in Quantitative Aptitude
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13 votes
13 votes

Find the area bounded by the lines $3x + 2y=14, 2x - 3y = 5$ in the first quadrant.

  1. $14.95$
  2. $15.25$
  3. $15.70$
  4. $20.35$
in Quantitative Aptitude
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4 Comments

15.25 sq units
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moved by
Bt how it will come 15.25 square units can someone expkain plz
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solve for x and y, observe that two given lines are perpendicular. so it will form something like rectangle in first quadrant.

now coordinates of rectangle are (0,0) (x,0) (0,y) (x,y)

find area = length * breadth
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2 Answers

12 votes
12 votes
Best answer
  • $3x+2y=14\rightarrow(1)$
  • $2x−3y=5\rightarrow(2)$

To get the intersection point, we solve the equations and get $x=4,y=1$

We can draw the required area like this 

 

From the above diagram,

Area $=\triangle ABC +\Box OBCE-\triangle CDE$

Area $=\left(\dfrac{1}{2}\times 4\times 6\right)+\left(1\times 4\right)-\left(\dfrac{1}{2}\times\dfrac{3}{2}\times 1\right)$

Area $=12+4-\dfrac{3}{4}=16-0.75=15.25$

So, the correct answer is $(B).$

edited by

4 Comments

Yeah, In question they should mention which axis. B’coz if we take first lower part then upper part, this will be a time-consuming problem.
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In the figure why △ DFC area is not taken??

Question is :

Find the area bounded by the lines 3x+2y=14 , 2x−3y=5 in the first quadrant.

And this area alone satisfies all conditions bounded by two lines given and also in the first quadrant. Isn’t it ??

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@ASNR1010 @Deepak Poonia Sir @Lakshman Patel RJIT . In the question it is given that we need to find the area in first quadrant then why are we finding the area bounded by Y-axis ??

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6 votes
6 votes

ans may be like this

1 comment

Hello can the picture be more clear please?
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Answer:

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