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# In the xy-plane, the points (a, 0) and (0, b) are on the lin

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Founder
Joined: 18 Apr 2015
Posts: 6901
Followers: 114

Kudos [?]: 1343 [0], given: 6316

In the xy-plane, the points (a, 0) and (0, b) are on the lin [#permalink]  21 Nov 2017, 13:02
Expert's post
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Question Stats:

77% (00:40) correct 22% (01:07) wrong based on 40 sessions
In the xy-plane, the points (a, 0) and (0, b) are on the line whose equation is $$y = \frac{1}{2} x + 10$$

 Quantity A Quantity B a b

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

_________________
Director
Joined: 07 Jan 2018
Posts: 644
Followers: 7

Kudos [?]: 602 [1] , given: 88

Re: In the xy-plane, the points (a, 0) and (0, b) are on the lin [#permalink]  19 Jan 2018, 01:28
1
KUDOS
Explanation

to get the value for point a
we replace 0 for y in the given equation
this gives 0=1/2 x + 10
-10=1/2 x>>>-20 = x

for b
replace 0 for x to get
y= 10
(B)
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Intern
Joined: 24 Jan 2018
Posts: 31
Followers: 0

Kudos [?]: 4 [1] , given: 8

Re: In the xy-plane, the points (a, 0) and (0, b) are on the lin [#permalink]  26 Jan 2018, 09:12
1
KUDOS
We replace:
(a,0)--> 0=1/2(a)+10; so we get a=-20
(0,b)--> b=1/2(0)+10 we get b=10
b>a
GRE Instructor
Joined: 10 Apr 2015
Posts: 1981
Followers: 60

Kudos [?]: 1803 [1] , given: 9

Re: In the xy-plane, the points (a, 0) and (0, b) are on the lin [#permalink]  24 May 2019, 15:56
1
KUDOS
Expert's post
Carcass wrote:
In the xy-plane, the points (a, 0) and (0, b) are on the line whose equation is $$y = \frac{1}{2} x + 10$$

 Quantity A Quantity B a b

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Since the points (a, 0) and (0, b) lie ON the line $$y = \frac{1}{2} x + 10$$, we know that the coordinates of those point must satisfy the equation of the line

That is, x = a and y = 0 satisfies the equation $$y = \frac{1}{2} x + 10$$
And x = 0 and y = b satisfies the equation $$y = \frac{1}{2} x + 10$$

Let's test each of these.
(a, 0)
replace x with a and y with 0 to get: $$0 = \frac{1}{2} a + 10$$
Subtract 10 from both sides: $$-10 = \frac{1}{2} a$$
Divide both sides by 1/2 to get: $$-20 = a$$

(0, b)
replace x with 0 and y with b to get: $$b = \frac{1}{2}(0) + 10$$
Simplify: $$b = 10$$

We get:
QUANTITY A: -20
QUANTITY B: 10

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

Re: In the xy-plane, the points (a, 0) and (0, b) are on the lin   [#permalink] 24 May 2019, 15:56
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