Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample from Mensuration that is important for all the competitive exams. We have included Some questions that are repeatedly asked in exams !!
- The length of a rectangle is reduced by 30%. By what percent would the width have to be increased to maintain the original area?
A)25%
B)42.86%
C)32.76%
D)35.50%
E)None of theseAnswer – B)42.86%
Explanation :
Width = 30*100/100-30
= 3000/70 = 42.86% - A circular wire of radius 49 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 4:7. The smaller side of the rectangle is
A)56cm
B)48cm
C)35cm
D)42cm
E)None of theseAnswer – A)56cm
Explanation :
circumference = 2*(22/7)*49 = 308 cm
length of rectangle sides are 4x, 7x.
circumference = 2*(4x+7x)
308 = 22x
X= 308/22 = 14
smaller side of rectangle = 4x= 4*14=56 cm - One of sides of a right-angled triangle is thrice the other, and the hypotenuse is 12 cm. The area of the triangle is
A) 25.4 cm2
B)6 cm2
C)5 cm2
D)6 cm2
E)None of theseAnswer – D) 21.6 cm2
Explanation :
x2 + (3x)2 = (12)2
10 x2= 144
x2 = 14.4 cm2
Area = 1/2 (x *3x)
=3(14.4)/2
= 21.6 cm2 - A bottle in the shape of a right circular cone of height 21cm is filled with water. The water is poured in right circular cylindrical bottle whose radius is one third of the radius of the circular cone. Then the height of the water in the bottle is
A)72cm
B)54cm
C)63cm
D)48cm
E)None of theseAnswer – C)63cm
Explanation :
1/3(3.14*r2*h1) = 3.14(r/3)2*h2
(r2*21)/3 = (r2/9)*h2
h2 =7*9 = 63cm - The base of an isosceles triangle is 40cm and its area is 160cm2. The length of one of its equal side is
A)22cm
B)20cm
C)40cm
D)35cm
E)None of theseAnswer – A)22cm
Explanation :
Area = [h*(40/2)]/2
160/2 = h*10
80 = h*10
h = 80/10 = 8cm
x = √82+202 = √(64+400)
x = 21.5 = 22cm - The circumference of a circle is twice the perimeter of a rectangle. The area of the circle is 3850sq cm. What is the area of the rectangle if the length of the rectangle is 30cm ?
A)750sq cm
B)620sq cm
C)700sq cm
D)650sq cm
E)None of theseAnswer – A)750sq cm
Explanation :
Area of circle = (22/7)*r2
3850 = (22/7)*r2
r2= 3850*7/22 = 175*7 = 1225
r = 35
circumference of the circle = 2*(22/7)*35 = 220cm
perimeter of the rectangle = 2(30+b)
220/2 = 60+2b
110 = 60+2b
2b = 50
b = 25cm
Area of the rectangle = 25*30 = 750sq cm - Inside a square plot a circular garden is developed which exactly fits in the square plot and the diameter of the garden is equal to the side of the square plot which is 24m. What is the area of the space left out in the square plot after developing the garden ?
A) 120m2
B) 110m2
C) 125m2
D) 150m2
E)None of theseAnswer – A) 120m2
Explanation :24*24 – 22/7 (12*12) = 576 – 456 = 120m2 - The ratio of the side of the room is 5:2, the cost of whitewashing the ceiling of the room at 50 paise per square metre is Rs.2500 and the cost of preparing the walls at 30 paise per square metre is Rs.2100. The height of the room is
A)31m
B)36m
C)23m
D)20m
E)None of theseAnswer – C)23m
Explanation :
Area = 2500/0.50 = 5000sq metres
5x*2x = 5000
10x2 = 5000
X2 =500
X=22.36 = 22
Area of wall = 2100/0.30 = 7000sq units
H = 7000/2(110+44) = 7000/308 = 22.72 = 23m - There are 2 rectangular tanks A and B with lengths 12m and 15m in a square field. If the total area of the square field excluding the rectangular tanks is 360sq m and the breadth of both the rectangular tanks is 1/3 of the side of the square field, find the perimeter of the square field?
A)100m
B)75m
C)82m
D)96m
E)None of theseAnswer – D)96m
Explanation :
Area A = 12*a/3 = 4a
Area B = 15*a/3 = 5a
a2-9a-360 = 0
(a-24)(a+15) = 0
A=24
Perimeter of the square = 4a = 4*24 = 96m - Meena wishes to start a 78sq m triangular flower garden. Since she has only 20m barbed wire, she fences three sides of the garden letting her house compound wall act as the fourth side fencing. The dimension of the garden is
A)12m*3m
B)26m*13m
C)16m*13m
D)20m*11m
E)None of theseAnswer – B)26m*3m
Explanation :
L =20 – 2b
area of the garden = 78 sq m,
L * b =78
b*(20-2b) = 78
20b – 2b2 = 78
b2 -10b+ 39 =0
b= 3 b=13
b = 3 then L = 26
b=13 then L = 6
Dimension = 26m * 3 m
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