Maths RD Sharma Solutions Class 10

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.

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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.

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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.

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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.

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 4.12

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.

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