Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample from **Mensuration** that are important for all the competitive exams. We have included some questions that are repeatedly asked in exams !!

**The length and the breadth of a rectangular door are increased by 1 m each and due to this the area of the door increased by 21 sq. m. But if the length is increased by 1 m and breadth decreased by 1 m, area is decreased by 5 sq. m. Find the perimeter of the door.**

A) 25 m

B) 20 m

C) 40 m

D) 60 m

E) 24 m**C) 40 m**

Explanation:

Let original length = l, breadth = b, so area = lb

When l and b increased by 1:

(l+1)(b+1) = lb + 21

Solve, l + b = 20

When l increased by 1, b decreased by 1:

(l+1)(b-1) = lb – 5

Solve, l – b = 6

Now solve both equations, l = 13, b = 7

Perimeter = 2(13+7)**The perimeter of a rectangular plot is 340 m. Find the cost of gardening 1 m broad boundary around it at the rate of Rs 10 per sq. m.**

A) Rs 3450

B) Rs 3400

C) Rs 3480

D) Rs 3440

E) Rs 3880**D) Rs 3440**

Explanation:

Given 2(l+b) = 340

1 m broad boundary means increase in l and b by 2 m

So area of the boundary will be [(l+2)(b+2) – lb] = 2(l+b) + 4 = 340 + 4 = 344

So cost of gardening = 344*10**The sides of a triangle are in the ratio 3 : 4 : 5 whose area is 216 sq. cm. What will be the perimeter of this triangle?**

A) 58 cm

B) 64 cm

C) 28 cm

D) 36 cm

E) 72 cm**E) 72 cm**

Explanation:

Sides 3x, 4x, 5x

So semi-perimeter, s = (3x+4x+5x)/2 = 6x

Area = √s(s-a)(s-b)(s-c)

= √6x*3x*2x*x = 6x^{2}cm^{2}

So 6x^{2}= 216, this gives x = 6

Perimeter = 12x = 12*6**If the base of a triangle is increased by 50% and its height is decreased by 50%, then what will be the effect on its area?**

A) 50% decrease

B) 75% increase

C) No effect

D) 25% decrease

E) 25% increase**D) 25% decrease**

Explanation:

Area of triangle = (1/2) * base * height

So effect on area = +50 + (-50) + (50)(-50)/100 = -25%**A rectangle whose sides are in the ratio 6 : 5 is formed by bending a circular wire of radius 42 cm. Find the largest side of the rectangle.**

A) 60 cm

B) 72 cm

C) 66 cm

D) 78 cm

E) 84 cm**B) 72 cm**

Explanation:

Length of wire = 2ᴨr = 264 which should be equal to the perimeter of rectangle in which it is bent.

So 2(6x + 5x) = 264

Solve, x= 12

Largest side = 6x = 6*12**A rectangular sheet of 0.5 cm thickness is made from an iron cube of side 10 cm by hammering it down. The sides of the sheet are in the ratio 1 : 5. Find the largest side of the sheet.**

A) 100 cm

B) 72 cm

C) 20 cm

D) 70 cm

E) 88 cm**A) 100 cm**

Explanation:

Sides = x and 5x

Now vol. of rectangle = vol. of cube

x * 5x * (0.5) = 10*10 *10

Solve, x = 20

Largest side = 5x = 5*20**The area of the inner part of a cylinder is 616 sq. cms and its radius is half its height. Find the inner volume of the cylinder.**

A) 1577.5 cm^{3}

B) 1768.2 cm^{3}

C) 1538.5 cm^{3}

D) 1435.8 cm^{3}

E) 1238.5 cm^{3}**B) 1538.5 cm**^{3}

Explanation:

Given 2ᴨrh + ᴨr^{2}= 616 and r = (1/2) * h

So 2ᴨ × (1/2)h × h + ᴨ × (1/4)h^{2}= 616

Solve, h = 28/√5

Volume = ᴨr^{2}h = (22/7) * (1/4) * h^{2}* h

Put h = 28/√5, vol. ≈ 1538.5**A cylinder and a cone have equal base and equal height. The ratio of the radius of base to height is 5 : 12. Find the ratio of the total surface area of the cylinder to that of the cone.**

A) 7 : 15

B) 16 : 9

C) 17 : 9

D) 9 : 17

E) 15 : 7**C) 17 : 9**

Explanation:

Let radius = 5x and height = 12x

Then slant height = √[(5x)^{2}+ (12x)^{2}]= 13x

Required ratio = 2ᴨr(h+r) : ᴨr(l+r)**A cone of radius 12 cm and height 5 cm is mounted on a cylinder of radius 12 cm and height 19 cm. Find the total surface area of the figure thus formed.**

A) 2498 cm^{2}

B) 2400 cm^{2}

C) 2476 cm^{2}

D) 2376 cm^{2}

E) 2546 cm^{2}**D) 2376 cm**^{2}

Explanation:

Slant height of cone, l = √(12^{2}+ 5^{2}) = 13 cm

Total surface area of final figure = curved surface area of cone + curved surface area of cylinder + area of base

= ᴨrl + 2ᴨrh + ᴨr^{2}

= ᴨr (l + 2h + r)

= (22/7) * 12 (13 + 2*19 +12)**How many spherical balls whose radius is half that of cylinder can be formed by melting a cylindrical iron rod whose height is eight times its radius?**

A) 44

B) 48

C) 60

D) 56

E) Cannot be determined**B) 48**

Explanation:

Let radius of rod = r, then height = 8r

Radius of 1 spherical ball = r/2

So number of balls = Vol. of cylindrical rod/Vol. of 1 spherical ball

= ᴨ × r^{2}× 8r / (4/3) × ᴨ × (r/2)^{3}

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