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Alan’s regular hourly wage is 1.5 times Barney’s regular [#permalink]
01 Dec 2010, 05:34

2

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00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

30% (02:23) correct
70% (01:24) wrong based on 214 sessions

Alan’s regular hourly wage is 1.5 times Barney’s regular hourly wage, but Barney gets paid at twice his regular wage for any hours he works on Saturday. Both men work an integer number of hours on any given day. If Alan and Barney each worked for the same total non-zero number of hours last week, and earned the same total in wages, which of the following must be true?

I. Alan worked fewer hours Monday through Friday than did Barney. II. Barney worked at least one hour on Saturday. III. Barney made more money on Saturday than did Alan.

A.I only B. II only C. I and II only D. I and III only E. II and III only

I would just like to see the equation written out, so I can visualize the problem. I am having trouble conceptualizing the problem. I understand the a=1.5b but I would like to see how the second part is written when they have equal hours and equal pay. Thanks,

Re: MGMAT CAT1 Question 11 [#permalink]
01 Dec 2010, 07:26

3

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Expert's post

2

This post was BOOKMARKED

mmcooley33 wrote:

Alan’s regular hourly wage is 1.5 times Barney’s regular hourly wage, but Barney gets paid at twice his regular wage for any hours he works on Saturday. Both men work an integer number of hours on any given day. If Alan and Barney each worked for the same total non-zero number of hours last week, and earned the same total in wages, which of the following must be true?

I. Alan worked fewer hours Monday through Friday than did Barney. II. Barney worked at least one hour on Saturday. III. Barney made more money on Saturday than did Alan.

I only

II only

I and II only

I and III only

II and III only

I would just like to see the equation written out, so I can visualize the problem. I am having trouble conceptualizing the problem. I understand the a=1.5b but I would like to see how the second part is written when they have equal hours and equal pay. Thanks,

Although I don't think that algebraic way is the best for this problem, here you go:

Let Barney's regular hourly wage be x, then his Saturday wage will be 2x and Alan's hourly wage will be 1.5x;

Let the # of hours Barney worked Monday through Friday be m and on Saturday be n and the # of hours Alan worked Monday through Friday be p and on Saturday be q;

Given: xm+2xn=1.5x(p+q) and m+n=p+q.

xm+2xn=1.5x(p+q) --> m+2n=1.5(m+n) --> m=n --> Barney worked the equal # of hours Monday-Friday and on Saturday.

The above directly tells us that II must be true (as Barney worked total non-zero # of hours and he worked an integer # of hours on any given day then he must have been worked at least one hour on Saturday.)

As for I: Alan may have worked ALL his hours Monday through Friday so in this case this statement is not true (p=total>m). Alan also may have worked all his hours on Saturday. Or algebraically: there are any distribution possible between p and q, p=0 and q=total or p=total and q=0 or any other;

The above means that III is also not always true: if Alan worked all his hours on Saturday then he made all his money on Saturday thus he made more money on Saturday than Barney did.

Answer: B (II only).

But the above can also be done with much less algebra:

As Alan and Barney worked the same # of hours and earned the same amount of money, then their hourly average wages must have been the same: (average wage)=(total amount earned)/(# of hours worked). Now, Alan has constant hourly wage which is 1.5*x and Barney's average (\frac{xm+2xn}{m+n}) to be equal to this he must have been worked the equal # of hours Monday-Friday and on Saturday, so m=n.

Re: MGMAT CAT1 Question 11 [#permalink]
25 May 2013, 04:10

Bunuel wrote:

mmcooley33 wrote:

Alan’s regular hourly wage is 1.5 times Barney’s regular hourly wage, but Barney gets paid at twice his regular wage for any hours he works on Saturday. Both men work an integer number of hours on any given day. If Alan and Barney each worked for the same total non-zero number of hours last week, and earned the same total in wages, which of the following must be true?

I. Alan worked fewer hours Monday through Friday than did Barney. II. Barney worked at least one hour on Saturday. III. Barney made more money on Saturday than did Alan.

I only

II only

I and II only

I and III only

II and III only

I would just like to see the equation written out, so I can visualize the problem. I am having trouble conceptualizing the problem. I understand the a=1.5b but I would like to see how the second part is written when they have equal hours and equal pay. Thanks,

Although I don't think that algebraic way is the best for this problem, here you go:

Let Barney's regular hourly wage be x, then his Saturday wage will be 2x and Alan's hourly wage will be 1.5x;

Let the # of hours Barney worked Monday through Friday be m and on Saturday be n and the # of hours Alan worked Monday through Friday be p and on Saturday be q;

Given: xm+2xn=1.5x(p+q) and m+n=p+q.

xm+2xn=1.5x(p+q) --> m+2n=1.5(m+n) --> m=n --> Barney worked the equal # of hours Monday-Friday and on Saturday.

The above directly tells us that II must be true (as Barney worked total non-zero # of hours and he worked an integer # of hours on any given day then he must have been worked at least one hour on Saturday.)

As for I: Alan may have worked ALL his hours Monday through Friday so in this case this statement is not true (p=total>m). Alan also may have worked all his hours on Saturday. Or algebraically: there are any distribution possible between p and q, p=0 and q=total or p=total and q=0 or any other;

The above means that III is also not always true: if Alan worked all his hours on Saturday then he made all his money on Saturday thus he made more money on Saturday than Barney did.

Answer: B (II only).

But the above can also be done with much less algebra:

As Alan and Barney worked the same # of hours and earned the same amount of money, then their hourly average wages must have been the same: (average wage)=(total amount earned)/(# of hours worked). Now, Alan has constant hourly wage which is 1.5*x and Barney's average (\frac{xm+2xn}{m+n}) to be equal to this he must have been worked the equal # of hours Monday-Friday and on Saturday, so m=n.

Hope it's clear.

Bunuel , u humble anyone's approach to logical and or mathematical problems. respect

Re: MGMAT CAT1 Question 11 [#permalink]
14 Jan 2014, 10:24

Bunuel wrote:

mmcooley33 wrote:

Alan’s regular hourly wage is 1.5 times Barney’s regular hourly wage, but Barney gets paid at twice his regular wage for any hours he works on Saturday. Both men work an integer number of hours on any given day. If Alan and Barney each worked for the same total non-zero number of hours last week, and earned the same total in wages, which of the following must be true?

I. Alan worked fewer hours Monday through Friday than did Barney. II. Barney worked at least one hour on Saturday. III. Barney made more money on Saturday than did Alan.

I only

II only

I and II only

I and III only

II and III only

I would just like to see the equation written out, so I can visualize the problem. I am having trouble conceptualizing the problem. I understand the a=1.5b but I would like to see how the second part is written when they have equal hours and equal pay. Thanks,

Although I don't think that algebraic way is the best for this problem, here you go:

Let Barney's regular hourly wage be x, then his Saturday wage will be 2x and Alan's hourly wage will be 1.5x;

Let the # of hours Barney worked Monday through Friday be m and on Saturday be n and the # of hours Alan worked Monday through Friday be p and on Saturday be q;

Given: xm+2xn=1.5x(p+q) and m+n=p+q.

xm+2xn=1.5x(p+q) --> m+2n=1.5(m+n) --> m=n --> Barney worked the equal # of hours Monday-Friday and on Saturday.

The above directly tells us that II must be true (as Barney worked total non-zero # of hours and he worked an integer # of hours on any given day then he must have been worked at least one hour on Saturday.)

As for I: Alan may have worked ALL his hours Monday through Friday so in this case this statement is not true (p=total>m). Alan also may have worked all his hours on Saturday. Or algebraically: there are any distribution possible between p and q, p=0 and q=total or p=total and q=0 or any other;

The above means that III is also not always true: if Alan worked all his hours on Saturday then he made all his money on Saturday thus he made more money on Saturday than Barney did.

Answer: B (II only).

But the above can also be done with much less algebra:

As Alan and Barney worked the same # of hours and earned the same amount of money, then their hourly average wages must have been the same: (average wage)=(total amount earned)/(# of hours worked). Now, Alan has constant hourly wage which is 1.5*x and Barney's average (\frac{xm+2xn}{m+n}) to be equal to this he must have been worked the equal # of hours Monday-Friday and on Saturday, so m=n.

Hope it's clear.

Hi Bunuel,

I believe the best approach is a mixture of conceptual and number picking, but I'm having a hard time getting to make work fast in under 2 minutes. Would you please show us how you deal with this problem in such way?

Re: Alan’s regular hourly wage is 1.5 times Barney’s regular [#permalink]
04 Feb 2014, 08:07

2

This post received KUDOS

Statement I. Alan worked fewer hours Monday through Friday than did Barney, well let's see. Maybe Alan did work fewer hours and then worked those missing hours on Saturday, or maybe not, we really can't tell the split of Alan's hours.

Statement II, what we do know is that Barney must have worked at least 1 hour on Saturday, how else then could he compensate for a lower salary if they work the same number of total hours?

Statement III what if Barney worked fewer hours during the week and then worked on Saturdays, then Alan could compensate by working more hours during the week, and maybe just working a smaller number of hours on Saturday (>=1). In that case we leave the possibility open of Barney earning a salary that is lower than Barney's but still making more money than Barney on Saturday. This can be possible because of the higher amount of hours worked by Alan.

Re: MGMAT CAT1 Question 11 [#permalink]
10 May 2014, 11:04

Bunuel wrote:

mmcooley33 wrote:

Alan’s regular hourly wage is 1.5 times Barney’s regular hourly wage, but Barney gets paid at twice his regular wage for any hours he works on Saturday. Both men work an integer number of hours on any given day. If Alan and Barney each worked for the same total non-zero number of hours last week, and earned the same total in wages, which of the following must be true?

I. Alan worked fewer hours Monday through Friday than did Barney. II. Barney worked at least one hour on Saturday. III. Barney made more money on Saturday than did Alan.

I only

II only

I and II only

I and III only

II and III only

I would just like to see the equation written out, so I can visualize the problem. I am having trouble conceptualizing the problem. I understand the a=1.5b but I would like to see how the second part is written when they have equal hours and equal pay. Thanks,

Although I don't think that algebraic way is the best for this problem, here you go:

Let Barney's regular hourly wage be x, then his Saturday wage will be 2x and Alan's hourly wage will be 1.5x;

Let the # of hours Barney worked Monday through Friday be m and on Saturday be n and the # of hours Alan worked Monday through Friday be p and on Saturday be q;

Given: xm+2xn=1.5x(p+q) and m+n=p+q.

xm+2xn=1.5x(p+q) --> m+2n=1.5(m+n) --> m=n --> Barney worked the equal # of hours Monday-Friday and on Saturday.

The above directly tells us that II must be true (as Barney worked total non-zero # of hours and he worked an integer # of hours on any given day then he must have been worked at least one hour on Saturday.)

As for I: Alan may have worked ALL his hours Monday through Friday so in this case this statement is not true (p=total>m). Alan also may have worked all his hours on Saturday. Or algebraically: there are any distribution possible between p and q, p=0 and q=total or p=total and q=0 or any other;

The above means that III is also not always true: if Alan worked all his hours on Saturday then he made all his money on Saturday thus he made more money on Saturday than Barney did.

Answer: B (II only).

But the above can also be done with much less algebra:

As Alan and Barney worked the same # of hours and earned the same amount of money, then their hourly average wages must have been the same: (average wage)=(total amount earned)/(# of hours worked). Now, Alan has constant hourly wage which is 1.5*x and Barney's average (\frac{xm+2xn}{m+n}) to be equal to this he must have been worked the equal # of hours Monday-Friday and on Saturday, so m=n.

Hope it's clear.

Hi Bunuel,

I'm having a hard time figuring out why statement III is wrong when NOT done algebraically.

I approached this problem conceptually: II is correct is rather easy to see. When it comes to III, the statement says that Barney made more money on Saturday than Alan. Correct?

Doesn't that HAVE to be true? What I mean by that is -- if Barney worked AT LEAST 1 hour on saturday, his salary is 2x vs. Alan's which is 1.5x, so doesn't that inherently make III true?

I would go even further and say that Barney would need to make a ton more money on Saturday to compensate for his lack of pay during the week.

Re: MGMAT CAT1 Question 11 [#permalink]
11 May 2014, 04:27

Expert's post

russ9 wrote:

Bunuel wrote:

mmcooley33 wrote:

Alan’s regular hourly wage is 1.5 times Barney’s regular hourly wage, but Barney gets paid at twice his regular wage for any hours he works on Saturday. Both men work an integer number of hours on any given day. If Alan and Barney each worked for the same total non-zero number of hours last week, and earned the same total in wages, which of the following must be true?

I. Alan worked fewer hours Monday through Friday than did Barney. II. Barney worked at least one hour on Saturday. III. Barney made more money on Saturday than did Alan.

I only

II only

I and II only

I and III only

II and III only

I would just like to see the equation written out, so I can visualize the problem. I am having trouble conceptualizing the problem. I understand the a=1.5b but I would like to see how the second part is written when they have equal hours and equal pay. Thanks,

Although I don't think that algebraic way is the best for this problem, here you go:

Let Barney's regular hourly wage be x, then his Saturday wage will be 2x and Alan's hourly wage will be 1.5x;

Let the # of hours Barney worked Monday through Friday be m and on Saturday be n and the # of hours Alan worked Monday through Friday be p and on Saturday be q;

Given: xm+2xn=1.5x(p+q) and m+n=p+q.

xm+2xn=1.5x(p+q) --> m+2n=1.5(m+n) --> m=n --> Barney worked the equal # of hours Monday-Friday and on Saturday.

The above directly tells us that II must be true (as Barney worked total non-zero # of hours and he worked an integer # of hours on any given day then he must have been worked at least one hour on Saturday.)

As for I: Alan may have worked ALL his hours Monday through Friday so in this case this statement is not true (p=total>m). Alan also may have worked all his hours on Saturday. Or algebraically: there are any distribution possible between p and q, p=0 and q=total or p=total and q=0 or any other;

The above means that III is also not always true: if Alan worked all his hours on Saturday then he made all his money on Saturday thus he made more money on Saturday than Barney did.

Answer: B (II only).

But the above can also be done with much less algebra:

As Alan and Barney worked the same # of hours and earned the same amount of money, then their hourly average wages must have been the same: (average wage)=(total amount earned)/(# of hours worked). Now, Alan has constant hourly wage which is 1.5*x and Barney's average (\frac{xm+2xn}{m+n}) to be equal to this he must have been worked the equal # of hours Monday-Friday and on Saturday, so m=n.

Hope it's clear.

Hi Bunuel,

I'm having a hard time figuring out why statement III is wrong when NOT done algebraically.

I approached this problem conceptually: II is correct is rather easy to see. When it comes to III, the statement says that Barney made more money on Saturday than Alan. Correct?

Doesn't that HAVE to be true? What I mean by that is -- if Barney worked AT LEAST 1 hour on saturday, his salary is 2x vs. Alan's which is 1.5x, so doesn't that inherently make III true?

I would go even further and say that Barney would need to make a ton more money on Saturday to compensate for his lack of pay during the week.

What am I missing here?

We got that Barney worked the same number of hours from Monday to Friday and on Saturday. Thus his wage is split into two parts 1 part is for the work done from Monday to Friday and 1.5 parts for the work done on Saturday.

Now, if Alan worked all his hours on Saturday then he made all his money on Saturday thus he made more money on Saturday than Barney did.

Re: MGMAT CAT1 Question 11 [#permalink]
15 May 2014, 15:35

Bunuel wrote:

russ9 wrote:

Hi Bunuel,

I'm having a hard time figuring out why statement III is wrong when NOT done algebraically.

I approached this problem conceptually: II is correct is rather easy to see. When it comes to III, the statement says that Barney made more money on Saturday than Alan. Correct?

Doesn't that HAVE to be true? What I mean by that is -- if Barney worked AT LEAST 1 hour on saturday, his salary is 2x vs. Alan's which is 1.5x, so doesn't that inherently make III true?

I would go even further and say that Barney would need to make a ton more money on Saturday to compensate for his lack of pay during the week.

What am I missing here?

We got that Barney worked the same number of hours from Monday to Friday and on Saturday. Thus his wage is split into two parts 1 part is for the work done from Monday to Friday and 1.5 parts for the work done on Saturday.

Now, if Alan worked all his hours on Saturday then he made all his money on Saturday thus he made more money on Saturday than Barney did.