Category: Class 10 Maths Objective Questions

[CLASS 10] Mathematics MCQs on Geometry – Distance Formula

Mathematics Multiple Choice Questions & Answers on “Geometry – Distance Formula”.

1. The distance between the points (5, 7) and (8, -5) is ________
a) √153
b) √154
c) √13
d) √53
Answer: a
Clarification: Using distance formula,
Distance between (5, 7) and (8, -5) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(8-5)^2 + (-5-7)^2})
= ( sqrt {(3)^2 + (-12)^2} )
= ( sqrt {9 + 144})
= √153

2. The distance of the point (9, -12) from origin will be ___________
a) 13
b) 15
c) 14
d) 17
Answer: b
Clarification: Distance between (9, -12) and (0, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(0-9)^2 + (0 + 12)^2} )
= ( sqrt {(9)^2 + (-12)^2} )
= ( sqrt {81 + 144})
= √225 = 15

3. What will be the value of x, if the distance between the points (5, 11) and (2, x) is 10?
a) -11 + √91, -11 – √91
b) 11 + √91, 11 – √91
c) 11 + √91, 11 + √91
d) -11 + √91, 11 – √91
Answer: b
Clarification: Distance between (5, 11) and (2, x) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(2-5)^2 + (x-11)^2} )
= ( sqrt {x^2-22x + 121 + (-3)^2} )
= ( sqrt {x^2-22x + 121 + 9} )
= ( sqrt {x^2-22x + 130} )
The distance between (5, 11) and (2, x) is 10
∴ ( sqrt {x^2-22x + 130} ) = 10
Squaring on both sides we get,
x2 – 22x + 130 = 100
x2 – 22x + 130 – 100 = 0
x2 – 22x + 30 = 0
x = 11 + √91, 11 – √91

4. What will be the point of x-axis which will be equidistant from the points (9, 8) and (3, 2)?
a) (10, 0)
b) (13, 0)
c) (11, 0)
d) (12, 0)
Answer: c
Clarification: Let the point on x-axis be (x, 0)
Distance between (9, 8) and (x, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(x-9)^2 + (0-8)^2} )
= ( sqrt {x^2-18x + 81 + (-8)^2} )
= ( sqrt {x^2-18x + 81 + 64} )
= ( sqrt {x^2-18x + 145} )
Distance between (3, 2) and (x, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(x-3)^2 + (0-2)^2} )
= ( sqrt {x^2-6x + 9 + (-2)^2} )
= ( sqrt {x^2-6x + 9 + 4} )
= ( sqrt {x^2-6x + 13} )
Since, the point ( x, 0) is equidistant to (3, 2) and (9, 8)
The distances will be equal
∴ ( sqrt {x^2-18x + 145} = sqrt {x^2-6x + 13} )
Squaring on both sides we get,
x2 – 18x + 145 = x2 – 6x + 13
-18x + 145 = -6x + 13
-18x + 6x = -145 + 13
-12x = -132
x = ( frac {132}{12} ) = 11
The point is (11, 0)

5. What will be the point of y-axis which will be equidistant from the points (-1, 0) and (3, 9)?
a) (5, ( frac {89}{18} ))
b) (1, ( frac {89}{18} ))
c) (16, ( frac {89}{18} ))
d) (0, ( frac {89}{18} ))
Answer: d
Clarification: Let the point on y-axis be (0, y)
Distance between (-1, 0) and (0, y) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(0 + 1)^2 + (y-0)^2} )
= ( sqrt {y^2 + (1)^2} )
= ( sqrt {y^2 + 1} )
Distance between (3, 9) and (0, y) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(0-3)^2 + (y-9)^2} )
= ( sqrt {y^2-18y + 81 + (-3)^2} )
= ( sqrt {y^2-18y + 81 + 9} )
= ( sqrt {y^2-18y + 90} )
Since, the point (0, y) is equidistant from (-1, 0) and (3, 9)
The distances will be equal
∴ ( sqrt {y^2 + 1} = sqrt {y^2-18y + 90} )
Squaring on both sides we get,
y2 + 1 = y2 – 18y + 90
1 – 90 = -18y
-89 = -18y
y = ( frac {89}{18} )
The point is (0, ( frac {89}{18} ))

6. If the point P(a, b) is equidistant from the points (3, 1) and (2, 0) then ____________
a) a + b = -3
b) a – b = -3
c) a + b = 3
d) a – b = 3
Answer: a
Clarification: The point is (a, b)
Distance between (3, 1) and (a, b) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(a-3)^2 + (b-1)^2} )
= ( sqrt {a^2-6a + 9 + b^2-2b + 1} )
= ( sqrt {a^2-6a + 10 + b^2-2b} )
Distance between (2, 0) and (a, b) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(a-2)^2 + (b-0)^2} )
= ( sqrt {a^2-4a + 4 + b^2 } )
Since, the point (a, b) is equidistant from (-1, 0) and (3, 9)
The distances will be equal
∴ ( sqrt {a^2-6a + 10 + b^2-2b} = sqrt {a^2-4a + 4 + b^2 } )
Squaring on both sides we get,
a2 – 6a + 10 + b2 – 2b = a2 – 4a + 4 + b2
-6a + 10 – 2b = -4a + 4
-2a – 6 = 2b
-a – b = 3
a + b = -3

7. The point on y-axis which is at a distance 5 unit from the point (-5, 7) is ___________
a) (7, 0)
b) (0, 7)
c) (1, 7)
d) (7, 7)
Answer: b
Clarification: Let the point on y-axis be (0, y)
Distance between (-5, 7) and (0, y) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(0 + 5)^2 + (y-7)^2} )
= ( sqrt {y^2-14y + 49 + 25} )
= ( sqrt {y^2-14y + 74} )
The distance between (-5, 7) and (0, y) is 5
∴ ( sqrt {y^2-14y + 74} ) = 5
Squaring on both sides, we get,
y2 – 14y + 74 = 25
y2 – 14y + 49 = 0
y = 7, 7
Hence, the point is (0, 7)

8. The point on x-axis which is at a distance 12 unit from the point (4, 6) is ___________
a) (-4 + ( sqrt {11i} ), 0), (-4 – ( sqrt {11i} ), 0)
b) (-4 – ( sqrt {11i} ), 0), (4 – ( sqrt {11i} ), 0)
c) (4 – ( sqrt {11i} ), 0), (4 – ( sqrt {11i} ), 0)
d) (4 + ( sqrt {11i} ), 0), (4 – ( sqrt {11i} ), 0)
Answer: d
Clarification: Let the point on x-axis be (x, 0)
Distance between (4, 6) and (x, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(x-4)^2 + (0-6)^2} )
= ( sqrt {x^2-8x + 16 + 36} )
= ( sqrt {x^2-8x + 52} )
The distance between (4, 6) and (x, 0) is 12
∴ ( sqrt {x^2-8x + 52} ) = 12
Squaring on both sides, we get,
x2 – 8x + 52 = 25
x2 – 8x + 27 = 0
x = 4 + ( sqrt {11i} ), 4 – ( sqrt {11i} )

9. If A(0, 3), B(5, 0) and C(-5, 0) are the vertices of ∆ABC, then the triangle is __________
a) Right-angled
b) Isosceles
c) Scalene
d) Equilateral
Answer: b
Clarification: Distance between (0, 3) and (5, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(5-0)^2 + (0-3)^2} )
= ( sqrt {5^2 + -3^2 } )
= ( sqrt {25 + 9} )
= √34
Distance between (5, 0) and (-5, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(-5-5)^2 + (0-0)^2} )
= ( sqrt {-10^2} )
= 10
Distance between (0, 3) and (-5, 0) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(-5-0)^2 + (0-3)^2} )
= ( sqrt {-5^2 + (-3)^2} )
= ( sqrt {25 + 9} )
= √34
Since, the two sides of the triangle are equal.
Hence, the triangle will be isosceles triangle.

10. The area of the triangle if A (-1, -1), B(-1, 3) and C (2, -1) are the vertices of the triangle is ____________
a) 8 units
b) 4 units
c) 6 units
d) 5 units
Answer: c
Clarification: Distance between A (-1, -1) and B (-1, 3) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(-1 + 1)^2 + (3 + 1)^2} )
= ( sqrt {4^2} )
= √16
= 4
Distance between B (-1, 3) and C(2, -1) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(2 + 1)^2 + (-1-3)^2} )
= ( sqrt {3^2 + (-4)^2} )
= ( sqrt {9 + 16} )
= 5
Distance between A (-1, -1) and C (2, -1) = ( sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} )
= ( sqrt {(2 + 1)^2 + (-1 + 1)^2} )
= ( sqrt {3^2 + 0^2} )
= 3
Now, AC2 + AB2 = 42 + 32 = 16 + 9 = 25
BC2 = 52 = 25
Hence, it is a right-angled triangle, right-angled at A.
Area of triangle = ( frac {1}{2}) × base × height = ( frac {1}{2}) × 4 × 3 = 6 units

[CLASS 10] Mathematics MCQs on Irrational and Rational Numbers

Mathematics Multiple Choice Questions & Answers on “Irrational and Rational Numbers”.

1. If x is a number whose simplest form is (frac {p}{q}), where p and q are integers and q≠0, then x is a terminating decimal only when q is of the form _________
a) 3m×5n
b) 2m×6n
c) 2m×5n
d) 7m×5n
Answer: b
Clarification: Let’s, take a number where q is of the form 2m×5n, say 250×510and p can be any integer
(frac {p}{2^{50}times 5^{10}} = frac {p times 5^{40}}{10^{50}})
The number (frac {p times 5^{40}}{10^{50}}) will terminate after 50 decimal places.
Hence, if q is of the form 2m×5n, it will terminate after some decimal places.

2. If x is a number whose simplest form is (frac {p}{q}), where p and q are integers and q ≠ 0, then x is a non-terminating repeating decimal only when q is not of the form ________
a) 2m×2n
b) 5m×5n
c) 2m×5n
d) 3m×4n
Answer: c
Clarification: Let’s, take a number where q is of the form 2m×5n, say 250×510and p can be any integer
(frac {p}{2^{50}times 5^{10}} = frac {p times 5^{40}}{10^{50}})
The number (frac {p times 5^{40}}{10^{50}}) will terminate after 50 decimal places.
Hence, if q is of the form 2m×5n, it will terminate after some decimal places.

3. Which of the following rational is non-terminating repeating decimal?
a) 0.25
b) (frac {4}{5})
c) (frac {4}{55})
d) (frac {2}{5})
Answer: c
Clarification: The value of (frac {4}{55}) is 0.07272727272…., which is non-terminating repeating decimal.
The other numbers terminate after few places of decimal.

4. The terminating rational number from the following numbers is _________
a) (frac {4}{9})
b) (frac {4}{3})
c) 0.146
d) (frac {4}{5})
Answer: d
Clarification: The value of (frac {4}{5}) is 0.8, which is terminating decimal.

5. The simplest form of the rational number 0.196 is ________
a) (frac {1}{6})
b) (frac {3}{6})
c) (frac {13}{66})
d) (frac {2}{5})
Answer: c
Clarification:
10x = 1.969696…..(1)
1000x = 196.9696…(2)
Subtracting (1) from (2)
We get,
990x=195
x = (frac {195}{990} = frac {13}{66})

6. The numbers of the form (frac {p}{q}) are integers, and q≠0 are called irrational number.
a) True
b) False
Answer: b
Clarification:
Irrational numbers cannot be written in the form of (frac {p}{q}).
For example, ∛4 cannot be written in a fraction form as it has non-terminating and non-repeating decimals.

7. After how many places of decimal, will the decimal expansion of the rational number (frac {57}{2^4 5^6}) terminate?
a) 4
b) 6
c) 7
d) 8
Answer: b
Clarification:
We have,
(frac {57}{2^4 5^6} = frac {57 times 2^2}{2^6 5^6} = frac {228}{10^6}) = 0.000228
The number (frac {57}{2^4 5^6}) will terminate after 6 decimal places.

8. From the following numbers, which number is not a rational number?
a) π
b) (frac {22}{7})
c) (frac {3}{4})
d) 0.666666…..
Answer: a
Clarification: A rational number has terminating or non-terminating but repeating decimals.
In case of π, it has a non-terminating as well as non-repeating decimal.
The other three numbers have terminating or non-terminating but repeating decimal, therefore, they are rational numbers.
Hence, it is an irrational number.

9. An irrational number has ________
a) Non-terminating decimal
b) Non-repeating decimal
c) Non-terminating and non-repeating decimal
d) Terminating decimal
Answer: c
Clarification: An irrational number has both non-terminating as well as non-repeating decimals.
For example, the number 1.353353335… has non-terminating as well as non-repeating decimals.

10. Which of the following numbers is not an irrational number?
a) π
b) (frac {22}{7})
c) 1.5353353335….
d) 2.7878878887….
Answer: b
Clarification:
An irrational number is expressible in the decimal form as non-terminating and non-repeating decimals.
From the given options,
π, 1.5353353335…, 2.7878878887… are non-terminating and non-repeating decimal.
Whereas, (frac {22}{7}) is non-terminating but repeating decimal.

11. The product of (frac {33}{2}) and (frac {5}{4}) is an irrational number.
a) True
b) False
Answer: b
Clarification:
(frac {33}{2} times frac {5}{4} = frac {165}{8})
(frac {165}{8}) is a rational number

12. The product of a rational and an irrational number is rational number.
a) True
b) False
Answer: b
Clarification: Take a rational and an irrational number, say 2 and 3√3
Product of 2 × 3√3 = 6√3.
6√3 is an irrational number
Hence, the product of a rational and an irrational number is a irrational number.

13. The product of two irrational numbers is an irrational number.
a) True
b) False
Answer: b
Clarification: Consider an irrational number, say √10
√10 × √10=10
10 is a rational number. Hence, the product of two irrational numbers is not always irrational.

14. The sum of two rational numbers is a rational number.
a) False
b) True
Answer: b
Clarification: Consider two rational numbers, say (frac {8}{9}, frac {3}{5})
Sum of these number = (frac {8}{9} + frac {3}{5} = frac {67}{45}), which is rational number.
Hence, the sum of two rational numbers is a rational number.

15. The sum of two irrational numbers is a rational number.
a) False
b) True
Answer: a
Clarification: Consider two irrational numbers, say, √2 and √5
Sum of these number = √2 + √3 = 3.14626… which is an irrational number.
Hence, the sum of two irrational numbers is an irrational number.

[CLASS 10] Mathematics MCQs on Number of Tangents from a Point on the Circle

Mathematics Online Quiz for Class 10 on “Number of Tangents from a Point on the Circle”.

1. How many tangents can a circle have?
a) Zero
b) Infinity
c) Estimated on the value of the radius
d) Fixed for every kind of circle
Answer: b
Clarification: A circle is a set of points on a plane that is equidistant from a fixed point and an infinite number of tangents can be drawn for any given circle.

2. Find the number of tangents that can be drawn for the given figure.

a) Estimated by using the Pythagorean theorem
b) Estimated on the value of the angle
c) Cannot be drawn
d) Estimated on the value of the diameter
Answer: c
Clarification: Tangent is part of a circle. It cannot be drawn for any other figure other than a circle. Hence, tangents cannot be drawn to the given figure which is a right – angled triangle.

3. How many tangents can be drawn at one point on a circle?
a) Only one
b) Three
c) Zero
d) Two
Answer: a
Clarification: A circle is a set of points on a plane that is equidistant from a fixed point and we can draw only one tangent at one point on any circle given.

4. A tangent touches a circle at a single point.
a) False
b) True
Answer: b
Clarification: A line that touches/intersects a circle at exactly one point of a circle is called a tangent and an infinite number of tangents are drawn to a circle whereas secant is a line that intersects two distinct points on a circle.

5. Number of tangents passing through a circle is _____
a) 2
b) 3
c) 1
d) 0
Answer: d
Clarification: A line that touches/intersects a circle at exactly one point of a circle is called a tangent and a tangent to a circle doesn’t pass through the circle.

6. What happens to the length of the chord when the chord comes closer to the center?
a) Decreases
b) Becomes an arc
c) Increases
d) Becomes a segment
Answer: c
Clarification: The length of the chord of a circle increases when it comes closer and closer to the center of the circle. Hence, the longest chord becomes the diameter.

7. Find the area of the sector if the radius is 6 cm and with an angle of 60°.
a) 18.35 cm
b) 18.85 cm
c) 18.00 cm
d) 18.05 cm
Answer: b
Clarification: The area of the sector = (frac {x^{circ }}{360^{circ }}) × πr2
= (frac {60^{circ }}{360^{circ }}) x (frac {22}{7}) × 62
= 18.85 cm

8. The area of the sector is _____
a) (frac {x^{circ }}{360^{circ }}) × πr2
b) (frac {x^{circ }}{360^{circ }}) – πr2
c) (frac {x^{circ }}{360^{circ }}) + πr3
d) (frac {x^{circ }}{360^{circ }}) × πr3
Answer: a
Clarification: The area of the sector is (frac {x^{circ }}{360^{circ }}) × πr2
Where x° is the degree measure of the angle at the center and r is the radius of the circle.

9. Find the radius of a circle if 2 m is the length of the tangent, 6 m is the distance between the center of the circle the external point.
a) 7 m
b) 5 m
c) √32 m
d) √38 m
Answer: c
Clarification: Length of the tangent = (sqrt {d^2 – r^2} )
2 = (sqrt {6^2 – r^2} )
r2 = 62 – 22
r = (sqrt {6^2 – 2^2} )
r = (sqrt {36 – 4} )
r = √32 m

10. Line C is secant to the circle.

a) False
b) True
Answer: a
Clarification: Tangent is a line that touches the circle at a single point. Hence, line C is a tangent, not a secant. Line A and line B are secants of the circle because they’re touching the circle at two distinct points.

To practice Mathematics Online Quiz for Class 10,

[CLASS 10] Mathematics MCQs on Pythagoras Theorem

Mathematics Multiple Choice Questions & Answers on “Pythagoras Theorem”.

1. ∆ABC is a right angled triangle, where AB = 5cm, BC = 10cm, AC = 15cm.
a) False
b) True

Answer: a
Clarification: If ∆ABC is a right angled triangle, then it should satisfy the Pythagoras Theorem.
AB = 5cm, BC = 10cm, AC = 15cm
AB2 + BC2 = 52 + 102 = 25 + 100 = 125
AC2 = 152 = 225
Since, AC2 ≠ AB2 + BC2
Hence, ABC is not a right-angled triangle.

2. What will be the distance of the foot of ladder from the building, if the ladder of 12 m high reaches the top of a building 35 m high from the ground?
a) 32.65 m
b) 32.87 m
c) 31.87 m
d) 32.85 m
Answer: b
3. The heights of two vertical lamp posts are 33 m and 24 m high. If the distance between them is 40 m, then what will be the distance between their tops?
a) 47.89m
b) 56.56m
c) 32.81m
d) 41m
Answer: d

4. ∆ABC is a right-angled triangle, right angled at B and BD ⊥ AC. If BD = 10cm, AB = 5 cm and BC = 5 cm then AC will be?
a) 44.72 cm
b) 5.59 cm
c) 18.11 cm
d) 22.36 cm

Answer: d

5. Which of triangle whose sides are given below are right angled?
a) AB = 89, AC = 80, BC = 39
b) AB = 57, AC = 50, BC = 45
c) AC = 34, AB = 20, BC = 21
d) AC = 50, AB = 32, BC = 20

Answer: a
Clarification: In (a), applying Pythagoras Theorem,
AC2 + BC2 = 802 + 392 = 7921 = AB2
Hence, this is a right angled triangle.
Now, in (b) applying Pythagoras Theorem,
AC2 + BC2 = 502 + 452 = 4525 ≠ AB2
Hence, this is not a right angled triangle.
Now, in (c) applying Pythagoras Theorem,
AB2 + BC2 = 202 + 212 = 841 ≠ AC2
Hence, this is not a right angled triangle.
Now, in (d) applying Pythagoras Theorem,
AB2 + BC2 = 322 + 202 = 1424 ≠ AC2
Hence, this is not right angled triangle.

6. The lengths of diagonals of a rhombus are 10 cm and 8 cm. What will be the length of the sides of rhombus?
a) 6.40 cm
b) 5.25 cm
c) 2.44 cm
d) 3.29 cm
Answer: a
7. If the side of rhombus is 13 cm and one of its diagonals is 24 cm, then what will be length of the other diagonal?
a) 8.4 cm
b) 4 cm
c) 11 cm
d) 10 cm
Answer: d
8. What will be the length of the altitude of an equilateral triangle whose side is 9 cm?
a) 4.567 cm
b) 7.794 cm
c) 8.765 cm
d) 4.567 cm
Answer: b
9. What will be the length of the square inscribed in a circle of radius 5 cm?
a) 2.34 cm
b) 3.45 cm
c) 5√2 cm
d) 2.45 cm
Answer: c

[CLASS 10] Mathematics MCQs on Real Numbers – Arithmetic Fundamental Theorem

Mathematics Multiple Choice Questions & Answers on “Real Numbers – Arithmetic Fundamental Theorem”.

1. Which of the following is a composite number?
a) 2
b) 3
c) 9
d) 7
Answer: c
Clarification: A prime number has two factors the number itself and 1. In case of 9 there are three factors i.e. 3 × 3 × 1. Hence, it is not a prime number.

2. Which of the following is a prime number?
a) 31
b) 52
c) 21
d) 32
Answer: a
Clarification: A prime number has two factors the number itself and 1. In case of 31 there are two factors i.e. 31 and 1. Hence, it is a prime number.

3. The fundamental theorem of arithmetic states that, every composite number can be factorized as product of primes and this factorization is unique.
a) False
b) True
Answer: b
Clarification: Let us consider a composite number, say 25
25 can be factorized as 5 × 5 × 1. This factorization is unique for 25 and no other number can be represented in the same manner.

4. The least common multiple of 135 and 24 is _________
a) 90
b) 360
c) 240
d) 1080
Answer: d
Clarification: 135 can be written as 3 × 3 × 3 × 5 × 1 and 24 can be written as 2 × 2 × 2 × 3 × 1.
LCM is the product of greatest power of each prime factor involved in the numbers.
Therefore, LCM = 2 × 2 × 2 × 3 × 3 × 3 × 5 = 1080

5. The highest common factor of 21 and 90 is _________
a) 3
b) 2
c) 1
d) 4
Answer: a
Clarification: 21 can be written as 3 × 7 × 1 and 90 can be written as 3 × 3 × 2 × 5.
HCF is the product of smallest power of each prime factor involved in the numbers.
Therefore, HCF = 3 × 1 = 3

6. The LCM of two numbers is 7991 and the two numbers are 61 and 131. What will be their HCF?
a) 2
b) 1
c) 3
d) 4
Answer: b
Clarification: For two numbers a and b, we know that
(a × b) = HCF of (a, b) × LCM of (a, b)
Here a = 61 and b = 131, and LCM is 7991
61 × 131 = HCF × 7991
HCF = (frac {7991}{7991}) = 1

7. What will be the largest number that divides 100 and 25, and leaves 3 as remainder in each case?
a) 7
b) 5
c) 1
d) 4
Answer: c
Clarification: The required number divides (100-3) i.e. 97 and (25-3) i.e. 22 exactly.
Now, 97 = 97 × 1 and 22 = 2 × 11
HCF of 97 and 22 is 1.
Hence, the required number is 1.

8. A bakery sells cookies in three boxes. The three boxes contain 60, 84 and 108 number of cookies. The baker wants to sells all the cookies in any of the three boxes. The least number of cookies that he can bake, in a day, so that he is able to sell all his cookies in any of the three boxes is _______
a) 5467
b) 2243
c) 1123
d) 3780
Answer: d
Clarification: The three boxes contain 60, 84 and 108 number of cookies.
60 can be factorized as 2 × 2 × 3 × 5, 84 as 2 × 2 × 3 × 7 and 108 as 2 × 2 × 3 × 3 × 3
To find the least number of cookies that can be filled in the container, we have to find the LCM of the three numbers
LCM of 60, 84 and 108 = 2 × 2 × 3 × 3 × 3 × 5 × 7 = 3780
Hence, the least number of cookies that can be filled in the container is 3780.

9. Two buckets contain 546 and 764 liters of water respectively. What will be maximum capacity of container which can measure the water of either buckets exact number of times?
a) 108
b) 54
c) 34
d) 456
Answer: a
Clarification: The two buckets contain 546 and 764 liters of water.
546 can be factorized as 2 × 2 × 3 × 3 × 3 × 5 and 764 as 2 × 2 × 3 × 3 × 3 × 7.
To find the maximum capacity of the container which can measure the water of either buckets exact number of times, we have to find the HCF of the two numbers.
HCF of 546 and 764 = 22 × 33=108
Hence, the maximum capacity of the container is 108 liters.

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[CLASS 10] Mathematics MCQs on Perimeter and Area of a Circle

Mathematics Multiple Choice Questions & Answers on “Perimeter and Area of a Circle”.

1. What is the formula for the circumference of a circle?
a) C = 2πd
b) C = 2πr
c) C = 2πa
d) C = 2πs
Answer: b
Clarification: The circumference of a circle (C) = 2πr.
Where C is the circumference of the circle and r is the radius of the given circle.
The value of π is taken as (frac {22}{7}).

2. What is the formula for the area of a circle?
a) A = πd2
b) A = πs2
c) A = πr2
d) A = πa2
Answer: c
Clarification: The area of a circle (A) = πr2.
Where A is the area of the circle and r is the radius of the given circle.
The value of π is taken as (frac {22}{7}).

3. Find the area of the circle if the radius is 3.14 cm.
a) 30.98 cm
b) 30.48 cm
c) 30.68 cm
d) 30.58 cm
Answer: a
Clarification: The area of a circle (A) = πr2
A = (frac {22}{7}) × (3.14)2
A = 30.98 cm

4. The perimeter of a circle is also called the circumference.
a) False
b) True
Answer: b
Clarification: The path that surrounds the outline of a two-dimensional shape is called the perimeter. The perimeter of a circle is also called the circumference.

5. The area of a semicircle is _____
a) (frac {2pi r}{2})
b) (frac {2pi r}{2})
c) (frac {pi r}{2})
d) (frac {pi r^2}{2})
Answer: d
Clarification: A semicircle is half of a full circle. Hence, the area of a semicircle is also half of a circle. Half of πr2 is (frac {pi r^2}{2}). Therefore, the area of a semicircle is (frac {pi r^2}{2}).

6. What is the circumference of a circle if the radius is 7 m?
a) 8 m
b) 2 m
c) 44 m
d) 22 m
Answer: c
Clarification: The circumference of a circle (C) = 2πr.
C = 2 × (frac {22}{7}) × 7
C = 44 m

7. Find the area of a semicircle if the radius is 6 cm.
a) 1.35 m
b) 6.54 m
c) 18.00 m
d) 8.05 m
Answer: b
Clarification: The area of the semicircle = πr2/2
= ((frac {22}{7}) × 62)/2
= 56.54 m

8. Find the radius of the circle if the circumference is 12 m.
a) 1.90 m
b) 1.09 m
c) 7.90 m
d) 1.40 m
Answer: a
Clarification: The circumference of a circle (C) = 2πr.
12 = 2 × (frac {22}{7}) × r
r = (frac {12}{2} times frac {7}{22})
r = 1.90 m

9. Find the radius of a circle if 2 m is the area of the circle.
a) √0.83 m
b) 5 m
c) √0.63 m
d) √38 m
Answer: c
Clarification: The area of the circle (A) = πr2
2 = (frac {22}{7}) × r2
r2 = (frac {7}{22}) × 2
r = √0.63 m

10. Find the radius of the circle if the area of the circle is 22 cm.
a) 1521.14 m
b) 1511.14 m
c) 1021.14 m
d) 1520.14 m
Answer: a
Clarification: The area of the circle (A) = πr2
A = (frac {22}{7}) (22)2
A = 1521.14 m

11. What is the circumference of the circle if the radius is 121 cm?
a) 760.00 cm
b) 765.57 cm
c) 750.57 cm
d) 760.57 cm
Answer: d
Clarification: The circumference of the circle (C) = 2πr
C = 2 × (frac {22}{7}) × 121
A = 760.57 cm

12. Find the radius of the wheel if the wheel rotates 100 times to cover 500 m.
a) 0.07 m
b) 0.47 cm
c) 0.79 m
d) 0.57 cm
Answer: c
Clarification: One rotation of the wheel = circumference of the wheel
100 rotations = 500 m
1 rotation = 5 m
So, circumference = 5 m
2πr = 5 m
r = 5 × (frac {7}{22} times frac {1}{2})
r = 0.79m

13. The difference between circumference and diameter of a ring is 10 cm. Find the radius of the ring.
a) 3.07 cm
b) 0.37 cm
c) 2.33 cm
d) 4.57 cm
Answer: c
Clarification: Circumference – Diameter = 10 cm (∵ Diameter = 2 × radius)
2πr – 2r = 10 cm
2r(π – 1) = 10 cm
2r((frac {22}{7}) – 1) = 10 cm
2r((frac {15}{7})) = 10 cm
r = 10 × (frac {7}{15} times frac {1}{2})
r = 2.33 cm

14. Find the diameter of the circle if the area of the circle is 6 m.
a) 3.07 m
b) 2.74 m
c) 2.33 m
d) 4.57 m
Answer: b
Clarification: The area of the circle (A) = πr2
6 = (frac {22}{7}) × r2
r2 = 1.90
r = √1.90
r = 1.37 m
Diameter of the circle = 2 × radius
= 2 × 1.37
= 2.74 m

15. Find the diameter of the circle if the circumference of the circle is (frac {11}{5})m.
a) 23.07 cm
b) 20.74 cm
c) 72.33 cm
d) 70 cm
Answer: d
Clarification: The circumference of the circle = (frac {11}{5})m
2πr = (frac {11}{5})m
πd = (frac {11}{5})m     (∵ Diameter = 2 × radius)
d = (frac {11}{5} times frac {7}{22})m
d = 0.7 m
d = 0.7 × 100 cm
d = 70 cm