In the xy-plane, the straight-line graphs of the three equations above

For more check Ultimate GMAT Quantitative Megathread
_________________

Solution provided by : mcelroytutoring

Let's start by substituting the point (p,r) into all equations in place of (x,y) which will make step #2 a bit easier to comprehend but is not necessary to solve the question. Then, let's consider the number of variables left in each equation.

#1: r = ap - 5 (3 variables R,A,P)
#2: r = p + 6 (2 variables R,P)
#3: r = 3p + b (3 variables R,B,P)

1) a = 2 allows us to reduce equation #1 to the variables r and p, which are the same two variables as equation #2. Thus we have simultaneous equations. As soon as we verify that the equations are different, we know that we can solve for both variables. Once we know r and p, we can substitute in equation #3 to solve for b. Sufficient.

2) r = 17 allows us to do the same thing, more or less. It reduces equation #2 to only one variable, allowing us to solve for p. Once we have p (and r), we can use equation #3 to solve for b. Sufficient.

_________________

if each of the 3 equations contains points (p,r) this means that they intersect in that point
1. a=2
Find the intercept Intercept for three simultaneous equations
y=2x-5
y=x+6
y=3x+b
Let's use the first 2 equations: plug y=x+6 in the secod equation
x+6=2x-5 -> x=11, y=17 we can use the values to calulate b in the 3rd equation
17=33+b -> b=-16 SUFFICIENT

2. Here we have directly the value for Y, let's plug it in the 2nd equation
y=x+6 -> 17=x+6 -> x=11, y=17; We can plug these values in the 3rd equation and find b as we did above
17=33+b -> b=-16 SUFFICIENT

Answer (D) Most important point is here to catch the hint about intersection of 3 lines at one point

I think it's D.

Keeping point (p,r) in all the equations we get :

p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)

Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.

Similarly for (2) we can get the values for r and p and hence can get the value for b.

So both statements individually are correct to answer the question.

Approach if right but the values you derived are wrong. According to me r = 17 and that is what stmt b also states.

y = ax - 5 ... eq 1
y = x + 6 ... eq 2
y = 3x + b ...eq 3

Total of 4 variables are present.

Statement 1 : a = 2

Insert in eq 1

We have y = 2x -5 and y = x+6

Solving we get x = 11 and y = 17

Substitute in eq 3 and we get value of b

Statement 2: r=17

This means the y co- ordinate is 17
Substitute in eq 2 we get x as 11

Again can find value of b from equation 3

Hence D

(eq1) \(y = ax - 5\)
(eq2) \(y = x + 6\)
(eq3) \(y = 3x + b\)

In the xy-plane, the straight-line graphs of the three equations above each contain the point (p, r). If a and b are constants, what is the value of b?

1) \(a = 2\)
2) \(r = 17\)

Solution:

1) \(a = 2\)
- Putting the value of a in eq1, we get: \(y = 2x - 5\)
- At this point you can solve for (x, y), plug (x, y) in (eq3) and solve for (b) [though this approach might take few seconds]
(alternatively, faster method)
- you can skip solving for (x, y) and deduce that given 3 equations and 3 unknowns (since a is given in statement 1) we can solve for all of them (including b), since the lines are have different slopes i.e. different lines. Hence, we can get single value of 'b', proving the condition SUFFICIENT.
NOTE: 3 equations and 3 unknowns does not ALWAYS mean that we can find 3 unknown. We have to make sure that 2 of them or all of them are not the same line.

2) \(r = 17\)
- Since, point (p, r) lie on all the line, we can plugin the point in above equation
\(r = ap - 5 => 17 = ap - 5\)
\(r = p + 6 => 17 = p + 6\)
\(r = 3p + b => 17 = 3p + b\)
- Again, we do not need to solve for all the variables and just recognize that the above equations will lead to single value of b. Hence, SUFFICIENT.

Answer: (D)

Would it be correct to simply say that we have 4 variables with 3 equations so eliminating any one variable gets us to three equations and three variables and is therefore sufficient? Is that logic sound?

what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",

i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables

The highlighted statement in effect says that all these 3 lines meet each other at one point and so there is a single value of (x,y) that satisfies these 3 equations. It is only because of this highlighted statement you can solve this set of equations for a unique value of x,y, a and b.

Hope this clarifies your doubt.
_________________

Hi All,

We're given the equations for 3 lines (and those equations are based on 4 unknowns: 2 variables and the 2 'constants' A and B):

Y = (A)(X) - 5
Y = X + 6
Y = 3X + B

We're told that the three lines all cross at one point on a graph (p,r). We're asked for the value of B. While this question looks complex, it's actually built around a 'system' math "shortcut" - meaning that since we have 3 unique equations and 4 unknowns, we just need one more unique equation (with one or more of those unknowns) and we can solve for ALL of the unknowns:

1) A =2

With this information, we now have a 4th equation, so we CAN solve for B.
Fact 1 is SUFFICIENT

2) R = 17

This information tell us the x co-ordinate where all three lines will meet, so it's the equivalent of having X=17 to work with. This 4th equation also allows us to solve for B.
Fact 2 is SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

We can begin by substituting p and r for x and y, respectively, in the three given equations.

1) r = ap – 5

2) r = p + 6

3) r = 3p + b

Statement One Alone:

a = 2

We can substitute 2 for a in the equation r = ap – 5. Thus, we have:

r = 2p – 5

Next we can set equations 1 and 2 equal to each other.

2p – 5 = p + 6

p = 11

Since p = 11, we see that r = 11 + 6 = 17

Finally, we can substitute 11 for p and 17 for r in equation 3. This gives us:

17 = 3(11) + b

17 = 33 + b

-16 = b

Statement one alone is sufficient to answer the question.

Statement Two Alone:

r = 17

We can substitute r into all three equations and we have:

1) 17 = ap – 5

2) 17 = p + 6

3) 17 = 3p + b

We see that p = 11. Now we can substitute 11 for p in equation 3 to determine a value for b.

17 = 3(11) + b

-16 = b

Statement two alone is also sufficient to answer the question.

Answer: D
_________________

Video solution for the same

https://gmatquantum.com/official-guides ... ial-guide/
_________________

Video solution from Quant Reasoning starts at 11:16
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

_________________

Answer: Option D

Video solution by GMATinsight



Get TOPICWISE: Concept Videos | Practice Qns 100+ | Official Qns 50+ | 100% Video solution CLICK.
Two MUST join YouTube channels : GMATinsight (1000+ FREE Videos) and GMATclub
_________________

y = ax - 5
y = x + 6
y = 3x + b



State 1 : a = 2

In eq 1

We have y = 2x -5 and y = x+6

Solving we get x = 11 and y = 17

From eq 3 and we get value of b

State 2: r=17

Gibes us y co- ordinate ias 17
Substitute in eq 2 we get x as 11

Again can find value of b from equation 3

Hence D

(x, y) => (p, r)

r = ap - 5
r = p + 6
r = 3p + b

1)
a = 2
r = 2*p

p + 6 = 2p - 5
P = 11

r = p + 6
r = 11 + 6 = 17

r = 3p + b
17 = 3 (11) + b
b = -16

SUFF

2) r = 17

r = p + 6
17 = p + 6
p = 11

r = 3p + b
17 = 3*11 + b
b = -16
SUFF

Statement (1) a = 2
We now have 4 equations and 4 unknowns(x,y,a,b) to solve for b.
(Sufficient)

Statement (2)r = 17
Substitute the value in the equations given in the question stem.
17 = a p – 5..........(i)

17 = p + 6...........(ii)

17 = 3p + b..........(iii)

p=11 from (ii)

b=17 - 3(11)

=>b= -16
(sufficient)
(option d)

Devmitra Sen
GMAT SME

GMATNinja, Bunuel, can you please clarify whether st1 and st2 are even necessary to solve this question?

I approached this question in the following way:

y = ax - 5----> r=ap-5;
y = x + 6---> r=p+6
y = 3x + b---> r=3p+b
3p+b=p+6--> p+b/3=p+6---> b/3=6--> b=18