{"id":38933,"date":"2019-12-06T16:32:37","date_gmt":"2019-12-06T11:02:37","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=38933"},"modified":"2019-12-06T16:32:37","modified_gmt":"2019-12-06T11:02:37","slug":"mensuration-questions-for-ibps-cerk-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/mensuration-questions-for-ibps-cerk-set-2-pdf\/","title":{"rendered":"Mensuration Questions For IBPS Cerk Set-2 PDF"},"content":{"rendered":"

Mensuration Questions For IBPS Clerk Set-2 PDF<\/span><\/strong><\/span><\/h1>\n

Download important Mensuration Inequalities PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Mensuration Inequalities for IBPS Clerk Exam.<\/p>\n

Download Mensuration Questions For IBPS Cerk Set-2 PDF<\/a><\/p>\n

Get 25 IBPS Clerk mocks for Rs. 149. Enroll here<\/a><\/p>\n

Take Free IBPS Clerk Mock Test<\/a><\/p>\n

Download IBPS Clerk Previous papers PDF<\/a><\/p>\n

Go to Free Banking Study Material<\/a> (15,000 Solved Questions)<\/p>\n

Question 1:\u00a0<\/b>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 ?<\/p>\n

a)\u00a0160.16<\/p>\n

b)\u00a06.16<\/p>\n

c)\u00a01122.66<\/p>\n

d)\u00a0Data Inadequate<\/p>\n

e)\u00a0None of these<\/p>\n

Question 2:\u00a0<\/b>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?<\/p>\n

a)\u00a05 meters<\/p>\n

b)\u00a08 meters<\/p>\n

c)\u00a010 meters<\/p>\n

d)\u00a012 meters<\/p>\n

e)\u00a0None of these<\/p>\n

Question 3:\u00a0<\/b>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)?<\/p>\n

a)\u00a0160<\/p>\n

b)\u00a040<\/p>\n

c)\u00a020<\/p>\n

d)\u00a0Cannot be determined<\/p>\n

e)\u00a0None of these<\/p>\n

IBPS Clerk Online Mock Test<\/a><\/p>\n

Question 4:\u00a0<\/b>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)<\/p>\n

a)\u00a014<\/p>\n

b)\u00a01.4<\/p>\n

c)\u00a038<\/p>\n

d)\u00a021<\/p>\n

e)\u00a03.5<\/p>\n

Question 5:\u00a0<\/b>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?<\/p>\n

a)\u00a0142 cms.<\/p>\n

b)\u00a0140 cms.<\/p>\n

c)\u00a0132 ems.<\/p>\n

d)\u00a0150 cms.<\/p>\n

e)\u00a0None of these<\/p>\n

Question 6:\u00a0<\/b>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.)?<\/p>\n

a)\u00a03250<\/p>\n

b)\u00a02445<\/p>\n

c)\u00a01350<\/p>\n

d)\u00a02210<\/p>\n

e)\u00a02940<\/p>\n

IBPS Clerk Previous Papers<\/a><\/p>\n

Question 7:\u00a0<\/b>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.\/m\u00b2)<\/p>\n

a)\u00a018<\/p>\n

b)\u00a011<\/p>\n

c)\u00a016<\/p>\n

d)\u00a015<\/p>\n

e)\u00a013<\/p>\n

Question 8:\u00a0<\/b>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 ?<\/p>\n

a)\u00a02116 square metres<\/p>\n

b)\u00a02047 square metres<\/p>\n

c)\u00a02090 square metres<\/p>\n

d)\u00a02178 square metres<\/p>\n

e)\u00a0None of these<\/p>\n

Question 9:\u00a0<\/b>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)<\/p>\n

a)\u00a092<\/p>\n

b)\u00a0108<\/p>\n

c)\u00a096<\/p>\n

d)\u00a088<\/p>\n

e)\u00a084<\/p>\n

IBPS Clerk Important Questions PDF<\/a><\/p>\n

Free Banking Study Material (15,000 Solved Questions)<\/a><\/p>\n

Question 10:\u00a0<\/b>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 ?<\/p>\n

a)\u00a0$17\\sqrt{2}m$<\/p>\n

b)\u00a0$21\\sqrt{2}m$<\/p>\n

c)\u00a0$22\\sqrt{2}m$<\/p>\n

d)\u00a0$23\\sqrt{2}m$<\/p>\n

e)\u00a0None of these<\/p>\n

Question 11:\u00a0<\/b>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 m\u00b2).<\/p>\n

a)\u00a0580<\/p>\n

b)\u00a0784<\/p>\n

c)\u00a0637<\/p>\n

d)\u00a0516<\/p>\n

e)\u00a0616<\/p>\n

Question 12:\u00a0<\/b>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 )<\/p>\n

a)\u00a0239<\/p>\n

b)\u00a0253<\/p>\n

c)\u00a0241<\/p>\n

d)\u00a0257<\/p>\n

e)\u00a0247<\/p>\n

Question 13:\u00a0<\/b>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)<\/p>\n

a)\u00a0112<\/p>\n

b)\u00a096<\/p>\n

c)\u00a072<\/p>\n

d)\u00a056<\/p>\n

e)\u00a060<\/p>\n

Question 14:\u00a0<\/b>A rectangular garden of length 12m is surrounded by a 2m wide path. If the area of the garden is 84 m\u00b2 and the cost of gravelling is Rs.8 per m\u00b2, what is the total cost of gravelling the path ?( in Rs.)<\/p>\n

a)\u00a0Rs.780\/-<\/p>\n

b)\u00a0Rs.742\/-<\/p>\n

c)\u00a0Rs.724\/-<\/p>\n

d)\u00a0Rs.775\/-<\/p>\n

e)\u00a0Rs.736\/-<\/p>\n

Question 15:\u00a0<\/b>The length of rectangular plot is thrice its breadth. If the area of the rectangular plot is 6075 sq. metres, what is its length ?<\/p>\n

a)\u00a0145 metres<\/p>\n

b)\u00a0130 metres<\/p>\n

c)\u00a075 metres<\/p>\n

d)\u00a045 metres<\/p>\n

e)\u00a0None of these<\/p>\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

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

=> Radius of outer circle\u00a0(ground+garden), R = 17.5 + 1.4 = 18.9 m<\/p>\n

Area of garden = Area of outer circle – Area of inner circle<\/p>\n

= $\\pi R^2-\\pi r^2 = \\pi(R-r)(R+r)$<\/p>\n

= $\\frac{22}{7}(18.9-17.5)(18.9+17.5)$<\/p>\n

= $\\frac{22}{7} \\times 1.4 \\times 36.4$<\/p>\n

= $22 \\times 0.2 \\times 36.4 = 160.16$ $m^2$<\/p>\n

=> Ans – (A)<\/p>\n

2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/figure>\n

AD is the tower = 15 m and CE is the electric pole<\/p>\n

Let AB = $x$ m and DE = BC = $y$ m<\/p>\n

Also, $\\angle$ AED = 60\u00b0 and\u00a0$\\angle$\u00a0ACB = 30\u00b0<\/p>\n

In\u00a0$\\triangle$\u00a0ADE, => $tan(\\angle AED)=\\frac{AD}{DE}$<\/p>\n

=> $tan(60)=\\sqrt{3}=\\frac{15}{y}$<\/p>\n

=> $y=\\frac{15}{\\sqrt{3}}=5\\sqrt{3}$ m<\/p>\n

In\u00a0$\\triangle$\u00a0ABC, => $tan(\\angle ACB)=\\frac{AB}{BC}$<\/p>\n

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

=> $x=5$ m<\/p>\n

$\\therefore$ CE = AD – AB = 15 – 5 = 10 meters<\/p>\n

=> Ans – (C)<\/p>\n

3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let the length of the plot be $l$ meters and breadth = $b$ meters<\/p>\n

New length = $4l$ meters<\/p>\n

Perimeter = $2(4l+b)=480$<\/p>\n

=> $4l+b=\\frac{480}{2}=240$<\/p>\n

=> $b=240-4l$ ———(i)<\/p>\n

Area = $(4l \\times b)=12800$<\/p>\n

=> $lb=\\frac{12800}{4}=3200$<\/p>\n

Substituting value of $b$ from equation (i)<\/p>\n

=> $l(240-4l)=3200$<\/p>\n

=> $240l-4l^2=3200$<\/p>\n

=> $l^2-60l+800$<\/p>\n

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

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

=> $l=\\frac{60 \\pm 20}{2}$<\/p>\n

=> $l=\\frac{60+20}{2},\\frac{60-20}{2}$<\/p>\n

=> $l=\\frac{80}{2},\\frac{40}{2}$<\/p>\n

=> $l=40,20$ meters<\/p>\n

=> Ans – (D)<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

<\/figure>\n

Diameter of circle = side of square = $d$ cm<\/p>\n

Area of square = $d^2$ sq. cm<\/p>\n

Area of 1 quadrant = $\\frac{1}{4} \\times \\pi \\times (\\frac{d}{2})^2$<\/p>\n

= $\\pi \\times \\frac{d^2}{16}$<\/p>\n

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

Area of shaded region = 168<\/p>\n

=> $d^2 – \\frac{\\pi d^2}{4} = 168$<\/p>\n

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

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

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

=> $d = \\sqrt{28 \\times 28} = 28$ cm<\/p>\n

$\\therefore$ Radius = $\\frac{28}{2} = 14$ cm<\/p>\n

5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Radius of circle = 21 cm<\/p>\n

=> Area of circle = $\\pi r^2$<\/p>\n

= $\\frac{22}{7} \\times 21 \\times 21 = 1386 cm^2$<\/p>\n

Thus, area of rectangle = $1386 cm^2$<\/p>\n

Let length and breadth of rectangle be $14x$ and $11x$ respectively.<\/p>\n

=> Area = $14x \\times 11x = 1386$<\/p>\n

=> $x^2 = \\frac{1386}{154} = 9$<\/p>\n

=> $x = \\sqrt{9} = 3$ cm<\/p>\n

$\\therefore$ Perimeter = $2 (14x + 11x)$<\/p>\n

= $2 \\times 25x = 50x$<\/p>\n

= $50 \\times 3 = 150$ cm<\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the dimension of the room be $3x , 2x , x$ metres<\/p>\n

Acc. to ques, => $3x + 2x + x = 18$<\/p>\n

=> $x = \\frac{18}{6} = 3$ metres<\/p>\n

Curved surface area of the room = $2 h (l + b)$<\/p>\n

= $2 \\times x \\times (3x + 2x) = 2x \\times 5x$<\/p>\n

= $10 (x)^2 = 10 \\times (3)^2$<\/p>\n

= $10 \\times 9 = 90 m^2$<\/p>\n

$\\therefore$ Total cost incurred on painting only the four walls of the room\u00a0= $15 \\times 90$<\/p>\n

= $Rs. 1,350$<\/p>\n

7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Side of square lawn = 18 m<\/p>\n

Part of the lawn that requires grass = $1 – \\frac{1}{4} = \\frac{3}{4}$<\/p>\n

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

= $3 \\times 9 \\times 9 = 243 m^2$<\/p>\n

Total amount to plant grass = Rs. 3159<\/p>\n

$\\therefore$ Rate that the gardener charges for planting synthetic grass = $\\frac{3159}{243} = 13$<\/p>\n

8)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Let length of rectangle = $x$\u00a0metre<\/p>\n

and breadth = $(x – 8)$ metre<\/p>\n

=> Perimeter = $2 (x + x – 8) = 180$<\/p>\n

=> $2x – 8 = 90$<\/p>\n

=> $x = \\frac{98}{2} = 49$<\/p>\n

=> Breadth = 49 – 8 = 41 metre<\/p>\n

$\\therefore$ Area of rectangle = 49 * 41 = 2009 sq. metre<\/p>\n

9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Length of rectangle = Side of square<\/p>\n

= $\\sqrt{729} = 27$ metre<\/p>\n

=> Breadth of rectangle = $27 – 6 = 21$ meters<\/p>\n

$\\therefore$ Perimeter of rectangle<\/p>\n

= $2 (27 + 21) = 2 \\times 48$<\/p>\n

= $96$ metre<\/p>\n

10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Perimeter of rectangular plot = $2 (l + b)$<\/p>\n

= 2\u00a0(23 + 19) = 2 * 42<\/p>\n

= 84 metre = Perimeter of square plot<\/p>\n

=> Side of square plot = 84\/4 = 21 metre<\/p>\n

$\\therefore$ Diagonal = $\\sqrt{2}$ * side<\/p>\n

= $21\\sqrt{2}$ metre<\/p>\n

Get 25 IBPS Clerk mocks for Rs. 149. Enroll here<\/a><\/p>\n

11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Area of the lawn = $\\pi r^2 = 154$<\/p>\n

=> $\\frac{22}{7} r^2 = 154$<\/p>\n

=> $r^2 = \\frac{154 \\times 7}{22} = 49$<\/p>\n

=> $r = \\sqrt{49} = 7$<\/p>\n

Now, radius of lawn including path = 7 + 7 = 14 metre<\/p>\n

$\\therefore$ Required area = $\\pi R^2$<\/p>\n

= $\\frac{22}{7} \\times 14^2$<\/p>\n

=\u00a0$22 \\times 28 = 616 m^2$<\/p>\n

12)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Let length of rectangle = $x$ m<\/p>\n

Breadth = $(x – 7)$ m<\/p>\n

=> Perimeter of rectangle = $2 (x + x – 7) = 90$<\/p>\n

=> $2x – 7 = \\frac{90}{2} = 45$<\/p>\n

=> $2x = 45 + 7 = 52$<\/p>\n

=> $x = \\frac{52}{2} = 26$<\/p>\n

=> Breadth = 26 – 7 = 19 m<\/p>\n

=> Height of triangle = 26 m and Base of triangle = 19 m<\/p>\n

$\\therefore$ Area of triangle = $\\frac{1}{2} \\times 26 \\times 19$<\/p>\n

= $13 \\times 19 = 247 m^2$<\/p>\n

13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Area of square shaped tile = $2 \\times 2 = 4 m^2$<\/p>\n

No. of tiles used = 49<\/p>\n

=> Area of floor = $49 \\times 4 = 196 m^2$<\/p>\n

Let side of floor = $x$ m<\/p>\n

=> $x^2 = 196$<\/p>\n

=> $x = \\sqrt{196} = 14$ m<\/p>\n

$\\therefore$ Perimeter of hall = $4 \\times 14 = 56$ m<\/p>\n

14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Length of rectangular garden = 12 m<\/p>\n

Area of garden = $12 \\times b = 84$<\/p>\n

=> $b = \\frac{84}{12} = 7$ m<\/p>\n

<\/figure>\n

Length of rectangular garden including path = $12 + 2 + 2 = 16$ m<\/p>\n

Breadth\u00a0of rectangular garden including path =\u00a0$7 + 2 + 2 = 11$ m<\/p>\n

Area\u00a0of rectangular garden including path =\u00a0$16 \\times 11 = 176 m^2$<\/p>\n

=> Area of path = $176 – 84 = 92 m^2$<\/p>\n

$\\therefore$ Costt of gravelling the path = $92 \\times 8$<\/p>\n

= Rs.\u00a0$736$<\/p>\n

15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Let Breadth = $x$ m<\/p>\n

=> Length of rectangle = $3x$ m<\/p>\n

=> Area of rectangle<\/p>\n

=> $x \\times 3x = 6075$<\/p>\n

=> $x^2 = \\frac{6075}{3} = 2025$<\/p>\n

=> $x = \\sqrt{2025} = 45$<\/p>\n

$\\therefore$ Length of rectangle = $3 \\times 45 = 135$ m<\/p>\n

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We hope this Mensuration for IBPS Clerk preparation will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

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. Take Free IBPS Clerk Mock Test Download IBPS Clerk Previous papers PDF Go to Free Banking Study Material (15,000 Solved Questions) Question 1:\u00a0A circular 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