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Mensuration Worked out Problem for Bank Exam

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 Mensuration Worked out Problem

Mensuration Worked out Problem for Bank Exam, SSC, SBI PO and other government  exam.

  1. The area of a rectangular field is 460 square meters. If the length is 15 percent more than the breadth, what is breadth of the rectangular field ?

              a) 15                                      b) 26                            c) 34.5                                    d) Cannot be determined                                             e) None of these

Answer : e) none of these

Solution: Let breadth of a rectangle be x

Then length = x + 15x/100

                   = x + 3x/20

                   = 23x/20

Area of a rectangle = 460

x × 23x/20 = 460

x × x = 400

x = 20

  1. If the length of the diagonal AC of a square ABCD is 5.2 cm, then the area of the square is :

    a) 15.12 sq. cm                     b) 13..52 sq.cm                      c) 12.62 sq.cm                         d) 10.00 sq. cm

Answer: b) 13.52 sq.cm

Solution: since the square is a rhombus and the diagonals of a square are equal

Area of a rhombus =  1/2 × product of diagonals

                              = 1/2× 5.2×5.2

                              = 2.6×5.2

                              = 13.52

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3.if the diagonal of two squares are in the ratio of 2:5, their area will be in the ratio of

   a) √2:√5                                  b) 2:5                                  c) 4:25                                      d) 4:5 

Answer: c) 4:25

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Solution: let the diagonals of the square be 2x and 5x respectively

Area of one square =  1/2 × 2x × 2x

                          = 4x2/2

Area of other square =  1/2 × 5x × 5x

                                 = 25x2/2

Ratio of areas of both the square = 4x2/2 25x2/2

                                                     = 4/25

                                                    = 4:25

  1. From four corners of a square sheet of side 4 cm, four pieces, each in the shape of arc of a circle with radius 2 cm, are cut out. The area of the remaining portion is

    a) 8 – π                                     b) 16 – 4 π                                          c) 16 – 8 π                                                        d) 4 – 2 π

Answer: b) 16 – 4 π

Solution: area of a square sheet = 4 × 4 = 16 sq. cm

Since the angle between two sides of a square is 90 degree and the radius of arc is 2 cm

S, area of 4 arc will be = 4 × 90/360 π22

                                    = 4π

Area of remaining portion = 16 – 4π

Mensuration Worked out Problem for Bank Exam

  1. The cost of building a fence around a circular field is rupee 7700 at the rate of  rupee 14 per foot . what is the area of the circular field ?

    a) 24062.5                                     b) 23864.4                                                c) 24644.5                                    d) Cannot be determined

    e) None of these

Answer: a) 24062.5

Solution: length of a fence = 7700/14

                                         = 550

Circumference of a circle field = 2πr

              2πr = 550

              2×22/r = 550

             r = 175/2

Area of circular field = π r2

                               = 22/7 × 175/2 × 175/2

                               = 24062.5

  1. The sides of a triangle are 6 cm, 8 cm and 10 cm. The area (in cm2) of the triangle formed by joining the mid_points of this triangle is:

     a) 12                                                     b) 24                                       c) 6                                        d) 18

Answer: c) 6

Solution: a = 10, b = 8, c = 6

    s= (6 + 8 + 10)/2

    S = 12

Area of a triangle ABC = √12(12-10) (12-6) (12-8)

                                         = √12(2) (6)) (4)

                                        = 24

All the four triangles ADE, BDF,CEF and DEF are congruent, so their area are equal

Hence, area of triangle DEF = 24/4

                                             = 6 sq cm.

  1. If D and E are the mid_points of the side AB and AC respectively of the ΔABC in the figure given here, the shaded region of the triangle is what percent of the whole triangular region?

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    a) 50%                                               b) 25%                                      c) 80%                                          d) 75%

Answer: d) 75%

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Solution: Since D and E are the mid_ points of AB and AC respectively

So, DE is parallel to BC and 2DE = BC

ΔADE∼ΔABC

area ΔADE/area ΔABC = (DE)2/(BC)2     

               = (DE)2/(2DE)2   

               = 1/4

area of shaded part /area ΔABC = ¾

% = 3/4100

   = 75%

  1. If the length of a rectangle is increased by 20% and breadth is decreased by 20% . then its area

    a) Increases by 4%                                    b) Decreases by 4%                                        c) Decreases by 1%                                         d) Remains unchanged

Answer: b) decreased by 4%

Solution: let the length of rectangle be 10

And its breadth be 10

Then, area = 100

When the length is increased by 20%, new length = 12

When its breadth is decreased by 20%, new breadth = 8

then , new area = 96

So, area decreased by 4%

  1. If the surface area of a cube is 294 sq. cm, then its volume(in sq. cm) is

    a) 216                                    b) 125                                          c) 343                                              d) 512

Answer  c) 343

Solution surface area of a cube = 294

                     6a2= 294

                      a2= 49

                      a = 7

Volume of a cube = a3

                            = 7 × 7 × 7

                            = 343

  1. A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side 3 cm to form a cone, the volume of the cone so formed is

    a) 16 π                                            b) 12 π                                         c) 15 π                                               d) 20 π

Answer: a) 16 π

Solution:

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when a triangle of side 3cm, 4cm and 5cm is rotated about the side 3 cm, then a cone of radius 4 cm and height 3 cm is formed

Volume of a cone = 1/3 π rh

                            =  1/3 π × 4 × 4 × 3

                            = 16 π

  1. A wooden box measures 20 cm by 12 cm by 10 cm. Thickness of the wood is 1 cm. Volume of wood to make the box (in cubic cm) is

    a) 960                                  b) 519                                                c) 2400                                                   d) 1120

Answer a) 960

Solution outer dimension of wooden box = 20 cm , 12 cm and 10 cm

Outer volume of box = 20 × 12 × 10

                                = 2400 cc ( cubic centimeter)

Thickness of wood = 1 cm

Inner dimension of wooden box = 20 – 2 = 18 cm, 12 – 2 = 10 cm and 10 – 2 = 8 cm

Inner volume of box = 18 × 10 × 8

                               = 1440 cc

Volume of wood in the box = 2400 – 1440  = 960 cc

  1. Water flows into a tank which is 200m long and 150m wide, through a pipe of cross-section 0.3m × 0.2m at 20 km/hour. Then the time (in hours) for the water level in the tank to reach 8m is

     a) 50                                b) 120                                   c) 150                                       d) 200

Answer  d) 200

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Solution: speed of water = 20 km/hour

                                      = 20000 m/hour

Area of base = 0.3m × 0.2m

                    = 0.06 sq. meter

Volume of water flows in 1 hour = area of base× speed of water in 1 hour

                                                  = 0.06 × 20000

                                                  = 1200 cubic meter

Volume of tank up to 8 m height = 200 × 150 × 8

                                                 = 240000 cubic meter

Time to reach the water 8 m high = 240000/1200

                                                     = 200 hours  

  1. The base of a right prism is a quadrilateral ABCD. given that AB = 9 cm, BC = 14 cm, CD = 13 cm, DA = 12 cm and angle DAB = 90 degree. If the volume of the prism be 2070 cm3, then the area of the lateral surface is

     a) 720                            b) 810                                   c) 1260                                    d) 2070

Answer a) 720

Solution Area of quadrilateral ABCD = area of ΔABD + area of  ΔBCD

Since  ΔABD is a right angled triangle

BD = √AB2 + AD2

      = √92 + 122

     = √225

     = 15 cm

Area of Δ ABD = 1/2 × 9 × 12

                           = 54 sq.cm

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For ΔBCD, s = (13 + 15 + 14)/2

                     = 21

area of ΔBCD =  √21(21 – 13) (21 – 14) (21 – 15)                                                               

                        =  √21 × 8 × 7 × 6

                        = 84 sq. cm

Area of quadrilateral ABCD = 54 + 84

                                            = 138 sq. cm  

Volume of prism = area of base×height

                 2070 = 138 × h

                    Height = 2070/138                                                                

                               = 15 cm

Perimeter of quadrilateral ABCD = 9 + 14 + 13 + 12

                                                    = 48 cm

Area of lateral surface = perimeter of base×height

                                    = 48 × 15

                                   = 720 sq. cm

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