Category: CBSE Class 10

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Question 1.
Write the first five terms of each of the following sequences whose nth terms are:

Solution:

Question 2.
Find the indicated terms in each of the following sequences whose nth terms are:

Solution:

Question 3.
Find the next five terms of each of the following sequences given by :

Solution:

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3

Other Exercises

• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.1
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS

Question 1.
A bucket has top and bottom diameters of 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 12 cm. Also, find the cost of tin sheet used for making the bucket at the rate of ?1.20 per dm2. (Use % = 3.14)
Solution:
Upper diameter = 40 cm
and lower diameter = 20 cm
∴  Upper radius (r₁) = \((\frac { 40 }{ 2 } )\) = 20 cm
and lower radius (r₂) = \((\frac { 20 }{ 2 } )\) = 10 cm
Depth or height (h) = 12 cm
Volume of the bucket

Question 2.
A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm, and height 3 cm. Find the area of its whole surface and volume.
Solution:
Base diameter of frustum = 20 cm 20
∴  Radius (r₁) = \((\frac { 20 }{ 2 } )\) = 10 cm
and diameter of top = 12 cm

Question 3.
The slant height of the frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface of the frustum.
Solution:
Perimeter of the top of frustum = 18 cm

Question 4.
The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.
Solution:
Perimeter of the top of frustum = 44 cm

Question 5.
If the radii of the circular ends of a conical bucket which is 45 cm high be 28 cm and 7 cm, find the capacity of the bucket. (Use π = 22/7). (C.B.S.E. 2000)
Solution:

Question 6.
The height of a cone is 20 cm. A small cone is cut off from the top by a plane parallel to the base. If its volume be \((\frac { 1 }{ 125 } )\)  of the volume of the original cone, determine at what height above the base the section is made.
Solution:
Total height of the cone (h₁) = 20 cm
Let a cone whose height is h₂ is cut off Then height of the remaining portion (frustum)

Question 7.
If the radii of the circular ends of a bucket 24 cm high are 5 cm and 15 cm respectively, find the surface area of the bucket.
Solution:
Height of the bucket (frustum) (A) = 24 cm
Radius of the top (r₁) = 15 cm 1
and radius of the bottom (r₂) = 5 cm

Question 8.
The radii of the circular bases of a frustum of a right circular cone are 12 cm and 3 cm and the height is 12 cm. Find the total surface area and the volume of the frustum.
Solution:
Height of the frustum (A) = 12 cm
Radius of the top (r₁) = 12 cm
and radius of the bottom (r₂) = 3 cm

Question 9.
A tent consists of a frustum of a cone capped by a cone. If the radii of the ends of the frustum be 13 m and 7 m, the height of the frutum be 8 m and the slant height of the conical cap be 12 m, find the canvas required for the tent. (Take : π = 22/7)
Solution:
Radius of the bottom of the tent (r₁) = 13 m
and radius of the top (r₂) = 7 m
Height of frustum portion (h₁) = 8 m
Slant height of the conical cap (l₂) = 12 m
Let l₁ be the slant height of the frustum portion, then

Question 10.
A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of ₹44 per litre which the container can hold. [NCERT Exemplar]
Solution:
Given that, height of milk container (h) = 16 cm
Radius of lower end of milk container (r) = 8 cm
and radius of upper end of milk container (R) = 20 cm

∴ Volume of the milk container made of metal sheet in the form of a frustum of a cone

Question 11.
A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity and surface area of the bucket. Also, find the cost of milk which can completely fill the container, at the rate of ₹25 per litre. (Use π = 3.14) [NCERT Exemplar]
Solution:

Question 12.
A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm³ of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of the metal sheet used in its making. (Use π = 3.14). [CBSE 2006C]
Solution:
Volume of frustum (bucket) = 12308.8 cm³
Upper radius (r₁) = 20 cm
and lower radius (r₂) = 12 cm

Question 13.
A bucket made of aluminium sheet is of height 20 cm and its upper and lower ends are of radius 25 cm and 10 cm respectively. Find the cost of making the bucket if the aluminium sheet costs Rs. 70 per 100 cm². (Use π = 3.14) (C.B.S.E. 2006C)
Solution:
Height of bucket (frustum) (h) = 20 cm
Upper radius (r₁) = 25 cm
and lower radius (r₂) = 10

Question 14.
The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Find its total surface area. (C.B.S.E. 2005)
Solution:
Upper radius of frustum (r₁) = 3.3 cm
and lower radius (r₂) = 27 cm
Slant height (l) = 10 cm

Question 15.
A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the volume of the bucket. Also, find the cost of the bucket if the cost of metal sheet used is Rs. 20 per 100 cm². (Use π = 3.14) (CBSE 2008)
Solution:
Lower radius of bucket (r) = \((\frac { 16 }{ 2 } )\) = 8 cm
and upper radius (R) = \((\frac { 40 }{ 2 } )\) = 20 cm
Height (h) = 16

Question 16.
A solid is in the shape of a frustum of a cone. The diameter of the two circular ends are 60 cm and 36 cm and the leight is 9 ³cm. Find the area of its whole surface and the volume. [CBSE 2010]
Solution:
In a solid frustum upper diameter = 60 cm
∴ Radius (r₁) = \((\frac { 60 }{ 2 } )\) = 30 cm
Lower diameter = 36 cm 36
∴ Radius (r₂) = \((\frac { 36 }{ 2 } )\) = 18 cm
Height (h) = 9 cm

Question 17.
A milk container is made of metal sheet in the shape of frustum of a cone whose volucrn is 10459\((\frac { 3 }{ 7 } )\) cm³. The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs. 1.40 per cm². (Use π = 22.7) [CBSE 2010]
Solution:
Volume of frustum = 10459\((\frac { 3 }{ 7 } )\) cm3 73216
= \((\frac { 73216 }{ 7 } )\) cm³
Lower radius (r₂) = 8 cm
and upper radius (r₁) = 20 cm

Question 18.
A solid cone of base radius 10 cm is cut into two parts through the mid-pint of its height, by a plane parallel to its base. Find the ratio in the volumes of two parts of the cone. [CBSE 2013]
Solution:
Radius of solid cone (r) = 10 cm
Let total height = h
In ΔAOB,
C is mid point of AO and CD || OB

Question 19.
A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24 cm and the diameters of its upper and lower circular ends are 30 cm and 10 cm respectively. Find the cost of metal sheet used in it at the rate of ₹10 per 100 cm². (Use π = 3.14). [CBSE 2013]
Solution:

Question 20.
In the given figure, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use π = 22/7 and \(\sqrt { 5 } \) = 2.236).  [ CBSE 2015]

Solution:
Total height of cone = 12 cm
Radius of its base = 6 cm
A cone of height 4 cm is cut out
Height of the so formed frustum = 12 – 4 = 8 cm
Let r be the radius of the cone cut out

Question 21.
The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.
Solution:
Let the height and radius of the given cone be H and R respectively.
The cone is divided into two parts by drawing a plane through the mid point of its axis and parallel to the base.
Upper part is a smaller cone and the bottom part is the frustum of the cone.

Question 22.
A bucket, made of metal sheet, is in the form of a cone whose height is 35 cm and radii of circular ends are 30 cm and 12 cm. How many litres of milk it contains if it is full to the brim? If the milk is sold at ₹40 per litre, find the amount received by the person. [CBSE 2017]
Solution:
Radii of the bucket in the form of frustum of cone = 30 cm
and 12 cm Depth of the bucket = 35 cm

Question 23.
A reservoir in the form of the frustum of a right circular cone contains 44 x 10⁷ litres of water which fills it completely. The radii of the bottom and top of the reservoir are 50 metres and 100 metres respectively. Find the depth of water and the lateral surface area of the reservoir. (Take : π = 22/7)
Solution:
A reservoir is a frustum in shape and its upper radius (r₁) = 100 m
Lower radius (r₂) = 50 m
and capacity of water in it = 44 x 10⁷ litres

P.Q. A metallic right circular cone 20 cm high and whose vertical angle is 90° is cut into two parts at the middle point of its axis by a plane parallel to the base. If the frustum so obtained be drawn into a wire  \((\frac { 1 }{ 16 } )\)cm, find the length of the wire.

Solution: In the cone ABC, ∠A = 90°, AL ⊥ BC and = 20 cm
It is cut into two parts at the middle point M on the axis AL
AL bisects ∠A also

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.1
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS

Question 1.
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of canvas required for the tent.
Solution:
Diameter of the base of the tent = 24 m
∴ Radius (r)= \((\frac { 24 }{ 2 } )\)  = 12m
Total height of the tent = 16 m
Height of the cylindrical portion (h₁) = 11 m
Height of the conical portion (h₂) =16-11 = 5 m

Question 2.
A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of radius 2.5 m and height 21 m and the cone has the slant height 8 m. Calculate the total surface area and the volume of the rocket.
Solution:
Radius of the base of the rocket (r) = 2.5 m
Height of cylindrical portion (h₁) = 21 m
Slant height of the conical portion (l) = 8 m
Let height of conical portion = h₂

Question 3.
A tent of height 77 dm is in the form of a right circular cylinder of diameter 36 m and height 44
dm surmounted by a right circular cone. Find the cost of the canvas at Rs. 3.50 per m² . (Use π = 22/7).
Solution:
Total height of the tent = 77 dm
Height of cylindrical part (h₁) = 44 dm
= 4.4 m
Height of conical part (h₂) = 7.7 – 4.4 = 3.3 m
Diameter of the base of the tent = 36 m

Question 4.
A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm, respectively. Determine the surface area of the toy. (Use π = 3.14).
Solution:
Diameter of the base of the toy = 6 cm
∴ Radius (r) = \((\frac { 6 }{ 2 } )\)  = 3 cm
Height (h) = 4 cm

Total surface area of the toy = curved surface area of the conical part + surface area of the hemispherical part
= πrl + 2πr² = πr (l + 2r)
= 3.14 x 3 (5 + 6) = 3.14 x 3 x 11 cm²
= 3.14 x 33 = 103.62 cm²

Question 5.
A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end. The radius of the common base is 3.5 cm and the ~ heights of the cylindrical and conical portions are 10 cm and 6 cm, respectively. Find the total surface area of the solid. (Use π = 22/7)
Solution:
Radius of the common base (r) = 3.5 m
Height of cylindrical part (h₁) = 10 cm
Height of conical part (h₂) = 6 cm

Now total surface area of the solid = curved surface of conical part + curved surface of cylindrical part + curved surface of hemispherical part

Question 6.
A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 30 cm. (C.B.S.E. 2002)
Solution:
Radius of the base of the cylindrical part (r) = 5 cm

Height of cylindrical part (h₁) = 13 cm
Height of the conical part (h₂) = 30 – (13 + 5) = 30- 18 = 12 cm

Question 7.
A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemi-sphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of the water left in the tub. (Take π = 22/7) (C.B.S.E. 2000C)
Solution:
Radius of the cylindrical tub (R) = 5 cm
and height (h₁) = 9.8 cm
Radius of the solid (r) = 3.5 cm
and height of cone (h₁) = 5 cm

= 38.5 x 4 = 154 cm3
∴ Water flowed out of the tub = 154 cm³
Remaining water in the tub = 770 – 154
= 616 cm³

Question 8.
A circus tent has cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20 m. The heights of the cylindrical and conical portions are 4.2 m and 2.1 m respectively. Find the volume of the tent.
Solution:
Radius of the tent (r) = 20 m
Height of the conical part (h₁) = 2.1 m
and height of the cylindrical part (h₂) = 4.2 m

Question 9.
A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with conical ends each of axis length 9 cm. Determine the capacity of the tank.
Solution:
Diameter of cylindrical part = 21 cm 21
∴ Radius (r) = \((\frac { 21 }{ 2 } )\) cm
Height of cylindrical part (h₁) = 18 cm
and height of each conical part (h₂) = 9 cm

Question 10.
A conical hole is drilled in a circular cylinder of height 12 cm and base radius 5 cm. The height and the base radius of the cone are also the same. Find the whole surface and volume of the remaining cylinder.
Solution:
Base radius of the cylinder (r) = 5 cm
and height (h) = 12 cm
∴ Volume = πr²h = n (5)2 x 12 cm3 = 300π cm³
∵ The base and height of the cone drilled are the same as those of the cylinder
∴ Volume of cone = \((\frac { 1 }{ 3 } )\)πr²h

Question 11.
A tent is in the form of a cylinder of diameter 20 m and height 2.5 m, surmounted by a cone of equal base and height 7.5 m. Find the capacity of the tent and the cost of the canvas at Rs. 100 per square metre.
Solution:
Diameter of the base of the tent = 20 m
∴ Radius (r) =\((\frac { 20 }{ 2 } )\) = 10 m
Height of cylindrical part (h₁) = 2.5 m
and height of conical part (h₂) = 7.5 m
Slant height of the conical part (l)

Question 12.
A boiler is in the form of a cylinder 2 m long with hemispherical ends each of 2 metre diameter. Find the volume of the boiler.
Solution:
Diameter of the cylinder = 2 m
∴ Radius (r) = \((\frac { 1 }{ 3 } )\) = 1 m
Height (length) of cylindrical part (h) = 2 m

Question 13.
A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylinder is \((\frac { 14 }{ 3 } )\) m and the diameter of hemisphere is 3.5 m. Calculate the volume and the internal surface area of the solid.
Solution:
Diameter of the cylindrical part = 3.5 m
∴Radius (r) = \((\frac { 3.5 }{ 2 } )\) = 1.75 = \((\frac { 7 }{ 4 } )\) m
and height (h) = \((\frac { 14 }{ 3 } )\) m

Question 14.
A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each of the hemispherical ends is 7 cm, find the cost of polishing its surface at the rate of Rs. 10 per dm² . (C.B.S.E. 2006C)
Solution:
Total height = 104 cm
Radius of hemispherical part (r) = 7 cm
Height of cylinder (h) = 104 cm – 2 x 7 cm = 104- 14 = 90 cm

Total outer surface area = curved surface area of the cylindrical part + 2 x curved surface area of each hemispherical part

Question 15.
A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16 cm and height 42 cm. The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimeters of cork dust will be required ?
Solution:
Diameter of inner cylinder = 14 cm
∴Radius (r) = \((\frac { 14 }{ 2 } )\) = 7 cm
Diameter of outer cylinder = 16 cm
∴ Radius (R) = \((\frac { 16 }{ 2 } )\) = 8 cm
Height (h) = 42cm

Question 16.
A cylindrical road roller made of iron is 1 m long. Its internal diameter is 54 cm and the thickness of the iron sheet used in making the roller is 9 cm. Find the mass of the roller, if 1 cm³ of iron has 7.8 gm mass. (Use π = 3.14)
Solution:
Length of roller (h) = 1 m = 100 cm
Inner diameter = 54 cm
Thickness of iron sheet = 9 cm
∴ Inner radius (r) = \((\frac { 52 }{ 2 } )\) = 27 cm
and outer radus (R) = 27 + 9 = 36 cm
∴ Volume of the mass = πR²h – πr²h
= πh (R²- r²)
= 3.14 x 100 (36² – 27²) cm³
= 314 x (36 + 27) (36- 27) cm³
= 314 x 63 x 9 cm^c
= 178038 cm³
Weight of 1 cm³ of iron = 7.8 gm
∴ Total weight = 178038 x 7.8 gm
= 1388696.4 gm
= 1388.6964 kg
= 1388.7 kg

Question 17.
A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
Diameter of hemisphere = 14 cm
Total height =13 cm
Radius of hemisphere = \((\frac { 14 }{ 2 } )\) = 7 cm
∴ Height of cylindrical part =13-7 = 6 cm

∴  Inner surface area of the vessel = inner surface area of cylindrical part + inner surface area of hemispherical part

Question 18.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius of cone (r) – 3.5 cm
Total height of the toy = 15.5 cm

Height of the conical part (h) = 15.5 – 3.5 = 12 cm
∴ Slant height of the cone (l)

Question 19.
The difference between outside and inside surface areas of cylindrical metallic pipe 14 cm long is 44 m². If the pipe is made of 99 cm³ of metal, find the outer and inner radii of the pipe.
Solution:
In cylindrical metallic pipe,
length of pipe = 14 cm

Difference between outside and inside
surface area = 44 m²
Volume of pipe material = 99 cm³
Let R and r be the outer and inner radii of the pipe respectively, then Outer surface area – inner surface area = 44 cm²

Question 20.
A right circular cylinder having diameter 12 cm and height 15 cm is full ice-cream. The ice-cream is to be filled in 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.
Solution:
Height of cylinder (H) = 15 cm
and diameter =12 cm

Question 21.
A solid iron pole having cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that the mass of 1 cm3 of iron is 8 gm.
Solution:
Diamter of the base = 12 cm
∴ Radius (r) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height of the cylindrical portion (h₁)= 110 cm
and height of conical portion (h₂) = 9 cm

Question 22.
A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover.
Solution:
Height of conical part (h) = 2 cm
Diameter of base = 4 cm

Now volume of the cylinder which circum scribes the toy = πr²h
= π (2)² x 4 = 16π cm³
∴  Difference of their volumes = 16π – 8π = 8πcm³

Question 23.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottoms. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Radius of conical part = 60 cm
and height (h) = 120 cm

Question 24.
A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the value of water (i) displaced out of the cylinder. (ii) left in the cylinder. (C.B.S.E. 2009)
Solution:
Internal diameters of cylindrical vessel = 10 cm
∴ Radius (r) = \((\frac { 10 }{ 2 } )\) =5 cm
and height (h) = 10.5 cm
∴ Volume of water filled in it

Question 25.
A hemispherical depression is cut out from one face of a cubical wooden block of edge 21 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Determine the volume and total surface area of the remaining block. [CBSE 2010]
Solution:
Edge of cube = 21 cm
∴ Diameter of the hemisphere curved out of

Question 26.
A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of the base of the cone is 21 cm and its volume is 2/3 of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy. (Use π = 22/7).
Solution:
Radius of base of the conical part (r) = 21 cm

Question 27.
A solid is in the shape of a cone surmounted on a hemi-sphere, the radius of each of them is being 3.5 cm and the total height of solid is 9.5 cm. Find the volume of the solid. [CBSE 2012]
Solution:

Question 28.
An wooden toy is made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius
3.5 cm, find the volume of wood in the toy. (Use π = 22/7). [CBSE 2013]
Solution:
Height of cylindrical part (h) = 10 cm
Radius of the base (r) = 3.5 cm

Question 29.
The largest possible sphere is carved out of a wooden solid cube of side 7 cm. Find the volume of the wood left. (Use π = 22/7). [CBSE 2014]
Solution:

Question 30.
From a solid cylinder of height 2.8 cm and diameter 4.2 cm, conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid, (take π = 22/7). [CBSE 2014]
Solution:
Diameter of solid cylinder= 4.2 cm
∴ Radius (r) =\((\frac { 4.2 }{ 2 } )\) = 2.1 cm
Height (h) = 2.8 cm

Question 31.
The largest cone is curved out from one face of solid cube of side 21 cm. Find the volume of the remaining solid. [CBSE 2015]
Solution:
Side of a solid cube (a) = 21 cm
∴ Volume = a³, = (21 )³ cm³
= 9261 cm³
Diameter of the base of cone = 21 cm
Now radius of cone curved from it (r) =\((\frac { 21 }{ 2 } )\) cm
and height (h) = 21 cm

Question 32.
A solid wooden toy is in the form of a hemisphere surmounted by a Cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is 166 \((\frac { 5 }{ 6 } )\) cm³ . Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of ₹10 per cm². (Take π  = 22/7). [CBSE 2015]
Solution:

Question 33.
In the given figure, from a cuboidal solid metalic block, of dimensions 15 cm x 10 cm x 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block. (Take  π = 22/7) [CBSE 2015]
Solution:

Radius of hole = \((\frac { 7 }{ 2 } )\)cm and height = 5 cm
Length of block (l) = 15 cm
Breadth (b) = 10 cm and height = 5 cm
∴ Surface area = 2(lb + bh + hl)
= 2(15 x 10 + 10 x 5 + 5 x 15) cm²
= 2(150 + 50 + 75) = 2 x 275 = 550 cm²
Area of circular holes of both sides = 2 x πr²

Question 34.
 A building is in the form of a cylinder surmounted by a hemi-spherical vaulted dome and
contains 41 \((\frac { 19 }{ 21 } )\) m³ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building? [NCERT Exemplar]
Solution:
Let total height of the building = Internal diameter of the dome = 2rm

Radius of building (or dome) = \((\frac { 2r }{ 2 } )\) = r m
Height of cylinder = 2r-r = rm

Question 35.
A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10 cm x 5 cm x 4 cm. The radius of each of the conical depression is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand. [NCERT Exemplar]
Solution:
Given that, length of cuboid pen stand (l) = 10 cm
Breadth of cuboid pen stand (b) = 5 cm
and height of cuboid pen stand (h) = 4 cm

∴ Volume of cuboid pend stand = l x b x h= 10 x 5 x 4 = 200 cm³
Also, radius of conical depression (r) = 0.5 cm
and height (depth) of a conical depression (h₁) = 2.1 cm
∴ Volume of a conical depression = πrh₁

Question 36.
 A building is in the form of a cylinder surmounted by a hemispherical dome. The base diameter of the dome is equal to \((\frac { 2 }{ 3 } )\) of the total height of the building. Find the height of the building, if it contains 67 \((\frac { 1 }{ 21 } )\) m³ of air.
Solution:
Let the radius of the hemispherical dome be r metres and the total height of the building be h metres.
Since the base diameter of the dome is equal to \((\frac { 2 }{ 3 } )\) of the total height, therefore
2 r = \((\frac { 2 }{ 3 } )\)h. This implies r = \((\frac { h }{ 3 } )\). Let H metres be the height of the cylindrical portion.
Therefore, H = h – \((\frac { h }{ 3 } )\) = \((\frac { 2 }{ 3 } )\)h metres.
Volume of the air inside the building = Volume of air inside the dome + Volume of the air inside the
cylinder = \((\frac { 2 }{ 3 } )\) πr³ + πr²H, where H is the height of the cylindrical portion

Question 37.
A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of cube and the toy. Also, find the total surface area of the toy. [NCERT Exemplar]
Solution:
Let r be the radius of the hemisphere and the cone and h be the height of the cone.
Volume of the toy=Volume of the hemisphere + Volume of the cone

= \((\frac { 1408 }{ 7 } )\) cm³
A cube circumsrcibes the given solid. Therefore, edge of the cube should be 8 cm. Volume of the cube = 83 cm3 = 512 cm3 Difference in the volume of the cube and

Question 38.
A circus tent is in the shape of a cylinder surmounted by a conical top of same diameter. If their common diameter is 56 m, the height of the cylindrical part is 6 m and the total height of the tent above the ground is 27 m, find the area of the canvas used in making the tent. [ICBSE 2017]
Solution:
We have, diameter of base of cylinder = d = 56 m
Radius of base of cylinder = r=  \((\frac { d }{ 2 } )\)=  \((\frac { 52 }{ 2 } )\)= 28 m
Height of tent = 27 m
Height of cylinder = 6 m

Height of conical portion = 27 – 6 = 21 m
Radius of conical portion, r = 28 m

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Mark the correct alternative in each of the following :
Question 1.
If the equation x² + 4x + k = 0 has real and distinct roots, then
(a) k < 4
(b) k > 4
(c) k ≥ 4
(d) k ≤ 4
Solution:
(a) In the equation x² + 4x + k = 0
a = 1, b = 4, c = k
D = b² – 4ac = (4)² – 4 x 1 x k = 16 – 4k
Roots are real and distinct
D > 0
=> 16 – 4k > 0
=> 16 > 4k
=> 4 > k
=> k < 4

Question 2.
If the equation x² – ax + 1 = 0 has two distinct roots, then
(a) |a| = 2
(b) |a| < 2
(c) |a| > 2
(d) None of these
Solution:
(c) In the equation x² – ax + 1 = 0
a = 1, b = – a, c = 1
D = b² – 4ac = (-a)² – 4 x 1 x 1 = a² – 4
Roots are distinct
D > 0
=> a² – 4 > 0
=> a² > 4
=> a² > (2)²
=> |a| > 2

Question 3.
If the equation 9x² + 6kx + 4 = 0, has equal roots, then the roots are both equal to
(a) ± \(\frac { 2 }{ 3 }\)
(b) ± \(\frac { 3 }{ 2 }\)
(c) 0
(d) ± 3
Solution:
(a)

Question 4.
If ax² + bx + c = 0 has equal roots, then c =

Solution:
(d) In the equation ax² + bx + c = 0
D = b² – 4ac
Roots are equal
D = 0 => b² – 4ac = 0
=> 4ac = b²
=> c = \(\frac { { b }^{ 2 } }{ 4a }\)

Question 5.
If the equation ax² + 2x + a = 0 has two distinct roots, if
(a) a = ±1
(b) a = 0
(c) a = 0, 1
(d) a = -1, 0
Solution:
(a) In the equation ax² + 2x + a = 0
D = b² – 4ac = (2)² – 4 x a x a = 4 – 4a²
Roots are real and equal
D = 0
=> 4 – 4a² = 0
=> 4 = 4a²
=> 1 = a²
=> a² = 1
=> a² = (±1)²
=> a = ±1

Question 6.
The positive value of k for which the equation x² + kx + 64 = 0 and x² – 8x + k = 0 will both have real roots, is
(a) 4
(b) 8
(c) 12
(d) 16
Solution:
(d) In the equation x² + kx + 64 = 0

Question 7.

Solution:
(b)

Which is not possible
x = 3 is correct

Question 8.
If 2 is a root of the equation x² + bx + 12 = 0 and the equation x² + bx + q = 0 has equal roots, then q =
(a) 8
(b) – 8
(c) 16
(d) -16
Solution:
(c)

Question 9.
If the equation (a² + b²) x² – 2 (ac + bd) x + c² + d² = 0 has equal roots, then
(a) ab = cd
(b) ad = bc
(c) ad = √bc
(d) ab = √cd
Solution:
(b)

Question 10.
If the roots of the equation (a² + b²) x² – 2b (a + c) x + (b² + c²) = 0 are equal, then ;
(a) 2b = a + c
(b) b² = ac
(c) b = \(\frac { 2ac }{ a + c }\)
(d) b = ac
Solution:
(b)

Question 11.
If the equation x² – bx + 1 = 0 does not possess real roots, then
(a) -3 < b < 3
(b) -2 < b < 2
(c) b > 2
(d) b < -2
Solution:
(b)

Question 12.
If x = 1 is a common root of the equations ax² + ax + 3 = 0 and x² + x + b = 0, then ab =
(a) 3
(b) 3.5
(c) 6
(d) -3
Solution:
(a) In the equation
ax² + ax + 3 = 0 and x² + x + b = 0
Substituting the value of x = 1, then in ax² + ax + 3 = 0

Question 13.
If p and q are the roots of the equation x² – px + q + 0, then
(a) p = 1, q = -2
(b) p = 0, q = 1
(c) p = -2, q = 0
(d) p = -2, q = 1
Solution:
(a)

Question 14.
If a and b can take values 1, 2, 3, 4. Then the number of the equations of the form ax² + bx + 1 = 0 having real roots is
(a) 10
(b) 7
(c) 6
(d) 12
Solution:
(b)
ax² + bx + 1 = 0
D = b² – 4a = b² – 4a
Roots are real
D ≥ 0
=> b² – 4a ≥ 0
=> b² ≥ 4a
Here value of b can be 2, 3 or 4
If b = 2, then a can be 1,
If b = 3, then a can be 1, 2
If b = 4, then a can be 1, 2, 3, 4
No. of equation can be 7

Question 15.
The number of quadratic equations having real roots and which do not change by squaring their roots is
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(c) There can be two such quad, equations whose roots can be 1 and 0
The square of 1 and 0 remains same
No. of quad equation are 2

Question 16.
If (a² + b²) x² + 2(ab + bd) x + c² + d² = 0 has no real roots, then
(a) ad = bc
(b) ab = cd
(c) ac = bd
(d) ad ≠ bc
Solution:
(d)

Question 17.
If the sum of the roots of the equation x² – x = λ (2x – 1) is zero, then λ =
(a) -2
(b) 2
(c) – \(\frac { 1 }{ 2 }\)
(d) \(\frac { 1 }{ 2 }\)
Solution:
(c)

Question 18.
If x = 1 is a common root of ax² + ax + 2 = 0 and x² + x + b = 0 then, ab =
(a) 1
(b) 2
(c) 4
(d) 3
Solution:
(b)

Question 19.
The value of c for which the equation ax² + 2bx + c = 0 has equal roots is

Solution:
(a)

Question 20.
If x² + k (4x + k – 1) + 2 = 0 has equal roots, then k =

Solution:
(b)

Question 21.
If the sum and product of the roots of the equation kx² + 6x + 4k = 0 are equal, then k =

Solution:
(b)

Question 22.
If sin α and cos α are the roots of the equations ax² + bx + c = 0, then b² =
(a) a² – 2ac
(b) a² + 2ac
(b) a² – ac
(d) a² + ac
Solution:
(b)

Question 23.
If 2 is a root of the equation x² + ax + 12 = 0 and the quadratic equation x² + ax + q = 0 has equal roots, then q =
(a) 12
(b) 8
(c) 20
(d) 16
Solution:
(d)

Question 24.
If the sum of the roots of the equation x² – (k + 6) x + 2 (2k – 1) = 0 is equal to half of their product, then k =
(a) 6
(b) 7
(c) 1
(d) 5
Solution:
(b) In the quadratic equation
x² – (k + 6) x + 2 (2k – 1) = 0
Here a = 1, b = – (k + 6), c = 2 (2k – 1)

Question 25.
If a and b are roots of the equation x² + ax + b = 0, then a + b =
(a) 1
(b) 2
(c) -2
(d) -1
Solution:
(d) a and b are the roots of the equation x² + ax + b = 0
Sum of roots = – a and product of roots = b
Now a + b = – a
and ab = b => a = 1 ….(i)
2a + b = 0
=> 2 x 1 + b = 0
=> b = -2
Now a + b = 1 – 2 = -1

Question 26.
A quadratic equation whose one root is 2 and the sum of whose roots is zero, is
(a) x² + 4 = 0
(b) x² – 4 = 0
(c) 4x² – 1 = 0
(d) x² – 2 = 0
Solution:
(b) Sum of roots of a quad, equation = 0
One root = 2
Second root = 0 – 2 = – 2
and product of roots = 2 x (-2) = – 4
Equation will be
x² + (sum of roots) x + product of roots = 0
x² + 0x + (-4) = 0
=> x² – 4 = 0

Question 27.
If one root of the equation ax² + bx + c = 0 is three times the other, then b² : ac =
(a) 3 : 1
(b) 3 : 16
(c) 16 : 3
(d) 16 : 1
Solution:
(c)

Question 28.
If one root of the equation 2x² + kx + 4 = 0 is 2, then the other root is
(a) 6
(b) -6
(c) -1
(d) 1
Solution:
(d) The given quadratic equation 2x² + kx + 4 = 0
One root is 2
Product of roots = \(\frac { c }{ a }\) = \(\frac { 4 }{ 2 }\) = 2
Second root = \(\frac { 2 }{ 2 }\) = 1

Question 29.
If one root of the equation x² + ax + 3 = 0 is 1, then its other root is
(a) 3
(b) -3
(c) 2
(d) -2
Solution:
(a) The quad, equation is x² + ax + 3 = 0
One root =1
and product of roots = \(\frac { c }{ a }\) = \(\frac { 3 }{ 1 }\) = 3
Second root = \(\frac { 3 }{ 1 }\) = 3

Question 30.
If one root of the equation 4x² – 2x + (λ – 4) = 0 be the reciprocal of the other, then λ =
(a) 8
(b) -8
(c) 4
(d) -4
Solution:
(a)

Question 31.
If y = 1 is a common root of the equations ay² + ay + 3 = 0 and y² + y + b = 0, then ab equals
(a) 3
(b) – \(\frac { 1 }{ 2 }\)
(c) 6
(d) -3 [CBSE 2012]
Solution:
(a)

Question 32.
The values of k for which the quadratic equation 16x² + 4kx + 9 = 0 has real and equal roots are
(a) 6, – \(\frac { 1 }{ 6 }\)
(b) 36, -36
(c) 6, -6
(d) \(\frac { 3 }{ 4 }\) , – \(\frac { 3 }{ 4 }\) [CBSE 2014]
Solution:
(c) 16x² + 4kx + 9 = 0
Here a = 16, b = 4k, c = 9
Now D = b² – 4ac = (4k)² – 4 x 16 x 9 = 16k² – 576
Roots are real and equal
D = 0 or b² – 4ac = 0
=> 16k² – 576 = 0
=> k² – 36 = 0
=> k² = 36 = (± 6)²
k = ± 6
k = 6, -6

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Answer each of the following questions either in one word or one sentence or as per requirement of the question :
Question 1.
Write the value of k for which the quadratic equation x² – kx + 4 = 0 has equal roots.
Solution:
x² – kx + 4 = 0
Here a = 1, b = – k, c = 4
Discriminant (D) = b² – 4ac
= (-k)² – 4 x 1 x 4 = k² – 16
The roots are equal
D = 0 => k² – 16 = 0
=> (k + 4) (k – 4) = 0.
Either k + 4 = 0, then k = – 4
or k – 4 = 0, then k = 4
k = 4, -4

Question 2.
What is the nature of roots of the quadratic equation 4x² – 12x – 9 = 0 ?
Solution:
4x² – 12x – 9 = 0
Here a = 4, b = -12, c = – 9
Discriminant (D) = b² – 4ac = (-12)² – 4 x 4 x (-9)
= 144 + 144 = 288
D > 0
Roots are real and distinct

Question 3.
If 1 + √2 is a root of a quadratic equation with rational co-efficients, write its other root.
Solution:
The roots of the quadratic equation with rational co-efficients are conjugate
The other root will be 1 – √2

Question 4.
Write the number of real roots of the equation x² + 3 |x| + 2 = 0.
Solution:

Question 5.
Write the sum of the real roots of the equation x² + |x| – 6 = 0.
Solution:

Question 6.
Write the set of values of ‘a’ for which the equation x² + ax – 1 = 0, has real roots.
Solution:
x² + ax – 1=0
Here a = 1, b = a, c = -1
D = b² – 4ac = (a)² – 4 x 1 x (-1) = a² + 4
Roots are real
D ≥ 0 => a² + 4 ≥ 0
For all real values of a, the equation has real roots.

Question 7.
In there any real value of ‘a’ for which the equation x² + 2x + (a² + 1) = 0 has real roots ?
Solution:
x² + 2x + (a² + 1) = 0
D = (-b)² – 4ac = (2)² – 4 x 1 (a² + 1) = 4 – 4a² – 4 = – 4a²
For real value of x, D ≥ 0
But – 4a² ≤ 0
So it is not possible
There is no real value of a

Question 8.
Write the value of λ, for which x² + 4x + λ is a perfect square.
Solution:
In x² + 4x + λ
a = 1, b = 4, c = λ
x² + 4x + λ will be a perfect square if x² + 4x + λ = 0 has equal roots
D = b² – 4ac = (4)² – 4 x 1 x λ = 16 – 4λ
D = 0
=> 16 – 4λ = 0
=> 16 = 4A
=> λ = 4
Hence λ = 4

Question 9.
Write the condition to be satisfied for which equations ax² + 2bx + c = 0 and bx² – 2√ac x + b = 0 have equal roots.
Solution:
In ax² + 2bx + c = 0

Question 10.
Write the set of values of k for which the quadratic equation has 2x² + kx – 8 = 0 has real roots.
Solution:
In 2x² + kx – 8 = 0
D = b²- 4ac = (k)² – 4 x 2 x (-8) = k² + 64
The roots are real
D ≥ 0
k² + 64 ≥ 0
For all real values of k, the equation has real roots.

Question 11.
Write a quadratic polynomial, sum of whose zeros is 2√3 and their product is 2.
Solution:
Sum of zeros = 2√3
and product of zeros = 2
The required polynomial will be

Question 12.
Show that x = – 3 is a solution of x² + 6x + 9 = 0 (C.B.S.E. 2008)
Solution:
The given equation is x² + 6x + 9 = 0
If x = -3 is its solution then it will satisfy it
L.H.S. = (-3)² + 6 (-3) + 9 = 9 – 18 + 9 = 18 – 18 = 0 = R.H.S.
Hence x = – 3 is its one root (solution)

Question 13.
Show that x = – 2 is a solution of 3x² + 13x + 14 = 0. (C.B.S.E. 2008)
Solution:
The given equation is 3x² + 13x + 14 = 0
If x = – 2 is its solution, then it will satisfy it
L.H.S. = 3(-2)² + 13 (- 2) + 14 =3 x 4 – 26 + 14
= 12 – 26 + 14 = 26 – 26 = 0 = R.H.S.
Hence x = – 2 is its solution

Question 14.
Find the discriminant of the quadratic equation 3√3 x² + 10x + √3 =0. (C.B.S.E. 2009)
Solution:

Question 15.
If x = \(\frac { -1 }{ 2 }\), is a solution of the quadratic equation 3x² + 2kx – 3 = 0, find the value of k. [CBSE 2015]
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
A piece of cloth costs Rs. 35. If the piece were 4 m longer and each metre costs Re. one less, the cost would remain unchanged. How long is the piece ?
Solution:
Let the length of piece of cloth = x m
Total cost = Rs. 35
Cost of 1 m cloth = Rs. \(\frac { 35 }{ x }\)
According to the condition,

=> x (x + 14) – 10 (x + 14) = 0
=> (x + 14) (x – 10) = 0
Either x + 14 = 0, then x = – 14 which is not possible being negative
or x – 10 = 0, then x = 10
Length of piece of cloth = 10 m

Question 2.
Some students planned a picnic. The budget for food was Rs. 480. But eight of these failed to go and thus the cost of food for each member increased by Rs. 10. How many students attended the picnic ?
Solution:
Let the number of students = x
and total budget = Rs. 480
Share of each students = Rs. \(\frac { 480 }{ x }\)
According to the condition,

=> x (x + 16) – 24 (x + 16) = 0
=> (x + 16) (x – 24) = 0
Either x + 16 = 0, then x = -16 which is not possible being negative
or x – 24 = 0, then x = 24
Number of students = 24
and number of students who attended the picnic = 24 – 8 = 16

Question 3.
A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.
Solution:
Let cost price ofjfie article = Rs. x
Selling price = Rs. 24
Gain = x %
According to the condition,

Question 4.
Out of a group of swans, \(\frac { 7 }{ 2 }\) times the square root of the total number are playing on the share of a pond. The two remaining ones are swinging in water. Find the total number of swans.
Solution:
Let the total number of swans = x
According to the condition,

Number of total swans = 16

Question 5.
If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys mope for Rs. 360. Find the original price of the toy. (C.B.S.E. 2002C)
Solution:
List price of the toy = Rs. x
Total amount = Rs. 360
Reduced price of each toy = (x – 2)
According to the condition,

=> x (x – 20) + 18 (x – 20) = 0
=> (x – 20) (x + 18) = 0
Either x – 20 = 0, then x = 20
or x + 18 = 0, then x = -18 which is not possible being negative
Price of each toy = Rs. 20

Question 6.
Rs. 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less. Find the original number of persons.
Solution:
Total amount = Rs. 9000
Let number of persons = x
Then each share = Rs. \(\frac { 9000 }{ x }\)
Increased persons = (x + 20)
According to the condition,

Either x + 45 = 0, then x = -45 which is not possible being negative
or x – 25 = 0, then x = 25
Number of persons = 25

Question 7.
Some students planned a picnic. The budget for food was Rs. 500. But 5 of them failed to go and thus the cost of food for each number increased by Rs. 5. How many students attended the picnic? (C.B.S.E. 1999)
Solution:
Let number of students = x
Total budget = Rs. 500
Share of each student = Rs. \(\frac { 500 }{ x }\)
No. of students failed to go = 5
According to the given condition,

Question 8.
A pole has to be erected at a point on the boundary of a. circular park of diameter 13 metres in such a way that the difference, of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it the possible to do so ? If yes, at what distances from the two gates Should the pole be erected ?
Solution:
In a circle, AB is the diameters and AB = 13 m

Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 5 = 0, then x = 5
P is at a distance of 5 m from B and 5 + 7 = 12 m from A

Question 9.
In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 marks more in Mathematics and 4 marks less in Science. The product of their marks, would have been 180. Find his marks in the two subjects. (C.B.S.E. 2008)
Solution:
Sum of marks in Mathematics and Science = 28
Let marks in Math = x
Then marks in Science = 28 – x
According to the condition,
(x + 3) (28 – x – 4) = 180
=> (x + 3) (24 – x) = 180
=> 24x – x² + 72 – 3x = 180
=> 21x – x² + 72 – 180 = 0
=> – x² + 21x – 108 = 0
=> x² – 21x + 108 = 0
=> x² – 9x – 12x + 108 = 0
=> x (x – 9) – 12 (x – 9) – 0
=> (x – 9)(x – 12) = 0
Either x – 9 = 0, then x = 9
or x – 12 = 0, then x = 12
(i) If x = 9, then Marks in Maths = 9 and marks in Science = 28 – 9 = 19
(ii) If x = 12, then Marks in Maths = 12 and marks in Science = 28 – 12 = 16

Question 10.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in two subjects. [NCERT]
Solution:
Sum of marks in Mathematics and English = 30
Let marks obtained in Mathematics = x
Then in English = 30 – x
According to the condition,
(x + 2) (30 – x – 3) = 210
=> (x + 2) (27 – x) = 210
=> 27x – x² + 54 – 2x – 210 = 0
=> – x² + 25x – 156 = 0
=> x² – 25x + 156 = 0
=> x² – 12x – 13x +156 = 0
=> x (x – 12) – 13 (x – 12) = 0
=> (x – 12) (x – 13) = 0
Either x – 12 = 0, then x = 12
or x – 13 = 0, then x = 13
(i) If x = 12, then
Marks in Maths =12 and in English = 30 – 12 = 18
(ii) If x = 13, then
Marks in Maths = 13 and in English = 30 – 13 = 17

Question 11.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, And the number of articles produced and the cost of each article. [NCERT]
Solution:
Total cost = Rs. 90
Let number of articles = x
Then price of each articles = 2x + 3
x (2x + 3) = 90
=> 2x² + 3x – 90 = 0
=> 2x² – 12x + 15x – 90 = 0
=> 2x (x – 6) + 15 (x – 6) = 0
=> (x – 6) (2x + 15) = 0
Either x – 6 = 0, then x = 6
or 2x + 15 = 0 then 2x = -15 => x = \(\frac { -15 }{ 2 }\) which is not possible being negative
x = 6
Number of articles = 6
and price of each article = 2x + 3 = 2 x 6 + 3 = 12 + 3 = 15

Question 12.
At t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than \(\frac { { t }^{ 2 } }{ 4 }\) minutes. Find t.
Solution:
We know that, the time between 2 pm to 3 pm = 1 h = 60 minutes
Given that, at t minutes past 2 pm, the time needed by the min. hand of a clock to show 3 pm was found to be 3 min. less than \(\frac { { t }^{ 2 } }{ 4 }\) min.
i.e., t = (\(\frac { { t }^{ 2 } }{ 4 }\) – 3) = 60
=> 4t + t² – 12 = 240
=> t² + 4t – 252 = 0
=> t² + 18t – 14t – 252 = 0 [by splitting the middle term]
=> t (t + 18) – 14 (t + 18) = 0
=> (t + 18) (t – 14) = 0 [since, time cannot be negative, so t ≠ -18]
t = 14 min.
Hence, the required value of t is 14 minutes

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

 

Question 1.
A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work. .
Solution:
Let B can do the work in = x days
A will do the same work in = (x – 10) days
A and B both can finish the work in = 12 days
According to the condition,

=> x (x – 4) – 30 (x – 4) = 0
=> (x – 4) (x – 30) = 0
Either x – 4 = 0, then x = 4
or x – 30 = 0, then x = 30
But x = 4 is not possible
B can finish the work in 30 days

Question 2.
If two pipes function simultaneously, a reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours will the second pipe take to fill the reservoir ?
Solution:
Two pipes can fill the .reservoir in = 12 hours
Let first pipe can fill the reservoir in = x hrs
Then second pipe will fill it in = (x – 10) hours
Now according to the condition,

=> x² – 10x = 24x – 120
=> x² – 10x – 24x + 120 = 0
=> x² – 34x + 120 = 0
=> x² – 30x – 4x + 120 = 0
=> x (x – 30) – 4 (x – 30) = 0
=> (x – 30) (x – 4) = 0
Either x – 30 = 0, then x = 30
or x – 4 = 0 but it is not possible as it is < 10
The second pipe will fill the reservoir in = x – 10 = 30 – 10 = 20 hours

Question 3.
Two water taps together can fill a tank in 9\(\frac { 3 }{ 8 }\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Two taps can fill the tank in = 9\(\frac { 3 }{ 8 }\) = \(\frac { 75 }{ 8 }\) hr
Let smaller tap fill the tank in = x hours
Then larger tap will fill it in = (x – 10) hours
According to the condition,


Smaller tap can fill the tank in = 25 hours
and larger tap can fill the tank in = 25 – 10 = 15 hours

Question 4.
Tw o pipes running together can fill a tank in 11\(\frac { 1 }{ 9 }\) minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately. [CBSE 2010]
Solution:

=> 9x (x – 20) + 25 (x – 20) = 0
=> (x – 20) (9x + 25) = 0
Either x = – 20 = 0, then x = 20 or 9x + 25 = 0 then 9x = -25
=> x = \(\frac { -25 }{ 9 }\) but it is not possible being negative
x = 20
Time taken by the two pipes = 20 minutes and 20 + 5 = 25 minutes

Question 5.
To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool? [CBSE 2015]
Solution:
Let pipe of larger diameter can fill the tank = x hrs
and pipe of smaller diameter can fill in = y hrs

=> 26x + 80 = x² + 10x
=> x² + 10x – 26x – 80 = 0
=> x² – 16x – 80 = 0
=> x² – 20x + 4x – 80 = 0
=> x (x – 20) + 4 (x – 20) = 0
=> (x – 20) (x + 4) = 0
Either x – 20 = 0, then x = 20
or x + 4 = 0, then x = – 4 which is not possible
x = 20 and y = 10 + x = 10 + 20 = 30
Larger pipe can fill the tank in 20 hours and smaller pipe can fill in 30 hours.

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
The perimeter of a rectangular field is 82 m and its area is 400 m². Find the breadth of the rectangle.
Solution:
Perimeter of a rectangle field = 82 m
Length + Breadth = \(\frac { 82 }{ 2 }\) = 41 m
Let breadth = x m
Length = (41 – x) m
According to the condition,
Area = Length x breadth
400 = x (41 – x)
=> 400 = 4x – x²
=> x² – 41x + 400 = 0
=> x² – 16x – 25x + 400 = 0
=> x (x – 16) – 25 (x – 16) = 0
=> (x – 16) (x – 25) = 0
Either x – 16 = 0, then x = 16
or x – 25 = 0 then x = 25
25 > 16 and length > breadth
Breadth = 16 m

Question 2.
The length of a hall is 5 m more than its breadth. If the area of the floor of the hall is 84 m², what are the length and breadth of the hall ?
Solution:
Let breadth of the hall = x m
Then length = x + 5
Area of the floor = 84 m2
Now according to the condition,
x (x + 5) = 84
=> x² + 5x – 84 = 0
=> x² + 12x – 7x – 84 = 0
=> x (x + 12) – 7 (x + 12) = 0
=> (x + 12) (x – 7) = 0
Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 7 = 0, then x = 7
Breadth of the hall = 7 m and length = 7 + 5 = 12 m

Question 3.
Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm². Find the sides of the squares.
Solution:
Side of first square = x cm
and side of the second square = (x + 4) cm
According to the condition,
x² + (x + 4)² = 656
=> x² + x² + 8x + 16 = 656
=> 2x² + 8x + 16 – 656 = 0
=> 2x² + 8x – 640 = 0
=> x² + 4x – 320 = 0 (Dividing by 2)
=> x² + 20x – 16x – 320 = 0
=> x (x + 20) – 16 (x + 20) 0
=> (x + 20) (x – 16) = 0
Either x + 20 = 0, then x = -20 which is not possible being negative
or x – 16 = 0, then x = 16
Side of first square = 16 cm
and side of second square = 16 + 4 = 20 cm

Question 4.
The area of a right angled triangle is 165 m². Determine its base and altitude if the latter exceeds the former by 7 m.
Solution:
Area of a right angled triangle = 165 m²
Let its base = x m
Then altitude = (x + 7) m
According to the condition,
\(\frac { 1 }{ 2 }\) x (x + 7) = 165
=> \(\frac { 1 }{ 2 }\) (x² + 7x) = 165
=> x² + 7x = 330
=> x² + 7x – 330 = 0
=> x² + 22x – 15x – 330 = 0
=> x (x + 22) – 15 (x + 22) = 0
=> (x + 22) (x – 15) = 0
Either x + 22 = 0, then x = -22 which is not possible being negative
or x – 15 = 0, then x = 15
Base = 15 m
and altitude = 15 + 7 = 22 m

Question 5.
Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m² ? If so, find its length and breadth.
Solution:
Area of rectangular mango grove = 800 m²
Let breadth = x m
Then length = 2x m
According to the condition,
2x x x = 800
=> 2x² = 800
=> x² = 400 = (±20)²
Yes, it is possible,
x = 20, -20
But x = -20 is not possible being negative
Breadth = 20 m
and length = 20 x 2 = 40 m

Question 6.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m² ? If so, find its length and breadth:
Solution:
Perimeter of rectangular park = 80 m
Length + Breadth = \(\frac { 80 }{ 2 }\) = 40 m
Let length = x m
Them breadth = 40 – x
According to the condition,
Area = Length x Breadth
x (40 – x) = 400
=> 40x – x² = 400
=> x² – 40x + 400 = 0
=> (x – 20)² = 0
=> x – 20 = 0
=> x = 20
Yes, it is possible
Length = 20 m
and breadth = 40 – x = 40 – 20 = 20 m

Question 7.
Sum of the areas of two squares is 640 m². If the difference of their perimeters is 64 m, find the sides of the two squares. [CBSE 2008]
Solution:
Let side of first square = x m
and of second squares = y m
According to the given conditions,
4x – 4y = 64
=> x – y = 16 ….(i)
and x² + y² = 640 ….(ii)
From (i), x = 16 + y
In (ii)
(16 + y)² + y² = 640
=> 256 + 32y + y² + y² = 640
=> 2y² + 32y + 256 – 640 = 0
=> y² + 16y – 192 = 0 (Dividing by 2)
=> y² + 24y – 8y – 192 = 0
=> y (y + 24) – 8 (y + 24) = 0
=> (y + 24)(y – 8) = 0
Either y + 24 = 0, then y = -24, which is not possible as it is negative
or y – 8 = 0, then y = 8
x = 16 + y = 16 + 8 = 24
Side of first square = 24 m
and side of second square = 8m

Question 8.
Sum of the areas of two squares is 400 cm². If the difference of their perimeters is 16 cm, find the sides of two squares. [CBSE 2013]
Solution:
Let perimeter of the first square = x cm
Then perimeter of second square = (x + 16) cm
Side of first square = \(\frac { x }{ 4 }\) cm
and side of second square = (\(\frac { x }{ 4 }\) + 4) cm
Sum of areas of these two squares = 400 cm²

Question 9.
The area of a rectangular plot is 528 m². The length of the plot (in metres) is one metre more then twice its breadth. Find the length and the breadth of the plot. [CBSE 2014]
Solution:
Area of a rectangular plot = 528 m²
Let breadth = x m
Then length = (2x + 1) m
x (2x + 1) = 528 (∴ Area = l x b)
2x² + x – 528 = 0
=> 2x² + 33x – 32x² – 528 = 0
=> x (2x + 33) – 16 (2x + 33) = 0
=> (2x + 33) (x – 16) = 0
Either 2x + 33 = 0 then 2x = – 33 => x = \(\frac { -33 }{ 2 }\) but it is not possible being negative
or x – 16 = 0, then x = 16
Length = 2x + 1 = 16 x 2 + 1 = 33 m
and breadth = x = 16 m

Question 10.
In the centre of a rectangular lawn of dimensions 50 m x 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m². Find the length and breadth of the pond. [NCERT Exemplar]
Solution:
Given that a rectangular pond has to be constructed in the centre of a rectangular lawn of dimensions 50 m x 40 m. So, the distance between pond and lawn would be same around the pond. Say x m.

Now, length of rectangular lawn (l₁) = 50 m
and breadth of rectangular lawn (b₁) = 40 m
Length of rectangular pond (l₂) = 50 – (x + x) = 50 – 2x
Also, area of the grass surrounding the pond = 1184 m²
Area of rectangular lawn – Area of rectangular pond = Area of grass surrounding the pond
l₁ x b₁ – l₂ x b₂= 1184 [∵ area of rectangle = length x breadth]
=> 50 x 40 – (50 – 2x) (40 – 2x) = 1184
=> 2000 – (2000 – 80x – 100x + 4x²) = 1184
=> 80x + 100x – 4x² = 1184
=> 4x² – 180x + 1184 = 0
=> x² – 45x + 296 = 0
=> x² – 21x – 8x + 296 = 0 [by splitting the middle term]
=> x (x – 37) – 8 (x – 37) = 0
=> (x – 37) (x – 8) = 0
∴ x = 8
[At x = 37, length and breadth, of pond are -24 and -34, respectively but length and breadth cannot be negative. So, x = 37 cannot be possible]
Length of pond = 50 – 2x = 50 – 2(8) = 50 – 16 = 34 m
and breadth of pond = 40 – 2x = 40 – 2(8) = 40 – 16 = 24 m
Hence, required length and .breadth of pond are 34 m and 24 m, respectively.

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.
Solution:
Length of the hypotenuse of a let right ∆ABC = 25 cm
Let length of one of the other two sides = x cm
Then other side = x + 5 cm

According to condition,
(x)² + (x + 5)² = (25)² (Using Pythagoras Theorem)
=> x² + x² + 10x + 25 = 625
=> 2x² + 10x + 25 – 625 = 0
=> 2x² + 10x – 600 = 0
=> x² + 5x – 300 = 0 (Dividing by 2)
=> x² + 20x – 15x – 300 = 0
=> x (x + 20) – 15 (x + 20) = 0
=> (x + 20) (x – 15) = 0
Either x + 20 = 0, then x = -20, which is not possible being negative
or x – 15 = 0, then x = 15
One side = 15 cm
and second side = 15 + 5 = 20 cm

Question 2.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let shorter side of the rectangular field = x m
Then diagonal = (x + 60) m
and longer side = (x + 30) m
According to the condition,
(Diagonal)² = Sum of squares of the two sides

=> (x + 60)² = x² + (x + 30)²
=> x² + 120x + 3600 = x² + x² + 60x + 900
=> 2x² + 60x + 900 – x² – 120x – 3600 = 0
=> x² – 60x – 2700 = 0
=> x² – 90x + 30x – 2700 = 0
=> x (x – 90) + 30 (x – 90) = 0
=> (x – 90) (x + 30) = 0
Either x – 90 = 0, then x = 90
or x + 30 = 0, then x = – 30 which is not possible being negative
Longer side (length) = x + 30 = 90 + 30= 120
and breadth = x = 90 m

Question 3.
The hypotenuse of a right triangle is 3√10 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 9√5 cm. How long are the legs of the triangle ?
Solution:
Let the smaller leg of right triangle = x cm
and larger leg = y cm
Then x² + y² = (3√10)² (Using Pythagoras Theorem)
x² + y² = 90 ….(i)
According to the second condition,
(3x)² + (2y)² = (9√5)²
=> 9x² + 4y² = 405 ….(ii)
Multiplying (i) by 9 and (ii) by 1

y = 9
Substituting the value of y in (i)
x² + (9)² = 90
=> x² + 81 = 90
=> x² = 90 – 81 = 9 = (3)²
x = 3
Length of smaller leg = 3 cm
and length of longer leg = 9 cm

Question 4.
A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that .the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected ?
Solution:
In a circle, AB is the diameters and AB = 13 m
Let P be the pole on the circle Let PB = x m,
then PA = (x + 7) m
Now in right ∆APB (P is in a semi circle)
AB² = AB² + AP² (Pythagoras Theorem)

(13)² = x² + (x + 7)²
=> x² + x² + 14x + 49 = 169
=> 2x² + 14x + 49 – 169 = 0
=> 2x²+ 14x – 120 = 0
=> x2 + 7x – 60 = 0 (Dividing by 2)
=> x² + 12x – 5x – 60 = 0
=> x (x + 12) – 5 (x + 12) = 0
=> (x + 12) (x – 5) = 0
Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 5 = 0, then x = 5
P is at a distance of 5 m from B and 5 + 7 = 12 m from A

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
Ashu is x years old while his mother Mrs. Veena is x² years old. Five years hence Mrs. Veena will be three times old as Ashu. Find their present ages.
Solution:
Present age of Ashu = x years
and age of his mother = x² years
5 years hence,
age of Ashu will be = (x + 5) years
and age of his mother = (x² + 5) years
According to the question,
x² + 5 = 3 (x + 5)
=> x² + 5 = 3x + 15
=> x² + 5 – 3x – 15 = 0
=> x² – 3x – 10 = 0
=> x² – 5x + 2x – 10 = 0
=> x (x – 5) + 2 (x – 5) = 0
=> (x – 5) (x + 2) = 0
Either x – 5 = 0, then x = 5
or x + 2 = 0, then x = -2 which is not possible being negative
Present age of Ashu = 5 years
and age of his mother = x² = (5)² = 25 years

Question 2.
The sum of the ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man’s age at that time. Find their present ages.
Solution:
Sum of ages of a man and his son = 45 years
Let the present age of the man = x years
Then age of his son = (45 – x) years
5 years ago,
Age of the man was = (x – 5) years
and age of his son = (45 – x – 5) years = (40 – x) years
According to the condition,
(x – 5) (40 – x) = 4 (x – 5)
=> 40 – x = 4 [Dividing by (x – 5)]
=> x = 40 – 4 = 36
Age of the man = 36 years
and age of his son = 45 – 36 = 9 years

Question 3.
The product of Shikha’s age five years ago and her age 8 years later is 30, her age at both times being given in years. Find her present age.
Solution:
Let present age of Shikha = x years
5 years ago, her age was = (x – 5) years
and 8 years later, her age will be = (x + 8) years
According to the condition,
(x – 5) (x + 8) = 30
=> x² + 3x – 40 = 30
=> x² + 3x – 40 – 30 = 0
=> x² + 3x – 70 = 0
=> x² + 10x – 7x – 70 = 0
=> x (x + 10) – 7 (x + 10) = 0
=> (x + 10)(x – 7) = 0
Either x + 10 = 0, then x = -10 which is not possible being negative
or x – 7 = 0, then x = 7
Her present age = 7 years

Question 4.
The product of Ramu’s age (in years) five years ago and his age (in years) nine years later is 15. Determine Ramu’s present age.
Solution:
Let present age of Ramu = x years
5 years ago his age was = (x – 5) years
and 9 years later his age will be = (x + 9) years
According to the condition,
(x – 5) (x + 9) = 15
=> x² + 9x – 5x – 45 = 15
=> x² + 4x – 45 – 15 = 0
=> x² + 4x – 60 = 0
=> x² + 10x – 6x – 60 = 0
=> x (x + 10) – 6 (x + 10) = 0
=> (x + 10) (x – 6) = 0
Either x + 10 = 0, then x = -10 but it is not possible being negative
or x – 6 = 0, then x = 6
Present age of Ramu = 6 years

Question 5.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Sum of ages of two friends = 20 years
Let present age of one friend = x years
Age of second friend = (20 – x) years
4 years ago,
Age of first friend = x – 4
and age of second friend = 20 – x – 4 = 16 -x
According to the condition,
(x – 4) (16 – x) = 48
=> 16x – x² – 64 + 4x = 48
=> – x² + 20x – 64 – 48 = 0
=> – x² + 20x – 112 = 0
=> x² – 20x + 112 = 0
Here a = 1, b = – 20, c = 112
Discriminant(D) = b² – 4ac = (-20)² – 4 x 1 x 112
= 400 – 448 = – 48
∴ D < 0
Roots are not real
It is not possible

Question 6.
A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Find their present ages. [CBSE2010]
Solution:
Let age of sister = x years
Then age of girl = 2x years
4 years hence,
Girl’s age = 2x + 4
and sister’s age = x + 4
According to the condition,
(2x + 4) (x + 4) = 160
=> 2x² + 8x + 4x + 16 = 160
=> 2x² + 12x + 16 – 160 = 0
=> 2x²+ 12x – 144 = 0
=> x² + 6x – 12 = 0
=> x² + 12x – 6x – 72 = 0
=> x (x + 12) – 6 (x + 12) = 0
=> (x + 12) (x – 6) = 0
Either x + 12 = 0, then x = – 12 which is not possible being negative
or x – 6 = 0, then x = 6
Age of sister = 6 years
and age of girl = 2x = 2 x 6 = 12 years

Question 7.
The sum of the reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is \(\frac { 1 }{ 3 }\). Find his present age. [NCERT]
Solution:
Let age of Rehman = x years
His age 3 years ago = x – 3
and age 5 years hence = x + 5
According to the condition,

=> x² + 2x – 15 = 6x + 6
=> x² + 2x – 15 – 6x – 6 = 0
=> x² – 4x – 21 =0
=> x² – 7x + 3x – 21 = 0
=> x (x – 7) + 3 (x – 7) = 0
=> (x – 7)(x + 3) = 0
Either x – 7 = 0, then x = 7
or x + 3 = 0 then x = – 3 which is not possible being negative
x = 7
His present age = 7 years

Question 8.
If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age. What is her age now? [NCERT Exemplar]
Solution:
Let the actual age of Zeba = x year .
Her age when she was 5 years younger = (x – 5) years
Now, by given condition,
Square of her age = 11 more than 5 times her actual age
(x – 5)² = 5 x actual age + 11
=> (x – 5)² = 5x + 11
=> x² + 25 – 10x = 5x + 11
=> x² – 15x + 14 = 0
=> x² – 14x – x + 14 = 0 [by splitting the middle term]
=> x (x – 14) – 1 (x – 14) = 0
=> (x – 1) (x – 14) = 0
=> x = 14
[Here, x ≠ 1 because her age is x – 5. So, x – 5 = 1 – 5= -4 i.e., age cannot be negative]
Hence, required Zeba’s age now is 14 years.

Question 9.
At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha. [NCERT Exemplar]
Solution:
Let Nisha’s present age be x year.
Then, Asha’s present age = x² + 2 [by given condition]
Now, when Nisha grows to her mother’s present age i.e., after {(x² + 2) – x} years.
Then, Asha’s age also increased by [(x² + 2) – x] year.
Again by given condition,
Age of Asha = One years less than 10 times the present age of Nisha
(x² + 2) + {(x² + 2) – x} = 10x – 1
=> 2x² – x + 4 = 10x – 1
=> 2x² – 11x + 5 = 0
=> 2x² – 10x – x + 5 = 0
=> 2x (x – 5) – 1(x – 5) = 0
=> (x – 5) (2x – 1) = 0
∴ x = 5
[Here, x = \(\frac { 1 }{ 2 }\) cannot be possible, because at x = \(\frac { 1 }{ 2 }\), Asha’s age is 2\(\frac { 1 }{ 4 }\) years which is not possible]
Hence, required age of Nisha = 5 years
and required age of Asha = x² + 2 = (5)² + 2 = 25 + 2 = 27 years

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
The speed of a boat in still water is 8 km/hr. It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream.
Solution:
Let the speed of stream = x km/h
and speed of boat in still water = 8 km/h
Distance covered up stream = 15 km
and downstream = 22 km
Total time taken = 5 hours
According to the conditions,

Question 2.
A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train. [NCERT Exemplar]
Solution:
Let the original speed of the train = x km/h
Then, the increased speed of the train = (x + 5) km/h [by given condition]
and distance = 360 km
According to the question,

Question 3.
A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.
Solution:
Total journey = 200 km
Let the speed of fast train = x km/hr
Then speed of slow train = (x – 10) km/hr
According to the condition,

=> x (x – 50) + 40 (x – 50) = 0
=> (x – 50) (x + 40) = 0
Either x – 50 = 0, then x = 50
or x + 40 = 0, then x = -40 but it is not possible being negative
Speed of the fast train = 50 km/hr
and speed of the slow train = 50 – 40 = 10 km/hr

Question 4.
A passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.
Solution:
Total journey = 150 km
Let the usual speed of the train = x km/hr
According to the condition,

=> x (x + 30) – 25 (x + 30) = 0
=> (x + 30) (x – 25) = 0
Either x + 30 = 0, then x = -30 but it is not possible being negative
or x – 25 = 0, then x = 25
Usual speed of the train = 25 km/hr

Question 5.
The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction ?
Solution:
Distance = 150 km
Let the speed of the person while going = x km/hr
Then the speed while returning = (x + 10) km/hr
According to the condition,


=> x (x + 30) – 20 (x + 30) = 0
=> (x + 30) (x – 20) = 0
Either x + 30 = 0, then x = -30 which is not possible being negative
or (x – 20) = 0 then x = 20
Usual speed of the man while going = 20 km/hr

Question 6.
A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.
Solution:
Distance = 1600 km
Let usual speed of the plane = x km/hr
Increased speed = (x + 400) km/hr
According to the condition,

is not possible being negative or x – 800 = 0, then x = 800
Usual speed of the plane = 800 km/hr

Question 7.
An aeroplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Find its usual speed.
Solution:
Distance = 1200 km
Let usual speed of the aeroplane = x km/hr
Increased speed = (x + 100) km/hr
According to the condition,

=> x (x + 400) – 300 (x + 400) = 0
=> (x + 400) (x – 300) = 0
Either x – 300 = 0, then x = 300
or x + 400 = 0, then x = -400 which is not possible being negative
Usual speed of the plane = 300 km/hr

Question 8.
A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed. If it takes 3 hours to complete^total journey, what is its original average speed? [NCERT Exemplar]
Solution:
Let its original average speed be x km/h. Therefore
\(\frac { 63 }{ x }\) + \(\frac { 72 }{ x + 6 }\) = 3

Question 9.
A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train. [CBSE 2006C]
Solution:
Distance to be covered = 90 km
Let uniform-original speed = x km/h
Increased speed = (x + 15) km/hr
According to the condition,

=> x² + 60x – 45x – 2700 = 0
=> x (x + 60) – 45 (x + 60) = 0
=> (x + 60)(x – 45) = 0
Either x + 60 = 0, then x = -60 which is not possible being negative
or x – 45 = 0, then x = 45
Original speed of the train = 45 km/hr

Question 10.
A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Total distance = 360 km
Let uniform speed of the train = x km/hr
Increased speed = (x + 5) km/hr
According to the condition,

=> x (x + 45) – 40 (x + 45) = 0
=> (x + 45) (x – 40) = 0
Either x + 45 = 0, then x = -45 but it is not possible being negative
or x – 40 = 0, then x = 40
Speed of the train = 40 km/hr

Question 11.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speeds of the two trains.
Solution:
Distance between Mysore and Bangalore = 132 km
Let the speed of the passenger train=x km/hr
Then speed of express train = x + 11
According to the condition,

Either x + 44 = 0, then x = -44 but it is not possible being negative
or x – 33 = 0, then x = 33
Speed of passenger train = 33 km/hr
and speed of express train = 33 + 11 = 44 km/hr

Question 12.
An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed. (CBSE 2010)
Solution:
Distance = 1250 km
Usual speed = x km/hr
Increased speed = (x + 250) km/hr
According to the condition,

Question 13.
While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away, in time, the pilot increased the speed by 100 km/hr. Find the original speed / hour of the plane. [CBSE 2013]
Solution:
Distance = 1500 km
Let the original speed of the aeroplane = x km/hr
Then increased speed = (x + 100) km/hr
According to the condition,

=> x² + 100x = 300000
=> x²+ 100x – 300000 = 0
=> x² + 600x – 500x – 300000 = 0
=> x (x + 600) – 500(x + 600) = 0
=> (x + 600) (x – 500) = 0
Either x + 600 = 0, then x = -600 which is not possible being negative
or x – 500 = 0, then x = 500
Original speed = 500 km/hr

Question 14.
A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. [CBSE 2014]
Solution:
Let speed of the stream be x km/hr,
Speed of the boat upstream = (18 – x) km/hr
and Speed of the boat downstream = (18 + x) km/hr
Distance = 24 km

48x = 324 – x²
x² + 48x – 324 = 0
x² + 54x – 6x – 324 = 0
x(x + 54) – 6(x + 54) = 0
(x – 6) (x + 54) = 0
x – 6 = 0 or x + 54 = 0
x = 6 or x = – 54
Since speed cannot be negative
Speed of stream, x = 6 km/hr

Question 15.
A car moves a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/ hour. Find the time taken to cover the distance. [CBSE 2017]
Solution:
Distance = 2592 km
Let the speed of the car = x km/hr
and time taken = \(\frac { x }{ 2 }\) hour
We have, Distance = Speed x Time
2592 = x x \(\frac { x }{ 2 }\)
=> 2592 = \(\frac { { x }^{ 2 } }{ 2 }\)
=> x² = 2592 x 2
=> x = √5184
=> x = 72 km/hr
and thus time taken = \(\frac { x }{ 2 }\) h = \(\frac { 72 }{ 2 }\) = 36 hour

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.