DSQ: Understanding work problems - GMAT Quantitative

Think of this in terms of "greenhouses per minute", not "minutes per greenhouse". Converting hours and minutes to just minutes, Steve can paint greenhouses per minute; Phil can paint greenhouses per minute.

If we let be the time in minutes that it takes to paint the greenhouse, then Steve and Phil paint and greenhouses, respectively; since one greenhouse total is painted, then we can add the labor and set up this equation:

Simplify and solve:

or 2 hours, 38 minutes.

A work problem is actually a rate problem in disguise.

If you know that an object working alone can do a job in hours, then you know that the object works at a rate of jobs per hour. After hours, the object accomplishes of a job. Similarly, the other object working alone does a job in hours, and therefore does of a job. Together, the objects do one whole job, so solve this equation

for .

Statement 1 alone gives us half the picture; , but is unknown.

Statement 2 alone tells us that is 75% of . But is unknown.

From Statement 1 and 2 together, we know

is 75% of - this allows us to calculate :

75% of is , so . Since we have both and , we have the complete equation

and we can calculate .

A work problem is actually a rate problem in disguise.

If you know that an object working alone can do a job in hours, then you know that the object works at a rate of jobs per hour. After hours, the object accomplishes of a job. Similarly, the other object working alone does a job in hours, and therefore does of a job. Together, the objects do one whole job, so solve this equation

for .

Statement 1 alone tells us that , and Statement 2 alone tells us that .

Therefore, each statement alone gives us only half the picture, but together, they give us the equation

,

which can be solved to yield the answer.

A work problem is actually a rate problem in disguise.

If you know that David working alone can do a job in hours, then you know that the object works at a rate of jobs per hour. After hours, the object accomplishes of a job. Similarly, Eddie and Floyd, working without David, do a job in hours, and therefore does of a job. Together, the objects do one whole job, so solve this equation

for .

Statement 1 alone tells us that , and Statement 2 alone tells us that ; each one alone leaves the other value unknown. However, if both statements are given, the equation

can be solved for to yield the correct answer.

Since the group is now without the help of Mrs. Smith, we look at Mrs. Smith's contribution to the work as a whole, and the sum of the other two ladies' contribution as a whole; Statement 2, which deals with Mrs. Hume alone, is irrelevant and unhelpful.

Since the three ladies together wrote 400 invitations in 120 minutes, we can infer that they would have spent three-fourths of this time, or 90 minutes, writing 300 invitations.

If Statement 1 is true, then Mrs. Smith, who can write 100 invitations in 90 minutes, would take three times this, or 270 minutes, to write 300 invitations.

A work problem is a rate problem in disguise. Think in terms of "jobs per hour", and take the reciprocal of each "hours per job". The three ladies together would have done job in one hour, and Mrs. Smith alone would have done job in one hour.

Therefore, in one hour today, the two ladies will do

jobs, and in hours today, they will do

job.

Solve for in this equation.

This proves that Statement 1 alone allows us to find the answer - but not Statement 2.

For statement (1), she spent

on rent, and

on clothes.

So she spent in total.

Therefore, she has left.

For statement (2), we can set Clara’s salary to be , then we have

Then simplify the equation:

Therefore, .

Now we can just follow what we did for the first statement to calculate the money he had left.

Think of this in terms of "greenhouses per minute", not "minutes per greenhouse". Converting hours and minutes to just minutes, Steve can paint greenhouses per minute; Phil can paint greenhouses per minute.

If we let be the time in minutes that it takes to paint the greenhouse, then Steve and Phil paint and greenhouses, respectively; since one greenhouse total is painted, then we can add the labor and set up this equation:

Simplify and solve:

or 2 hours, 38 minutes.

A work problem is actually a rate problem in disguise.

If you know that an object working alone can do a job in hours, then you know that the object works at a rate of jobs per hour. After hours, the object accomplishes of a job. Similarly, the other object working alone does a job in hours, and therefore does of a job. Together, the objects do one whole job, so solve this equation

for .

Statement 1 alone gives us half the picture; , but is unknown.

Statement 2 alone tells us that is 75% of . But is unknown.

From Statement 1 and 2 together, we know

is 75% of - this allows us to calculate :

75% of is , so . Since we have both and , we have the complete equation

and we can calculate .

A work problem is actually a rate problem in disguise.

If you know that an object working alone can do a job in hours, then you know that the object works at a rate of jobs per hour. After hours, the object accomplishes of a job. Similarly, the other object working alone does a job in hours, and therefore does of a job. Together, the objects do one whole job, so solve this equation

for .

Statement 1 alone tells us that , and Statement 2 alone tells us that .

Therefore, each statement alone gives us only half the picture, but together, they give us the equation

,

which can be solved to yield the answer.

A work problem is actually a rate problem in disguise.

If you know that David working alone can do a job in hours, then you know that the object works at a rate of jobs per hour. After hours, the object accomplishes of a job. Similarly, Eddie and Floyd, working without David, do a job in hours, and therefore does of a job. Together, the objects do one whole job, so solve this equation

for .

Statement 1 alone tells us that , and Statement 2 alone tells us that ; each one alone leaves the other value unknown. However, if both statements are given, the equation

can be solved for to yield the correct answer.

Since the group is now without the help of Mrs. Smith, we look at Mrs. Smith's contribution to the work as a whole, and the sum of the other two ladies' contribution as a whole; Statement 2, which deals with Mrs. Hume alone, is irrelevant and unhelpful.

Since the three ladies together wrote 400 invitations in 120 minutes, we can infer that they would have spent three-fourths of this time, or 90 minutes, writing 300 invitations.

If Statement 1 is true, then Mrs. Smith, who can write 100 invitations in 90 minutes, would take three times this, or 270 minutes, to write 300 invitations.

A work problem is a rate problem in disguise. Think in terms of "jobs per hour", and take the reciprocal of each "hours per job". The three ladies together would have done job in one hour, and Mrs. Smith alone would have done job in one hour.

Therefore, in one hour today, the two ladies will do

jobs, and in hours today, they will do

job.

Solve for in this equation.

This proves that Statement 1 alone allows us to find the answer - but not Statement 2.

For statement (1), she spent

on rent, and

on clothes.

So she spent in total.

Therefore, she has left.

For statement (2), we can set Clara’s salary to be , then we have

Then simplify the equation:

Therefore, .

Now we can just follow what we did for the first statement to calculate the money he had left.

Think of this in terms of "greenhouses per minute", not "minutes per greenhouse". Converting hours and minutes to just minutes, Steve can paint greenhouses per minute; Phil can paint greenhouses per minute.

If we let be the time in minutes that it takes to paint the greenhouse, then Steve and Phil paint and greenhouses, respectively; since one greenhouse total is painted, then we can add the labor and set up this equation:

Simplify and solve:

or 2 hours, 38 minutes.

A work problem is actually a rate problem in disguise.

If you know that an object working alone can do a job in hours, then you know that the object works at a rate of jobs per hour. After hours, the object accomplishes of a job. Similarly, the other object working alone does a job in hours, and therefore does of a job. Together, the objects do one whole job, so solve this equation

for .

Statement 1 alone gives us half the picture; , but is unknown.

Statement 2 alone tells us that is 75% of . But is unknown.

From Statement 1 and 2 together, we know

is 75% of - this allows us to calculate :

75% of is , so . Since we have both and , we have the complete equation

and we can calculate .

A work problem is actually a rate problem in disguise.

If you know that an object working alone can do a job in hours, then you know that the object works at a rate of jobs per hour. After hours, the object accomplishes of a job. Similarly, the other object working alone does a job in hours, and therefore does of a job. Together, the objects do one whole job, so solve this equation

for .

Statement 1 alone tells us that , and Statement 2 alone tells us that .

Therefore, each statement alone gives us only half the picture, but together, they give us the equation

,

which can be solved to yield the answer.

A work problem is actually a rate problem in disguise.

If you know that David working alone can do a job in hours, then you know that the object works at a rate of jobs per hour. After hours, the object accomplishes of a job. Similarly, Eddie and Floyd, working without David, do a job in hours, and therefore does of a job. Together, the objects do one whole job, so solve this equation

for .

Statement 1 alone tells us that , and Statement 2 alone tells us that ; each one alone leaves the other value unknown. However, if both statements are given, the equation

can be solved for to yield the correct answer.

Since the group is now without the help of Mrs. Smith, we look at Mrs. Smith's contribution to the work as a whole, and the sum of the other two ladies' contribution as a whole; Statement 2, which deals with Mrs. Hume alone, is irrelevant and unhelpful.

Since the three ladies together wrote 400 invitations in 120 minutes, we can infer that they would have spent three-fourths of this time, or 90 minutes, writing 300 invitations.

If Statement 1 is true, then Mrs. Smith, who can write 100 invitations in 90 minutes, would take three times this, or 270 minutes, to write 300 invitations.

A work problem is a rate problem in disguise. Think in terms of "jobs per hour", and take the reciprocal of each "hours per job". The three ladies together would have done job in one hour, and Mrs. Smith alone would have done job in one hour.

Therefore, in one hour today, the two ladies will do

jobs, and in hours today, they will do

job.

Solve for in this equation.

This proves that Statement 1 alone allows us to find the answer - but not Statement 2.

For statement (1), she spent

on rent, and

on clothes.

So she spent in total.

Therefore, she has left.

For statement (2), we can set Clara’s salary to be , then we have

Then simplify the equation:

Therefore, .

Now we can just follow what we did for the first statement to calculate the money he had left.

Think of this in terms of "greenhouses per minute", not "minutes per greenhouse". Converting hours and minutes to just minutes, Steve can paint greenhouses per minute; Phil can paint greenhouses per minute.

If we let be the time in minutes that it takes to paint the greenhouse, then Steve and Phil paint and greenhouses, respectively; since one greenhouse total is painted, then we can add the labor and set up this equation:

Simplify and solve:

or 2 hours, 38 minutes.