 + 1 = 2 -> reject

b. Unit digit = 9 - 1 = 8 -> reject

c. Unit digit = 1 - 1 = 0 -> hold

d. Unit digit = 1 - 1 = 0 -> hold

e. Unit digit = 1 + 1 = 2 -> reject

At this point, I'm not

...

]]>

I have another possible approach but would need help from the experts. Notice that the first term is 10 so the entire expression will end in a 0. Now, let's use concept of units digits and check each options.

a. Unit digit = 1 + 1 = 2 -> reject

b. Unit digit = 9 - 1 = 8 -> reject

c. Unit digit = 1 - 1 = 0 -> hold

d. Unit digit = 1 - 1 = 0 -> hold

e. Unit digit = 1 + 1 = 2 -> reject

At this point, I'm not

...]]>

Ranges for x: x >= 1; -1 <= x <= 0

1/2 does not fall in either.

Answer (C).

]]>

Ranges for x: x >= 1; -1 <= x <= 0

1/2 does not fall in either.

Answer (C).]]>

= 53! * 55

Number of trailing 0s in 53! = number of 5s in the expansion of 53!

= 10 + 2 = 12

There is 1 more 5 in 55.

Hence, total number of trailing 0s = 12 + 1 = 13

Answer (B).

In most cases, when we are adding multiple terms, all of which have trailing 0s, the sum will have as many trailing 0s as that in the lowest term. Example: 20 + 2300 + 34210000 = 34212320 -> one

...

]]>

= 53! * 55

Number of trailing 0s in 53! = number of 5s in the expansion of 53!

= 10 + 2 = 12

There is 1 more 5 in 55.

Hence, total number of trailing 0s = 12 + 1 = 13

Answer (B).

In most cases, when we are adding multiple terms, all of which have trailing 0s, the sum will have as many trailing 0s as that in the lowest term. Example: 20 + 2300 + 34210000 = 34212320 -> one

...]]>

I. x > z -> always true since hypotenuse is the longest side

II. x = y + z -> never true since the length of any side of a triangle is less than the sum of length of other two sides

III. y^2 + z^2 is a positive integer -> true in this case since it is given that y and z are positive integers

Answer (D).

]]>

I. x > z -> always true since hypotenuse is the longest side

II. x = y + z -> never true since the length of any side of a triangle is less than the sum of length of other two sides

III. y^2 + z^2 is a positive integer -> true in this case since it is given that y and z are positive integers

Answer (D).]]>

A certain company employs 6 senior officers and 4 junior officers. If a committee is to be created, that is made up of 3 senior officers and 1 junior officer, how many different committee are possible?

A. 8

B. 24

C. 58

D. 80

E. 210

We need to determine how many different committees are possible with 3 senior officers and 1 junior officer from 6 senior officers and 4 junior officers. Let’s first determine the number of ways to select 3 senior officers.

number of ways to select 3 senior officers

...

]]>

A certain company employs 6 senior officers and 4 junior officers. If a committee is to be created, that is made up of 3 senior officers and 1 junior officer, how many different committee are possible?

A. 8

B. 24

C. 58

D. 80

E. 210

We need to determine how many different committees are possible with 3 senior officers and 1 junior officer from 6 senior officers and 4 junior officers. Let’s first determine the number of ways to select 3 senior officers.

number of ways to select 3 senior officers

...]]>

Sent from my Lenovo A7010a48 using GMAT Club Forum mobile app

]]>

Sent from my Lenovo A7010a48 using GMAT Club Forum mobile app]]>

a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?

A. 3/8

B. 1/2

C. 11/16

D. 5/7

E. 3/4

Another option is toPLUG in numbers .

We're told thatthe median of set Q is (7/8)c

Let's letc = 8

This means the median = (7/8)(8 =7

For

...

]]>

a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?

A. 3/8

B. 1/2

C. 11/16

D. 5/7

E. 3/4

Another option is toPLUG in numbers .

We're told thatthe median of set Q is (7/8)c

Let's letc = 8

This means the median = (7/8)(8 =7

For

...]]>

Let p = the product of all the odd integers between 500 and 598, and let q = the product of all the odd integers between 500 and 602. In terms of q, what is the value of\(\frac{1}{p}\) +\(\frac{1}{q}\) ?

A.\(\frac{1}{600q}\)

B.\(\frac{1}{359,999q}\)

C.\(\frac{1,200}{q}\)

D.\(\frac{360,000}{q}\)

E.\(359,999q\)

...

]]>

Let p = the product of all the odd integers between 500 and 598, and let q = the product of all the odd integers between 500 and 602. In terms of q, what is the value of\(\frac{1}{p}\) +\(\frac{1}{q}\) ?

A.\(\frac{1}{600q}\)

B.\(\frac{1}{359,999q}\)

C.\(\frac{1,200}{q}\)

D.\(\frac{360,000}{q}\)

E.\(359,999q\)

...]]>

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For each positive integer\(n\) ,\(p(n)\) is defined to be the product of the digits of\(n\) . For example,\(p(724) = 56\) since\(7 * 2 * 4 =56\) .

Which of the following statements must be true?

I.\(p(10n) = p(n)\)

II.\(p(n+1) > p(n)\)

III.\(p(2n) = 2p(n)\)

--

A. None

B. I and II only

C. I and III only

D. II and III only

E. I, II, and III

...

]]>

For each positive integer\(n\) ,\(p(n)\) is defined to be the product of the digits of\(n\) . For example,\(p(724) = 56\) since\(7 * 2 * 4 =56\) .

Which of the following statements must be true?

I.\(p(10n) = p(n)\)

II.\(p(n+1) > p(n)\)

III.\(p(2n) = 2p(n)\)

--

A. None

B. I and II only

C. I and III only

D. II and III only

E. I, II, and III

...]]>

Buneul, here's my doubt:

# of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) --> total of 6*5=30 trailing zeros for these 5 terms;

# of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) --> total of 7*4=28 trailing zeros for these 5 terms;

for calculating trailing zeros up til 24! you did just 20/5=4.

but above those numbers i.e., from 25! on wards you did (25/5+25/5^2=6) and (30/5+30/5^2=7)

Suppose I want # of trailing zeros in 310!

using

...

]]>

Buneul, here's my doubt:

# of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) --> total of 6*5=30 trailing zeros for these 5 terms;

# of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) --> total of 7*4=28 trailing zeros for these 5 terms;

for calculating trailing zeros up til 24! you did just 20/5=4.

but above those numbers i.e., from 25! on wards you did (25/5+25/5^2=6) and (30/5+30/5^2=7)

Suppose I want # of trailing zeros in 310!

using

...]]>

Number of fives => two

number of two's are greater but we don't have more fives to generate 10's more than 2

hence number of trailing zeroes = 2

Smash that C

]]>

Number of fives => two

number of two's are greater but we don't have more fives to generate 10's more than 2

hence number of trailing zeroes = 2

Smash that C]]>

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]]>

hence the decimal expansion would be 0.54545454545454

Clearly the 57th digit is 5=> Smash that C

Why did we write 6/11 as 54/99 => if denominator and numerator have same number of digits and if the denominator can be written as 10^n-1 ie 9,99,999,9999 etc then we need not carry any division process . The result would be given by the numerator

E.g => 69/99 = 0.69696969...

3/9=0.33333...

158/999=0.158158158...

...

]]>

hence the decimal expansion would be 0.54545454545454

Clearly the 57th digit is 5=> Smash that C

Why did we write 6/11 as 54/99 => if denominator and numerator have same number of digits and if the denominator can be written as 10^n-1 ie 9,99,999,9999 etc then we need not carry any division process . The result would be given by the numerator

E.g => 69/99 = 0.69696969...

3/9=0.33333...

158/999=0.158158158...

...]]>

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SMASH THAT E

]]>

SMASH THAT E]]>

SO total possibility is 2 to the power 5 ( one coin can be either H or T SO five times tossed = 2*2*2*2*2*2= 32

Kate should get between 10 and 15 dollar 10<k<15

if either HHTTT ( OVERALL ONE DOLLAR PLUS MEANS 11 DOLLARS

if 02 head and 3 tails total possibility ( here order matters so that is why i said it is not a combination question= factorial 5/ factorial 2 * factorial 3 = 10

second scenario for four tails and one head- means

...

]]>

SO total possibility is 2 to the power 5 ( one coin can be either H or T SO five times tossed = 2*2*2*2*2*2= 32

Kate should get between 10 and 15 dollar 10<k<15

if either HHTTT ( OVERALL ONE DOLLAR PLUS MEANS 11 DOLLARS

if 02 head and 3 tails total possibility ( here order matters so that is why i said it is not a combination question= factorial 5/ factorial 2 * factorial 3 = 10

second scenario for four tails and one head- means

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We need 5's as number of two's will off-course be sufficient to make a 10

5

5

5

5

5^2

5

5

5

5

5^2

5

5

5

5

5^2

5

5

5

5

5^2

5

5

5

5

5^3

5

5

5

5

5^2

5

5

5

5

5^2

5

5

5

5

5^2

Add them up => 49

SMASH THAT E

]]>

We need 5's as number of two's will off-course be sufficient to make a 10

5

5

5

5

5^2

5

5

5

5

5^2

5

5

5

5

5^2

5

5

5

5

5^2

5

5

5

5

5^3

5

5

5

5

5^2

5

5

5

5

5^2

5

5

5

5

5^2

Add them up => 49

SMASH THAT E]]>

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.]]>

x/z=12/20=60%

]]>

x/z=12/20=60%]]>

Sent from my iPhone using GMAT Club Forum mobile app

]]>

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

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.]]>

Please Merge your Question with this one.=>

if-y-is-80-greater-than-x-than-x-is-what-less-than-y-163031.html?fl=similar

*** REPEATED QUESTION. IT has already been Discussed

...

]]>

Please Merge your Question with this one.=>

if-y-is-80-greater-than-x-than-x-is-what-less-than-y-163031.html?fl=similar

*** REPEATED QUESTION. IT has already been Discussed

...]]>

Kudos will follow!

]]>

Kudos will follow!]]>

Bunuel wrote:

vigneshpandi wrote:

Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A?

1. 1

2. 3

3. 4

4. 6

5. 8

1. 1

2. 3

3. 4

4. 6

5. 8

Bag A:

R/W=1/3=2/6;

W/B=2/3=6/9;

So R/W/B=2/6/9 --> # of marbles in bag A would be\(2x\) ,\(6x\) , and\(9x\) , for somepositive integer multiple \(x\) , where\(6x\) corresponds to the # of white

...

]]>

Bunuel wrote:

vigneshpandi wrote:

Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A?

1. 1

2. 3

3. 4

4. 6

5. 8

1. 1

2. 3

3. 4

4. 6

5. 8

Bag A:

R/W=1/3=2/6;

W/B=2/3=6/9;

So R/W/B=2/6/9 --> # of marbles in bag A would be\(2x\) ,\(6x\) , and\(9x\) , for somepositive integer multiple \(x\) , where\(6x\) corresponds to the # of white

...]]>

Now we are asked if x rounded to tens would be greater than x or not.

For that to happen the unit digit must be greater than 4 i.e 5,6,7,8,9

Its a YES/NO question .

Lets look at the statements =>

Statement 1 => the unit digit can be 0,2,4,6,8 .

Remember that digit constraint allows only 10 values for any digit .

Hence here

...

]]>

Now we are asked if x rounded to tens would be greater than x or not.

For that to happen the unit digit must be greater than 4 i.e 5,6,7,8,9

Its a YES/NO question .

Lets look at the statements =>

Statement 1 => the unit digit can be 0,2,4,6,8 .

Remember that digit constraint allows only 10 values for any digit .

Hence here

...]]>