Line M has a y-intercept of 4, and its slope must be an integer-multi

Question

Line M has a y-intercept of –4, and its slope must be an integer-multiple of 1/7. Given that Line M passes below (4, –1) and above (5, –6), how many possible slopes could Line M have?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

khrystal · Accepted Answer

The slope of the line M has to be in between the slope of line A (passing through (0, -4) and (4, -1)) and line B(passing through (5, -6) and (0, -4))

Now, as we know slope of line with two given points = (y2-y1)/(x2-x1)
Therefore ,
slope of line A => 3/4
slope of line B => -2/5

Therefore slope of line M => -2/5 < m < 3/4

As given in question, m is integer multiple of 1/7 therefore m =k/7 (where k is any integer)
-2/5 < k/7 < 3/4
-14/5 < k < 21/4
-2.8 < k < 5.1

and as we know k is integer
Therefore possible values of k would be -2, -1, 0, 1, 2, 3 ,4 ,5
Total = 8 values (C Answer)

Bunuel · Answer

MAGOOSH OFFICIAL SOLUTION:

Well, for starters, zero is a multiple of every number, and a line with slope zero, the horizontal line y = –4 passes below (4, –1) and above (5, –6). That’s horizontal line is our starting point.

The point (4, –1) is over 4, up 3 from the y-intercept (0, –4). A line with a slope of +3/4 would go straight from (0, –4) to (4, –1). Thus, we need a slope that is less than +3/4. Notice that 3/4 = 21/28. Notice that 5/7 = 20/28, so this would be less than 3/4. Therefore, +1/7 through +5/7 will all slope up, obviously above (5, –6), and all will pass below (4, –1). That’s five upward sloping lines.

The point (5, –6) is over 5, down 2, from the y-intercept (0, –4). A line with a slope of –2/5 would go straight from (0, –4) to (5, –6). Thus, we need a slope that is more than –2/5; another way to say that is, we need a negative slope whose absolute value is less than +2/5. Well, 2/5 = 14/35, while 2/7 = 10/35 and 3/7 = 15/35, so (2/7) < (2/5) < (3, 7). The negatively sloping lines obviously pass below (4, –1), but only two of them, –1/7 and –2/7, pass above (5, –6).

That’s one horizontal line, five upward sloping lines, and two downward sloping lines, for a total of eight.

Answer = (C).

sterling19 · Answer

Let m = slope of Line M
Using the y-intercept and the given points, we can find that  -\frac{2}{5} < m < \frac{3}{4} 
Since we need to think in multiples of  \frac{1}{7} , let's convert this to  -\frac{14}{35} < m < \frac{21}{28}

Multiplying  \frac{1}{7}  by the consecutive integers between -2 and 5, inclusive, will give us values that fall in this range.
The correct answer is C.

balamoon · Answer

Please elaborate the last sentence....how can values faill in the range after multiplying 1/7 between -2 and 5.

ravi2107 · Answer

Hi Sterling,

Thanks for your answer.Could be please elaborate it ?

Regards

aniketm.87@gmail.com · Answer

Trying to make sense of it :

-14/35 < m < 21/28

so m must be less than 21/28
20/28 = 5/7 so we can take values from 20/28 or 5/7 to 0
ie 5/7, 4/7, 3/7, 2/7, 1/7 and 0 (6 values)

now m must be > -14/35 ie -2/5
take m as -10/35 ie - 2/7
so two values -2/7 and -1/7

total values = 6 + 2 (8) (C)

OreoShake · Answer

But if the slope must be a integer multiple of 1/7 then how can there be 8 slopes? this would mean that slope can only be +/- 1,2,3 and so on.. Can anyone resolve this?

praneet87 · Answer

I am confused here too. I got to the point of -2/5 < m < 3/4. After that I couldn't figure how to get  integer  multiple of 1/7.

OreoShake · Answer

what is meant by 'integer multiple' of 1/7? does that mean multiples such as 1 2 3 4, or -1, -2, -3, -4 or something else? Thanks

goraya · Answer

Step 1.

Find range for possible slope values i.e., -0.4(-2/5) to 0.75(3/4)

Step 2.

We know from question stems that the limits of the above range are not included in our target values and that the values of slope have to be multiples of 0.14(1/7). Hence, only possible values are 
 
-0.28, -0.14, 0, 0.14, 0.18, 0.42, 0.56 and 0.70

a total of 8 values.

rohan994 · Answer

Line M has y-intercept = -4, which means one of the points is (0,-4),

And it is given that the slope m=n \frac{1}{7}, where n is an integer. 
Therefore the equation of the line M is (y+4)=(n/7)(x-0)
This can be simplified for n, n=7(y+4)/x

Given that the line is below (4,-1) and above (5,-6) implies that the values of n should lie between -14/5 and 21/4 or -56/20 and 105/20

Since n is an integer, there are only 8 possible values -> -40/20, -20/20, 0/20, 20/20, 40/20, 60/20, 80/20 and 100/20

Hence Answer  C

Fdambro294 · Answer

SO Close. Completely Forgot the Horizontal Line with Slope = 0 at Y = -(4).

I see lots of others did as well.

GOAL: Find the UPPER and LOWER Boundaries of the Possible Slopes of Lines that could "FIT" within the Points Given.

(1st) The Line must Pass ABOVE the Point (5 , -6) and Pass through Y-Intercept (0 , -4)

If the Line Actually DID Pass through Point (5 , -6), the Slope of the that Line would be:

(-4 - -6)/ (0 - 5) = +2 / -5 = -(2/5)

The Upper Boundary of our Slope (Let's call it M) must be: M > -(2/5)

(2nd) The Line must Pass BELOW the Point of (4 , -1) and Pass through Y-Intercept (0 , -4)

The UPPER Boundary of the Possible Values Slope M can take would be the Slope that actually DOES Pass through Point (4 , -1)

(-4 - -1) / (0 - 4) = (-3) / (-4) = +(3/4)

Out Slope M must be: M < +(3/4)

(3rd) Find the "Integer-Multiples" of 1/7 that would Satisfy the Conditions

The Given Range that our Slope M can take is the following:

-(2/5) < M < +(3/4)

and we are told that M must be an "Integer-Multiple" of 1/7

I took this to mean that the Slope must have an INTEGER in the NUM and the DEN must be 7

Examples: +1/7 , -1/7 , +2/7 , -2/7, etc.

However, do NOT forget that 0 is an INTEGER Also. Thus, the Slope could be: M = 0/7 ---- or the Slope of a Horizontal Line Parallel to the X-Axis

(4th) Find the L.C.D. so we can compare the Boundaries with the Possible Values Slope - M can Take:

-(2/5) = -(56/140)

+(3/4) = +(105/140)

1/7 = 20/140
2/7 = 40/140
3/7 = 60/140
4/7 = 80/140
5/7 = 100/140

Given the Range that our M-Slope must fall within:

-(56/140) < M < +(105/140)

M can take the Following Values:

-(40/140)
-(20/140)

0 ----- (Horizontal Line: y = -(4) )

20/140
40/140
60/140
80/140
100/140

Solution:

There are 8 Possible Slopes of a Line that would meet the Conditions in the Question

-C-

Kinshook · Answer

Line M has a y-intercept of –4, and its slope must be an integer-multiple of 1/7.
Given that Line M passes below (4, –1) and above (5, –6), how many possible slopes could Line M have?

The equation of the line M :
y = mx/7 - 4; m is an integer

Since line M passes below (4,-1)
y = 4m/7 - 4 < - 1
4m < 21
m < 21/4 = 5.25

Since line M passes above (5,-6)
y = 5m/7 - 4 > -6
5m > -14
m > -14/5 = - 2.8

-2.8 < m < 5.25
m = {-2,-1,0,1,2,3,4,5} : 8 possible values of m

Possible Slopes of line M = m/7 = {-2/7,-1/7,0,1/7,2/7,3/7,4/7,5/7}: 8 possible values of slope of line M

IMO C

Line M has a y-intercept of 4, and its slope must be an integer-multi

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Line M has a y-intercept of 4, and its slope must be an integer-multi

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards