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Another Cuckoo prep Question

Expert replies
by Sanjeev k Sexena » Tue Jan 21, 2014 11:44 am
In an xy-coordinate plane, a line is defined by y = kx + 1. If (4, b), (a, 4), and (a, b
+1) are three points on the line, where a and b are unknown, then k = ?

(A) 1/2

(B) 1

(C) 1(1/2) Mixed number

(D) 2

(E) 2(1/2) Mixed number
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Source: — Problem Solving |

by [email protected] » Tue Jan 21, 2014 1:50 pm
Hi Sanjeev,

I think that this question has a typo in it. Two of the three points on the line are defined as (a,4) and (a,b+1), but since the x-coordinate is the same, then you either don't have a line or you don't have 2 different coordinates. If you can double-check and update this question, then I'll be happy to help you solve it.

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by Sanjeev k Sexena » Wed Jan 22, 2014 2:35 am
[email protected] wrote:Hi Sanjeev,

I think that this question has a typo in it. Two of the three points on the line are defined as (a,4) and (a,b+1), but since the x-coordinate is the same, then you either don't have a line or you don't have 2 different coordinates. If you can double-check and update this question, then I'll be happy to help you solve it.

GMAT assassins aren't born, they're made,
Rich
Hi there,

Check page 186, Question:77 of the VERITAS PREP Geometry Guide. The question is so written ! Don't know whether they made the typo themselves.
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by [email protected] » Wed Jan 22, 2014 2:22 pm
Hi Sanjeev,

I don't own that book, but I can still offer a suggestion. Check the wording in the explanation; it's unlikely that a typo would show up in both spots. If that doesn't work, then post in the Veritas Forum (on this site); someone from that company should be able to address the issue.

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by sarathought777 » Tue Feb 11, 2014 12:38 pm
answer is A
PUT the 3 inputs of x and y in line
and get
b = 4k+1 ...(1)
4 = ak+1 ...(2)
b+1 = ak+1 ...(3)

now compare (2) & (3)
and get
b+1 = 4
so b= 3

put the value of b in (1)
3 = 4k+1
2=4k
k = 1/2
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by GMATGuruNY » Tue Feb 11, 2014 1:21 pm
Sanjeev k Sexena wrote:In an xy-coordinate plane, a line is defined by y = kx + 1. If (4, b), (a, 4), and (a, b
+1) are three points on the line, where a and b are unknown, then k = ?

(A) 1/2

(B) 1

(C) 1(1/2) Mixed number

(D) 2

(E) 2(1/2) Mixed number
(a, 4) and (a, b+1) must represent the SAME point on the line.
Thus, the y-value in each case must be the same:
4 = b+1
b=3.

Since b=3, (4,b) = (4,3).
Since the equation of the line is y = kx + 1, the y-intercept = (0,1).
Since both (4,3) and (0,1) are on the line, the slope = (3-1)/(4-0) = 2/4 = 1/2.

The correct answer is A.
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by vipulgoyal » Wed Feb 12, 2014 11:18 pm
there is another way to prove it

b+1-b/a-4 = 4-b/a-4
b=3
Since b=3, (4,b) = (4,3)
(3-1)/(4-0) = 2/4 = 1/2.
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by Brent@GMATPrepNow » Fri Jan 10, 2020 5:32 am
Sanjeev k Sexena wrote:In an xy-coordinate plane, a line is defined by y = kx + 1. If (4, b), (a, 4), and (a, b
+1) are three points on the line, where a and b are unknown, then k = ?

(A) 1/2

(B) 1

(C) 1(1/2) Mixed number

(D) 2

(E) 2(1/2) Mixed number
There's a HUGE CLUE in the fact that (a, 4) and (a, b+1) are both on the same line. Notice that the x-coordinates are the same. If the x-coordinates are the same, then there are two possible scenarios:
scenario #1: The points (a, 4) and (a, b+1) are DIFFERENT points, in which case the line is vertical (with undefined slope)
scenario #2: The points (a, 4) and (a, b+1) define the SAME point

IMPORTANT: If a line is defined by y = kx + 1, then k represents the slope. So, the question is really asking us to find the slope of the line.

In scenario #1, the slope would be undefined. Since none of the answer choices are undefined, we can rule out scenario #1, which means (a, 4) and (a, b+1) define the SAME point. So, we can be certain that b+1 = 4, which means b = 3

Now that we know that b = 3, we can use the fact that the point (4,b) is on the line.
This means that the point (4,3) is on the line y = kx + 1.
When we plug x=4 and y=3 into the equation, we get 3 = (k)(4) + 1
Solve to get k = 1/2

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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