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# The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2

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The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2  [#permalink]

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14 Feb 2020, 00:55
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15% (low)

Question Stats:

86% (01:34) correct 14% (01:37) wrong based on 49 sessions

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The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2). What is the value of a + b?

A. 0
B. 3/4
C. 1
D. 9/4
E. 3

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Re: The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2  [#permalink]

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14 Feb 2020, 01:16
1
(1) x = 1/4*y + a
(2) y = 1/4*x + b

(1)+(2)
(x+y) = (x+y)/4 + (a+b)
3.(x+y)/4 = (a+b)

(x,y) =(1,2) then
3.(1+2)/4=(a+b)
(a+b)=9/4

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Re: The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2  [#permalink]

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14 Feb 2020, 01:17
1
Quote:
The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2). What is the value of a + b?

A. 0
B. 3/4
C. 1
D. 9/4
E. 3

For intersection of x = 1/4*y + a i.e. y = 4*(x-a) and y = 1/4*x + b

4*(x-a) = 1/4*x + b

i.e. 16(x-a) = x+4b
i.e. 16x - 16a = x + 4b
i.e. 15x = 4b + 16a
i.e. x = (1/15)*(4b + 16a) = 1 [i.e. x coordinate of point (1,2)]
i.e. 4b + 16a = 15

and y = 4*(x-a) = 4*(x-a) = 2 [i.e. y coordinate of point (1,2)]

i.e. x - a = 1/2
i.e. (1/15)*(4b + 16a) - a = 1/2
i.w. 4b + 16a - 15a = 15/2
i.e. 4b + a = 15/2

Taking the difference of the two equations in bold letters

(4b + 16a) - (4b + a) = 15 - (15/2)

i.e. 15 a = 15/2
i.e. a = 1/2

Since, 4b + a = 15/2
i.e. 4b = (15/2) - (1/2) = 7
i.e. b = 7/4

i.e. a+b = (1/2)+(7/4) = 9/4

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Re: The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2  [#permalink]

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14 Feb 2020, 01:20
1
as both the lines pass through point (1,2)
substitute 1,2 in each of the given equation we get values of a and b as 1/2 and 7/4 which adds to 9/4

IMO ans is D
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Re: The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2  [#permalink]

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14 Feb 2020, 03:37
1
Quote:
The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2). What is the value of a + b?

A. 0
B. 3/4
C. 1
D. 9/4
E. 3

x = 1/4*y + a… 1=1/4(2)+a…a=1-1/2=1/2=2/4
y = 1/4*x + b…2=1/4+b…b=2-1/4=8-1/4=7/4
a+b=(2+7)/4=9/4

Ans (D)
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Re: The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2  [#permalink]

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14 Feb 2020, 08:34
for given two eqn of line substitute value of x & y
we get
a= 1/2 and b = 7/4
a+b ; 9/4
IMO D

The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2). What is the value of a + b?

A. 0
B. 3/4
C. 1
D. 9/4
E. 3
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Re: The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2  [#permalink]

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14 Feb 2020, 12:05
1
The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2). What is the value of a + b?

point (1,2) would satisfy both equations then
or
$$x = \frac{1}{4}*y + a$$
$$1 =\frac{2}{4}+a$$ or a =$$\frac{1}{2}$$
$$y = \frac{1}{4}*x + b$$
$$2= \frac{1}{4}+b$$or b = $$\frac{7}{4}$$
$$a+b = \frac{1}{2}+\frac{7}{4} = \frac{9}{4}$$
D
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The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2  [#permalink]

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Updated on: 19 Feb 2020, 09:57
.

Originally posted by madgmat2019 on 15 Feb 2020, 01:16.
Last edited by madgmat2019 on 19 Feb 2020, 09:57, edited 1 time in total.
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Re: The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2  [#permalink]

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16 Feb 2020, 13:33
1
The lines $$x = \frac{1}{4}*y + a$$ and $$y = \frac{1}{4}*x + b$$ intersect at the point (1, 2). What is the value of a + b?

--> $$1= (\frac{1}{4})*2+a$$ --> $$a=\frac{1}{2}$$

--> $$2=(\frac{1}{4})*1+ b$$ --> $$b= \frac{7}{4 }$$

$$a+b = \frac{1}{2}+ \frac{7}{4} = \frac{(2+7)}{4}= \frac{9}{4}$$

Re: The lines x = 1/4*y + a and y = 1/4*x + b intersect at the point (1, 2   [#permalink] 16 Feb 2020, 13:33
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