Aptitude - Area - Discussion

Discussion Forum : Area - General Questions (Q.No. 5)
5.
A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
2.91 m
3 m
5.82 m
None of these
Answer: Option
Explanation:

Area of the park = (60 x 40) m2 = 2400 m2.

Area of the lawn = 2109 m2.

Area of the crossroads = (2400 - 2109) m2 = 291 m2.

Let the width of the road be x metres. Then,

60x + 40x - x2 = 291

x2 - 100x + 291 = 0

(x - 97)(x - 3) = 0

x = 3.

Video Explanation: https://youtu.be/R3CtrAKGxkc

Discussion:
75 comments Page 3 of 8.

Raji said:   1 decade ago
@Aysooh superb yar

Aaru said:   1 decade ago
Ayoosh your explaination is nice. I understood well now,

Praveen said:   1 decade ago
Thanks ayoosh very nice explanation.

Manogaram said:   1 decade ago
Hats off to ayoosh and prajwal for nice explanation.

Shruti said:   1 decade ago
Thank you ayoosh.

Rkbm said:   1 decade ago
X be the width. (60-x) (40-x) is area of remaining field.

Surendar sundaram said:   1 decade ago
40 X is width of the road.....
_________ We have already find out the area of the road by
| | | | subtracting (total area -lawn area)...
|___| |___| road area == 291 sq units
|___| |___|60 i.e Area of two triangle(roads)-(overlap btwn them)
| | | | over lap is width^2====x^2
|___|_|___| area of two triangle (60*x) and (40*x)
|x| so (60x+40x)-x^2===291
| | by solving tis v can find x===3

Ankanna said:   1 decade ago
See road is a rect. and they are intercrossing. let the width be x, then for one road area will be 40x and for the other road will be 60x since 60 and 40 will be the lengths of road.

Now, we area of roads is 291 m2, So

area(road1) + area(road2) - area(intercrossing).

60x + 40x - x^2 = 291 and then on you can find the answer.

Imran Shaikh said:   1 decade ago
How is this calculated ?

x2 - 100x + 291 = 0 ----------> (x - 97)(x - 3) = 0

Hashmuddin said:   1 decade ago
x2 -100x +291=0
x2 -97x -3x +291=0
x(x -97)-3(x-97)=0
(x-97)(x-3)=0


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