Aptitude Test - Area - PrashnaPatra

Aptitude Topics

Area

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

A.	15360
B.	153600
C.	30720
D.	307200

Answer: Option B

Explanation:

Perimeter = Distance covered in 8 min. =		12000	x 8	m = 1600 m.
		60

Let length = 3x metres and breadth = 2x metres.

Then, 2(3x + 2x) = 1600 or x = 160.

Length = 480 m and Breadth = 320 m.

Area = (480 x 320) m² = 153600 m².

An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

A.	2%
B.	2.02%
C.	4%
D.	4.04%

Answer: Option D

Explanation:

100 cm is read as 102 cm.

A₁ = (100 x 100) cm² and A₂ (102 x 102) cm².

(A₂ - A₁) = [(102)² - (100)²]

= (102 + 100) x (102 - 100)

= 404 cm².

Percentage error =		404	x 100	%	= 4.04%
		100 x 100

The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

A.	16 cm
B.	18 cm
C.	24 cm
D.	Data inadequate
E.	None of these

Answer: Option B

Explanation:

2(l + b)	=	5
b	=	1

2l + 2b = 5b

3b = 2l

b =	2	l
	3

Then, Area = 216 cm²

l x b = 216

l x	2	l	= 216
	3

l² = 324

l = 18 cm.

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

A.	40%
B.	42%
C.	44%
D.	46%

Answer: Option C

Explanation:

Let original length = x metres and original breadth = y metres.

Original area = (xy) m².

New length =		120	x	m	=		6	x	m.
		100					5

New breadth =		120	y	m	=		6	y	m.
		100					5

New Area =		6	x x	6	y	m²	=		36	xy	m².
		5		5					25

The difference between the original area = xy and new-area 36/25 xy is

= (36/25)xy - xy

= xy(36/25 - 1)

= xy(11/25) or (11/25)xy

Increase % =		11	xy x	1	x 100	%	= 44%.
		25		xy

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?

A.	2.91 m
B.	3 m
C.	5.82 m
D.	None of these

Answer: Option B

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.

The diagonal of the floor of a rectangular closet is 7 feet. The shorter side of the closet is 4 feet. What is the area of the closet in square feet?

5	1
	4

13	1
	2

Answer: Option C

Explanation:

Other side

=√[

	15		2	-		9		2
	2					2

]ft

=√[

225	-	81
4	-	4

]ft

=√[

144

]ft

6 ft.

Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.

A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:

A.	10%
B.	10.08%
C.	20%
D.	28%

Answer: Option D

Explanation:

Let original length = x and original breadth = y.

Decrease in area

= xy -		80	x	x	90	y
		100			100

=		xy -	18	xy
			25

=	7	xy.
	25

Decrease % =		7	xy x	1	x 100	%	= 28%.
		25		xy

A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

A.	20
B.	24
C.	30
D.	33

Answer: Option C

Explanation:

Let the side of the square(ABCD) be x metres.

Then, AB + BC = 2x metres.

AC = √2x = (1.41x) m.

Saving on 2x metres = (0.59x) m.

Saving % =		0.59x	x 100	%	= 30% (approx.)
		2x

The diagonal of a rectangle is √41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

A.	9 cm
B.	18 cm
C.	20 cm
D.	41 cm

Answer: Option B

Explanation:

√(l² + b² )= √41.

Also, lb = 20.

(l + b)² = (l² + b²) + 2lb = 41 + 40 = 81

(l + b) = 9.

Perimeter = 2(l + b) = 18 cm.

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

A.	814
B.	820
C.	840
D.	844

Answer: Option A

Explanation:

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 x 41) cm².

Required number of tiles =		1517 x 902		= 814.
		41 x 41

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

A.	1520 m²
B.	2420 m²
C.	2480 m²
D.	2520 m²

Answer: Option D

Explanation:

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.

Solving the two equations, we get: l = 63 and b = 40.

Area = (l x b) = (63 x 40) m² = 2520 m².

The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

A.	25% increase
B.	50% increase
C.	50% decrease
D.	75% decrease

Answer: Option B

Explanation:

Let original length = x and original breadth = y.

Original area = xy.

New length =	x	.
	2

New breadth = 3y.

New area =		x	x 3y		=	3	xy.
		2				2

Increase % =		1	xy x	1	x 100	%	= 50%.
		2		xy

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

A.	40
B.	50
C.	120
D.	Data inadequate
E.	None of these

Answer: Option E

Explanation:

Let breadth = x metres.

Then, length = (x + 20) metres.

Perimeter =		5300		m = 200 m.
		26.50

2[(x + 20) + x] = 200

2x + 20 = 100

2x = 80

x = 40.

Hence, length = x + 20 = 60 m.

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

A.	34
B.	40
C.	68
D.	88

Answer: Option D

Explanation:

We have: l = 20 ft and lb = 680 sq. ft.

So, b = 34 ft.

Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:

A.	Rs. 456
B.	Rs. 458
C.	Rs. 558
D.	Rs. 568

Answer: Option C

Explanation:

Area to be plastered	= [2(l + b) x h] + (l x b)
	= {[2(25 + 12) x 6] + (25 x 12)} m²
	= (444 + 300) m²
	= 744 m².

Cost of plastering = Rs.		744 x	75		= Rs. 558.
			100