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 !!

**After measuring 100m of a rope, it was discovered that the metre rod was 2cm longer. The true length of the rod is**

A)95m

B)98m

C)96m

D)93m

**B)98m**

Explanation :

100*2 = 200cm measured long

Correct length = 100 – (200/100) = 100 -2 = 98m

**A circle is drawn circumscribing a rectangle of sides 8cm and 6cm, find the area of the circle**

A)78.5cm^{2}

B)82.7cm^{2}

C)65.4cm^{2}

D)54.9cm^{2}

**A)78.5cm**^{2}

Explanation :

Diagonal of rectangle = Diameter of circle

Diameter = √ (64+36)^{ }= 10

Radius = 5cm

Area of circle = 3.14*5*5 = 78.5cm^{2}

**If the sides of a equilateral triangle is increased by 10% , 30%and 60%then a new triangle is formed. By what % perimeter of the triangle is increased ?**

A)40.50%

B)32.45%

C)33.33%

D)35.67%

**C)33.33%**

Explanation :

Let a = side of the triangle

Perimeter = 3a

New perimeter = a*110/100 + a*130/100 + a*140/100

= a(11+13+16)/10 = 4a

% increased in perimeter = (4a – 3a/3a)*100

= 100/3 = 33.33%

**If length, breadth and height of a hall is 14m,8m,8m then find the diagonal of the hall ?**

A)15m

B)17m

C)19m

D)18m

**D)18m**

Explanation :

Diagonal = √ (14^{2}+8^{2}+8^{2})

= √ (196+64+64)

= √324

= 18m

**The base and the other side of an isosceles triangle is 8cm and 12cm. Find its triangle ?**

A)45.2 cm^{2}

B)54.8 cm^{2}

C)40.5 cm^{2}

D)38.7 cm^{2}

**A)45.2 cm**^{2}

Explanation :

Area = b/4√ (4a^{2}–b^{2})

= 8/4√ (4*144 – 64)

= 2(22.6) = 45.2cm^{2}

**Find the area of sheet required to prepare an 60m height cone with base radius 32cm.**

A) 138625cm^{2}

B) 158510cm^{2}

C) 149607cm^{2}

D) 128612cm^{2}

**D) 128612cm**^{2}

Explanation :

L = √(60^{2}+ 32^{2}) = 68

Area = 3.14*r(h+l)

Area of sheet = 3.14*32(60+68) = 12861.44 = 128612cm^{2}

**A hall is 80m long 65m wide. It has to be designed by tiles of 15m*10m. Find no of tiles required ?**

A)41

B)37

C)35

D)33

**C)35**

Explanation :No of tiles = 80*65/15*10 = 5200/150 = 34.7 = 35

**Find the cost of carpenting a room 12 m long and 7 m broad with a carpet 10m long and 6m wide and at the rate of Rs.150 per meter.**

A)Rs.320

B)Rs.210

C)Rs.200

D)Rs.235

**B)Rs.210**

Explanation :

Area of the room = 12 × 7 => 84 m^{2}

Area of carpet = 10*6 = 60 m^{2}

Cost = (84/60)*150 = 210

**The side of a cube is 13 cm, Find its surface area?**

A) 1014cm^{2}

B) 1040cm^{2}

C) 1010cm^{2}

D) 1024cm^{2}

**A) 1014cm**^{2}

Explanation :Surface area = 6a^{2}= 6*13*13 = 1014cm^{2}

**The length of box 12 Cm long, 6 Cm breadth and 4Cm height. Find the maximum length of scale in that box?**

A)11

B)14

C)12

D)9

**B)14**

Explanation :

Diagonal = √l^{2}+b^{2}+h^{2}

= √12^{2}+6^{2}+4^{2}

= √196 => 14 Cm

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