__Question No. 01__**A rectangular lawn is 80 meters long and 60 meters wide. The time taken by a man to walk along its diagonal at the speed of 18 km per hour is**

(A) 25 sec.

(B) 20 sec.

(C) 18 sec.

(D) 10 sec.

Answer:
Option B

__Explanation:__
Diagonal, d
= √(80² + 40²) = √10000 = 100

Speed, v =
18 km/hr = 18 × (5/18) m/sec = 5 m/sec

∴ Time
taken, t = d/v = 100/5 = 20 sec.

__Question No. 02__**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 × 41) cm

^{2}
∴ Required
number of tiles = (1517 × 902)/ (41 × 41) = 814

__Question No. 03__**A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paisa 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*) ×*h*] + (*l*×*b*)
= [{2(25 +
12) × 6} + (25 × 12)] m

^{2}
= (444 + 300)
m

^{2}
= 744 m

^{2}.
∴ Cost
of plastering = Rs. 744 × (75/100) = Rs. 558

__Question No. 04__**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) 15,360

(B) 1,53,600

(C) 30,720

(D) 3,07,200

Answer:
Option B

__Explanation:__
Perimeter =
Distance covered in 8 min. = (12,000/60) × 8 m = 1600 m.

Let length =
3

*x*meters and breadth = 2*x*meters.
Then, 2(3

*x*+ 2*x*) = 1600 or*x*= 160
∴ Length
= 480 m and Breadth = 320 m.

∴ Area
= (480 × 320) m

^{2}= 153600 m^{2}.

__Question No. 05__**The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot @ 26.50 per meter is Rs. 5300, what is the length of the plot in meters?**

(A) 40

(B) 50

(C) 120

(D) None of
these

Answer:
Option D

__Explanation:__
Let breadth
=

*x*meters.
Then, length
= (

*x*+ 20) meters.
Perimeter =
(5300/26.50) m = 200 m.

∴ 2[(

*x*+ 20) +*x*] = 200
=> 2

*x*+ 20 = 100
=> 2

*x*= 80
=>

*x*= 40
Hence,
length =

*x*+ 20 = 60 m.

__Question No. 06__**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.

∴ A

_{1}= (100 x 100) cm^{2}and A_{2}(102 x 102) cm^{2}.
(A

_{2}- A_{1}) = [(102)^{2}- (100)^{2}]
= (102 +
100) x (102 - 100)

= 404 cm

^{2}.
∴ Percentage
error = [{404/(100 × 100)} × 100] % = 4.04%

__Question No. 07__**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*+ 2*b*) = (20 + 68) ft = 88 ft.

__Question No. 08__**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) None of
these

Answer:
Option B

__Explanation:__
According to question; [2(

*l*+*b*)]/b = 5/1
=> 2

*l*+ 2*b*= 5*b*
=> 3

*b*= 2*l*
∴ b = 2

*l*/3
Then, Area =
216 cm

^{2}
=>

*l*×*b*= 216
=>

*l*× (2*l*/3) = 216
=>

*l*² = 324
=>

*l*= 18 cm.

__Question No. 09__**What is the total surface area of a right circular cone of height 14 cm and base radius 7 cm?**

(A) 344.35
cm

^{2}
(B) 462 cm

^{2}
(C) 498.35
cm

^{2}
(D) None of
these

Answer:
Option C

__Explanation:__*h*= 14 cm,

*r*= 7 cm.

So,

*l*= √[(7)^{2}+ (14)^{2}]= √245 = 7√5 cm.
∴ Total
surface area =

*π rl*+*π r*^{2}
= [{(22/7)
× 7 × 7√5} + {(22/7) × 7 × 7}] cm

^{2}
= [154(√5 +
1)] cm

^{2}
= (154 ×
3.236) cm

^{2}
= 498.35
cm

^{2}.

__Question No. 10__**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 × 40) m

^{2}= 2400 m^{2}.
Area of the
lawn = 2109 m

^{2}.
∴ Area
of the crossroads = (2400 - 2109) m

^{2}= 291 m^{2}.
Let, the
width of the road be

*x*meters.
Then, 60

*x*+ 40*x*-*x*^{2}= 291
=>

*x*^{2}- 100*x*+ 291 = 0
=> (

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

*x*= 3
∴ The
width of the road is 3 meters.

**Area of plane Surfaces:**

**Formula: Area Aptitude Formulas**

**Solved Examples: Solved Examples: Set 01**

**Practice Test: Practice Test: 01**

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