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Line M has a yintercept of –4, and its slope must be an integermulti
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10 Mar 2015, 06:16
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46% (02:28) correct 54% (02:55) wrong based on 257 sessions
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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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15 Mar 2015, 22:22
Bunuel wrote: Line M has a yintercept of –4, and its slope must be an integermultiple 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
Kudos for a correct solution. 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 yintercept (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 yintercept (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).
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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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17 Apr 2017, 10:04
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 = (y2y1)/(x2x1) 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)




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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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14 Mar 2015, 14:07
Let m = slope of Line M Using the yintercept 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.



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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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02 Jun 2015, 11:43
sterling19 wrote: Let m = slope of Line M Using the yintercept 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. Please elaborate the last sentence....how can values faill in the range after multiplying 1/7 between 2 and 5.
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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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14 Jul 2016, 11:08
sterling19 wrote: Let m = slope of Line M Using the yintercept 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. Hi Sterling, Thanks for your answer.Could be please elaborate it ? Regards



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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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27 Jul 2016, 23:01
ravi2107 wrote: sterling19 wrote: Let m = slope of Line M Using the yintercept 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. Hi Sterling, Thanks for your answer.Could be please elaborate it ? Regards 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)
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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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28 Oct 2016, 09:55
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?



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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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10 Jan 2017, 19:44
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.



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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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08 Apr 2017, 12:12
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



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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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18 Apr 2017, 00:40
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.



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Re: Line M has a yintercept of –4, and its slope must be an integermulti
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03 Jun 2018, 07:10
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