# Quantitative Aptitude: Mensuration Questions Set 6

1. Poles are to be fixed along the boundary of a rectangular field in such a way that distance between any two adjacent poles is 2 m.The perimeter of the field is 70m and length and the breadth of the field are in the ratio 4:3 resp. How many poles will be required?
A) 42
B) 40
C) 35
D) 38
E) 45
Option C
Solution:
Required between the two poles = (Perimeter/Dist.between any two adjacent poles) = 70 / 2 = 35
2. The circumference of  a circular garden is 1320m.Find the area. Outside the garden , a road of 2m width runs around it .What is the area of this road and calculate the cost of gravelling it at the rate of 50 paise per sq. m .
A) 2500.15 m2, Rs.1500.15
B) 2652.57 m2, Rs.1326.285
C) 2541.14 m2, Rs.1600.47
D) 3245.78 m2,Rs.2000
E) 4157.12 m2,Rs.1452.11
Option B
Solution:
Circumference of the garden = 2*pi*R = 1320
R= 210m
Outer radius = 210 +2= 212 m
Area of the road = pi*(212)^a  – pi*(210)^2
= pi*422*2 = 2652.57 m^2
Therefore ,
cost of gravelling = 2652.57 * 0.5 = Rs.1326.285
3. A square shape of park of area 23,104 sq. m is to be enclosed with wire placed at heights 1,2,3,4 m above the ground . Find required length of the wire ,,if its length required for each circuit is 10% greater than the perimeter of the field ?
A) 2675.2m
B) 2145.12m
C) 2750m
D) 2478.11m
E) 2400.5m
Option A
Solution:
Perimeter = √23,104  * 4 = (152 * 4)m
Length of each circuit = 152 * 4 *(110/100)
The wire goes around 4 times ,so the total length of the wire required =  152 * 4 *(110/100) * 4 = 2675.2 m
4. Area of a hexagon is 54√3 cm^2. What is its side ?
A) 7cm
B) 5cm
C) 4cm
D) 6cm
E) 8cm
Option C
Solution:
(6√3/4) *a^2 = 54√3
=> a^2 = 36
=> a = 6 cm
5. Smallest side of a right angled triangle is 8 cm less than the side of a square of perimeter 64cm . Second largest side of the right angled triangle is 4 cm less than the length of rectangle of area 112 sq. cm and breadth 8 cm .What is the largest side of the right angled triangle?
A) 9.2cm
B) 7.75cm
C) 10.50cm
D) 14cm
E) 12.80cm
Option E
Solution:
Side of a square = (perimeter /4) = 64/4 = 16 cm
smallest side  = 16 – 8   = 8cm
Length of the rectangle = Area/Breadth = 112/8 = 14cm
Second side of triangle = 14 – 4 = 10cm
Hypotenuse of the right angled triangle = √(8)^2+(10)^2 = 12.80 cm
6. If the radius of the circular field is equal to the side of a square field .If the difference between the area of the circular field and area of the square field is 5145 sq. m ,then calculate the perimeter of the circular field?
A) 421m
B) 315m
C) 310m
D) 308m
E) 300m
Option D
Solution:
Let the radius of the circular field and the side of the square field be r
Then,
pi*r^2 – r^2 = 5145
=> r^2[(22-7)/7] = 5145
=> r = 49 m
Therefore ,
circumference of the circular field = 2*pi*r = 308m
7. A rectangular plot has a concrete path running in the middle of the plot parallel to the parallel to the breadth of the plot. The rest of the plot is used as a lawn ,which has an area of 240sq. m. If the width of the path is 3m and the length of the plot is greater than its breadth by 2m ,what is the area of the rectangular plot(in m )?
A) 410m
B) 288m
C) 250m
D) 300m
E) 320m
Option B
Solution:
Let width be x m
and length be (x+2)m
Area of path = 3x sq. m
x(x+2) – 3x = 240
=> x^2 – x – 240 = 0
=> x(x – 16) +15 (x – 16) = 0
=>(x – 16)(x + 15) = 0
=>x = 16
Length =  16 + 2 = 18m
Therefore ,
Area of  plot = 16 * 18 = 288sq. m
8. A solid spherical ball  of radius r is converted into a solid circular cylinder of radius R. If the height of the cylinder is twice the radius of the sphere ,then find the relation between these two with respect to radius.
A) R = r√(3/4)
B) R = r√(3/2)
C) R = r√(1/2)
D) R = r√(2/3)
E) R = r√(1/3)
Option D
Solution:
Since one object is converted into another so the volume will remain the same .
Therefore ,
(4/3)*pi*r^3 = pi*R^2*H
=>R = r√(2/3)
9. A rectangular tank of length 37 (1/3) m internally , 12 m in breadth and 8 m in depth is full of water .Find the weight of water in metric tons, given that one cubic metre of water weighs 1000kg.
A) 3584 metric tons
B) 4500 metric tons
C) 4101 metric tons
D) 3870 metric tons
E) 5721 metric tons
Option A
Solution:
Volume of  water = 37(1/3)*12*8 m^3
Weight of water = (112/3)*12*8*1000 = 3584metric tons.
10. An equilateral triangle and a regular hexagon have equal perimeters. The ratio of the area of the triangle and that of the hexagon is :
A) 3:4
B) 4:9
C) 1:2
D) 2:3
E) 4:5
Option D
Solution:
Let side of triangle be x  and the side of regular hexagon be y .
3x = 6y
=>x = 2y
Area of triangle = (√3/4)x2
Area of hexagon = 6*(√3/4) * y2 = (3√3/8)*x2
Required ratio = 2 : 3