The equation of line k is y = mx + b, where m and b are some constants

Stmnt 1:

If by statement 1 you mean that the point (b, 2b) lies on the line k, you can plug in the point in the equation of the line y=mx+b to get 2b = m*b + b.
2b = b(m + 1)
b = mb
(m-1)b = 0
Either m is 1 or b = 0 or both. In case we prove that b is not 0, then m must be 1. If b is 0, m could be anything including 1. Not sufficient.

Stmnt 2:
If line passes through (2, 2), we get 2 = m*2 + b
We have 2 unknowns and only one equation. Not enough to get the value of m.

Using both together, we have
2m + b = 2 .... (I)
such that either b is 0 or m is 1 or both.

If b is 0, put b = 0 in (I) which gives us m = 1
If b is not 0, m must be 1.

So in any case, m is always 1.

Answer (C)

I will give it a try

St1 says (b,2b) belong to k so, we have
2b=mb+b or b=mb or (m-1)*b=0 ----->This implies either m=1 or b=0. St1 is not sufficient. So A and D ruled out

St 2: 2=2m+b or \(m= (2-b)/2\) or m =1-(b/2).. But we don't know anything about b. So not sufficient. Option B ruled out

Combining we get that b=0 then m=1. Alternatively if m=1 then b=0.

Therefore ans is C
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Can anybody explain why m cant be 0 ?

Also when 2b = mb + b, why can i not cancel b to get m+1 = 2 and m =1?

From (1): m = 1 or b = 0 or both.
From (2): 2m + b = 2.

If m is not 1, then b must be 0. But if b = 0, then from 2m + b = 2, we get that m = 1. So, in any case, m = 1.

Hope it's clear.
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If you divide (reduce) \(2b = mb + b\) by b you assume, with no ground for it, that b does not equal to zero thus exclude a possible solution (notice that both b = 0 AND m = 1 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Hope it's clear.
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Thank you Bunuel, much appreciated

Solution -

Stmt1 - (b, 2b) belongs to k --> Let’s plug in the equation of the line. 2b = mb + b --> b = mb --> b(m – 1) = 0. So, either b = 0 or m = 1. If b=0, we may not find definite value of m. Not Sufficient.

Note - From b = mb above, please do not conclude m=1 after dividing b both sides. We do not know the value of b.

Stmt2 - (2, 2) belongs to k --> Let’s plug in the equation of the line. 2 = m*2 + b. Without knowing value of b, we can not find the value of m. Not Sufficient.

1+2 --> We can derive below equations from both statements.
b(m – 1) = 0 ----1
2 = 2m + b ------2

If b = 0 from 1, then second equation is 2 = 2m --> m = 1.
If m = 1 from 1, then second equation is 2 = 2 + b --> b = 0. Thus the value of m = 1. Sufficient.

ANS C.

For line k, y = mx + b. What is the value of m?

I took more of a conceptual approach. In order for us to know the slope m, we must be given 2 points that lie on line k.

(1) (b, 2b) belongs to k.
we only know 1 point, which is ambiguous given b.
insufficient

(2) (2, 2) belongs to k.
we only know 1 point
insufficient

combined, b=mb and 2=2m+b. We get b=0 and m=1
Sufficient
Answer: C

800score Official Solution:

Statement (1) tells us that point (b, 2b) belongs to line k. Let’s plug it in the equation of the line. 2b = mb + b. This yields b(m – 1) = 0. So either b = 0 or m = 1. We do NOT have a definite value of m if b = 0. Therefore statement (1) by itself is NOT sufficient.

Statement (2) tells us that (2, 2) belongs to k. Let’s plug it in the equation of the line. 2 = m × 2 + b. Clearly, not knowing the value of b, we can NOT find the value of m. Therefore statement (2) by itself is NOT sufficient.

When we use the both statements together, we have the following system of two equations:
b(m – 1) = 0
2 = 2m + b
If the solution of the first equation is b = 0, then the second one yields 2 = 2m. So in this case m = 1.
If the solution of the first equation is m = 1, then the second one yields 2 = 2 + b. So in this case b = 0. Thus the solution of the system is always b = 0, m = 1. (The equation of the line is y = x). We have the definite value of m. Therefore statements (1) and (2) taken together are sufficient to answer the question.

The correct answer is C.
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y=mx+b, in this equation if we put x=0, then we get y=b.
So we have a point (0,b)
We know that above equation is in the standard form that is y=mx+c, where m=slope, c=y intercept of the line
hence ,
m=(y1-y2)/(x1-x2)
m = 2b-b/b-0
m=1
Is there something wrong in this method? cause according to this the answer is A

You are ignoring the fact that b=0 is a possible solution.

Remember to not eliminate variables in DS question when you dont know what values they can take.

When you plug in b,2b in the equation y=mx+b ---> 2b=mb+b ---> mb-b=0 --> b(m-1)=0 (do not cancel out b as you will eliminate 1 solution)--> either b=0 or m=1.

So if b=0 , then m can take all possible values. This makes this statement not sufficient.

Hope this helps.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

The equation of line k is y=mx+b, where m and b are constants. What is the value of m?

1) (b,2b) belongs to k.
2) (2,2) belongs to k.

In the original condition, we need to know the gradient and the y-intercept, so we need 2 equations
There are 2 equations given by the 2 conditions, so there is high chance (C) will be our answer.
Looking at the conditions together, from 2b=mb+b, b=mb, b(m-1)=0, we get b=0 or m=1, and from 2=2m+b, if b=0, m=1. This is sufficient, and the answer becomes (C).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Bunuel
This question has been discussed here
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Merged the topics. Thank you.
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