If the total cost of a housing project with f units is $1,750,000, what was the cost per unit for the housing project?
(1) If half as many units were built for the same total price, the cost per unit would be $250,000.
(2) If six more units had been built for the same total price, the cost per unit would be $3,750 less.
The OA is D .
Experts how can I use each statement to get an answer? I don't know how to solve it.
If the total cost of a housing project with f units is. . .
This topic has expert replies
GMAT/MBA Expert
- ErikaPrepScholar
- Legendary Member
- Posts: 503
- Joined: Thu Jul 20, 2017 9:03 am
- Thanked: 86 times
- Followed by:15 members
- GMAT Score:770
Let x equal the cost per unit for the housing project. Then the question stem gives:
$$f\cdot x=1,750,000$$
Statement 1
This tells us that we can build f/2 units for 1,750,000, where x = 250,000. In other words:
$$\frac{f}{2}\cdot250,000\ =\ 1,750,000$$
Solving for f gives 14 units. We can then plug into our initial equation:
$$14\cdot x=1,750,000$$
Solving for x gives $125,000 per unit. Sufficient.
Statement 2
This tells us that we can build f + 6 units for 1,750,000 at x - 3,750 per unit. In other words:
$$\left(f+6\right)\cdot\left(x-3,750\right)\ =\ 1,750,000$$
We can rearrange our original equation to get
$$f\ =\ \frac{1,750,000}{x}$$
Then we substitute that in for f
$$\left(\frac{1,750,000}{x}+\ 6\right)\left(x-3750\right)=1,750,000$$
At this point, we could expand, giving a quadratic equation that we can then factor, but we should recognize at this point that we CAN do that, hence we CAN solve for x, and statement 2 is sufficient. These numbers aren't pretty, so continuing to solve the quadratic is a waste of time.
$$f\cdot x=1,750,000$$
Statement 1
This tells us that we can build f/2 units for 1,750,000, where x = 250,000. In other words:
$$\frac{f}{2}\cdot250,000\ =\ 1,750,000$$
Solving for f gives 14 units. We can then plug into our initial equation:
$$14\cdot x=1,750,000$$
Solving for x gives $125,000 per unit. Sufficient.
Statement 2
This tells us that we can build f + 6 units for 1,750,000 at x - 3,750 per unit. In other words:
$$\left(f+6\right)\cdot\left(x-3,750\right)\ =\ 1,750,000$$
We can rearrange our original equation to get
$$f\ =\ \frac{1,750,000}{x}$$
Then we substitute that in for f
$$\left(\frac{1,750,000}{x}+\ 6\right)\left(x-3750\right)=1,750,000$$
At this point, we could expand, giving a quadratic equation that we can then factor, but we should recognize at this point that we CAN do that, hence we CAN solve for x, and statement 2 is sufficient. These numbers aren't pretty, so continuing to solve the quadratic is a waste of time.
Erika John - Content Manager/Lead Instructor
https://gmat.prepscholar.com/gmat/s/
Get tutoring from me or another PrepScholar GMAT expert: https://gmat.prepscholar.com/gmat/s/tutoring/
Learn about our exclusive savings for BTG members (up to 25% off) and our 5 day free trial
Check out our PrepScholar GMAT YouTube channel, and read our expert guides on the PrepScholar GMAT blog
GMAT/MBA Expert
- ErikaPrepScholar
- Legendary Member
- Posts: 503
- Joined: Thu Jul 20, 2017 9:03 am
- Thanked: 86 times
- Followed by:15 members
- GMAT Score:770
Another trick we can use on statement 2: if we have n variables, we need n distinct equations to solve for them.
In statement 2, we have 2 variables (f and x) and 2 distinct equations (the one from the question stem and the one from statement 2). So we should know even without substituting in for f that we can solve for x using these two equations together.
In statement 2, we have 2 variables (f and x) and 2 distinct equations (the one from the question stem and the one from statement 2). So we should know even without substituting in for f that we can solve for x using these two equations together.
Erika John - Content Manager/Lead Instructor
https://gmat.prepscholar.com/gmat/s/
Get tutoring from me or another PrepScholar GMAT expert: https://gmat.prepscholar.com/gmat/s/tutoring/
Learn about our exclusive savings for BTG members (up to 25% off) and our 5 day free trial
Check out our PrepScholar GMAT YouTube channel, and read our expert guides on the PrepScholar GMAT blog