Aptitude - Area - Discussion

@Rkbm your answer best answer.

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.

There is no intercrossing left. When we considered 60x +40x, aren't we considering the intercrossing too? Otherwise the lengths couldn't have been 60 and 40.

The road is in the shape of plus mark.

So area of 1st road is 60x.

So area of 2nd road is 40x.

But in the middle the square has been calculated twice above. So subtract that area x^2.

Cosidering Road parallel to 60m side Area :- 60x.
Cosidering Road parallel to 40m side Area :- 40x.

Common area :- x*x.

Hence :- 291=60x+40x-x*x.

40*3=120
60*3=180
TOTAL=300
LESS=291
BALANCE=9 SQCM
THAT IS 3*3
ANS=3.

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

@Puskar,

You may go through Diego's answer for better understanding.

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.

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