Concise Mathematics Class 10 ICSE 2018 Solutions Pdf

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24D.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E

Question 1.
Find the mode of the following data:
(i) 7,9,8,7,7,6,8,10,7 and 6
(ii) 9,11,8,11,16,9,11,5,3,11,17 and 8
Solution:
(i) Mode = 7
because it occurs 4 times
(ii) Mode =11
because it occurs 4 times

Question 2.
The following table shows the frequency distribution of heights of 50 boys:

Find the mode of heights.
Solution:
Mode is 122 because it occurs maximum times i.e its., frequency is 18.

Question 3.
Find the mode of following data, using a histogram:

Solution:
Mode class = 20 – 30
Mode = 24

We see in the histogram that line AD and CB intersect at P. Draw perpendicular Q to the horizontal x-axis. Which is the value of the mode = 24

Question 4.
The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:

Solution:
Model class is = 30 – 35
and Mode = 34

We see in the histogram that line AD and CB intersect at P. Draw perpendicular Q to the horizontal axis. Which is the value of the mode.

Question 5.
Find the median and mode for the set of numbers 2,2,3,5,5,5,6,8 and 9.
Solution:
Median = \(\frac { 9 +1 }{ 2 }\) = 5th term which is 5
Mode = 5, because it occurs in maximum times.

Question 6.
A boy scored following marks in various class tests during a term, each test being marked out of 20.
15,17,16,7,10,12,14,16,19,12,16.
(i) What are his modal marks ?
(ii) What are his median marks ?
(iii) What are his total marks ?
(iv) What are his mean marks ?
Solution:
Arranging the given data in ascending order : 7, 10,12, 12,14, 15,16,16, 16, 17,19.
(i) Mode = 16 as it occurs in maximum times.
(ii) Median= \(\frac { 11 +1 }{ 2 }\) = 6th term which is 15
(iii) Total marks = 7 + 10+ 12+ 12+ 14+ 15+ 16 + 16+ 16+ 17+ 19= 154

Question 7.
Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks.
0,0,2,2,3,3,3,4,5,5,5,5,6, 6,7,8
Solution:

(ii) Median = Mean of 8th and 9th term
= \(\frac { 4 +5 }{ 2 }\) = \(\frac { 9 }{ 2 }\) = 4.5
(iii) Mode = 5 as it occurs in maximum times.

Question 8.
At a shooting competition the score of a com-petitor were as given below :

(i) What was his modal score ?
(ii) What was his median score ?
(iii) What was his total score ?
(iv) What was his mean score ?
Solution:

(i) Modal score =4 as its frequency is 7, the maximum.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F

Question 1.
Find three numbers in A.P. whose sum is 24 and whose product is 440.
Solution:
Let three numbers be a – d, a, a + d
a – d + a + a + d = 24

Question 2.
The sum of three consecutive terms of an A.P. is 21 and the slim of their squares is 165. Find these terms.
Solution:
Let three consecutive numbers in A.P. are
a – d, a, a + d
a – d + a + a + d = 21

Question 3.
The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.
Solution:
Let the angles of a quadrilateral are
a, a + d, a + 2d, a + 3d
d= 20°
a + a + d + a + 2d + a + 3d = 360°
(Sum of angles of a quadrilateral)
=> 4a + 6d = 360°
=> 4a + 6 x 20° = 360°
=> 4a + 120° = 360°
=> 4a = 360 – 120° = 240°
a = \(\\ \frac { 240 }{ 4 } \) = 60°
Angles are 60°, 80°, 100°, 120°

Question 4.
Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.
Solution:
Number = 96
Let its four parts be a, a + d, a + 2d, a + 3d
a + a + d + a + 2d + a + 3d = 96
=> 4a + 6d = 96
=> 2a + 3d = 48 …(i)
Product of means : Product of extremes = 15 : 7

Question 5.
Find five numbers in A.P. whose sum is \(12 \frac { 1 }{ 2 } \) and the ratio of the first to the last terms is 2 : 3.
Solution:
Let 5 numbers in A.P. be
a, a + d, a + 2d, a + 2d, a + 4d
a + a + d + a + 2d + a + 3d + a + 4d =

Question 6.
Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.
Solution:
Number = 207
Let part be a – d, a, a + d
a – d + a + a + d = 207

Question 7.
The sum of three numbers in A.P. is 15 the sum of the squares of the extreme is 58. Find the numbers
Solution:
Let three numbers in A.P. be a – d, a, a + d
a – d + a + a + d = 15
=> 3a = 15

Question 8.
Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
Solution:
Let four numbers in A.P. be
a – 3d, a – d, a + d, a + 3d
a – 3d + a – d + a + d + a + 3d = 20

Question 9.
Insert one arithmetic mean between 3 and 13.
Solution:
Let A be the arithmetic mean between 3 and 13
\(\left( A=\frac { a+b }{ 2 } \right) \)
A = \(\\ \frac { 3+13 }{ 2 } \)
= \(\\ \frac { 16 }{ 2 } \)
= 8

Question 10.
The angles of a polygon are in A.P. with common difference 5°. If the smallest angle is 120°, find the number of sides of the polygon.
Solution:
Angles of a polygon are in A.P.
and common difference (d) = 5°
Smallest angle (a) = 120°
Let n be the number of sides of the polygon then sum of angles = (2n – 4) x 90°
a = 120° and d = 5°

Question 11.
\(\\ \frac { 1 }{ a } \), \(\\ \frac { 1 }{ b } \) and \(\\ \frac { 1 }{ c } \) are in A.P
S.T : bc, ca and ab are also in A.P
Solution:
\(\\ \frac { 1 }{ a } \), \(\\ \frac { 1 }{ b } \) and \(\\ \frac { 1 }{ c } \) are in A.P
We have to show that

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems (Based on Quadratic Equations) Ex 6B

Question 1.
The sides of a right-angled triangle containing the right angle are 4x cm and (2x – 1) cm. If the area of the triangle is 30 cm²; calculate the lengths of its sides
Solution:


But -13 is not possible.
Sides are 5 cm, 12 cm and 13 cm.

Question 2.
The hypotenuse of a right-angled triangle is 26 cm and the sum of other two sides is 34 cm. Find the lengths of its sides.
Solution:

AC + BC = 34 cm
Let AC = x cm Then BC = (34 – x) cm
But AC² + BC² = AB² (By Pythagoras Theorem)
⇒ (x)² + (34 – x)² = (26)²
⇒ x² + 1156 + x² – 68x = 676
⇒ 2x² – 68x + 1156 – 676 = 0
⇒ 2x² – 68x + 480 = 0
⇒ x² – 34x + 240 = 0 (Dividing by 2)
⇒ x² – 24x – 10x + 240 = 0
⇒ x (x – 24) – 10 (x – 24) = 0
⇒ (x – 24) (x- 10) = 0
Either x – 24 = 0, then x = 24
or x – 10 = 0, then x = 10
If one side is 24 cm, then second will be 34 – 24 = 10 cm
If one side is 10 cm, then second side will be 34 – 10 = 24
Sides are 24 cm and 10 cm

Question 3.
The sides of a right-angled triangle are (x – 1) cm, 3x cm and (3x + 1) cm. Find :
(i) The value of x,
(ii) the lengths of its sides,
(iii) its area.
Solution:

AB = x – 1, BC = 3x and AC = 3x +1
According to Pythagoras Theorem,
AC² = AB² + BC²
⇒ (3x + 1)² = (x – 1)² + (3x)²
⇒ 9x² + 6x + 1 = x² – 2x + 1 + 9x²
⇒ 9x² + 6x + 1 – x² + 2x -1 – 9x² = 0
⇒ – x² + 8x = 0
⇒ x² – 8x = 0
⇒ x (x – 8) = 0
Either x = 0 but it is not possible.
or x – 8 = 0, then x = 8
(i) x = 8
(ii) Side AB = x – 1 = 8 – 1 = 7cm
BC = 3x = 3 x 8 = 24 cm
AC = 3x + 1 = 3 x 8 + 1 = 24 + 1 = 25 cm
(iii) Area = \(\frac { 1 }{ 2 }\) x AB x BC = \(\frac { 1 }{ 2 }\) x 7 x 24 cm² = 84 cm²

Question 4.
The hypotenuse of a right-angled triangle exceeds one side by 1 cm and the other side by 18 cm; find the lengths of the sides of the triangle.
Solution:

In right triangle ABC, ∠B = 90°
Let hypotenuse AC = x, then AB = x – 1 and BC = x – 18
Now, according to Pythagoras Theorem,
AC² = AB² + BC²
⇒ x² = (x – 1)² + (x – 18)²
⇒ x² = x² – 2x + 1 + x² – 36x + 324
⇒ x² = 2x² – 38x + 325
⇒ 2x² – 38x + 325 – x² = 0
⇒ x² – 38x + 325 = 0
⇒ x² – 13x – 25x + 325 =0
⇒ x (x – 13) – 25 (x – 13) = 0
⇒ (x – 13) (x – 25) = 0
Either x – 13 = 0, then x = 13 But it is not possible.
x – 18 = 13 – 18 = -5 cannot be possible.
or x – 25 = 0, then x = 25
AC = 25,
AB = x – 1 = 25 – 1 = 24
and BC = x – 18 = 25 – 18 = 7
Sides are 25, 24, 7

Question 5.
The diagonal of a rectangle is 60 m more than its shorter side and the larger side is 30 m more than the shorter side. Find the sides of the rectangle.
Solution:
Let shorter side of a rectangle (b) = x m
Longer side (l) = (x + 30) m
and length of diagonal = (x + 60) m
According to the condition,
(Diagonal)² = (Longer side)² + (Shorter side)² (Pythagoras Theorem)
⇒ (x + 60)² = (x + 30)² + x²
⇒ x² + 120x + 3600 = x² + 60x + 900 + x²
⇒ 2x² + 60x + 900 – x² – 120x – 3600 = 0
⇒ x² – 60x – 2700 = 0
⇒ x² – 90x + 30x – 2700 = 0 {-2700 = -90 x 30, -60 = -90 + 30}
⇒ x (x – 90) (x + 30) = 0
Either x – 90 = 0, then x = 90
or x + 30 = 0, then x = -30 but is not possible being negative
Length = x + 30 = 90 + 30 = 120 m
Breadth = x = 90 m

Question 6.
The perimeter of a rectangle is 104 m and its area is 640 m². Find its length and breadth.
Solution:

Perimeter = 104 m
⇒ 2 (l + b) = 104 m
l + b = 52 m
Let length of the rectangular plot = x m
Breadth = 52 – x
Area = l x b = x (52 – x)
But area of plot = 640 m²
x (52 – x) = 640
⇒ 52x – x² = 640
⇒ – x² + 52x – 640 = 0
⇒ x² – 52x + 640 = 0
⇒ x² – 20x – 32x + 640 = 0
⇒ x (x – 20) – 32 (x – 20) = 0
⇒ (x – 20) (x – 32) = 0
Either x – 20 = 0, then x = 20
or x – 32 = 0, then x = 32
Length = 32 m
and breadth = 52 – 32 = 20 m

Question 7.
A footpath of uniform width runs round the inside of a rectangular field 32 m long and 24 m wide. If the path occupies 208 m², find the width of the footpath.
Solution:

Length of field = 32m
and width = 24 m
Area of the field = 32 x 24 m² = 768m²
Area of path = 208 m²
Let width of path = x m
Inner length = 32 – 2x
and inner width = 24 – 2x
and inner area = (32 – 2x) (24 – 2x) m²
Area of path = 768 – (32 – 2x) (24 – 2x)
Now, according to the condition,
768 – (32 – 2x) (24 – 2x) = 208
⇒ 768 – (768 – 64x – 48x + 4x²) = 208
⇒ 768 – 768 + 64x + 48x – 4x² = 208
⇒ -4x² + 112x – 208 = 0
Dividing by -4 , we get:
⇒ x² – 28x + 52 = 0
⇒ x² – 26x – 2x + 52 = 0
⇒ x (x – 26) – 2 (x – 26) = 0
⇒ (x – 26 ) (x – 2) = 0
Either x – 26 = 0 , then x = 26 But it is not possible.
or x – 2 – 0 then x = 2
Width of path = 2 m

Question 8.
Two squares have sides x cm and (x + 4 ) cm. The sum of their areas is 656 sq. cm. Express this as an algebraic equation in x and solve the equation to find the sides of the squares.
Solution:
Side of first square = x cm
Area = x² cm²
Side of second square = (x + 4) cm
Area = (x + 4)² cm²
Sum of squares = 656 cm²
⇒ x² + (x + 4)² = 656
⇒ x² + x² + 8x + 16 – 656 = 0
⇒ 2x² + 8x – 640 = 0
⇒ x² + 4x- 320 = 0 (Dividing by 2)
⇒ x² + 20x – 16x – 320 = 0
⇒ x (x + 20) – 16 (x + 20) = 0
⇒ (x + 20) (x – 16) = 0
Either x + 20 = 0, then x = – 20 But it is not possible.
or x – 16 = 0, then x = 16
Side of first square = 16 cm
and side of second square = 16 + 4 = 20 cm

Question 9.
The dimensions of a rectangular field are 50 m by 40 m. A flower bed is prepared inside this field leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and gravelling the path at Rs. 30 and Rs. 20 per square metre, respectively, is Rs. 52,000. Find the width of the gravel path.
Solution:

Length of the field (l) = 50 m
and width (b) = 40 m
Area of rectangular field = l x b = 50 x 40 = 2000 m²
Let width of gravel path be x
Then inner length = 50 – 2x
and width = 40 – 2x
Area of inner flower bed = (50 – 2x) (40 – 2x)
= 2000 – 80x – 100x + 4x² = 4x² – 180x + 2000 sq m
and Area of gravel path = 2000 – (4x² – 180x + 2000)
= 2000 – 4x² + 180x – 2000 = -4x² + 180x
Now rate of gravelling = Rs. 20 per m²
and rate of laying flowers = Rs. 30 per m²
and total cost = Rs. 52000
According to the condition,
(-4x² + 180x) x 20 + (4x² -180x + 2000) x 30 = 52000
-80x² + 3600x + 120x² – 5400x + 60000 = 52000
⇒ 40x² – 1800x – 60000 – 52000 = 0
⇒ 40x² – 1800x + 8000 = 0
⇒ x² – 45x + 200 = 0 (Dividing by 40)
⇒ x² – 40x – 5x + 200 = 0 (200= -40 x (- 5), -45 = -40 – 5)
⇒ x (x – 40) – 5 (x – 40) = 0
⇒ (x – 40) (x – 5) = 0
Either x – 40 = 0, then x = 40
or x – 5 = 0, then x = 5
But x = 40 is not possible
x = 5
Width of gravel path = 5 m

Question 10.
An area is paved with square tiles of a certain size and the number required is 128. If the tiles had been 2 cm smaller each way, 200 tiles would have been needed to pave the same area. Find the size of the larger tiles.
Solution:
No. of square tiles = 128
Let the size of square tile = x cm
Area of one square tile = x² cm²
Area of total tiles = 128 x x² = 128x² cm²
If the size of square tile is reduced by 2 cm
Then size of square tile = (x – 2) cm
Area of one tile = (x – 2)² cm²
Now number of tiles will be = 200
⇒ 128x² = 200 (x – 2)²
⇒128x² = 200x² – 800x + 800
⇒ 200x² – 800x + 800 – 128x² = 0
⇒ 72x² – 800x + 800 = 0
⇒ 9x² – 100x + 100 = 0 (Dividing by 8)
⇒ 9x² – 90x – 10x + 100 = 0
⇒ 9x (x – 10) – 10 (x – 10) = 0
⇒ (x – 10) (9x – 10) = 0
Either x- 10 = 0, then x = 10
or 9x – 10 = 0, then 9x = 10 ⇒ x = \(\frac { 10 }{ 9 }\)
Which is not possible x = 10
Size of square tile = 10 cm

Question 11.
A farmer has 70 m. of fencing, with which he encloses three sides of a rectangular sheep pen; the fourth side being a wall. If the area of the pen is 600 sq. m, find the length of its shorter side.
Solution:

Length of fencing = 70 m.
Area of rectangular pen = 600 sq. m
(i) Let the length of shorter side = x m
Length of larger side = 70 – 2x
Area of rectangular pen = x ( 70 – 2x) ….(ii)
From (i) and (ii)
x (70 – 2x) = 600
⇒ 70x – 2x² = 600
⇒ -2x² + 70x – 600 = 0
⇒ x² – 35x + 300 = 0 (Dividing by -2)
⇒ x² – 15x – 20x + 300 = 0
⇒ x (x – 15) – 20 (x – 15) = 0
⇒ (x – 15) (x – 20) = 0
Either x – 15 = 0 then x = 15
or x – 20 = 0 then x = 20
Shorter side = 15 m

Question 12.
A square lawn is bounded on three sides by a path 4 m wide. If the area of the path is \(\frac { 7 }{ 8 }\) that of the lawn, find the dimensions of the lawn.
Solution:

Let the side of square lawn = x m
Area of lawn = x² m²
Width of path = 4 m.
Area of path = 4 x x + 4 x x + (x + 8) x 4 = 8x + 4x + 32 = 12x + 32
But 12x + 32 = \(\frac { 7 }{ 8 }\) x2
⇒ 96x + 256 = 7x²
⇒ 7x² – 96x – 256 = 0
⇒ 7x² – 112x + 16x – 256 = 0
⇒ 7x (x – 16) + 16 (x – 16) = 0
⇒ (x – 16 ) (7x + 16) = 0
Either x – 16 = 0, then x = 16
0r 7x + 16 = 0, then 7x = -16 ⇒ x = \(\frac { -16 }{ 7 }\)
But it is not possible.
Side of square lawn = 16 m

Question 13.
The area of a big rectangular room is 300 m². If the length were decreased by 5m and the breadth increased by 5 m; the area would be unaltered. Find the length of the room.
Solution:
Area of big rectangular room = 300 m²
Let length of the room = x m.
Width = m
In second case,
Length = (x – 5) m
and width = (\(\frac { 300 }{ x }\) + 5) m

⇒ (x – 5) ( 300 + 5x) = 300x
⇒ 300x + 5x² – 1500 – 25x = 300x
⇒ 5x² + 300x – 25x – 300x – 1500 = 0
⇒ 5x² – 25x – 1500 = 0
⇒ x² – 5x – 300 = 0 (Dividing by 5)
⇒ x² – 20x + 15x – 300 = 0
⇒ x (x – 20) + 15 (x – 20) = 0
⇒ (x – 20) (x + 15) = 0
Either x – 20 – 0 then x = 20
or x + 15 = 0 then x = -15 But it is not possible.
Length of room = 20 m
and width = \(\frac { 300 }{ 20 }\) = 15 m

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6B 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E

Question 1.
A student got the following marks in 9 questions of a question paper : 3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.
Solution:
Arranging the given data in descending order, we get:
8, 7, 6, 5,4,3, 3, 1,0
The middle term is 4 which is 5th terms
∴ Median = 4

Question 2.
The weights (in kg) of 10 students of a class are given below :
21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21, 24 Find the median of their weights.
Solution:
Arranging the given data in descending order.
We get 23.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5
the middle terms are 24 and 24, 5th and 6th terms.

Question 3.
The marks obtained by 19 students of a class are given below :
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35, 28. Find :
(i) Media
(ii) Lower quartile
(iii) Upper quartile
(iv) Inter – quartle range
Solution:
(i) Arranging in order say ascending, we get
22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 31, 32, 32, 33, 35, 35,36, 36, 37
Middle term is 10th term i.e. 29
∴ Median = 29

Question 4.
From the following data, find :
(i) Median
(ii) Upper quartile
(iii) Inter-quartile range
25,10, 40, 88, 45, 60, 77, 36,18, 95, 56, 65, 7, 0, 38 and 83.
Solution:
(i) Arrange in ascending order, we get
0,7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65 ,77, 83, 88, 95

Question 5.
The ages of 37 students in a class are given in the following table :

Find the median.
Solution:

Question 6.
The weight of 60 boys are given in the following distribution table :

Find : (i) Median
(ii) Lower quartile
(iii) Upper quartile
(iv) Inter-quartile range
Solution:

Question 7.
Estimate the median for the given data by drawing ogive :

Solution:

Through mark of 25.5 on the y-axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis, which meets x-axis at B.
∴ The value of B is the median which is 28.

Question 8.
By drawing an ogive; estimate the median for the following frequency distribution :

Solution:

Through mark of 28th on the y- axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular line segment to x- axis. Which meets it at B.

∴ The value of B is the median which is 18.4

Question 9.
From the following cumulative frequency table, draw ogive and then use it to find :
(i) Median,
(ii) Lower quartile,
(iii) Upper quartile.

Solution:

No. of terms = 80
Median = 40th term Through mark of 40 draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B
(ii) Lower quartile (Q1) = \(\frac { n }{ 4 }\) th term
= \(\frac { 80 }{ 4 }\) th term (Here n = 80 which is even)
= 20th term =18
(iii) Upper quartile (Q1) = \(\frac { 3 }{ 4 }\) nth term =\(\frac { 3 x80 }{ 4 }\) = 60th term = 66 .

∴ Value of B is the median which is 40.

Question 10.
In a school 100 pupils have heights as tabulated below:

Find the median height by drawing an ogive.
Solution:


Through mark 50, draw a line parallel to x- axis which meets the curve at A. From A, draw per-pendicular to x-axis which meets x-axis at B is the median which is 148 cm.

P.Q.
Attempt this question on a graph paper. The table shows the distribution of marks gained by a group of 400 students in an examination :


Using a scale of 2 cm to represent 10 marks and 2 cm to represent 50 students, plot these points and draw a smooth curve through the points
Estimate from the graph :
(i) Median marks
(ii) quartile marks. [1997]
Solution:
Plot the points (10, 5), (20, 10), (30, 30), (40, 60), (50,105), (60,180), (70,270), (80, 355), (90, 390), (100, 400) on the graph and join them with free hand to get an ogive (curve) as shown:
(i) Total students = 400

From 200 on y-axis draw a line parallel to x-axis meeting the curve at P. From P, draw PL perpendicular on x-axis then L is the median which is 62.
(ii) Lower quartile (Q1) = \(\frac { 1 }{ 4 }\) x n = \(\frac { 1 }{ 4 }\) x 400 =100
From 100 ony-axis, draw a line parallel to x-axis meeting the curve at Q, from Q, draw QM ⊥ x-axis.
M is the required lower quartile (Q1) which is 49 3 3
Upper quartile (Q3) = \(\frac { 3 }{ 4 }\) n = \(\frac { 3 }{ 4 }\) x 400 = 300
From 300 on y-axis, draw a line parallel to x-axis meeting the curve at R. From R draw RN perpendicular to x-axis N is the required upper quartile (Q3) which is = 74

P.Q.
Attempt this question on graph paper.

(i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10 years on one axis and 2cm = 10 casualities on the other, (ii) From your graph determine :
(a) Median
(b) Lower quartile. (1995)
Solution:

No. of terms = 83
∴ Median \(\frac { 83 }{ 2 }\) =41.5 th term .
Through marks 41.5,draw a line segment par allel to x-axis which meets the curve at A. From A, draw a line segment perpendicular to x-axis meeting it at B.


Through 21 on the y-axis draw a line parallel to x-axis meeting the curve at M
From M, draw a perpendicular on x-axis which meets it at N.
∴N is the lower quartile which is 29 (approx)

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24C 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.

 

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10E
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F

Question 1.
Find the sum of the first 22 terms of the A.P. : 8, 3, – 2,…..
Solution:
A.P. = 8, 3, – 2,…..
Here, a = 8, d = 3 – 8 = – 5, n = 22

Question 2.
How many terms of the A.P. : 24, 21, 18, must be taken so that their sum is 78 ?
Solution:
Let n term of the given A.P. be taken
and A.P. = 24, 21, 18……
Let n be the number of terms.
Here, a = 24, d = 21 – 24 = – 3, S_n = 78

Question 3.
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
Solution:
nth term (T_n) = 8n – 5
T₁ = 8 – 5 = 3
T₂ = 8 x 2 – 5 = 16 – 5 = 11
T₃ = 8 x 3 – 5 = 24 – 5 = 19
A.P. is 3, 11, 19,
Here, a = 3,d = 11 – 3 = 8 and n = 28

Question 4.
Find the sum of :
(i) all odd natural numbers less than 50.
(ii) first 12 natural numbers each of which is a multiple of 7.
Solution:
(i) Sum of all odd natural numbers less then 50
1 + 3 + 5 + 7 +…….+ 49
Here a = 1, d = 3 – 1 = 2
Let n be the number of term, then
49 = a + (n – 1)d
=> 49 = 1 + (n – 1) x 2
=> 49 – 1 = (n – 1) x 2

Question 5.
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Solution:
Sum of first 51 terms of an A.P. in which T₂ = 14 and T₃ = 18
d = T₃ – T₂ = 18 – 14 = 4
and T₂ = a + d
=> 14 = a + 4
=> a = 14 – 4 = 10
A.P. = 10, 14, 18, 22,….

Question 6.
The sum of first 7 terms of an A.P. is 49 and that of first 17 terms of it is 289. Find the sum of first tt terms.
Solution:
S₇ = 49,
S₁₇ = 289

Question 7.
The first term of an A.P. is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
Solution:
First term of an AP (a) = 5
Last term = 45
and S_n = 1000

Question 8.
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Solution:
All natural numbers between 250 and 1000 which are divisible by 9 are
252, 261, 270……999

Question 9.
The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
Solution:
In an A.P.
T₁ = a = 34, l = 700, d = 18
Let n be the number of terms, then
T_n = a + (n – 1 )d
=> 700 = 34 + (n – 1) x 18
=> 700 – 34 = 18(n – 1)
=> 180 – 0 = 666

Question 10.
In an A.P. the first term is 25, nth term is – 17 and the sum of n terms is 132. Find n and the common difference.
Solution:
In an A.P.
First term (a) = 25
nth term = – 17
and S_n = 132
Let d be the common difference

Question 11.
If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of this A.P.
Solution:
In an A.P.
8th term = 37
15th term = 12th term + 15
Let a be the first term and d be the common difference, then

Question 12.
Find the sum of all multiples of 7 lying between 300 and 700.
Solution:
Multiples of 7 lying between 300 and 700 are 301, 308, 315, 322,…., 693

Question 13.
The sum of n natural numbers is 5n² + 4n. Find its 8th term.
Solution:
Sum of n natural number = 5n² + 4n
S_n = 5n² + 4n
S₁(a) = 5 x (1)² + 4 x 1
= 5 + 4 = 9
S₂ = 5(2)² + 4 x 2 = 20 + 8 = 28
S₂ – S₁ = T₂ = 28 – 9 = 19
=> a + d = 19 => 9 + d = 19
d = 19 – 9 = 10
a = 9, d = 10
T₈ = a + (n – 1 )d = 9 + (8 – 1) x 10
= 9 + 7 x 10 = 9 + 70 = 79

Question 14.
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
Solution:
In an A.P.
T₄ = 11
T₈ = 2T₄ + 5
Now, a + 3d = 11
a + 7d = 2 x 11 + 5 = 22 + 5 = 27
Subtracting, 4d = 16
=> d = \(\\ \frac { 16 }{ 4 } \) = 4

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C 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.