Answers and Solutions

Answer

Refer to the answer below:

Answer

The  tent  can be divided  into two parts namely  the lower cylindrical part and upper conical part.

1. Lower  Cylindrical part--

Height of walls (ht) = 3.5 meters Radius of base (rb) = 2.8 meters

Area of  canvas  required for floor  =  pi*rb*rb       =  3.14*2.8*2.8        = 24.62 sq. metres

Area of  canvas required for walls   =  2*pi*rb*ht      = 2*3.14*2.8*3.5      = 61.54 sq. metres

2. Area of conical roof

Height of conical roof (cr) = 2.1 metres Conical length of roof = square root (rb*rb + cr*cr) = 3.5 metres Area = pi*rb*Conical length = pi*2.8*3.5 = 30.77 sq. metres

Total area of tarpaulin required for one tent    = Area of floor + Area of walls + Area of Conical top

       = 24.62 + 61.54 + 30.77     = 116.93 square metres

Cost of tarpaulin   = Rs. 120.00 per square metre Cost of tarpaulin for one tent  = 120 * 116.93 = Rs. 14,031.6 Cost of 1500 tents  = 1500*14,031.6 = 21,047,400.00

Contribution from each of the 50 schools   = 21,047,400.00/50     = 4,20,948.00      = Four Lakh twenty thousand     and nine hundred and forty eight  

Answer

ANSWER:

Given:  Radius of the cylindrical portion of tent= r =2.8 mtrs

           Ht of the cylindrical portion of the tent= h1 = 3.5 mtrs

           Radius of the conical top of the tent=      r = 2.8 mtrs

           Ht of the conical top of the tent=             h2=  2.1 mtrs  

          Therefore slant ht of  the conical top of the tent=  l = (r^2+h2^2)^0.5= { (2.8^2)+(2.1)^2}= 3.5 mtrs                 

The cloth required for the tent is equal to the lateral surface area of the cylinder &top cone.

Cloth required for 1 tent= 2πrh1 + πrl=   πr(2h1+l)= π*2.8*( 2*3.5+3.5)= (22/7)*2.8*10.5= 92.4 Sq.mtr

Cost of cloth per tent @Rs 120 per Sq.mtr= 92.4*120= Rs 11088

Total cost of 1500 tents=  11088*1500= Rs 1,66,32,000

No. of schools  bearing the cost=  50

Therefore average cost per school = 16632000/50=Rs 3,32,640 

Answer

Solution to the given problem

Answer

Total area of canvas required will be = 1500 * (curved surface area of cylinder + curved surface area of cone)

                                                          = 1500 * (2*pi*r*h + pi*r*l) 

                                                          = 1500 * (92.4)

                                                          = 138600 square metre

Total cost needed for canvas = 138600 * number of tents

                                              = 138600 * 1500

                                              = 16632000 Rs.

Amount shared by each school = 16632000/ 50

                                                  = 332640 Rs.

Answer

Answer

A teny has two surfaces cylindrical of height 3.5 m and above that conical of height 2.1 m. The slant height comes to sqrt(2.8x2.8 + 2.1x2.1) or 3.5 m

So the total surface area of a tent = (22/7) x 2.8x (2 x 2.8 + 3.5)= 92.4 m2

So the cost to each school = 92.4 x 120 x1500/5 = Rs 3,32,640

Answer

Providing you the outline of the solution, I would like you to solve it based on this.

Area of cylinder: 2 π r h     sq metres

Area of cone: π r l     sq metres

Total Area: 2 π r h + π r l                           [ l is the slant height. l = √ (r^2 + h^2)   ]

Hence find the area for 1500 tents and find total cost by multiplying the value by 120  [ given Rs 120 per sq meter is the cost of canvasing each tent ]

finally share of each school = total value obtained above divided by 50

hope this helps.

Answer

Answer

Here slant height of cone(l)  = 

TSA OF TENT  = LSA OF CYLINDER+LSA OF CONE

 TSA OF TENT = 2rh +

 TSA OF TENT = r (2h+l)            (As radius of cylinder = radius of cone)

TSA OF TENT =

TSA OF TENT = 92.4 sq m.

Total cost = 1500

Each school’s contribution =  =  = 332,640 rs

Answer

Area of Canvas = curved area of cone + curved surface area of the cylinder.

                            =( 22/7 r l (slant height) + 2* 22/7 r h) m2 = (30.8 + 61.6)m2   = 92.4 m2.

Total cost obtained to make 1500 canvas = 1500 * 120*  92.4 = Rs 16,632,000.

Amount share by each school = 16632000/50 = Rs 3,32,640.

Answer

Answer