It can be observed that the maximum class frequency is 10, belonging to 30 - 35. Therefore, Modal class = 30 - 35 Class size (h) = 5 Lower limit (l) of modal class = 30 Frequency (f₁) of modal class = 10 Frequency (f₀) of class preceding the modal class = 9 Frequency (f₂) of class succeeding the modRead more
It can be observed that the maximum class frequency is 10, belonging to 30 – 35.
Therefore, Modal class = 30 – 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f₁) of modal class = 10
Frequency (f₀) of class preceding the modal class = 9
Frequency (f₂) of class succeeding the modal class 3
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 30 + ((10 – 9)/(2 × 10 – 9 – 3)) × 5 = 30 + 1/8 × 5 = 30 + 0.625 = 30.625
It represent that most of the states/U.T have a teacher-student ratio as 30.6.
To find the class marks, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Taking 32.5 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows.
From the table, we obtain
∑fᵢ = 35, ∑fᵢuᵢ = -23, a = 32.5 and h = 5
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 32.5 + (-23/35) × 5 = 32.5 – 23/7 = 32.5 – 3.28 = 29.22
Therefore, mean of the data is 29.2.
It represents that on an average, teacher-student ration was 29.2
To find the class marks (xᵢ), the following relation is used. xᵢ = (Upper Limit + Lower Limit)/2 Taking 30 as assumed mean (a), dᵢ and fᵢdᵢ are calculated as follows. From the table, we obtain ∑fᵢ = 80, ∑fᵢdᵢ = 430 and a = 30 mean (X̄) = a + (∑fᵢdᵢ / ∑fᵢ) = 30 + (430/80) = 30 + 5.375 = 35.375 = 35.3Read more
To find the class marks (xᵢ), the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Taking 30 as assumed mean (a), dᵢ and fᵢdᵢ are calculated as follows.
From the table, we obtain
∑fᵢ = 80, ∑fᵢdᵢ = 430 and a = 30
mean (X̄) = a + (∑fᵢdᵢ / ∑fᵢ) = 30 + (430/80) = 30 + 5.375 = 35.375 = 35.38
Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years. It can be observed that the maximum class frequency is 23 belonging to class interval 35 -45.
Modal class = 35 – 45
Lower limit (l) of modal class 35
Frequency (f₁) of modal class = 23
Class size (h) = 10
Frequency (f₀) of class preceding the modal class = 21
Frequency (f₂) of class succeeding the modal class 14
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 35 + ((23 – 21)/(2 × 23 – 21 – 14)) × 10 = 35 + 2/11 × 10 = 35 + 1.81 = 36.81
Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.
Here is the video explanation of the above question😃
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 - 80. Therefore, Modal class = 60 - 80 Lower limit (l) of modal class = 60 Frequency (f₁) of modal class = 61 Frequency (f₀) of class preceding the modal class = 52 Frequency (f₂) ofRead more
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 – 80.
Therefore, Modal class = 60 – 80
Lower limit (l) of modal class = 60
Frequency (f₁) of modal class = 61
Frequency (f₀) of class preceding the modal class = 52
Frequency (f₂) of class succeeding the modal class 38
Class size (h) = 20
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 60 + ((61 – 52)/(2 × 61 – 52 – 38)) × 20 = 60 + 9/32 × 20 = 60 + 5.625 = 65.625
Therefore, modal lifeline of electrical components is 65.625 hours.
It can be observed that the maximum class frequency is 40, belonging to 1500 - 2000 intervals. Therefore, Modal class = 1500 - 2000 Lower limit (l) of modal class = 1500 Frequency (f₁) of modal class = 40 Frequency (f₀) of class preceding the modal class = 24 Frequency (f₂) of class succeeding the mRead more
It can be observed that the maximum class frequency is 40, belonging to 1500 – 2000 intervals.
Therefore, Modal class = 1500 – 2000
Lower limit (l) of modal class = 1500
Frequency (f₁) of modal class = 40
Frequency (f₀) of class preceding the modal class = 24
Frequency (f₂) of class succeeding the modal class 33
Class size (h) = 500
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 500 + ((40 – 24)/(2 × 40 – 24 – 33)) × 500 = 1500 + 16/23 × 500 = 1500 + 347.826 = 1847.83
Therefore, modal monthly expenditure was ₹1847.83
To find the class mark, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Class size (h) of the given data = 500.
Taking 2750 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows.
From the table, we obtain
∑fᵢ = 200, ∑fᵢuᵢ = -35, a = 2750 and h = 500
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 2750 + (-35/200) × 500 = 2750 – 87.5 = 2662.50
We know that each angle of equilateral triangle is 60°. Area of sector OCDE = 60°/360° × πr² = 1/6 × π(6)² = 1/6 × 22/7 × 6 × 6 = 132/7 cm² Area of equilateral triangle OAB = √3/4 (12)² = 36√3 cm² Area of circle = πr² = π(6)² = 22/7 × 6 × 6 = 792/7 cm² Area of shaded region = Area of circle + Area oRead more
We know that each angle of equilateral triangle is 60°.
Area of sector OCDE
= 60°/360° × πr² = 1/6 × π(6)² = 1/6 × 22/7 × 6 × 6 = 132/7 cm²
Area of equilateral triangle OAB
= √3/4 (12)² = 36√3 cm²
Area of circle
= πr² = π(6)² = 22/7 × 6 × 6 = 792/7 cm²
Area of shaded region
= Area of circle + Area of triangle OAB- Area of sector OCDE
= (792/7 + 36√3 – 132/7) cm²
= (36√3 + 660/7) cm²
Radius of circle = 32 Centroid O divides median AD into 2:1, therefore A0:0D = 2:1 = AO/OD = 2/1 ⇒ 32/OD = 2/1 ⇒ OD = 16 Therefore, AD = 32 + 16 = 48 cm In AABD, AB² = AD² + BD² = (48)² + (AB/2)² ⇒ 3/4 AB² = (48)² ⇒ AB = (48 × 2)/√3 = (96√3)/3 = 32√3 Area of equilateral triangle ABC = √3/4 (32√3)² =Read more
Radius of circle = 32
Centroid O divides median AD into 2:1, therefore A0:0D = 2:1
= AO/OD = 2/1 ⇒ 32/OD = 2/1 ⇒ OD = 16
Therefore, AD = 32 + 16 = 48 cm
In AABD,
AB² = AD² + BD² = (48)² + (AB/2)²
⇒ 3/4 AB² = (48)² ⇒ AB = (48 × 2)/√3 = (96√3)/3 = 32√3
Area of equilateral triangle ABC
= √3/4 (32√3)² = 768√3 cm²
Area of circle
= πr² = π(32)² = 22/7 × 32 × 32 = 22528/7 cm²
Area of design = Area of circle – Area of equilateral triangle ABC
= (22528/7 – 768√3) cm²
The circles drawn taking A, B, C and D form quadrants of radius 7 cm in square. Radius of each quadrant = 7 cm Area of each quadrant = 90°/360° × πr² = 1/4 × π(7)² = 1/4 × 22/7 × 7 × 7 = 77/2 cm² Area of square = (Side)² = (14)² = 196 cm² Area of shaded region = Area of square - Area of 4 quadrantsRead more
The circles drawn taking A, B, C and D form quadrants of radius 7 cm in square.
Radius of each quadrant = 7 cm
Area of each quadrant
= 90°/360° × πr² = 1/4 × π(7)²
= 1/4 × 22/7 × 7 × 7 = 77/2 cm²
Area of square = (Side)² = (14)² = 196 cm²
Area of shaded region = Area of square – Area of 4 quadrants
= 196 – 4 × 77/2 = 196- 154 = 42 cm²
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India.
It can be observed that the maximum class frequency is 10, belonging to 30 - 35. Therefore, Modal class = 30 - 35 Class size (h) = 5 Lower limit (l) of modal class = 30 Frequency (f₁) of modal class = 10 Frequency (f₀) of class preceding the modal class = 9 Frequency (f₂) of class succeeding the modRead more
It can be observed that the maximum class frequency is 10, belonging to 30 – 35.
Therefore, Modal class = 30 – 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f₁) of modal class = 10
Frequency (f₀) of class preceding the modal class = 9
Frequency (f₂) of class succeeding the modal class 3
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 30 + ((10 – 9)/(2 × 10 – 9 – 3)) × 5 = 30 + 1/8 × 5 = 30 + 0.625 = 30.625
It represent that most of the states/U.T have a teacher-student ratio as 30.6.
To find the class marks, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Taking 32.5 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows.
From the table, we obtain
See less∑fᵢ = 35, ∑fᵢuᵢ = -23, a = 32.5 and h = 5
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 32.5 + (-23/35) × 5 = 32.5 – 23/7 = 32.5 – 3.28 = 29.22
Therefore, mean of the data is 29.2.
It represents that on an average, teacher-student ration was 29.2
The following table shows the ages of the patients admitted in a hospital during a year:
To find the class marks (xᵢ), the following relation is used. xᵢ = (Upper Limit + Lower Limit)/2 Taking 30 as assumed mean (a), dᵢ and fᵢdᵢ are calculated as follows. From the table, we obtain ∑fᵢ = 80, ∑fᵢdᵢ = 430 and a = 30 mean (X̄) = a + (∑fᵢdᵢ / ∑fᵢ) = 30 + (430/80) = 30 + 5.375 = 35.375 = 35.3Read more
To find the class marks (xᵢ), the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Taking 30 as assumed mean (a), dᵢ and fᵢdᵢ are calculated as follows.
From the table, we obtain
∑fᵢ = 80, ∑fᵢdᵢ = 430 and a = 30
mean (X̄) = a + (∑fᵢdᵢ / ∑fᵢ) = 30 + (430/80) = 30 + 5.375 = 35.375 = 35.38
Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years. It can be observed that the maximum class frequency is 23 belonging to class interval 35 -45.
Modal class = 35 – 45
Lower limit (l) of modal class 35
Frequency (f₁) of modal class = 23
Class size (h) = 10
Frequency (f₀) of class preceding the modal class = 21
Frequency (f₂) of class succeeding the modal class 14
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 35 + ((23 – 21)/(2 × 23 – 21 – 14)) × 10 = 35 + 2/11 × 10 = 35 + 1.81 = 36.81
Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.
Here is the video explanation of the above question😃
See lessThe following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 - 80. Therefore, Modal class = 60 - 80 Lower limit (l) of modal class = 60 Frequency (f₁) of modal class = 61 Frequency (f₀) of class preceding the modal class = 52 Frequency (f₂) ofRead more
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 – 80.
See lessTherefore, Modal class = 60 – 80
Lower limit (l) of modal class = 60
Frequency (f₁) of modal class = 61
Frequency (f₀) of class preceding the modal class = 52
Frequency (f₂) of class succeeding the modal class 38
Class size (h) = 20
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 60 + ((61 – 52)/(2 × 61 – 52 – 38)) × 20 = 60 + 9/32 × 20 = 60 + 5.625 = 65.625
Therefore, modal lifeline of electrical components is 65.625 hours.
The following data gives the distribution of total monthly household expenditure of 200 families of a village.
It can be observed that the maximum class frequency is 40, belonging to 1500 - 2000 intervals. Therefore, Modal class = 1500 - 2000 Lower limit (l) of modal class = 1500 Frequency (f₁) of modal class = 40 Frequency (f₀) of class preceding the modal class = 24 Frequency (f₂) of class succeeding the mRead more
It can be observed that the maximum class frequency is 40, belonging to 1500 – 2000 intervals.
Therefore, Modal class = 1500 – 2000
Lower limit (l) of modal class = 1500
Frequency (f₁) of modal class = 40
Frequency (f₀) of class preceding the modal class = 24
Frequency (f₂) of class succeeding the modal class 33
Class size (h) = 500
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 500 + ((40 – 24)/(2 × 40 – 24 – 33)) × 500 = 1500 + 16/23 × 500 = 1500 + 347.826 = 1847.83
Therefore, modal monthly expenditure was ₹1847.83
To find the class mark, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Class size (h) of the given data = 500.
Taking 2750 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows.
From the table, we obtain
See less∑fᵢ = 200, ∑fᵢuᵢ = -35, a = 2750 and h = 500
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 2750 + (-35/200) × 500 = 2750 – 87.5 = 2662.50
Find the area of the shaded region in Figure.
We know that each angle of equilateral triangle is 60°. Area of sector OCDE = 60°/360° × πr² = 1/6 × π(6)² = 1/6 × 22/7 × 6 × 6 = 132/7 cm² Area of equilateral triangle OAB = √3/4 (12)² = 36√3 cm² Area of circle = πr² = π(6)² = 22/7 × 6 × 6 = 792/7 cm² Area of shaded region = Area of circle + Area oRead more
We know that each angle of equilateral triangle is 60°.
Area of sector OCDE
= 60°/360° × πr² = 1/6 × π(6)² = 1/6 × 22/7 × 6 × 6 = 132/7 cm²
Area of equilateral triangle OAB
= √3/4 (12)² = 36√3 cm²
Area of circle
= πr² = π(6)² = 22/7 × 6 × 6 = 792/7 cm²
Area of shaded region
= Area of circle + Area of triangle OAB- Area of sector OCDE
= (792/7 + 36√3 – 132/7) cm²
= (36√3 + 660/7) cm²
Here is the video explanation 😃🙌
See lessIn a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Figure. Find the area of the design.
Radius of circle = 32 Centroid O divides median AD into 2:1, therefore A0:0D = 2:1 = AO/OD = 2/1 ⇒ 32/OD = 2/1 ⇒ OD = 16 Therefore, AD = 32 + 16 = 48 cm In AABD, AB² = AD² + BD² = (48)² + (AB/2)² ⇒ 3/4 AB² = (48)² ⇒ AB = (48 × 2)/√3 = (96√3)/3 = 32√3 Area of equilateral triangle ABC = √3/4 (32√3)² =Read more
Radius of circle = 32
See lessCentroid O divides median AD into 2:1, therefore A0:0D = 2:1
= AO/OD = 2/1 ⇒ 32/OD = 2/1 ⇒ OD = 16
Therefore, AD = 32 + 16 = 48 cm
In AABD,
AB² = AD² + BD² = (48)² + (AB/2)²
⇒ 3/4 AB² = (48)² ⇒ AB = (48 × 2)/√3 = (96√3)/3 = 32√3
Area of equilateral triangle ABC
= √3/4 (32√3)² = 768√3 cm²
Area of circle
= πr² = π(32)² = 22/7 × 32 × 32 = 22528/7 cm²
Area of design = Area of circle – Area of equilateral triangle ABC
= (22528/7 – 768√3) cm²
In this Figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
The circles drawn taking A, B, C and D form quadrants of radius 7 cm in square. Radius of each quadrant = 7 cm Area of each quadrant = 90°/360° × πr² = 1/4 × π(7)² = 1/4 × 22/7 × 7 × 7 = 77/2 cm² Area of square = (Side)² = (14)² = 196 cm² Area of shaded region = Area of square - Area of 4 quadrantsRead more
The circles drawn taking A, B, C and D form quadrants of radius 7 cm in square.
Radius of each quadrant = 7 cm
Area of each quadrant
= 90°/360° × πr² = 1/4 × π(7)²
= 1/4 × 22/7 × 7 × 7 = 77/2 cm²
Area of square = (Side)² = (14)² = 196 cm²
Area of shaded region = Area of square – Area of 4 quadrants
= 196 – 4 × 77/2 = 196- 154 = 42 cm²
Here is the Video explanation 😀👇
See lessFigure depicts a racing track whose left and right ends are semicircular.
(i)The distance around the track along its inner edge = AB + arcBEC + CD + arcDFA = 106 + 1/2 × 2πr + 106 + 1/2 × 2πr = 106 + 1/2 × 2 × 22/7 × 30 + 106 + 1/2 × 2 × 22/7 × 30 = 212 + 2 × 22/7 × 30 = 212 + 1320/7 = 2804/7 (ii) Area of Track = (Area of GHIJ - Area of ABCD) + (Area of semicircle HKI - ARead more
(i)The distance around the track along its inner edge
See less= AB + arcBEC + CD + arcDFA
= 106 + 1/2 × 2πr + 106 + 1/2 × 2πr
= 106 + 1/2 × 2 × 22/7 × 30 + 106 + 1/2 × 2 × 22/7 × 30
= 212 + 2 × 22/7 × 30 = 212 + 1320/7 = 2804/7
(ii) Area of Track = (Area of GHIJ – Area of ABCD) + (Area of semicircle HKI – Area of semicircle BRC) + (Area of semicircle GLJ – Area of semicircle AFD)
= (106 × 80 – 106 × 60) + 1/2 × 22/7 × [(40)² – (30)²] + 1/2 × 22/7 × [(40)² – (30)²]
= 106(80 – 60) + 1/2 × 22/7 × (700) + 1/2 × 22/7 × (700) = 2120 + 22/7 × (700) = 2120 + 2200 = 4320