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Quants Questions : Mensuration Set 4

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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 !!

  1. 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 these
    Answer – B)42.86%
    Explanation :
    Width = 30*100/100-30
    = 3000/70 = 42.86%

  2. 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 these
    Answer – 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

  3. 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 these
    Answer – D) 21.6 cm2
    Explanation :
    x2 + (3x)= (12)2
    10 x2= 144
    x= 14.4 cm2
    Area = 1/2 (x *3x)
    =3(14.4)/2
    = 21.6 cm2

  4. 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 these
    Answer – 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

  5. 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 these
    Answer – 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

  6. 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 these
    Answer – 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

  7. 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 these
    Answer – A) 120m2
    Explanation :24*24 – 22/7 (12*12) = 576 – 456 = 120m2

  8. 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 these
    Answer – 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

  9. 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 these
    Answer – 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

  10. 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 these
    Answer – 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