Quantitative Aptitude · Mathematics

Mensuration and Costing

74 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 business mathematics and statistics applications of calculus marginal income and marginal cost to find the maximum profit if marginal revenue and marginal cost function are given: integral calculus – ii

The cost of cultivating a square field at the rate of $Rs.\,135$ per hectare is $Rs.\,1215$. The cost of putting a fence around it at the rate of $75\;paise$ per metre would be ...... .

  1. $Rs.\,360$
  2. $Rs.\,810$
  3. $Rs.\,900$
  4. $Rs.\,1800$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

rate of cultivating a square field =Rs.135 per hectare, 
cost of cultivating a square field=Rs.1215
area of square field=cost of cultivating a square field/rate of cultivating a square field
$A=1215/135$
$A=9 hectare$
1 hectare =10000m^2
$Area=a^2$
$90000=a^2$
$a=300$

cost of putting a fence around it at the rate of 75 paise per metre 
Total cost=0.75*4*300Rs
C=900Rs
Answer (C) 900RS

Multiple choice maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

A flooring tile has the shape of a parallelogram whose base is $24: cm$ and the corresponding height is $10: cm$. How many such tiles are required to cover a floor of area $1080$ $m^2$? 

  1. $20000$
  2. $35000$
  3. $45000$
  4. $65000$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Area of the parallelogram $=$ Base$\times $Height


So area of each tiles $=24\times 10=240 :cm^2$

Area of the floor $=1080 : m^2=(1080\times 100\times 100)  : cm^2$

$\therefore$ Required number of tiles $=\dfrac{\text{Area of the floor}}{\text{Area of each tiles}}=\dfrac{10800000}{240}=45000$

Multiple choice maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

A field is in the form of a parallelogram whose base is $420\ m$ and altitude is $3.6\ dam$. Find the cost of watering at $10$ paise per sq. m.

  1. Rs. $15.120$
  2. Rs. $1512$
  3. Rs. $151.20$

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Base  $= 420\ m$
Height $=  36\ m$
Area $= b \times h = 420 \times 36$
$= 15,120 \displaystyle \ m^{2}$
The cost of watering per sq m
$= 10$ paise
Cost  of watering the field $= 15120 \times 0.1$
Rs. $1512$

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

The inner circumference of a circular tracks is $220\ m$. The track is $7$ $m$ wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of Rs. $2$ per metre. Use $\pi=\displaystyle\frac{22}{7}$

  1. Rs. $947$
  2. Rs. $726$
  3. Rs. $612$
  4. Rs. $528$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Circumference of inner side = $220 m$


$\Rightarrow 2\pi r=220\Rightarrow r=\dfrac { 220\times 7 }{ 44 } =35m$


Now, width of track  $=7m$

$\therefore $ Outer radius $=35+7 = 42 m$

Therefore outer circumference  $=2\pi R=2\times \dfrac { 22 }{ 7 } \times 42=264m$

$\therefore $ Cost of fencing $=$ Rs. $\left( 264\times 2 \right) $ $=$ Rs. $528$

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

The diameter of semi circular field is $49$ m. What the cost of fencing the plot of Rs. $10$ per metre?

  1. Rs. $ 1259.3$
  2. Rs. $1260$
  3. Rs. $1250$

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The perimeter of a semi circle is $= \pi r+2r$

Here $r:$ radius of the semi circle
Given that $2r=49$
$\Rightarrow r= \dfrac { 49 }{ 2 } $
Perimeter of the semi circle $ = (\pi +2)r= (\pi +2)\dfrac { 49 }{ 2 } = 125.93$ m 
Cost of fencing the plot per metre $=$ Rs. $10$
Total cost of fencing $= 125.93\times10=$ Rs. $1259.3$.

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

A circular grassy plot of land, 42 m in diameter, has a path 3.5 m wide running around it on the outside. Find the cost graveling the path at Rs 4 per square meter.

  1. $Rs 2002$
  2. $Rs 2003$
  3. $Rs 2004$
  4. $Rs 2000$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given : Radius ($r _1$) $=21m$

             Radius ($r _2$) $=24.5m$

Required area $=\pi (r _2^{2}-r _1^{2})$

Required area $=\pi (24.5^{2}-21^{2})=500.5$

Cost $=500.5\times 4$

         $=2002Rs$

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

200 wooden balls each of diameter 70 mm are to be painted Find the cost of painting these balls at 10 paise/$\displaystyle cm^{2}$

  1. Rs.3080

  2. Rs.2771

  3. Rs.4000

  4. Rs.7000
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total surface area of 200 balls of $\displaystyle \frac{35}{10}$ cm radius will be $\displaystyle 200\times 4\times \frac{22}{7}\times \frac{35}{10}\times \frac{35}{10}$ mm
The cost will be Rs $\displaystyle \frac{200\times 4\times 22\times 35\times 35}{7\times 10\times 10\times 100}$ which simplifies to Rs 3080

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

The inner diameter of a circular well is $3.5$ m. It is $10$ m deep. Find the cost of plastering this curved surface at the rate of Rs. $40$ per m$^2$.

  1. Rs. $4000$
  2. Rs. $4400$
  3. Rs. $4500$
  4. Rs. $4800$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given diameter of well is $3.5$ m and depth is $10$ m and cost of plastering is Rs. $40$ per sq m.

Then radius of well $=\dfrac{3.5}{2}=1.75$ m
And height of well $=10$ m
Then curved surface area of well $=$ $2\pi rh=2\times \dfrac{22}{7}\times 1.75\times 10$
$=$ $2\times 22\times 0.25\times 10=110 m^{2}$
Then cost of plastering curved surface area $=$ $110\times 40=$ Rs. $4400$.

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

A magnet is in the form of a ring with inner diameter $4cm$ and outer diameter $6cm$. If the thickness of the magnet is $2cm$ . What is the cost of fabricating the surface of the magnet if the cost of fabrication per ${cm}^{2}$ is $Rs.10$

  1. $Rs.950$
  2. $Rs.945$
  3. $Rs.942$
  4. $Rs.1000$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Surface area of a ring-shaped magnet (hollow cylinder) includes inner CSA, outer CSA, and two annular ends. Inner r=2, outer R=3, h=2. Area = 2*pi*r*h + 2*pi*R*h + 2*(pi*R^2 - pi*r^2) = 2*pi*2*2 + 2*pi*3*2 + 2*pi*(9-4) = 8*pi + 12*pi + 10*pi = 30*pi. 30 * 3.14 = 94.2. Cost = 94.2 * 10 = 942.

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

A flower pot is in the form of a hollow cylinder with a closed base with inner radius $2cm$ and outer radius $4cm$ . The height of the flower pot is $10cm$ . If the pot has to be polished find the cost of polishing if the cost of polishing per ${cm}^{2}$ is $Rs.2$.

  1. $Rs.276.32$
  2. $Rs.275$
  3. $Rs.270$
  4. $Rs.278.64$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have to first find the total surface area of the flower pot.
Outer radius $=4cm$
Inner radius $=2cm$
Height of flower pot $=10cm$
Total surface area $=A=2\pi Rh+2\pi rh+\pi { { R }^{ 2 } }+\pi { { r }^{ 2 } }$
$A=2\pi (R+r)h+\pi ({ R }^{ 2 }+{ r }^{ 2 })\ A=2\pi (4+2)2+\pi (16+4)\ A=24\pi +20\pi =44\pi \ A=40\pi =44\times 3.14=138.16{ cm }^{ 2 }$
Cost of fabrication per ${cm}^{2}$ $=Rs.2$
Total cost of fabrication $==138.16\times 2=Rs.276.32$

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