Quantitative Aptitude
VOLUME AND SURFACE AREA MCQs
Total Questions : 820
| Page 79 of 82 pages
Answer: Option C. -> 2h units
Radius of the base = r units
Curved surface area of a right cylinder = $$4\pi {\text{r}}h$$
Curved surface area of cylinder = $$2\pi {\text{RH}}$$
∴ According to the question,
$$2\pi {\text{rH = }}4\pi {\text{rh}}$$
⇒ Height of cylinder = 2h units
Radius of the base = r units
Curved surface area of a right cylinder = $$4\pi {\text{r}}h$$
Curved surface area of cylinder = $$2\pi {\text{RH}}$$
∴ According to the question,
$$2\pi {\text{rH = }}4\pi {\text{rh}}$$
⇒ Height of cylinder = 2h units
Answer: Option B. -> 11 m, 5 m
Let the breadth and height of the room be b and h metres respectively.
Then,
Area of the floor $$ = \left( {14b} \right)\,{m^2}$$
$$\eqalign{
& \therefore 14b = 2.2 \times 70 \cr
& \Rightarrow b = \frac{{2.2 \times 70}}{{14}} \cr
& \Rightarrow b = 11 \cr} $$
Volume of the room :
$$\eqalign{
& = \left( {14 \times 11 \times h} \right){m^3} \cr
& = \left( {154h} \right){m^3} \cr} $$
$$\eqalign{
& \therefore 154h = 11 \times 70 \cr
& \Rightarrow h = \frac{{11 \times 70}}{{154}} \cr
& \Rightarrow h = 5 \cr} $$
Let the breadth and height of the room be b and h metres respectively.
Then,
Area of the floor $$ = \left( {14b} \right)\,{m^2}$$
$$\eqalign{
& \therefore 14b = 2.2 \times 70 \cr
& \Rightarrow b = \frac{{2.2 \times 70}}{{14}} \cr
& \Rightarrow b = 11 \cr} $$
Volume of the room :
$$\eqalign{
& = \left( {14 \times 11 \times h} \right){m^3} \cr
& = \left( {154h} \right){m^3} \cr} $$
$$\eqalign{
& \therefore 154h = 11 \times 70 \cr
& \Rightarrow h = \frac{{11 \times 70}}{{154}} \cr
& \Rightarrow h = 5 \cr} $$
Answer: Option B. -> 36
Volume of water displaced :
$$\eqalign{
& = \left( {24 \times 15 \times \frac{1}{{100}}} \right){m^3} \cr
& = \frac{{18}}{5}{m^3} \cr} $$
Volume of water displaced by 1 man = 0.1 m3
∴ Number of men :
$$\eqalign{
& = \left( {\frac{{\frac{{18}}{5}}}{{0.1}}} \right) \cr
& = \left( {\frac{{18}}{5} \times 10} \right) \cr
& = 36 \cr} $$
Volume of water displaced :
$$\eqalign{
& = \left( {24 \times 15 \times \frac{1}{{100}}} \right){m^3} \cr
& = \frac{{18}}{5}{m^3} \cr} $$
Volume of water displaced by 1 man = 0.1 m3
∴ Number of men :
$$\eqalign{
& = \left( {\frac{{\frac{{18}}{5}}}{{0.1}}} \right) \cr
& = \left( {\frac{{18}}{5} \times 10} \right) \cr
& = 36 \cr} $$
Question 784. A rectangular water tank is open at the top. Its capacity is 24 m3. Its length and breadth are 4 m and 3 m respectively. Ignoring the thickness of the material used for building the tank, the total cost of painting the inner and outer surface of the tank at the rate of Rs. 10 per m2 is :
Answer: Option D. -> Rs. 800
Depth of the tank :
$$\eqalign{
& = \left( {\frac{{24}}{{4 \times 3}}} \right)m \cr
& = 2\,m \cr} $$
Since the tank is open and thickness of material is to be ignored, we have
Sum of inner and outer surface :
$$\eqalign{
& = 2\left[ {\left\{ {2\left( {l + b} \right) \times h} \right\} + lb} \right] \cr
& = 2\left[ {\left\{ {2\left( {4 + 3} \right) \times 2} \right\} + 4 \times 3} \right]{m^2} \cr
& = 80\,{m^2} \cr} $$
∴ Cost of painting :
$$\eqalign{
& = {\text{Rs}}{\text{.}}\left( {80 \times 10} \right) \cr
& = {\text{Rs}}{\text{. 800}} \cr} $$
Depth of the tank :
$$\eqalign{
& = \left( {\frac{{24}}{{4 \times 3}}} \right)m \cr
& = 2\,m \cr} $$
Since the tank is open and thickness of material is to be ignored, we have
Sum of inner and outer surface :
$$\eqalign{
& = 2\left[ {\left\{ {2\left( {l + b} \right) \times h} \right\} + lb} \right] \cr
& = 2\left[ {\left\{ {2\left( {4 + 3} \right) \times 2} \right\} + 4 \times 3} \right]{m^2} \cr
& = 80\,{m^2} \cr} $$
∴ Cost of painting :
$$\eqalign{
& = {\text{Rs}}{\text{.}}\left( {80 \times 10} \right) \cr
& = {\text{Rs}}{\text{. 800}} \cr} $$
Answer: Option C. -> 7 cm
Length of rectangle paper = Circumference of the base of cylinder
If r is the radius of the cylinder :
$$\eqalign{
& \Rightarrow 44 = 2\pi r \cr
& \Rightarrow r = \frac{{44 \times 7}}{{2 \times 22}} \cr
& \Rightarrow r = 7\,cm \cr} $$
Length of rectangle paper = Circumference of the base of cylinder
If r is the radius of the cylinder :
$$\eqalign{
& \Rightarrow 44 = 2\pi r \cr
& \Rightarrow r = \frac{{44 \times 7}}{{2 \times 22}} \cr
& \Rightarrow r = 7\,cm \cr} $$
Answer: Option B. -> 1 dm
Let the thickness of the bottom be x cm
Then,
$$\left[ {\left( {330 - 10} \right) \times \left( {260 - 10} \right) \times \left( {110 - x} \right)} \right]$$ $$ = 8000 \times 1000$$
$$ \Rightarrow 320 \times 250 \times \left( {110 - x} \right) = $$ $$8000 \times 1000$$
$$\eqalign{
& \Rightarrow \left( {110 - x} \right) = \frac{{8000 \times 1000}}{{320 \times 250}} \cr
& \Rightarrow \left( {110 - x} \right) = 100 \cr
& \Rightarrow x = 10\,cm \cr
& \Rightarrow x = 1\,dm \cr} $$
Let the thickness of the bottom be x cm
Then,
$$\left[ {\left( {330 - 10} \right) \times \left( {260 - 10} \right) \times \left( {110 - x} \right)} \right]$$ $$ = 8000 \times 1000$$
$$ \Rightarrow 320 \times 250 \times \left( {110 - x} \right) = $$ $$8000 \times 1000$$
$$\eqalign{
& \Rightarrow \left( {110 - x} \right) = \frac{{8000 \times 1000}}{{320 \times 250}} \cr
& \Rightarrow \left( {110 - x} \right) = 100 \cr
& \Rightarrow x = 10\,cm \cr
& \Rightarrow x = 1\,dm \cr} $$
Answer: Option C. -> 10 cm
Let the length of each side of the cube be a cm
Then, Volume of the part of cube outside water = Volume of the mass placed on it
⇒ 2a2 = 0.2 × 1000
⇒ 2a2 = 200
⇒ a2 = 100
⇒ a = 10
Let the length of each side of the cube be a cm
Then, Volume of the part of cube outside water = Volume of the mass placed on it
⇒ 2a2 = 0.2 × 1000
⇒ 2a2 = 200
⇒ a2 = 100
⇒ a = 10
Answer: Option D. -> $$\sqrt 3 :\sqrt 2 $$
Let their heights be 2h and 3h and radii be r and R respectively
Then,
$$\eqalign{
& \pi {r^2}\left( {2h} \right) = \pi {R^2}\left( {3h} \right) \cr
& \Rightarrow \frac{{{r^2}}}{{{R^2}}} = \frac{3}{2} \cr
& \Rightarrow \frac{r}{R} = \frac{{\sqrt 3 }}{{\sqrt 2 }}\,i.e.,\sqrt 3 :\sqrt 2 \cr} $$
Let their heights be 2h and 3h and radii be r and R respectively
Then,
$$\eqalign{
& \pi {r^2}\left( {2h} \right) = \pi {R^2}\left( {3h} \right) \cr
& \Rightarrow \frac{{{r^2}}}{{{R^2}}} = \frac{3}{2} \cr
& \Rightarrow \frac{r}{R} = \frac{{\sqrt 3 }}{{\sqrt 2 }}\,i.e.,\sqrt 3 :\sqrt 2 \cr} $$
Answer: Option B. -> 5.5 cm
Volume of the cylinder :
= Volume of the cube
= (11)3 cm3
= 1331 cm3
Let the radius of the base be r cm
Then,
$$\eqalign{
& \frac{{22}}{7} \times {r^2} \times 14 = 1331 \cr
& \Rightarrow {r^2} = \frac{{1331}}{{44}} = \frac{{121}}{4} \cr
& \Rightarrow r = \frac{{11}}{2} \cr
& \Rightarrow r = 5.5 \cr} $$
Volume of the cylinder :
= Volume of the cube
= (11)3 cm3
= 1331 cm3
Let the radius of the base be r cm
Then,
$$\eqalign{
& \frac{{22}}{7} \times {r^2} \times 14 = 1331 \cr
& \Rightarrow {r^2} = \frac{{1331}}{{44}} = \frac{{121}}{4} \cr
& \Rightarrow r = \frac{{11}}{2} \cr
& \Rightarrow r = 5.5 \cr} $$
Answer: Option C. -> 6400
$$\eqalign{
& {\text{Number of bricks :}} \cr
& = \frac{{{\text{Volume of the wall}}}}{{{\text{Volume of 1 brick}}}} \cr
& = \left( {\frac{{800 \times 600 \times 22.5}}{{25 \times 11.25 \times 6}}} \right) \cr
& = 6400 \cr} $$
$$\eqalign{
& {\text{Number of bricks :}} \cr
& = \frac{{{\text{Volume of the wall}}}}{{{\text{Volume of 1 brick}}}} \cr
& = \left( {\frac{{800 \times 600 \times 22.5}}{{25 \times 11.25 \times 6}}} \right) \cr
& = 6400 \cr} $$