GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Oct 2019, 15:55

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# The equation of a straight line containing the points (10,100) and (15

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58402
The equation of a straight line containing the points (10,100) and (15  [#permalink]

### Show Tags

30 Oct 2017, 23:55
1
3
00:00

Difficulty:

15% (low)

Question Stats:

83% (01:25) correct 17% (01:26) wrong based on 138 sessions

### HideShow timer Statistics

The equation of a straight line containing the points (10,100) and (15, 60) is

(A) y = –8x + 180
(B) y = 8x – 180
(C) y = x/8 + 7.5
(D) y = –8x – 180
(E) y = –x/8 + 22.5

_________________
Intern
Joined: 05 Mar 2017
Posts: 11
The equation of a straight line containing the points (10,100) and (15  [#permalink]

### Show Tags

31 Oct 2017, 08:21
A wins.

Eq. is y = mx + b

m = (y2- y1)/(x2-x1) = (60-100)/(15-10) = -8

b is value of y when x = 0.
Here, x reduces from 15 to 10 when y increases from 60 to 100. So, x will become 0 when y increases by 80.
New y = 100 + 80 = 180

Hence, y = -8x + 180
Senior SC Moderator
Joined: 22 May 2016
Posts: 3548
The equation of a straight line containing the points (10,100) and (15  [#permalink]

### Show Tags

31 Oct 2017, 09:17
2
2
Bunuel wrote:
The equation of a straight line containing the points (10,100) and (15, 60) is

(A) y = –8x + 180
(B) y = 8x – 180
(C) y = x/8 + 7.5
(D) y = –8x – 180
(E) y = –x/8 + 22.5

We can use the slope-intercept form of a line equation:
$$y = mx + b$$
$$m$$ = slope
$$b$$ = y-intercept

The answers are all in this form.

1) Find the slope of line from coordinates of the two given points:

$$\frac{rise}{run}=\frac{y2-y1}{x2-x1}=\frac{(100-60)}{10-15}=\frac{40}{-5}= -8 =$$ slope

Insert the (-8) where $$m$$ is in $$y = mx+b$$

$$y = -8x + b$$

2. Now find $$b.$$
Use the (x,y) coordinates from one of the two given points.
Substitute them into what we have so far:
$$y = -8x + b$$

We can substitute because every point on a line (i.e., its coordinates) will satisfy the equation for the line.

I will use (x,y) = (10,100) to find b, the y-intercept:

$$100 = -8(10) + b$$
$$100 + 80=b$$
$$b = 180$$

$$b$$ is positive, so in the original equation, the + sign on the RHS stays the same

We found $$m$$ and $$b$$.
For this question, all we have to do now is put those two values back into the slope-intercept equation, the one we started with.

3) Final: $$y = -8x + 180$$

_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.

Choose life.
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4007
The equation of a straight line containing the points (10,100) and (15  [#permalink]

### Show Tags

31 Oct 2017, 10:00
1
Top Contributor
Bunuel wrote:
The equation of a straight line containing the points (10,100) and (15, 60) is

(A) y = –8x + 180
(B) y = 8x – 180
(C) y = x/8 + 7.5
(D) y = –8x – 180
(E) y = –x/8 + 22.5

Notice that all of the answers are expressed in slope y-intercept form (y = mx + b, where m represents the line's slope and b represents the line's y-intercept)
We can use this to our advantage.

First let's sketch the two points and connect them with a line...

First notice that the slope of the line is NEGATIVE
So, we can ELIMINATE answer choices B and C, since those equations represent lines with POSITIVE slopes

Next notice that the y-intercept will be POSITIVE
So, we can ELIMINATE answer choice D, since that equation represents a line with a NEGATIVE y-intercept

We're left with answer choices A and E
The slope of answer choice A (y = –8x + 180) is -8, and the slope of answer choice E (y = –x/8 + 22.5) is -1/8

From our sketch, we can see that the line is quite steep, so we can ELIMINATE answer choice E, since a slope of -1/8 is not very steep.

RELATED VIDEOS FROM OUR COURSE

_________________
Test confidently with gmatprepnow.com
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8085
Location: United States (CA)
Re: The equation of a straight line containing the points (10,100) and (15  [#permalink]

### Show Tags

07 May 2019, 18:41
Bunuel wrote:
The equation of a straight line containing the points (10,100) and (15, 60) is

(A) y = –8x + 180
(B) y = 8x – 180
(C) y = x/8 + 7.5
(D) y = –8x – 180
(E) y = –x/8 + 22.5

We know the equation has to be of the form y = mx + b where m and b are constants.

If we let y = 100 and x = 10, we get 10m + b = 100.

If we let y = 60 and x = 15, we get 15m + b = 60.

Subtracting the second equation from the first, we get -5m = 40 and thus m = -8. If we substitute either of the solutions to y = -8x + b, for instance, if we let x = 10 and y = 100, we get b = 180. Thus, the equation of the line is y = -8x + 180.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: The equation of a straight line containing the points (10,100) and (15   [#permalink] 07 May 2019, 18:41
Display posts from previous: Sort by