Quantitative Aptitude · Mathematics

Mensuration and Costing

76 Questions

Mensuration and costing questions combine geometric calculations with practical expenses like painting, fencing, and carpeting. They are highly relevant for quantitative aptitude tests in SSC and state exams. Solving these requires accuracy in applying area and volume formulas.

painting and plastering costfencing and levelingsurface area calculationvolume of solids

Mensuration and Costing Questions

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

The cost of canvas required for a conical tent of height 8 m and diameter of base 12 m at the rate of Rs 3.50 per $\displaystyle m^{2}$ is

  1. RS 620

  2. Rs 600

  3. Rs 640

  4. Rs 660
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that conical tent of canvas height is 8 cm and diameter of base is 12 m

Then radius of base =$\frac{12}{2}=6 m$
Then area of conical tent =$\pi r(\sqrt{r^{2}+h^{2}})= \frac{22}{7}\times 6\times \sqrt{(10)^{2}+(6)^{2}}\Rightarrow 60\times \frac{22}{7}cm^{2}$
If cost of canvas is 3.50 per sq cm 
Then cost of canvas=$\times 60\times \frac{22}{7}\times 3.5= Rs 660 $

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

A conical tent of radius of 12 m and height 16 m is to be made, then the cost of canvas required at the rate Rs 10 per $ \displaystyle   m^{2}   $ is

  1. RS 7445

  2. Rs 7543

  3. Rs 7550

  4. Rs 7500
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given the radius a of height of tent is 12 m and 16 m respt 

Then slant height of tent s=$\sqrt{r^{2}+h^{2}}=\sqrt{(12)^{2}+(16)^{2}}=\sqrt{144+256}=\sqrt{400}=20 m$
Then covered surface area of tent=$\pi rs=3.143\times 12\times 20=754.30 m^{2}$
If cost of canvas is Rs 10 per sq m
Then total cost =$754.30\times10=7543 RS

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

What length of tarpaulin $3$ m wide will be required to make conical tent of height $8$ m and base radius $6$ m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $20$ cm.

 (Use $\pi=3.14$)

  1. $63$ m
  2. $85$ m
  3. $74$ m
  4. $92$ m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

$h=8m$, $r=6m$

$\therefore l=\sqrt {r^2+h^2}$

 $=\sqrt {6^2+8^2}$

 $=\sqrt {36+64}$

 $=\sqrt {100}$

 $=10$ m

$\therefore$ width surface area $=$ $\pi(r)(l)$

                                    $=3.14(6)(10)$

                                    $=1883.4m^2$

Width of tarpaulin $= 3$ m

$\therefore$ length of tarpaulin $= \displaystyle \frac{188.4}{3}$
                                   $=62.8$ m

Extra length of material required $=20$ cm
                                                       $=0.2$ m

$\therefore$ actual length of tarpaulin required $= 62.8$ m $+0.2$ m
                                                              $=63$ m

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

The slant height and base diameter of a conical tomb are $25$ m and $14$ m respectively. Find the cost of white washing its curved surface at the rate of Rs. $210$ per $100{m}^{2}$.

  1. Rs. $5627$
  2. Rs. $4156$
  3. Rs. $1155$
  4. Rs. $964$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Curved surface area of a cone $= \pi rl$, where $r$ is the radius of the cone and $l$ is the slant height.
Radius of the tomb $ = \dfrac {\text{Diameter}}{2} = 7  m $
Hence, CSA of this conical tomb , $ = \dfrac {22}{7} \times 7 \times 25 = 550$  sq. m.
Cost of white washing the tomb at Rs. $210$ per $100 {m}^{2} = \dfrac {210}{100} \times 550 =  $ Rs. $  1155 $

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

The length of canvas $1.1$ m wide required to build a conical tent of height $14$ m and the floor area $346.5{m}^{2}$, is

  1. $65$ m
  2. $525$ m
  3. $490$ m
  4. $860$ m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $h = 14$ m and floor area $= 346.5$ $m^{ 2 }$
$\Rightarrow \pi { r }^{ 2 }=346.5\ \Rightarrow { r }^{ 2 }=346.5\times \dfrac { 7 }{ 22 } =110.25\ \Rightarrow r=10.5$
$\therefore$ $l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { \left( 10.5 \right)  }^{ 2 }+{ \left( 14 \right)  }^{ 2 } } =\sqrt { 110.25+196 } $
$=\sqrt { 306.25 } =17.5$ m
$\therefore$ curved surface area $=\pi rl$
$=\cfrac { 22 }{ 7 } \times 10.5\times 17.5=\cfrac { 4042.5 }{ 7 } { m }^{ 2 }$
Width of cloth $=1.1$ m
$\therefore$ length of cloth required $=\cfrac { \cfrac { 4042.5 }{ 7 }  }{ 1.1 } =\cfrac {4042.5 }{ 7.7 } = 525$ m
Hence, $525$ m length of canvas is required to build the conical tent.

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

A circus tent is in the form of a cone over a cylinder. The diameter of the base is $9$ m, the height of cylindrical part is $4.8$ m and the total height of the tent is $10.8$ m. The canvas required for the tent is ..........

  1. $24.184$ sq.m
  2. $2418.4$ sq.m
  3. $241.84$ sq.m

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The amount of canvas used to make the tent $ = $ Curved surface area of cylindrical part $ + $ Curved surface area of the conical part.
Curved Surface Area of a Cylinder of Radius "$R$" and height "$h$" $ = 2\pi Rh$
Radius of the cylindrical part $ = \dfrac {Diameter}{2} = \dfrac {9}{2} $ m 
Curved surface area of a cone $= \pi rl$, where $r$ is the radius of the cone and $l$ is the slant height.

Radius of the conical part $ = \dfrac {Diameter}{2} = \dfrac {9}{2}  m $

Height of the conical part $ = 10.8 - 4.8 = 6 $ m 

For a cone, $  l= \sqrt { { h }^{ 2 }+  {r}^{ 2 } } $, where $h$ is the height. 

Hence, $l = \sqrt { { 6 }^{ 2 }+  {4.5}^{ 2 } } $

$\therefore  l = 7.5  $ m 

Hence, area of canvas $ = 2 \times \pi \times 4.5 \times 4.8 + \pi \times 4.5 \times 7.5$

$ = 76.95 \pi  $

$= 241.84  {m}^{2} $

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

The radius of the base of a conical tent is $7\space m$. The tent is $24\space m$ high. Find the cost of the canvas required to make the tent, if one square meter of canvas costs $Rs. 180$ (Take $\pi = 3.14$)

  1. $Rs. 99000$
  2. $Rs. 98000$
  3. $Rs. 95000$
  4. $Rs. 97000$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Radius $= 7m$,  height $= 24m$

CSA of tent = $\pi rl$
Now, $l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { 7 }^{ 2 }+{ 24 }^{ 2 } } =\sqrt { 49+576 } =\sqrt { 625 } =25cm$
Now, CSA of tent = $\dfrac { 22 }{ 7 } \times 7\times 25=550m$
Therefore, cost of canvas = Rs.$(550$$\times $$180) = Rs.99000$

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

The cost of the canvas required to make a conical tent of base radius $8$ m at the rate of Rs. $40$ per $\displaystyle m^{2}$ is Rs. $10,048$. Find the height of the tent .$\displaystyle \left ( Take\  \pi =3.14 \right )$ 

  1. $6$ m
  2. $7$ m
  3. $8$ m
  4. $10$ m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the amount of canvas required to make a conical tent, we need to calculate the lateral surface area of the tent.
Lateral surface area of a cone of radius $ r $, height $ h = \pi r\sqrt{{h}^{2} + {r}^{2}} $
As the total cost to make the tent is Rs $10048 $ at the rate of Rs $ 40 $ per sq m, total area LSA $ = \dfrac {10048}{40} = 251.2 $ sq m 

So, $ \pi r \times \sqrt{{h}^{2} + {r}^{2}} = 251.2 $
$ \Rightarrow  3.14 \times 8 \times \sqrt{{h}^{2} + {8}^{2}} = 251.2 $
$ \Rightarrow  \sqrt{{h}^{2} + {8}^{2}} = 10 $
$ \Rightarrow  {h}^{2} + 64 = 100 $
$ \Rightarrow  {h}^{2}  = 36 $
$ \Rightarrow  h = 6 $ m 

Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

A tree, in each year, grows $5\ cm$ less than it grew in the previous year. If it grew half a meter per year, then the height of the tree (in meters), when it ceases to grow, is

  1. $2.50$
  2. $2.00$
  3. $3.00$
  4. $2.75$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The annual growth forms an arithmetic progression with a first term a = 0.50 meters (50 cm) and a common difference d = -0.05 meters (-5 cm). The tree ceases to grow when its annual growth becomes zero or negative, which happens after n terms where 0.50 + (n - 1)(-0.05) <= 0, giving n = 10 years. The sum of this finite arithmetic progression is given by S = n/2 * (2a + (n - 1)d), which calculates to 2.75 meters.

Multiple choice

A farmer is considering two different irrigation systems, A and B. System A has a fixed cost of \$10,000 and a variable cost of \$10 per acre-inch of water. System B has a fixed cost of \$15,000 and a variable cost of \$5 per acre-inch of water. At what level of water usage (in acre-inches) does System B become more cost-effective than System A?

  1. 1000 acre-inches

  2. 2000 acre-inches

  3. 3000 acre-inches

  4. 4000 acre-inches
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The total cost of System A is $10,000 + $10x, where x is the number of acre-inches of water used. The total cost of System B is $15,000 + $5x. System B becomes more cost-effective than System A when $10,000 + $10x < $15,000 + $5x. Solving this inequality, we get x > 2000. Therefore, System B becomes more cost-effective than System A at a level of water usage of 2000 acre-inches.

Multiple choice

A farmer is considering two different irrigation systems, A and B. System A has a fixed cost of \$10,000 and a variable cost of \$5 per acre-inch of water. System B has a fixed cost of \$15,000 and a variable cost of \$3 per acre-inch of water. At what level of water usage (in acre-inches) does System B become more cost-effective than System A?

  1. 2000 acre-inches

  2. 3000 acre-inches

  3. 4000 acre-inches

  4. 5000 acre-inches
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The total cost of System A is $10,000 + $5x, where x is the number of acre-inches of water used. The total cost of System B is $15,000 + $3x. System B becomes more cost-effective than System A when $10,000 + $5x < $15,000 + $3x. Solving this inequality, we get x > 3000. Therefore, System B becomes more cost-effective than System A at a level of water usage of 3000 acre-inches.