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# Mensuration Questions For IBPS Clerk Set-2 PDF

Download important Mensuration Inequalities PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Mensuration Inequalities for IBPS Clerk Exam.

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Question 1: A circular Ground whose diameter is 35 meters has a 1.4 meters broad garden around it.What is the area of the garden in square meters ?

a) 160.16

b) 6.16

c) 1122.66

e) None of these

Question 2: The top of 15 meters high tower makes an angle of depression of 60 degrees with the bottom of a electric pole and an angle of 30 degrees with the top of the pole. What is the height of the electric pole?

a) 5 meters

b) 8 meters

c) 10 meters

d) 12 meters

e) None of these

Question 3: When the length of the rectangular plot is increased by four times its perimeter becomes 480 meters and area becomes 12800 sq.m. What is its original length(in meters)?

a) 160

b) 40

c) 20

d) Cannot be determined

e) None of these

Question 4: Four circles having equal radii are drawn with center at the four corners of a square. Each circle touches the other two adjacent circle. If remaining area of the square is 168 cm, what is the size of the radius of the radius of the circle? (in centimeters)

a) 14

b) 1.4

c) 38

d) 21

e) 3.5

Question 5: Area of a rectangle is equal to the area of the circle whose radius is 21 cms. If the length and the breadth of the rectangle are in the ratio of 14 : 11 respectively, what is its perimeter?

a) 142 cms.

b) 140 cms.

c) 132 ems.

d) 150 cms.

e) None of these

Question 6: The sum of the dimensions of a room (i.e. length, breadth and height) is 18 metres and its length, breadth and height are in the ratio of 3 : 2 : 1 respectively. If the room is to be painted at the rate of Rs. 15 per m2, what would be the total cost incurred on painting only the four walls of the room (in Rs.)?

a) 3250

b) 2445

c) 1350

d) 2210

e) 2940

Question 7: It takes Rs. 3159 to plant synthetic grass in a square lawn, ${1 \over 4}$ of which is paved (and thus does not require grass). If each side of this lawn measures 18m, what is the rate that the gardener charges for planting synthetic grass? (in Rs./m²)

a) 18

b) 11

c) 16

d) 15

e) 13

Question 8: If the perimeter of a rectangle is 180 metres and the difference between the length and the breadth is 8 metres, what is the area of the rectangle ?

a) 2116 square metres

b) 2047 square metres

c) 2090 square metres

d) 2178 square metres

e) None of these

Question 9: The difference between the length and breadth of a rectangle is 6 metre. The length of rectangle is equal to the side of the square whose area is 729 sq. metre. What is the perimeter of rectangle ? (in metre)

a) 92

b) 108

c) 96

d) 88

e) 84

Question 10: The perimeter of a square plot is equal to the perimeter of a rectangular plot which is 23 metre long and 19 metre broad. What will be the diagonal of the square plot ?

a) $17\sqrt{2}m$

b) $21\sqrt{2}m$

c) $22\sqrt{2}m$

d) $23\sqrt{2}m$

e) None of these

Question 11: A circular lawn has an area of 154 m2. A path of 7 m width surrounds the lawn. What is the area of the lawn including the path? (in m²).

a) 580

b) 784

c) 637

d) 516

e) 616

Question 12: The height and base of a triangle are equal to the length and breadth of a rectangle respectively. If the perimeter of the rectangle is 90m and the difference between its length and breadth is 7m, what is the area of the triangle ? (in m )

a) 239

b) 253

c) 241

d) 257

e) 247

Question 13: The floor of a square hall is tiled completely with forty nine square shaped tiles. If the side of each tile measures 2 m, what was the perimeter of the hall ? (in m)

a) 112

b) 96

c) 72

d) 56

e) 60

Question 14: A rectangular garden of length 12m is surrounded by a 2m wide path. If the area of the garden is 84 m² and the cost of gravelling is Rs.8 per m², what is the total cost of gravelling the path ?( in Rs.)

a) Rs.780/-

b) Rs.742/-

c) Rs.724/-

d) Rs.775/-

e) Rs.736/-

Question 15: The length of rectangular plot is thrice its breadth. If the area of the rectangular plot is 6075 sq. metres, what is its length ?

a) 145 metres

b) 130 metres

c) 75 metres

d) 45 metres

e) None of these

Radius of circular ground = $r = \frac{35}{2}=17.5$ m and width of broad garden = 1.4 m

=> Radius of outer circle (ground+garden), R = 17.5 + 1.4 = 18.9 m

Area of garden = Area of outer circle – Area of inner circle

= $\pi R^2-\pi r^2 = \pi(R-r)(R+r)$

= $\frac{22}{7}(18.9-17.5)(18.9+17.5)$

= $\frac{22}{7} \times 1.4 \times 36.4$

= $22 \times 0.2 \times 36.4 = 160.16$ $m^2$

=> Ans – (A)

AD is the tower = 15 m and CE is the electric pole

Let AB = $x$ m and DE = BC = $y$ m

Also, $\angle$ AED = 60° and $\angle$ ACB = 30°

In $\triangle$ ADE, => $tan(\angle AED)=\frac{AD}{DE}$

=> $tan(60)=\sqrt{3}=\frac{15}{y}$

=> $y=\frac{15}{\sqrt{3}}=5\sqrt{3}$ m

In $\triangle$ ABC, => $tan(\angle ACB)=\frac{AB}{BC}$

=> $tan(30)=\frac{1}{\sqrt{3}}=\frac{x}{5\sqrt{3}}$

=> $x=5$ m

$\therefore$ CE = AD – AB = 15 – 5 = 10 meters

=> Ans – (C)

Let the length of the plot be $l$ meters and breadth = $b$ meters

New length = $4l$ meters

Perimeter = $2(4l+b)=480$

=> $4l+b=\frac{480}{2}=240$

=> $b=240-4l$ ———(i)

Area = $(4l \times b)=12800$

=> $lb=\frac{12800}{4}=3200$

Substituting value of $b$ from equation (i)

=> $l(240-4l)=3200$

=> $240l-4l^2=3200$

=> $l^2-60l+800$

=> $l=\frac{-(-60) \pm \sqrt{(-60)^2-(4 \times 1 \times 800)}}{2}$

=> $l=\frac{60 \pm \sqrt{3600-3200}}{2} = \frac{60 \pm \sqrt{400}}{2}$

=> $l=\frac{60 \pm 20}{2}$

=> $l=\frac{60+20}{2},\frac{60-20}{2}$

=> $l=\frac{80}{2},\frac{40}{2}$

=> $l=40,20$ meters

=> Ans – (D)

Diameter of circle = side of square = $d$ cm

Area of square = $d^2$ sq. cm

Area of 1 quadrant = $\frac{1}{4} \times \pi \times (\frac{d}{2})^2$

= $\pi \times \frac{d^2}{16}$

=> Area of 4 quadrants = $4 \times \pi \times \frac{d^2}{16} = \frac{\pi d^2}{4}$ sq. cm

Area of shaded region = 168

=> $d^2 – \frac{\pi d^2}{4} = 168$

=> $\frac{1}{4} [d^2 (4 – \pi)] = 168$

=> $d^2 = \frac{168 \times 4}{4 – \pi} = \frac{168 \times 4}{4 – \frac{22}{7}}$

=> $d^2 = \frac{168 \times 4 \times 7}{28 – 22} = \frac{168}{6} \times 28$

=> $d = \sqrt{28 \times 28} = 28$ cm

$\therefore$ Radius = $\frac{28}{2} = 14$ cm

Radius of circle = 21 cm

=> Area of circle = $\pi r^2$

= $\frac{22}{7} \times 21 \times 21 = 1386 cm^2$

Thus, area of rectangle = $1386 cm^2$

Let length and breadth of rectangle be $14x$ and $11x$ respectively.

=> Area = $14x \times 11x = 1386$

=> $x^2 = \frac{1386}{154} = 9$

=> $x = \sqrt{9} = 3$ cm

$\therefore$ Perimeter = $2 (14x + 11x)$

= $2 \times 25x = 50x$

= $50 \times 3 = 150$ cm

Let the dimension of the room be $3x , 2x , x$ metres

Acc. to ques, => $3x + 2x + x = 18$

=> $x = \frac{18}{6} = 3$ metres

Curved surface area of the room = $2 h (l + b)$

= $2 \times x \times (3x + 2x) = 2x \times 5x$

= $10 (x)^2 = 10 \times (3)^2$

= $10 \times 9 = 90 m^2$

$\therefore$ Total cost incurred on painting only the four walls of the room = $15 \times 90$

= $Rs. 1,350$

Side of square lawn = 18 m

Part of the lawn that requires grass = $1 – \frac{1}{4} = \frac{3}{4}$

=> Area of the lawn that requires grass = $\frac{3}{4} \times (18)^2$

= $3 \times 9 \times 9 = 243 m^2$

Total amount to plant grass = Rs. 3159

$\therefore$ Rate that the gardener charges for planting synthetic grass = $\frac{3159}{243} = 13$

Let length of rectangle = $x$ metre

and breadth = $(x – 8)$ metre

=> Perimeter = $2 (x + x – 8) = 180$

=> $2x – 8 = 90$

=> $x = \frac{98}{2} = 49$

=> Breadth = 49 – 8 = 41 metre

$\therefore$ Area of rectangle = 49 * 41 = 2009 sq. metre

Length of rectangle = Side of square

= $\sqrt{729} = 27$ metre

=> Breadth of rectangle = $27 – 6 = 21$ meters

$\therefore$ Perimeter of rectangle

= $2 (27 + 21) = 2 \times 48$

= $96$ metre

Perimeter of rectangular plot = $2 (l + b)$

= 2 (23 + 19) = 2 * 42

= 84 metre = Perimeter of square plot

=> Side of square plot = 84/4 = 21 metre

$\therefore$ Diagonal = $\sqrt{2}$ * side

= $21\sqrt{2}$ metre

Area of the lawn = $\pi r^2 = 154$

=> $\frac{22}{7} r^2 = 154$

=> $r^2 = \frac{154 \times 7}{22} = 49$

=> $r = \sqrt{49} = 7$

Now, radius of lawn including path = 7 + 7 = 14 metre

$\therefore$ Required area = $\pi R^2$

= $\frac{22}{7} \times 14^2$

= $22 \times 28 = 616 m^2$

Let length of rectangle = $x$ m

Breadth = $(x – 7)$ m

=> Perimeter of rectangle = $2 (x + x – 7) = 90$

=> $2x – 7 = \frac{90}{2} = 45$

=> $2x = 45 + 7 = 52$

=> $x = \frac{52}{2} = 26$

=> Breadth = 26 – 7 = 19 m

=> Height of triangle = 26 m and Base of triangle = 19 m

$\therefore$ Area of triangle = $\frac{1}{2} \times 26 \times 19$

= $13 \times 19 = 247 m^2$

Area of square shaped tile = $2 \times 2 = 4 m^2$

No. of tiles used = 49

=> Area of floor = $49 \times 4 = 196 m^2$

Let side of floor = $x$ m

=> $x^2 = 196$

=> $x = \sqrt{196} = 14$ m

$\therefore$ Perimeter of hall = $4 \times 14 = 56$ m

Length of rectangular garden = 12 m

Area of garden = $12 \times b = 84$

=> $b = \frac{84}{12} = 7$ m

Length of rectangular garden including path = $12 + 2 + 2 = 16$ m

Breadth of rectangular garden including path = $7 + 2 + 2 = 11$ m

Area of rectangular garden including path = $16 \times 11 = 176 m^2$

=> Area of path = $176 – 84 = 92 m^2$

$\therefore$ Costt of gravelling the path = $92 \times 8$

= Rs. $736$

Let Breadth = $x$ m

=> Length of rectangle = $3x$ m

=> Area of rectangle

=> $x \times 3x = 6075$

=> $x^2 = \frac{6075}{3} = 2025$

=> $x = \sqrt{2025} = 45$

$\therefore$ Length of rectangle = $3 \times 45 = 135$ m

We hope this Mensuration for IBPS Clerk preparation will be helpful to you.