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) cm2
∴ 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)] m2
= (444 + 300)
m2
= 744 m2.
∴ 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 =
3x meters and breadth = 2x meters.
Then, 2(3x +
2x) = 1600 or x = 160
∴ Length
= 480 m and Breadth = 320 m.
∴ Area
= (480 × 320) m2 = 153600 m2.
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
=> 2x + 20 = 100
=> 2x = 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.
∴ 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/(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 + 2b) = (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
=> 2l + 2b = 5b
=> 3b = 2l
∴ b = 2l/3
Then, Area =
216 cm2
=> l × b =
216
=> l × (2l/3) = 216
=> l² =
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
cm2
(B) 462 cm2
(C) 498.35
cm2
(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 r2
= [{(22/7)
× 7 × 7√5} + {(22/7) × 7 × 7}] cm2
= [154(√5 +
1)] cm2
= (154 ×
3.236) cm2
= 498.35
cm2.
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) 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 meters.
Then, 60x +
40x - x2 = 291
=> x2 - 100x +
291 = 0
=> (x - 97)(x -
3) = 0
=> x = 3
∴ The
width of the road is 3 meters.
Area of plane surfaces: Next Tests: