Category: Class 10 Maths Objective Questions

[CLASS 10] Mathematics MCQs on Surface Area and Volume of Combination of Solids

Mathematics Assessment Questions for Class 10 on “Surface Area and Volume of Combination of Solids – 2”.

1. Find the surface area of the given solid which is in the form of a cone mounted on a hemisphere. The radius and height of the cone are 5cm and 12cm.
a) 214.4 cm2
b) 279.53 cm2
c) 70 cm2
d) 72.5 cm2
Answer: b
Clarification: Slant height = (sqrt {h^2+r^2})
= (sqrt {12^2+5^2})
= √169
= 13cm
The surface area of the toy = C.S.A of the cone + C.S.A of the sphere
= πrl + 2πr2
= (3.14 × 3 × 13) + (2 × 3.14 × 52)
= 47.1 + 56.52
= 279.53cm2

2. A wooden box is in the shape of a cuboid with three conical depressions. 7 cm × 3 cm × 4 cm are the dimensions of the cuboid and the radius and depth of the conical depressions are 0.5 cm and 1.2 cm. Find the volume of the entire wooden box?
a) 109.4 cm3
b) 80.05 cm3
c) 150 cm3
d) 89.4 cm3
Answer: b
Clarification: Volume of the wooden box = volume of the cuboid – the volume of 3 conical depressions
= lbh – 3((frac {1}{3})πr2h)
= (7 × 3 × 4) – 3((frac {1}{3}) × 3.14 × 0.52 × 1.2)
= 80.05 cm3

3. Find the volume of the largest right circular cone that can be cut out of cube having 5 cm as its length of the side.
a) 32.72 cm3
b) 15 cm3
c) 25.4 cm3
d) 37.2 cm3
Answer: a
Clarification: Length of the side of the cube = height of the cone = 5 cm
The radius of the base of the cone = (frac {5}{2}) cm = 2.5 cm
The volume of the cone = (frac {1}{3})πr2h
= (frac {1}{3}) × 3.14 × 2.52 × 5
= 32.72 cm3

4. A toy is in the form of a cone mounted on a hemisphere and a cylinder. The radius and height of the cone are 3 m and 4 m. Find the volume of the given solid?
a) 193.21 m3
b) 207.30 m3
c) 184.21 m3
d) 282.21 m3
Answer: b
Clarification: Volume of the toy = volume of the cone + volume of the hemisphere + volume of the cylinder
= (frac {1}{3})πr2h + (frac {2}{3})πr3 + πr2h
= ((frac {1}{3}) × 3.14 × 32 × 4) + ((frac {2}{3}) × 3.14 × 33) + (3.14 × 32 × 4)
= 207.30 m3

5. What is the length of the resulting solid if two identical cubes of side 7 cm are joined end to end?
a) 26 cm
b) 16 cm
c) 21 cm
d) 14 cm
Answer: d
Clarification: Length of resulting cuboid = 2 × side of the cube
= 2 × 7 cm
= 14 cm

6. The length, breadth and height of the cuboid is 8 cm, 4 cm and 4 cm. Find the volume of the cuboid?
a) 152.76 cm3
b) 154 cm3
c) 128 cm3
d) 141.76 cm3
Answer: c
Clarification: The volume of the cuboid = lbh
= 8 × 4 × 4
= 128 cm3

7. What is the volume of an article which is made by digging out a hemisphere from each end of a solid cylinder?
a) πr2h + 2(2πr3)
b) 2πrh – 2(πr2)
c) 2πrh + 2((frac {2}{3}) πr2)
d) πr2h – 2((frac {2}{3})πr3)
Answer: d
Clarification: Volume of the article = volume of the cylinder – 2(volume of the hemisphere)
= πr2h – 2((frac {2}{3})πr3)

8. What is the formula to find the height of an iron pillar consisting of a cylinder and a cone mounted on it?
a) The radius of the cylinder + 2(height of the cone)
b) Height of the cylinder + 2(height of the cone)
c) The radius of the cylinder + radius of the cone
d) Height of the cylinder + height of the cone
Answer: d
Clarification: To find the height of an iron pillar consisting of a cylinder and a cone mounted on it, we require heights of both cone and cylinder.
Height of the pillar = height of the cylinder + height of the cone.

9. What is the volume of an article which is made by digging out a hemisphere from each end of a solid cylinder where the radius, height of the cylinder is 5 cm, 8 cm respectively and the radius of the hemisphere is 5 cm?
a) 104.40 cm3
b) 205.6 cm3
c) 168.23 cm3
d) 604 cm3
Answer: a
Clarification: Volume of the article = volume of the cylinder – 2(volume of the hemisphere)
= πr2h – 2((frac {2}{3})πr3)
= (3.14 × 52 × 8) – 2((frac {2}{3}) × 3.14 × 53)
= 104.40 cm3

10. What is the formula required to use for T.S.A of an article which is made by digging out a hemisphere from each end of a solid cylinder?
a) C.S.A of the cylinder – 2(C.S.A of the hemisphere)
b) C.S.A of the cylinder + C.S.A of the hemisphere
c) C.S.A of the cylinder + 2(C.S.A of the hemisphere)
d) C.S.A of the cylinder – C.S.A of the hemisphere
Answer: c
Clarification: To find the T.S.A of an article which is made by digging out a hemisphere from each end of a solid cylinder, we need C.S.A of the cylinder and C.S.A of the hemisphere.
T.S.A of the article = C.S.A of the cylinder + 2(C.S.A of the hemisphere).

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[CLASS 10] Mathematics MCQs on Nth Term of Arithmetic Progression

Mathematics Multiple Choice Questions & Answers on “Nth Term of Arithmetic Progression”.

1. What will be the nth term of the AP 1, 5, 9, 13, 17…….?
a) 4n – 3
b) 3n – 4
c) 4n + 3
d) 3n + 4
Answer: a
Clarification: Here a = 1 and d = 4
The nth term of the AP = a + (n – 1)d = 1 + (n – 1)4 = 1 + 4n – 4 = 4n – 3

2. If the nth term of the AP is 2n + 7, then the common difference will be _____
a) -2
b) 2
c) 3
d) -3
Answer: b
Clarification: nth term of the AP is 2n + 7
Tn = 2n + 7
Common difference = T2 – T1 = (2 × 2 + 7 – (2 × 1 + 7)) = 11 – 9 = 2

3. If the nth term of the AP is 7n – 9, then the first term is ________
a) 1
b) 6
c) 3
d) 4
Answer: b
Clarification: nth term of the AP is 7n – 9
Tn = 7n – 1
T1 = 7(1) – 1 = 6
The first term of AP is 6.

4. If the nth term of the AP is 8n + 1, then the 20th term will be ______
a) 160
b) 120
c) 161
d) 121
Answer: c
Clarification: nth term of the AP is 8n + 1
Tn = 8n + 1
T20 = 8(20) + 1
T20 = 161

5. What will be the 99th term from the end of the AP 500, 489, 478, 467… – 1139?
a) 1078
b) 1123
c) 12
d) 61
Answer: d
Clarification: In this case since we have to find the 99th term from the end.
We will consider the first term to be -1139 and the common difference will be 11
Now, a = -1139, d = 11 and n = 99
T99 = a + (n – 1)d
T99 = -1139 + (99 – 1)11
T99 = -1139 + 1078 = -61
The value of 99th term from the end is 61.

6. If the pth term of an AP is q and its qth term is p, then what will be the value of its (p + q)th term?
a) 1
b) p + q – 1
c) 0
d) 2(p + q – 1)
Answer: c
Clarification: pth term = q
a + (p – 1)d = q
a + pd – d = q     (1)
qth term = p
a + (q – 1)d = p
a + qd – d = p     (2)
Subtracting (2) from (1) we get,
a + qd – d – (a + pd – d) = p – q
qd – pd = p – q
d = -1
Substituting in equation 1, we get,
a = p + q – 1
(p + q)th term = a + (n – 1)d = p + q – 1 + (p + q – 1)(-1) = p + q – 1 – p – q + 1 = 0

7. If 5 times the 5th term of an AP is equal to 15 times its 15th term, then the value of its 20th term will be _______
a) 0
b) 1
c) 2
d) 3
Answer: a
Clarification: 5(5th term) = 15(15th term)
5th term = a + (5 – 1)d = a + 4d
15th term = a + (15 – 1)d = a + 14d
5(a + 4d) = 15(a + 14d)
5a + 20d = 15a + 210d
20d – 210d = 15a – 5a
-190d = 10a
a = -19d
Now, the 20th term = a + (20 – 1)d = a + 19d
But a = -19d
Hence, -19d + 19d = 0

8. If the 11th term of an AP is (frac {1}{13}) and its 13th term is (frac {1}{11}), then what will be the value of 143th term?
a) (frac {1}{143})
b) 1
c) 0
d) (frac {23}{143})
Answer: b
Clarification: Here 11th term = (frac {1}{13})
13thterm = (frac {1}{11})
Let the first term of the AP be a and common difference be d
T11 = a + (n – 1)d = (frac {1}{13})
T11 = a + (11 – 1)d = (frac {1}{13})
T11 = a + 10d = (frac {1}{13})     (1)
T13 = a + (n – 1)d = (frac {1}{11})
T13 = a + (13 – 1)d = (frac {1}{11})
T13 = a + 12d = (frac {1}{11})     (2)
Subtracting (1) from (2)
We get,
a + 12d – (a + 10d) = (frac {1}{11} – frac {1}{13})
2d = (frac {2}{143})
d = (frac {1}{143})
Now, substituting value of d in equation 1
We get,
T11 = a + 10((frac {1}{143})) = (frac {1}{13})
a = (frac {1}{143})
The 143th term = (frac {1}{143}) + 142(frac {1}{143} = frac {(1+142)}{143} ) = 1

9. If the 7th term of an AP is 20 and its 11th term is 40 then, what will be the common difference?
a) 3
b) 4
c) 5
d) 2
Answer: c
Clarification: Here 7th term = 20
11thterm = 40
Let the first term of the AP be a and common difference be d
T7 = a + (n – 1)d = 20
T7 = a + (7 – 1)d = 20
T7 = a + 6d = 20     (1)
T11 = a + (n – 1)d = 40
T11 = a + (11 – 1)d = 40
T11 = a + 10d = 40     (2)
Subtracting (1) from (2)
We get,
a + 10d – (a + 6d) = 40 – 20
4d = 20
d = 5

10. What will be the 14th term of the AP 5, 8, 11, 14, 17…….?
a) 30
b) 41
c) 40
d) 44
Answer: d
Clarification: Here a = 5, d = 8 – 5 = 3 and n = 14
T14 = a + (n – 1)d
T14 = 5 + (14 – 1)3
T14 = 5 + 13 × 3
T14 = 44

[CLASS 10] Mathematics MCQs on Trigonometric Ratios

Mathematics Exam Questions and Answers for Class 10 on “Trigonometric Ratios – 2”.

1. Trigonometric ratios are only applicable to which kind of triangles?
a) Right-angled triangles
b) Any type of triangles
c) Acute angled triangles
d) Obtuse angled triangles
Answer: a
Clarification: Trigonometric ratios are only applicable to right-angled triangles whose angle is 90° between two sides of a triangle.

2. Trigonometric ratios are ______
a) sine, cosine and cotangent
b) sine, tangent, cotangent and secant
c) sine, cosine, tangent, cotangent, secant and cosecant
d) tangent, cotangent and secant
Answer: c
Clarification: Trigonometric ratios are generally defined as the ratios of any two sides of a right-angled triangle and it gives the relation between sides of a right-angled triangle to its appropriate angle. Sine, cosine, tangent, cotangent, secant and cosecant are six trigonometric ratios used in trigonometry.

3. What is the inverse of cosecant?
a) Cosine
b) Secant
c) Sine
d) Tangent
Answer: c
Clarification: The inverse of cosecant is sine.
Cosec θ = (frac {1}{Sin theta })

4. (frac {Sin theta }{Cos theta}) equals to ______
a) tan θ
b) cot θ
c) cosec θ
d) sec θ
Answer: a
Clarification: Sin θ = (frac {Length , of , opposite , side , of , the , triangle}{Length , of , hypotenuse , of , the , triangle})
Cos θ = (frac {Length , of , adjacent , side , of , the , triangle}{Length , of , hypotenuse , of , the , triangle})
(frac {Sin theta }{Cos theta} = frac {Length , of , opposite , side , of , the , triangle}{Length , of , adjacent , of , the , triangle})
= Tan θ

5. Who is the ‘Father of Trigonometry’?
a) Hipparchus
b) Euclid
c) Aristotle
d) Archimedes
Answer: a
Clarification: Hipparchus was a Greek mathematician and is considered as the ‘Father of Trigonometry’. Hipparchus not only contributed to mathematics but also to astronomy.

6. The product of sin θ and Cosec θ is 1.
a) True
b) False
Answer: a
Clarification: The inverse of cosecant is sine.
Sin θ = (frac {1}{Cosec theta })
Sin θ.Cosec θ = 1

7. What is the cotangent of an angle θ if the side adjacent to θ is 8 units and the side opposite to θ is 3 units?
a) (frac {3}{8})
b) (frac {8}{3})
c) (frac {4}{3})
d) (frac {3}{4})
Answer: b
Clarification: Cotangent θ = (frac {Length , of , adjacent , side , of , the , triangle}{Length , of , opposite , of , the , triangle})
= (frac {8}{3})

8. Choose the correct reciprocal ratios.
a) Tan θ, Sec θ
b) Cosec θ, Sec θ
c) Sec θ, Sin θ
d) Tan θ, Cot θ
Answer: d
Clarification: Tan θ, Cot θ are the reciprocal ratios because they are inverse to each other.
Tanθ = (frac {1}{Cot theta })

9. The basic trigonometric ratios are _____
a) sine
b) tangent
c) sine, cosine and tangent
d) cosine
Answer: c
Clarification: The basic trigonometric ratios are sine, cosine and tangent and the other three trigonometric ratios secant, cosecant, cotangent are derived from these basic trigonometric ratios.

10. (frac {Length , of , hypotenuse}{Length , of , opposite , to , angle , A}) is _____
a) sine of angle A
b) cosec of angle A
c) tan of angle A
d) cot of angle A
Answer: b
Clarification: The formula for cosec A is (frac {Length , of , hypotenuse , of , the , triangle}{Length , of , opposite , side , to , angle , A}).

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[CLASS 10] Mathematics MCQs on Conversion of Solid from One Shape to Another

Mathematics Multiple Choice Questions & Answers on “Conversion of Solid from One Shape to Another”.

1. A metallic sphere whose radius is 5 cm is melted and cast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
a) 21.4 cm
b) 43.63 cm
c) 70 cm
d) 72.5 cm2
Answer: b
Clarification: Volume of the sphere = volume of the cylinder
(frac {4}{3})πr3 = πr2h
(frac {4}{3}) × 3.14 × 53 = 3.14 × 62 × h
h = (frac {4 times 3.14 times 3.14 times 125}{3.14 times 36})
h = 43.63 cm

2. Three metallic spheres of radius 3 cm, 6 cm, 9 cm are melted into a single sphere. Find the radius of the resulting sphere.
a) 109.4 cm3
b) 4071.48 cm3
c) 1520 cm3
d) 1869.4 cm3
Answer: b
Clarification: Volume of the resulting sphere = sum of the volumes of all three spheres
= (frac {4}{3})π33 + (frac {4}{3})π63 + (frac {4}{3})π93
= 113.09 + 904.77 + 3053.62
= 4071.48 cm3

3. A well of depth 25 m with a radius 4 m is dug from the earth forming a platform of length 28 m and a breadth of 16 m. Find the height of the platform.
a) 2.8 m3
b) 5 m3
c) 5.4 m3
d) 7.2 m3
Answer: a
Clarification: Volume of the well = volume of the platform
πr2h = lbh
3.14 × 42 × 25 = 28 × 16 × h
h = 2.8 m3

4. How many coins of 1 cm in diameter and thickness of 1.2 cm need to be melted to form a cuboid with dimensions of 5 cm × 10 cm × 4 cm?
a) 193
b) 213
c) 184
d) 282
Answer: b
Clarification: Volume of the cuboid = number of coins(volume of a coin)
lbh = number of coins(πr2h)
5 × 10 × 4 = number of coins(3.14 × 0.52 × 1.2)
200 = number of coins(0.94)
number of coins = 213

5. A sphere of radius 14 cm is melted and cast into a number of tiny cones of radius 2.33 cm each and height 6 cm. Find the number of cones that will be formed?
a) 726
b) 816
c) 721
d) 821
Answer: d
Clarification: Number of cones(volume of a cone) = volume of sphere
number of cones((frac {1}{3})πr2h) = (frac {4}{3})πr3
number of cones((frac {1}{3}) × 3.14 × 2.332 × 6) = (frac {4}{3}) × 3.14 × 143
number of cones(14) = 11488.21
number of cones = 821

6. How many cylinders having 2.1 cm of radius and 1.4 cm of height can be made out of a cuboid metal box having dimensions 33 cm, 21 cm, 10.5 cm?
a) 152
b) 154
c) 844
d) 841
Answer: c
Clarification: Number of cylinders(volume of a cylinder) = volume of a cuboid
Number of cylinders (πr2h) = lbh
Number of cylinders (3.14 × 1.42 × 1.4) = 33 × 21 × 10.5
Number of cylinders = 844

7. A sphere having a radius of 3 cm is melted and elongated into a wire having a circular cross-section of radius 0.1 cm. Find the length of the wire?
a) 2400 cm
b) 3100 cm
c) 1200 cm
d) 3600 cm
Answer: d
Clarification: Volume of the wire = volume of the sphere
πr2h = (frac {4}{3})πr3
3.14 × 0.12 × h = (frac {4}{3}) × 3.14 × 27
h = 3600 cm

8. A cylindrical hole of depth 20 m with a radius 5 m is dug from the earth forming a platform of length 14 m and a breadth of 12 m. Find the height of the platform.
a) 8.34 m3
b) 11.84 m3
c) 7.64 m3
d) 9.34 m3
Answer: d
Clarification: Volume of the cylindrical hole = volume of the platform
πr2h = lbh
3.14 × 52 × 20 = 14 × 12 × h
h = 9.34 m3

9. What is the formula to find the rise in the water level when ‘x’ spherical balls are dropped into a cylindrical beaker?
a) (frac {Volume , of , ‘x’ , spherical , balls}{Base , area , of , cylinder})
b) (frac {Volume , of , ‘x’ , spherical , balls}{2(Base , area , of , cylinder)})
c) The volume of the cylinder + volume of the beaker
d) The volume of the cylinder – 2(volume of the beaker)
Answer: a
Clarification: To find the rise in the water level we require the volume of ‘x’ spherical balls and the base area of the cylinder
Rise in the water level = (frac {Volume , of , ‘x’ , spherical , balls}{Base , area , of , cylinder})

10. A metallic sphere whose radius is 4 cm is melted and cast into the shape of a right circular cone of radius 7 cm. Find the height of the cylinder?
a) 14.48 cm
b) 22.36 cm
c) 16.40 cm
d) 20.32 cm
Answer: c
Clarification: Volume of the sphere = volume of the cone
(frac {4}{3}) πr3 = (frac {1}{3}) πr2h
(frac {4}{3}) × 3.14 × 43 = 3.14 × 72 × h
h = (frac {4 times 3.14 times 3.14 times 64}{3.14 times 49})
h = 16.40 cm

11. Three metallic spheres of radius 2 cm, 4 cm, 8 cm are melted into a single sphere. Find the radius of the resulting sphere.
a) 1009.4 cm3
b) 2446.25 cm3
c) 1520 cm3
d) 2869.4 cm3
Answer: b
Clarification: Volume of the resulting sphere = sum of the volumes of all three spheres
= (frac {4}{3})π23 + (frac {4}{3})π43 + (frac {4}{3})π83
= (frac {4}{3}) × 3.14(23 + 43 + 83)
= 2446.25 cm3

[CLASS 10] Mathematics MCQs on Arithmetic Progression Basics

Mathematics Multiple Choice Questions & Answers on “Arithmetic Progression Basics”.

1. A sequence in which each term differs from its preceding term by a constant is called arithmetic progression.
a) True
b) False
Answer: a
Clarification: Consider an example, 8, 11, 14, 17, 20, ….
Now, 11 – 8 = 3
14 – 11 = 3
Here the constant is 3. This constant is also termed as common difference.

2. If we add the terms of an AP, then we get a corresponding series.
a) True
b) False
Answer: a
Clarification: For example, consider an AP 2, 11, 20, 29……
Adding the terms of the AP we get the arithmetic series as 2 + 11 + 20 + 29 + ⋯

3. Sequences which follow a definite pattern called progressions.
a) True
b) False
Answer: a
Clarification: Let us consider an example,
2, 4, 6, 8, …
Here each term is twice the preceding term.
Since it follows a definite pattern it is called as progressions.
Progressions are of three types: Arithmetic progression, Geometric progression and Harmonic progression.

4. If the first term of an AP is a and its common difference is d then the nth term is given by Tn = a + (n – 1)d
a) True
b) False
Answer: a
Clarification: In the given AP we have, first term as a and common difference as d.
So, the given AP may be written as a, a + d, a + 2d, a + 3d, …
In this AP, we have,
First term = a = a + (1 – 1)d
Second term = a = a + (2 – 1)d
Third term = a = a + (3 – 1)d
And so, on
Hence, the nth term of the AP will be a + (n – 1)d.

5. Which term of the AP 2, 11, 20, 29…… is 290?
a) 32
b) 35
c) 30
d) 33
Answer: d
Clarification: Here a = 2 and d = 11 – 2 = 9
Let the given AP contain n terms.
Tn = 290
a + (n – 1)d = 290
2 + (n – 1)9 = 290
2 + 9n – 9 = 290
9n = 297
n = 33
Thus, the 33 term of the AP is 290.

6. How many terms are there in the AP 9, 16, 23, 30……………282?
a) 50
b) 45
c) 40
d) 42
Answer: c
Clarification: Here a = 9 and d = 16 – 9 = 7
Let the given AP contain n terms.
Tn = 282
a + (n – 1)d = 282
9 + (n – 1)7 = 282
9 + 7n – 7 = 282
7n = 280
n = 40
Thus, the given AP has 40 terms.

7. The middle term of the AP 11, 8, 5, … , -79 will be?
a) 30
b) -34
c) 34
d) -30
Answer: b
Clarification: Here a = 11 and d = 8 – 11 = -3
Let the given AP contain n terms.
Tn = -79
a + (n – 1)d = -79
11 + (n – 1) – 3 = -79
11 – 3n + 3 = -79
14 + 79 = 3n
n = 31
Thus, the given AP has 31 terms.
∴ its middle term = (frac {1}{2}) (31 + 1) = 16th term
∴ 16th term = 11 + (16 – 1) – 3 = 11 – 45 = -34

8. Which term of the AP 80, 76, 72, 68……. is the first negative term?
a) 22
b) 21
c) 23
d) 24
Answer: a
Clarification: Here a = 80 and d = 76 – 80 = – 4
Let Tn be the first negative term of the given AP.
Tn < 0
a + (n – 1)d < 0
80 + (n – 1) – 4 < 0
80 – 4n + 4 < 0
84 < 4n
n > 21
Hence the 22th term of the given AP will be its first negative term.

9. What is the arithmetic mean between a2 and b2 ?
a) (frac {1}{2}) (a – b)
b) (frac {1}{2}) (a + b)
c) (frac {1}{2}) (a2 – b2)
d) (frac {1}{2}) (a2 + b2)
Answer: d
Clarification: We know that the arithmetic mean between two numbers is (frac {1}{2}) (a + b)
∴ Arithmetic mean between a2 and b2 = (frac {1}{2}) (a2 + b2)

10. What will be the value of k so that (2k + 1), (5k – 3) and (-8k + 5) are three consecutive terms of an AP?
a) 4
b) 3
c) (frac {3}{4})
d) (frac {4}{3})
Answer: c
Clarification: Since, (2k + 1), (5k – 3) and (-8k + 5) are three consecutive terms of an AP.
The common difference will be same.
(5k – 3) – (2k + 1) = (-8k + 5) – (5k – 3)
3k – 4 = -13k + 8
16k = 12
k = (frac {3}{4})

11. The sum of three numbers is 30 and their product 750. What are the three numbers?
a) -2, -6 and -10
b) 2, 6, and 10
c) 5, 10 and 15
d) -5, -10 and -15
Answer: c
Clarification: Let the three numbers in AP be (a + d), a, and (a – d).
Sum of the five numbers is 30.
∴ (a + d) + a + (a – d) = 30
3a = 30
a = 10
Now, product of the numbers is 750.
∴ (a + d)a(a – d) = 750
a3 – d2 a = 750
Substituting a = 10, we get
103 – d2 (10) = 750
1000 – d2 (10) = 750
250 = 10d2
d2 = 25
d = 5, -5
The three numbers are 5, 10 and 15.

12. Four numbers whose sum is 40 and sum of their squares is 480. What are the four numbers?
a) 4, 8, 12, and -16
b) 4, 8, 12, and 16
c) -4, 8, 12, and 16
d) 4, -8, 12, and 16
Answer: b
Clarification: Let the four numbers be (a + 3d), (a + d), (a – 3d), (a – d).
Sum of the four numbers is 40.
a + 3d + a + d + a – 3d + a – d = 40
4a = 40
a = 10
Sum of their squares is 480.
(a + 3d)2 + (a + d)2 + (a – 3d)2 + (a – d)2 = 480
4a2 + 20d2 = 480
102 + 5d2 = 120
100 + 5d2 = 120
5d2 = 20
d2 = 4
d = 2, -2
Hence the four numbers are 4, 8, 12, and 16.

13. The sum of 5 numbers in AP is 10 and their product is 80. What are the five numbers?
a) 2 + 2√6, 2 + √6, 2, 2 – 2√6, 2 – √6
b) 2 + 2√6, 2 + √6, 2 – 2√6, 2 – √6
c) 2 + 2√6, 2 – √6, 2, 2 – 2√6, 2 – √6
d) 2 + 2√6, 2, 2 – 2√6, 2 – √6
Answer: a
Clarification: Let the five numbers in AP be (a + 2d), (a + d), a, (a – 2d) and (a – d).
Sum of the five numbers is 10.
∴ (a + 2d) + (a + d) + a + (a – 2d) + (a – d) = 10
5a = 10
a = 2
Now, product of the numbers is 80.
∴ (a + 2d)(a + d)a(a – 2d)(a – d) = 80
a5 – 5a3 d2 + 4d4 a = 80
Substituting a = 2, we get
25 – 5(23 d2) + 4d4 (2) = 80
32 – 40d2 + 8d4 = 80
8d4 – 40d2 – 48 = 0
Substitute d2 = t
t2 – 5t – 6 = 0
Solving the equation, we get,
t = 6, -1
Now, t = d2
Hence, d = ±√6, ±√-1
The five numbers are 2 + 2√6, 2 + √6, 2, 2 – 2√6, 2 – √6 or 2 – 2√6, 2 – √6, 2, 2 + 2√6, 2 + √6 or 2 + 2√-1, 2 + √-1, 2, 2 – 2√-1, 2 – √-1 or 2 – 2√-1, 2 – √-1, 2, 2 + 2√-1, 2 + √-1.

[CLASS 10] Mathematics MCQs on Trigonometric Ratios

Mathematics Question Papers for Class 10 on “Trigonometric Ratios – 3”.

1. If sin A = (frac {3}{5}), then find tan A.
a) (frac {3}{4})
b) (frac {3}{5})
c) (frac {5}{3})
d) (frac {4}{3})
Answer: a
Clarification: Sin A = (frac {Opposite , side}{Hypotenuse} = frac {3}{5})
From Pythagoras theorem, (Hypotenuse)2 = (Opposite side)2 + (Adjacent side)2
52 = 32 + (Adjacent side)2
(Adjacent side)2 = 52 – 32
Adjacent side = √16 = 4
Tan A = (frac {Opposite , side}{Adjacent , side} = frac {3}{4})

2. (frac {Cos A}{Sin A}) = ______
a) Tan A
b) Sin A
c) Cot A
d) Sec A
Answer: c
Clarification: Sin A = (frac {Length , of , the , opposite , side}{Length , of , the , hypotenuse}), Cos A = (frac {Length , of , the , adjacent , side}{Length , of , the , hypotenuse})
(frac {Cos A}{Sin A} = frac {Length , of , the , adjacent , side}{Length , of , the , opposite , side})
= Cot A

3. If sin B = (frac {3}{5}), then find sec B.
a) (frac {3}{4})
b) (frac {4}{5})
c) (frac {5}{4})
d) (frac {5}{3})
Answer: c
Clarification: Sin B = (frac {Opposite , side}{Hypotenuse} = frac {3}{5})
From Pythagoras theorem, (Hypotenuse)2 = (Opposite side)2 + (Adjacent side)2
52 = 32 + (Adjacent side)2
(Adjacent side)2 = 52 – 32
Adjacent side = √16 = 4
Sec B = (frac {Hypotenuse}{Adjacent , side} = frac {5}{4})

4. What is the value of cos2θ – sin2θ if the length of the opposite side is 20 units and the length of the hypotenuse is 29 units?
a) (frac {- 41}{841})
b) (frac {- 41}{840})
c) (frac {41}{841})
d) (frac {41}{840})
Answer: a
Clarification: From Pythagoras theorem, (Hypotenuse)2 = (Opposite side)2 + (Adjacent side)2
(Adjacent side)2 = (Hypotenuse)2 – (Opposite side)2
Adjacent side = √441 = 21
Cosθ = (frac {Length , of , the , adjacent , side}{Length , of , the , hypotenuse} = frac {21}{29}), Sinθ = (frac {Length , of , the , opposite , side}{Length , of , the , hypotenuse} = frac {20}{29} )
cos2θ – sin2θ = ((frac {21}{29}))2 + ((frac {20}{29} ))2
= (frac {- 41}{841})

5. If the length of the side opposite to angle A is 15 units and the length of the hypotenuse is 17 units then the length of the side adjacent to angle A is _____
a) 8 units
b) 7 units
c) 4 units
d) 5 units
Answer: a
Clarification: From Pythagoras theorem, (Hypotenuse)2 = (Opposite side)2 + (Adjacent side)2
(Adjacent side)2 = (Hypotenuse)2 – (Opposite side)2
Adjacent side = (sqrt {289 – 225}) = 8 units

6. The meaning of the word trigonometry is three angles measure.
a) True
b) False
Answer: a
Clarification: The word trigonometry is from the language Greek where ‘Tri’ means three and ‘Goria’ means ‘angle’ and ‘Metron’ means measure which gives the meaning as three angles measure.

7. Trigonometry is also applicable to obtuse angles triangles.
a) True
b) False
Answer: b
Clarification: Trigonometry is only applicable to right – angled triangles whose angle is 90° between two sides of a triangle and it gives the relation between the length of sides and angles of a right – angled triangle.

8. If the length of the opposite side is 20 units and the length of the hypotenuse is 29 units then find cosec A?
a) (frac {20}{29})
b) (frac {29}{21})
c) (frac {21}{20})
d) (frac {29}{20})
Answer: d
Clarification: Cosec A = (frac {Length , of , the , hypotenuse}{Length , of , the , opposite , side})
= (frac {29}{20})

9. If tan A = (frac {89}{17}), then cot A is _____
a) (frac {89}{17})
b) (frac {89}{16})
c) (frac {17}{89})
d) (frac {16}{89})
Answer: c
Clarification: Tan A and cot A are reciprocal ratios and cot is inverse of tan.
Tan A = (frac {1}{Cot A} = frac {1}{89/17})
= (frac {17}{89})

10. If cos C = (frac {8}{17}) and sec C = (frac {17}{8}), then 1 is the product of cos C and sec C.
a) False
b) True
Answer: b
Clarification: Cos and sine are reciprocal trigonometric ratios. These two ratios are inverse to each other.
Cos C = (frac {1}{Sec , C})
(Cos C)(Sec C) = 1
((frac {8}{17}) (frac {17}{8})) = 1

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