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Area Calculation - Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Area Calculation. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - A way of uniform width keeps running round within a rectangular Field 38m long and 32m wide. On the off chance that the way involves 600m², then The width of the way is:

Answer : B

Explanation

area of the field = (32*38)m²=  1216m²
Let the width of the path be x mtr.
Area without path= (38-2x) (32-2x) m²
= 1216+4x²-140x
Area of the path= 1216- (1216+4x²-140x) = 140x-4x²
∴ 140x-4x²=600
⇒4x²-140x+600=0
⇒x²-35x+150=0 ⇒ (x-30) (x-5) =0 ⇒x=5m

Q 2 - The edge of a rectangle and a square are 160m each. The range of the rectangle is not exactly that of the square by 100sq. meters. The length of the rectangle is:

Answer : C

Explanation

Each side of the square= 160/4m = 40m
2(L+b) =160 ⇒   (L+b) = 80
(40) ²-Lb= 100 ⇒    lb= (1600-100) = 1500
(L-b) ²= (L+b) ²- 4Lb= (80) ²-4*1500= (6400-6000) =400⇒L-b=20
∴L+b= 80, L-b= 20 ⇒   2L=100 ⇒ L= 50 m

Q 3 - The region of a square is 1/2 hectare. The length of its slanting is:

Answer : B

Explanation

area = (1/2*10000)m²= 5000m²
1/2*(diagonal) ²= 5000 ⇒ d²= 10000⇒d= √10000= 100m
∴ Length of diagonal = 100m

Q 4 - The length of a rectangular plot is twice its expansiveness. On the off chance that the length of its corner to corner is 9√5m, the border of the perimeter of the rectangle is:

Answer : B

Explanation

let breadth = x meter, Then, length = 2x meter
Diagonal = √ (2x) ²+ x² =√5x²= √5x meter
∴ √5x= 9√5 ⇒9m ⇒Perimeter = 2(18+9) m =54m

Q 5 - The territory of the four dividers of the room is 168m². The broadness and Length of the room is 8m and 6m. The length of the room is:

Answer : C

Explanation

2(L+8)*6= 168 ⇒L+8= 168/12= 14 ⇒L= (14-8) = 6m

Q 6 - Each side of an equilateral triangle measures 2√3cm.The length of its altitude is:

Answer : C

Explanation

Each side=  2√3
1/2 * base *altitude= √3/4* (2√3)² = √3/4 *12 = 3√3
⇒ 1/2 *2 √3* altitude= 3√3⇒altitude= 3cm

Q 7 - The area of a right angled triangle is 20cm² and one of the sides containing the right angle is 4cm. The altitude on the hypotenuse is:

Answer : D

Explanation

Let the altitude be x cm. Then,
1/2 *4*x = 20 ⇒ x= 10 cm
BC= Hypotenuse = √ (10)²+ (4)² =√116 =√4*29= 2√29
Let AD⏊ BC. Then,
1/2 * BC* AD = area of ∆ ABC ⇒1/2*2 √29* AD= 20
∴ AD = 20/√29 cm

Q 8 - The length of the middle of an equilateral triangle is x. The range of the triangle is:

Answer : D

Explanation

Let each side be a and height be x. then,
A²/4+ x² = a² ⇒ (a²-a²/4) = x²⇒   a²= 4x²/3
Area of the triangle = √ 3/4 a² = √3/4* 4x²/3= √3x²/3

Q 9 - If the side of a rhombus is 20cm and its shorter corner to corner is three-fourth of its more extended askew, then the range of the rhombus is:

Answer : C

Explanation

Let the longer diagonal be x cm , then shorter diagonal = (3/4)x cm
∴ AC= x cm and BD=(3/4 )x cm
AO =1/2*AC =x/2 cm, BO= 1/2 BD= (3/8) X cm and AB= 20 cm
In right ∆ AOB, we have AO² +BO² = AB²
(x/2) ²+ (3x/8) ²= (20) ²⇒x²/4+9x²/64=400 ⇒16x²+9x²=25600 ⇒25x²= 25600 ⇒x²=1024
⇒ x=√1024= 32 cm
∴ AO =32/2=16cm, BO= (3/8*32) cm=12cm
∴ AC=2*AO= 32 cm, BD= 2*BO= 24cm
Area of the rhombus = (1/2*32*24) cm² =384 cm²

Q 10 - A round greenery enclosure has an outline of 440 m. There is a 7m wide fringe inside the patio nursery along its outskirts. The territory of the fringe is:

Answer : D

Explanation

2πR =440 ⇒ 2*22/7*R= 440 ⇒R= (440* 7/44)=70 m
Outer radius = 70m, inner radius = (70-7) =63 m
Required area = π [(70)²-(63)²] m²= 22/7 *(70+63) (70-63) m²
= (22*133) m², = 2926m²

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