NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes - Exercise 13.3

1. A metallic sphere of radius $4.2$ cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Ans: Let the radius of the sphere be ${r_1}$. Thus, ${r_1} = 4.2\,{\text{cm}}$. Now, let the radius of the cylinder be ${r_2}$. Thus, ${r_2} = 6\,{\text{cm}}$. Let the height of the cylinder be $h$. 

The volume of sphere is given by, ${V_{{\text{sphere}}}} = \dfrac{4}{3}\pi {r_1}^3$ and volume of a cylinder is ${V_{{\text{cylinder}}}} = \pi {r_2}^2h$.

The volume of the cylinder and sphere will be the same.

$ \Rightarrow \dfrac{4}{3}\pi {r_1}^3 = \pi {r_2}^2h$

Substitute ${r_1}$ as $4.2$, and ${r_2}$ as 6 and solve for $h$.

$   \Rightarrow \dfrac{4}{3}\pi {\left( {4.2} \right)^3} = \pi {\left( 6 \right)^2}h $

$   \Rightarrow \dfrac{4}{3}\pi  \times \left( {4.2} \right) \times \left( {4.2} \right) \times \left( {4.2} \right) = \pi  \times 36 \times h $

Cancel $\pi $ from the sides.

$ \Rightarrow \dfrac{4}{3}\left( {4.2} \right) \times \left( {4.2} \right) \times \left( {4.2} \right) = 36 \times h$

Divide both sides by 36.

$   \Rightarrow \dfrac{4}{3} \times \dfrac{{\left( {4.2} \right) \times \left( {4.2} \right) \times \left( {4.2} \right)}}{{36}} = h $

$   \Rightarrow \dfrac{{\left( {1.4} \right) \times \left( {4.2} \right) \times \left( {4.2} \right)}}{9} = h $

$   \Rightarrow {\left( {1.4} \right)^3} = h $

$   \Rightarrow 2.74 = h$

Thus, $h = 2.74\;{\text{cm}}$.  Hence, the height of the cylinder formed after recasting it from a sphere is $2.74$ cm.

2. Metallic spheres of radii 6 cm, 8 cm, and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Ans: Let the radius of the first sphere be ${r_1}$, and volume ${V_1}$. Here, ${r_1} = 6\,{\text{cm}}$.

Let the radius of the second sphere be ${r_2}$, and volume ${V_2}$. Here, ${r_2} = 8\,{\text{cm}}$.

Let the radius of the third sphere be ${r_3}$, and volume \[{V_3}\]. Here, ${r_3} = 10\,{\text{cm}}$.

Let the radius of the sphere formed be $r$ and the volume be $V$.

The volume of the sphere is given by, $V = \dfrac{4}{3}\pi {r^3}$ .

The volume of the larger sphere formed will be equal to the sum of the volume of the three spheres. Thus, \[V = {V_1} + {V_2} + {V_3}\].

The volume of the larger sphere formed will be equal to the sum of the volume of the three spheres. Thus, \[V={{V}_{1}}+{{V}_{2}}+{{V}_{3}}\].

$\Rightarrow V={{V}_{1}}+{{V}_{2}}+{{V}_{3}}$

$\Rightarrow \dfrac{4}{3}\pi {{r}^{3}}=\dfrac{4}{3}\pi {{r}_{1}}^{3}+\dfrac{4}{3}\pi {{r}_{2}}^{3}+\dfrac{4}{3}\pi {{r}_{3}}^{3}$

Take $\dfrac{4}{3}\pi $ common from the terms of the right-hand side.

$\Rightarrow \dfrac{4}{3}\pi {{r}^{3}}=\dfrac{4}{3}\pi \left[ {{r}_{1}}^{3}+{{r}_{2}}^{3}+{{r}_{3}}^{3} \right]$

$\Rightarrow {\dfrac{4}{3}}{\pi }{{r}^{3}}={\dfrac{4}{3}}{\pi }\left[ {{r}_{1}}^{3}+{{r}_{2}}^{3}+{{r}_{3}}^{3} \right]$

$\Rightarrow {{r}^{3}}=\left[ {{r}_{1}}^{3}+{{r}_{2}}^{3}+{{r}_{3}}^{3} \right]$

Substitute ${{r}_{1}}$ as 6, ${{r}_{2}}$ as 8, and ${{r}_{3}}$ as 10 and solve for $r$.

$\Rightarrow {{r}^{3}}=\left[ {{\left( 6 \right)}^{3}}+{{\left( 8 \right)}^{3}}+{{\left( 10 \right)}^{3}} \right]$

$\Rightarrow {{r}^{3}}=\left[ 216+512+1000 \right]$

$\Rightarrow {{r}^{3}}=1728$

Take the cube root of both sides.

$\Rightarrow {{r}^{3}}={{\left( 12 \right)}^{3}}$

$\Rightarrow r=12$

Hence, the radius of the sphere formed is 12 cm.

3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform. Use $\pi  = \dfrac{{22}}{7}$.

Ans: The well will have a cylindrical shape.  The platform that is made will be cuboidal in shape. The length of the platform is 22 m, and breadth is 14 m. Assume the height of the platform as $H$.

Let $h$ be the depth of the well, $d$ be the diameter, and $r$ be the radius. Here,

Let $h$ be the depth of the well, $d$ be the diameter, and $r$ be the radius. Here,

$h=20\,\text{m}$

$d=7\,\text{m}$

The radius is always half of the diameter.

$\Rightarrow r=\dfrac{d}{2}$

Substitute $d$ as 7.

$\Rightarrow r=\dfrac{7}{2}$

So, the radius of the well is $\dfrac{7}{2}$ m.

We can find the height of the platform by equating the volume of soil from well with the volume of soil used to make the platform. We will use the formula of volume of cylinder, $V=\pi {{r}^{2}}h$ to find the volume of soil from well and the formula of volume of a cuboid, $V=lbh$ to find the volume of soil used to make the platform.

Thus, $\pi {{r}^{2}}h=l\times b\times H$,

Substitute $r$ as $\dfrac{7}{2}$, $l$ as 22, $b$ as 14, and solve for $H$.

$\Rightarrow \pi {{\left( \dfrac{7}{2} \right)}^{2}}\times 20=22\times 14\times H$

Take $\pi =\dfrac{22}{7}$.

$\Rightarrow \dfrac{22}{7}\times {{\left( \dfrac{7}{2} \right)}^{2}}\times 20=22\times 14\times H$

$\Rightarrow \dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}\times 20=22\times 14\times H$

Divide both sides by $22\times 14$.

$\Rightarrow \dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}\times 20\times \dfrac{1}{22\times 14}=\dfrac{22\times 14\times H}{22\times 14}$

$\Rightarrow \dfrac{{22}}{{7}}\times \dfrac{{7}}{2}\times \dfrac{7}{2}\times 20\times \dfrac{1}{{22}\times 14}=\dfrac{{22}\times {14}\times H}{{22}\times {14}}$

On simplifying the expression we get,

$\Rightarrow 5\times \dfrac{1}{2}=H$

$\Rightarrow 2.5=H$

Hence, the height of the platform formed will be $2.5$ m.

4. A well of diameter 3 m is dug 14 m deep.  The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment.  Find the height of the embankment.

Ans: The well will have a cylindrical shape.  The embankment that is made will be a circular ring in shape. 

Let ${h_1}$ be the depth of the well, $d$ be the diameter, and ${r_1}$ be the radius.Here,

$h=14\,\text{m}$

$d=3\,\text{m}$

The radius is always half of the diameter.

$ \Rightarrow {r_1} = \dfrac{d}{2}$

Substitute $d$ as 3.

$ \Rightarrow {r_1} = \dfrac{3}{2}$

So, the radius of the well is $\dfrac{3}{2}$ m.

The width of the embankment is 4 m. we observe from the figure that the embankment will be cylindrical. Also, notice that it will be a hollow cylinder. Assume the height of the embankment is ${h_2}$.The volume of embankment will be the difference of the volume of the inner cylinder and outer cylinder.

Let the outer radius be ${r_2}$.

$\Rightarrow {{r}_{2}}=4+\dfrac{3}{2}$

$\Rightarrow {{r}_{2}}=\dfrac{8+3}{2}$

$\Rightarrow {{r}_{2}}=\dfrac{11}{2}\,\text{m}$

Now, the volume of soil dug from a well will be equal to the volume of earth used to form an embankment.

$\Rightarrow \pi \times {{r}_{1}}^{2}\times {{h}_{1}}=\pi \times \left( {{r}_{2}}^{2}-{{r}_{1}}^{2} \right)\times {{h}_{2}}$

$\Rightarrow {{r}_{1}}^{2}\times {{h}_{1}}=\left( {{r}_{2}}^{2}-{{r}_{1}}^{2} \right)\times {{h}_{2}}$

Substitute ${{r}_{1}}$ as $\dfrac{3}{2}$, ${{r}_{2}}$ as $\dfrac{11}{2}$, ${{h}_{1}}$ as 14.

$\Rightarrow {{\left( \dfrac{3}{2} \right)}^{2}}\times 14=\left[ {{\left( \dfrac{11}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}} \right]\times {{h}_{2}}$

$\Rightarrow \dfrac{9}{4}\times 14=\left[ \dfrac{121}{4}-\dfrac{9}{4} \right]\times {{h}_{2}}$

$\Rightarrow \dfrac{9}{4}\times 14=\left[ \dfrac{121-9}{4} \right]\times {{h}_{2}}$

$\Rightarrow \dfrac{9}{4}\times 14=\left[ \dfrac{112}{4} \right]\times {{h}_{2}}$

On simplifying we get,

\[\Rightarrow 9\times 4=112\times {{h}_{2}}\]

Divide both sides by 112.

$\Rightarrow \dfrac{9\times 4}{112}={{h}_{2}}$

$\Rightarrow \dfrac{9}{8}={{h}_{2}}$

$\Rightarrow 1.125={{h}_{2}}$

Therefore, the height of the embankment will be \[1.125\] m.

5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream.  The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Ans: The container has cylindrical shape and the ice cream cones have a conical base surmounted by a hemisphere.

Let ${{h}_{1}}$ be the height of the well, $d$ be the diameter, and ${{r}_{1}}$ be the radius. Here,

${{h}_{1}}=15\,\text{cm}$

$d=12\,\text{cm}$

The radius is always half of the diameter.

$\Rightarrow {{r}_{1}}=\dfrac{d}{2}$

Substitute $d$ as 12.

$\Rightarrow {{r}_{1}}=\dfrac{12}{2}$

$\Rightarrow {{r}_{1}}=6$

So, the radius of the well is 6 m.

Thus, let ${{V}_{1}}$ be the volume of the container.

${{V}_{1}}=\pi {{r}_{1}}^{2}{{h}_{1}}$

Now, let ${{r}_{2}}$ be the radius of circular end of the ice cream cone and ${{h}_{2}}$ be the height of the conical part of ice-cream cone.

${{h}_{2}}=12\,\text{cm}$

${{d}_{2}}=6\,\text{cm}$

Now,

$\Rightarrow {{r}_{2}}=\dfrac{{{d}_{2}}}{2}$

Substitute ${{d}_{2}}$ as 6.

$\Rightarrow {{r}_{2}}=\dfrac{6}{2}$

$\Rightarrow {{r}_{2}}=3$

The volume of ice cream will be the sum of the volume of the cone and hemispherical top. Let $V$ be the volume of the ice-cream cone.

$\Rightarrow V=\text{Volume of 1 ice-cream cone+Volume of hemisphere on the top}$

$\Rightarrow V=\dfrac{1}{3}\pi {{r}_{2}}^{2}{{h}_{2}}+\dfrac{2}{3}\pi {{r}_{2}}^{3}$

Let the number of ice creams that can be filled with ice cream present in the cylinder be $n$.

$\Rightarrow n=\dfrac{\text{Volume of container}}{\text{Volume of the ice cream cone}}$

$\Rightarrow n=\dfrac{\pi {{r}_{1}}^{2}{{h}_{1}}}{\dfrac{1}{3}\pi {{r}_{2}}^{2}{{h}_{2}}+\dfrac{2}{3}\pi {{r}_{2}}^{3}}$

Substitute ${{h}_{1}}$ as 15, ${{r}_{1}}$ as 6, ${{r}_{2}}$ as 3, and ${{h}_{2}}$ as 12.

$\Rightarrow n=\dfrac{\pi \times {{\left( 6 \right)}^{2}}\times \left( 15 \right)}{\dfrac{1}{3}\times \pi \times {{\left( 3 \right)}^{2}}\times \left( 12 \right)+\dfrac{2}{3}\times \pi \times {{\left( 3 \right)}^{3}}}$

$\Rightarrow n=\dfrac{\pi \times 36\times \left( 15 \right)}{\dfrac{1}{3}\times \pi \times \left( 9\times \left( 12 \right)+2\times 27 \right)}$

$\Rightarrow n=\dfrac{36\times \left( 15 \right)}{\dfrac{1}{3}\left( 108+54 \right)}$

$\Rightarrow n=\dfrac{36\times \left( 15 \right)\times 3}{162}$

$\Rightarrow n=10$

Therefore, 10 ice cream cones can be filled with the ice cream present in the container.

6. How many silver coins, $1.75$ cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions, $5.5{\text{ cm}} \times 10\,{\text{cm}} \times {\text{3}}{\text{.5}}\,{\text{cm}}$. Use $\pi  = \dfrac{{22}}{7}$.

Ans: The shape of the silver coins will be cylindrical in shape which are too melted to form a cuboid.

The diameter of the coin is $1.75$ cm. 

The radius is always half of the diameter.

$\Rightarrow r=\dfrac{d}{2}$

Substitute $d$ as $1.75$.

$\Rightarrow r=\dfrac{1.75}{2}$

$\Rightarrow r=0.875\,\text{cm}$

The height of the coin will be its thickness. The height, $h$ is in millimeters. So, convert it to centimeters.

$\Rightarrow h=\dfrac{2}{100}\,\text{cm}$

$\Rightarrow h=0.02\,\text{cm}$

The volume of the cuboid will be $5.5\text{ cm}\times 10\,\text{cm}\times \text{3}\text{.5}\,\text{cm}$.

The volume of a cylinder is $\pi {{r}^{2}}h$.

The volume of $n$ coins will be equal to the cuboid. Hence, $n\times \pi \times {{r}^{2}}\times h=l\times b\times h$.

Substitute the values. 

$\Rightarrow n\times \pi \times {{\left( 0.875 \right)}^{2}}\times \left( 0.2 \right)=5.5\times 10\times \text{3}\text{.5}$

Divide both sides by $\pi \times {{\left( 0.875 \right)}^{2}}\times \left( 0.2 \right)$.

$\Rightarrow n=\dfrac{5.5\times 10\times \text{3}\text{.5}}{\pi \times {{\left( 0.875 \right)}^{2}}\times \left( 0.2 \right)}$

Take $\pi =\dfrac{22}{7}$.

$\Rightarrow n=\dfrac{7\times 5.5\times 10\times \text{3}\text{.5}}{22\times 0.875\times 0.875\times \left( 0.2 \right)}$

On simplifying the calculation, we get,

$\Rightarrow n=400$

Hence, the number of coins melted to form a cuboid are 400.

7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand.  This bucket is emptied on the ground and a conical heap of sand is formed.  If the height of the conical heap is 24 cm.  Find the radius and slant height of the heap.

Ans: The container has a cylindrical bucket and the heap is conical.

Let ${{h}_{1}}$ be the height of the bucket, and ${{r}_{1}}$ be the radius. Here,

${{h}_{1}}=32\text{cm}$

${{r}_{1}}=18\text{cm}$

The volume of the cylinder bucket is ${{V}_{1}}=\pi {{r}_{1}}^{2}{{h}_{1}}$.

Now, let ${{r}_{2}}$ be the radius of the conical heap, and ${{h}_{2}}$ be the height.

Here, ${{h}_{2}}=24\text{cm}$.

The volume of conical heap is \[{{V}_{2}}=\dfrac{1}{3}\pi {{r}_{2}}^{2}{{h}_{2}}\].

The volume of sand in the cylindrical bucket will be equal to the volume of sand present in the conical heap.

$\Rightarrow {{V}_{1}}={{V}_{2}}$

$\Rightarrow \pi {{r}_{1}}^{2}{{h}_{1}}=\dfrac{1}{3}\pi {{r}_{2}}^{2}{{h}_{2}}$

Substitute ${{h}_{1}}$ as 32, ${{r}_{1}}$ as 18, and ${{h}_{2}}$ as 24.

$\Rightarrow \pi \times {{\left( 18 \right)}^{2}}\times \left( 32 \right)=\dfrac{1}{3}\times \pi \times {{\left( {{r}_{2}} \right)}^{2}}\times \left( 24 \right)$

$\Rightarrow 324\times \left( 32 \right)={{\left( {{r}_{2}} \right)}^{2}}\times 8$

Divide both sides by 8.

$\Rightarrow \dfrac{324\times \left( 32 \right)}{8}={{r}_{2}}^{2}$

$\Rightarrow 324\times 4={{r}_{2}}^{2}$

$\Rightarrow 1296={{r}_{2}}^{2}$

Take square root on both sides.

$\Rightarrow \sqrt{1296}=\sqrt{{{r}_{2}}^{2}}$

$\Rightarrow \pm 36={{r}_{2}}$

Since radius cannot be negative then, the positive value will be considered.

\[\Rightarrow 36={{r}_{2}}\]

Now, find the slant height of the heap. Let $l$ be the slant height of the cone. The formula for slant height is $l=\sqrt{{{r}_{2}}^{2}+{{h}_{2}}^{2}}$.

Substitute ${{r}_{2}}$ as 36, and ${{h}_{2}}$ as 24.

$l=\sqrt{{{\left( 36 \right)}^{2}}+{{\left( 24 \right)}^{2}}}$

$l=\sqrt{1296+576}$

$l=\sqrt{1872}$

$l=12\sqrt{13}$

Hence, the slant height of the heap is $12\sqrt{13}$ cm and the height of the conical heap is 36 cm.

8. Water in the canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. how much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Ans: Let the area of the cross section of canal as a rectangle. The figure shows the cross section as $ABCD$. 

The area of cross-section will be the product of length and breadth.

$\Rightarrow 6\times 1.5=9\,{{\text{m}}^{\text{2}}}$

The speed of water is 10 km/hr. Convert it into meters/min.

$\Rightarrow 10\text{ km/hr}=\dfrac{10000}{60}\text{ meter/min}$

The water that flows in 1 minute from the canal is the product of area of cross section of canal and speed of water. Hence,

$ \Rightarrow 9\times \dfrac{10000}{60}=9\times \dfrac{1000}{6}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=1500\,\,{{\text{m}}^{3}}$

Multiply the above quantity with 30 to get the volume of water that flows in 30 minutes from the canal.

$\Rightarrow 30\times 1500=45000\,{{\text{m}}^{3}}$

Let the area which is to be irrigated be $A$. Now, the volume of water irrigating the required area will be equal to the volume of water that flows in 30 minutes from the canal.

$\Rightarrow 45000=\dfrac{\text{A}\times 8}{100}$

$\Rightarrow 45000\times 100=8\text{A}$

$\Rightarrow 4500000=8\text{A}$

Divide both sides by 8.

$\Rightarrow \dfrac{4500000}{8}=\text{A}$

$\Rightarrow 562500=\text{A}$

Therefore, the area will be irrigated in 30 minutes in 562500 square meter.

9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep.  If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Ans: The area of the cross section of the pipe will be cylindrical with a diameter 20 cm.

Assume the height of the pipe as ${{h}_{1}}$ and radius as ${{r}_{1}}$.

The diameter is twice the value of radius.

$\Rightarrow {{r}_{1}}=\dfrac{{{d}_{1}}}{2}$

Substitute ${{d}_{1}}$ as 20.

$\Rightarrow {{r}_{1}}=\dfrac{20}{2}$

$\Rightarrow {{r}_{1}}=10$

The radius of the pipe will be 10 cm. Convert the radius in meters. To convert any quantity from centimeters to meters we divide by 100.

$\Rightarrow {{r}_{1}}=\dfrac{10}{100}\,\text{m}$

$\Rightarrow {{r}_{1}}=\dfrac{1}{10}\,\text{m}$

The area of the cross section of the pipe is circular. So, we will use the area of the circle to find the area of the cross section.

$\Rightarrow A=\pi {{r}_{1}}^{2}$

$\Rightarrow A=\pi \times {{\left( 0.1 \right)}^{2}}$

$\Rightarrow A=0.01\pi \,{{\text{m}}^{2}}$

The speed of water is 3 km/hr. On converting it into metre/ minute we get,

$\Rightarrow 3\text{ km/hr}=\dfrac{3000}{60}$

$\Rightarrow 3\text{ km/hr}=50\,\text{meter/min}$

The volume of water that flows in 1 minute from the pipe is given by,

$\Rightarrow 50\times 0.01\pi =0.50\pi$

$=0.5\pi \,\,{{\text{m}}^{2}}$

In $t$ minutes from the pipe is $t\times 0.5\pi \,{{\text{m}}^{\text{3}}}$.

Now, for the tank, its height is 2 m and diameter is 10 m. Assume the height of the pipe as ${{h}_{2}}$ and radius as ${{r}_{2}}$.

The diameter is twice the value of radius.

$\Rightarrow {{r}_{2}}=\dfrac{{{d}_{2}}}{2}$

Substitute ${{d}_{2}}$ as 10.

$\Rightarrow {{r}_{2}}=\dfrac{10}{2}$

$\Rightarrow {{r}_{2}}=5\,\text{m}$

Let the time taken by the pipe to completely fill the tank be $t$ minutes. The volume of water filled in the tank in $t$ minutes is equal to the volume of water flowing in $t$ minutes is equal to the volume flowing in $t$ minutes from the pipe.

Volume of water that flows in $t$ minutes from the pipe is equal to the volume of water in the tank.

$\Rightarrow t\times 0.5\pi =\pi \times {{\left( {{r}_{2}} \right)}^{2}}\times {{h}_{2}}$

$\Rightarrow t\times 0.5={{\left( {{r}_{2}} \right)}^{2}}\times {{h}_{2}}$

Substitute ${{r}_{2}}$ as 5.

$\Rightarrow t\times 0.5={{\left( 5 \right)}^{2}}\times 2$

$\Rightarrow t\times 0.5=25\times 2$

$\Rightarrow t\times 0.5=50$

Divide both sides by $0.5$.

$\Rightarrow t=\dfrac{50}{0.5}$

$\Rightarrow t=100$

Therefore, the cylindrical tank will be filled in 100 minutes.

Topics Covered in Class 10 Chapter 13 Exercise 13.3

The questions in this exercise are based on the topic “Conversion of solid from one shape to another”.

How to Convert Solid From One Shape to Another in Math?

While converting one solid shape to another, its volume remains the same, regardless of the new shape that is formed. We will equate the volumes of both the shapes and find the solution. For example, if you melt one big cylindrical candle to 10  small cylindrical candles, the sum of the volumes of the smaller candles is equivalent to the volume of the big cylindrical candle that you have melted.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3

Opting for the NCERT solutions for Ex 13.3 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.3 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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NCERT Solutions for Class 10 Maths Chapter 13 Exercises