Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D.

Other Exercises

Question 1.
If a : b = 3 : 5, find : (10a + 3b) : (5a + 2b)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q1.1

Question 2.
If 5x + 6y : 8x + 5y = 8 : 9, find x : y
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q2.1

Question 3.
If k (3x – 4y) : (2x – 3y) = (5x – 6y) : (4x – 5y), find x : y.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q3.1

Question 4.
Find the :
(i) duplicate ratio of 2√2 : 3√5
(ii) triplicate ratio of 2a : 3b,
(iii) sub-duplicate ratio of 9x2 a4 : 25y6 b2
(iv) sub-triplicate ratio of 216 : 343
(v) reciprocal ratio of 3 : 5
(vi) ratio compounded of the duplicate ratio of 5 : 6, the reciprocal ratio of 25 : 42 and the sub-duplicate ratio of 36 : 49.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q4.1

Question 5.
Find the value of x, if :
(i) (2x + 3) : (5x – 38) is the duplicate ratio of √5 : √6
(ii) (2x + 1) : (3x + 13) is the sub-duplicate ratio of 9 : 25.
(iii) (3x -7): (4x + 3) is the sub-triplicate ratio of 8 : 27.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q5.2

Question 6.
What quantity must be added to each term of the ratio x : y so that it may become equal to c : d ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q6.1

Question 7.
A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 84 kg?
Solution:
Ratio in reduction of weight = 7 : 5
Originally weight of the woman = 84 kg
Reduced weight = \(\frac { 84 x 5 }{ 7 }\) = 60 kg

Question 8.
If 15 (2x² – y²) = 7xy, find x : y, if x and y both are positive.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q8.2

Question 9.
Find the :
(i) fourth proportional to 2xy, x² and y².
(ii) third proportional to a² – b² and a + b.
(iii) mean proportion to (x – y) and (x3 – x²y)
Solution:
(i) Let a be the fourth proportional
Then 2xy : x² :: y² : a
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q9.1

Question 10.
Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112.
Solution:
Let x and y be the two numbers.
Then 14 is the mean proportional between x and y
xy = 14² => xy = 196 ….(i)
and 112 is the third proportional to x, y
y² = 112 x ….(ii)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q10.1

Question 11.
If x and y be unequal and x : y is the duplicate ratio of x + z and y + z, prove that z is mean proportional between x and y.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q11.1

Question 12.
If q is the mean proportional between p and r, prove that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q12.3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q11.2

Question 13.
If a, b and c are in continued proportion, prove that: a : c = (a² + b²) : (b² + c²).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q13.1

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q14.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q14.2

Question 15.
If (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove that a : b = c : d.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q15.1

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q16.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q16.2

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q17.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q17.2

Question 18.
There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3 : 1. How many more girls should be added to the council so that the ratio of number of boys to the number of girls may be 9 : 5 ?
Solution:
Total number of members = 36
Ratio in boys and girls = 3 : 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q18.1

Question 19.
If 7x – 15y = 4x + y, find the value of x : y. Hence, use componendo and dividendo to find the values of:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q19.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q19.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q19.3

Question 20.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q20.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q20.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q20.3

Question 21.
If x, y, z arc in continued proportion, prove that
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q21.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q21.2

Question 22.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q22.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q22.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q22.3

Question 23.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q23.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q23.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q23.3

Question 24.
Using componendo and dividendo, find the value
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q24.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q24.2

Question 25.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q25.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q25.2

Question 26.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q26.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q26.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q26.3

Question 27.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q27.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q27.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D Q27.3

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D 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 7 Ratio and Proportion Ex 7C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C.

Other Exercises

Question 1.
If a : b = c : d, prove that:
(i) 5a + 7b : 5a – 7b = 5c + 7d : 5c – 7d.
(ii) (9a + 13b) (9c – 13d) = (9c + 13d) (9a – 13b).
(iii) xa + yb : xc + yd = b : d.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q1.2

Question 2.
If a : b = c : d, prove that: (6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q2.1

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q3.2

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q4.2

Question 5.
If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d), prove that a : b = c : d.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q5.1

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q6.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q6.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q6.4

Question 7.
If (a + b + c + d) (a – b – c + d) = (a + b – c – d) (a – b + c – d), prove that a : b = c : d.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q7.1

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q8.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q8.2

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q9.2

Question 10.
If a, b and c are in continued proportion, prove that
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q10.3

Question 11.
Using properties of proportion, solve for x:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q11.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q11.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q11.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q11.4
But x = – 1 does not satisfy it
x = 1

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q12.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q12.2

Question 13.
Using the properties of proportion, solve for x, given
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q13.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q13.2

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q14.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q14.2

Question 15.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q15.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C Q15.3

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C 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 22 Heights and Distances Ex 22B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B

Other Exercises

Question 1.
In the figure given below, it is given that AB is perpendicular to BD and is of length X metres. DC = 30 m. ∠ADB = 30° and ∠ACB = 45°. Without using tables, find X.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q1.2

Question 2.
Find the height of a tree when it is found that on walking away from it 20m, in a horizontal line through its base, the elevation of its top Changes from 60° to 30°.
Solution:
Let AB be the tree and its height be x DC = 20 m.
Now in right ∆ADB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q2.2

Question 3.
Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30° to 45°.
Solution:
Let AB be the building and its
height be x and DC =40m
Now in right ∆ADB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q3.2

Question 4.
From the top of a light house 100 m high, the angles of depression of two ships are observed as 48° and 36° respectively. Find the distance between the two ships (in the nearest metre) if :
(i) the ships are on the same side of the light house,
(ii) the ships are on the opposite sides of the light house.
Solution:
AD = 100 m
In right angled ∆ABD
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q4.2

Question 5.
Two pillars of equal heights stand 011 either side of a roadway, which is 150m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60° and 30”; find the height of the pillars and the position of the point.
Solution:
Let AB and CD be the two pillars which stand on either side of a road BD, then BD = 150m
Let AB = CD = h
Let P be the point on the road such that the angles of elevation from P to the top of the pillars are 60° and 30° respectively.
Let BP = x then PD = 150 – x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q5.2

Question 6.
From the figure given below, calculate the length of CD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q6.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q6.2

Question 7.
The angle of elevation of the top of a tower is observed to be 60°. At a point 30 m vertically above the first point of observation, the elevation is found to be 45°. Find :
(i) the height of the tower,
(ii) its horizontal distance from the points of observation.
Solution:
Let height of tower = x i.e. AB = x
In the right ∆ADB,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q7.3

Question 8.
From the top of a cliff. 60 metre high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°. Find the height of the tower.
Solution:
Height of the cliff AB = 60m
Let the height of tower = x
Draw TR \(\parallel\) SB. meeting AB in R
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q8.2

Question 9.
A man on a cliff observes a boat, at an angle of depression 30°, which is sailing towards the shore to the point immediately beneath him. Three minutes later the angle of depression of the boat is found to be 60°. Assuming that the boat sails at a uniform speed, determine :
(i) how much more time it will take to reach the shore.
(ii) the speed of the boat in metre per second, if the height of the cliff is 500 m.
Solution:
Height of cliff = 500 m.
In right ∆ACD,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q9.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q9.3

Question 10.
A man in a boat rowing away from a light house 150m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 45°. Find the speed of the boat.
Solution:
Height of the lighthouse AB = 150m
Now is right ∆ACB,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q10.2

Question 11.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find :
(i) the height of the tree, correct to 2 decimal places,
(ii) the width of the river.
Solution:
TR is the height of tree and RQ is the width of the river.
Let TR = x
Now in right ∆TQR
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q11.2

Question 12.
The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160m high, is 45°. Find the height of the first tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q12.1
Let the height of first tower CD = x
and height of second tower AB = 160 m
Distance between them DB = 75 m.
In right ∆ACE,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q12.2

Question 13.
The length of the shadow of a tower standing on level plane is found to be 2y metres longer when the sun’s altitude is 30″ than when it was
45°. Prove that the height of the tower is y (\(\sqrt { 3 }\) +1) metres.
Solution:
Let AB be the tower and AB = x
Distance CD = 2y
In right ∆ADB,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q13.2

Question 14.
An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°, find the uniform speed of the aeroplane in km per hour.
Solution:
Height of aeroplane = 1 km = 1000 m.
In right ∆ACB,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q14.2

Question 15.
From the top of a hill, the angles of depression of two consecutive kilometre stones, due east, are found to be 30° and 45° respectively. Find the distance of the two stones from the foot of the hill. (2007)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B Q15.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B 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 7 Ratio and Proportion Ex 7B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B.

Other Exercises

Question 1.
Find the fourth proportional to :
(i) 1.5, 4.5 and 3.5
(ii) 3a, 6a² and 2ab²
Solution:
(i) Let x be the fourth proportional to 1.5, 4.5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q1.1

Question 2.
Find the third proportional to :
(i) 2\(\frac { 2 }{ 3 }\) and 4
(ii) a – b and a² – b²
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q2.2

Question 3.
Find the mean proportional between :
(i) 6 + 3√3 and 8 – 4√3
(ii) a – b and a3 – a²b.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q3.2

Question 4.
If x + 5 is the mean proportion between x + 2 and x + 9 ; find the value of x.
Solution:
x + 5 is the mean proportion between x + 2 and x + 9
(x + 5)² = (x + 2) (x + 9) {b² = ac}
⇒ x² + 10x + 25 = x² + 11x + 18
⇒ x² + 10x – x² – 11x = 18 – 25
⇒ -x = -7
⇒ x = 7
Hence x = 7

Question 5.
If x², 4 and 9 are in continued proportion, find x.
Solution:
x², 4 and 9 are in continued proportion
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q5.1

Question 6.
What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional. (2005, 2013)
Solution:
Let x to be added to each number then 6 + x, 15 + x, 20 + x and 43 + x are in proportion.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q6.1

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q7.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q7.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q7.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q7.5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q7.6

Question 8.
What least number must be subtracted from each of the numbers 7,17 and 47 so that the remainders are in continued pro-portion ?
Solution:
Let x be subtracted from each of the numbers 7, 17 and 47.
Then 7 – x, 17 – x and 47 – x are in continued proportion.
7 – x : 17 – x : : 17 – x : 47 – x
⇒ (7 – x) (47 – x) = (17 – x) (17 – x)
⇒ 329 – 7x – 47x + x² = 289 – 17x – 17x + x²
⇒ -7x – 47x + x² + 17x + 17x – x² = 289 – 329
⇒ -54x + 34x = – 40
⇒ -20x = -40
⇒ x = 2
2 is to be the subtracted

Question 9.
If y is the mean proportional between x and z; show that xy + yz is the mean proportional between x² + y² and y² + z².
Solution:
y is the mean proportional between x and z.
y² = xz
Now, we have to prove that
xy + yz is the mean proportional between x² + y² and y² + z²
i.e., (xy + yz)² = (x² + y²) (y² + z²)
L.H.S. (xy + yz)² = [y(x + z)]² = y² (x + z)² = xz (x + z)²
R.H.S. (x² + y²) (y² + z²) = (x² + xz) (xz + z²) = x (x + z) z (x + z) = xz (x + z)²
L.H.S. = R.H.S.
Hence proved.

Question 10.
If q is the mean proportional between p and r, show that: pqr (p + q + r)3 = (pq + qr + pr)3.
Solution:
q is the mean proportional between p and r,
q² = pr
Now L.H.S. = pqr (p + q + r)3
= qq² (p + q + r)3
= q3 (p + q + r)3
= [q(p + q + r)]3
= (pq + q² + qr)3
= (pq + pr + qr)3
= (pq + qr + pr)3
= R.H.S.

Question 11.
If three quantities are in continued proportion; show that the ratio of the first to the third is the duplicate ratio of the first to the second.
Solution:
Let x, y and z are three quantities which are
in continued proportion
Then x : y : y : z ⇒ y² = zx
Now, we have to prove that
x : z = x² : y² or xy² = zx²
L.H.S. = xy² = x x zx (y² = zx)
= x² z = R.H.S.
Hence Proved.

Question 12.
If y is the mean proportional between x and z, prove that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q12.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q12.2

Question 13.
Given four quantities a, b, c and d are in proportion. Show that:
(a – c) b² : (b – d) cd = (a² – b² – ab) : (c² – d² – cd)
Solution:
a, b, c and d are in proportion Then a : b :: c : d
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q13.1

Question 14.
Find two numbers such that the mean proportional between them is 12 and the third proportional to them is 96.
Solution:
Let a and b be the two numbers, whose mean proportional is 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q14.1

Question 15.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q15.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q15.2

Question 16.
If p : q = r : s ; then show that: mp + nq : q = mr + ns : s.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q16.1

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q17.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q17.2

Question 18.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q18.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q18.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q18.3

Question 19.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q19.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q19.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q19.3

Question 20.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q20.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B Q20.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B 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 22 Heights and Distances Ex 22A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A

Other Exercises

Question 1.
The height of a tree is \(\sqrt { 3 }\) times the length of its shadow. Find the angle of elevation of the sun.
Solution:
Let AB be the tree and BC be its shadow.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q1.1
∴ θ = 60°
∵ Angle of elevation of the sun = 60°

Question 2.
The angle of elevation of the top of a tower, from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.
Solution:
Let AB be the tower and C is the point which is 160 m away from the foot of the tower,
i.e. CB = 160 m
Let height of the tower be x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q2.1

Question 3.
A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68“ with the ground. Find the height, upto which the ladder reaches.
Solution:
Let AB be the wall and
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q3.1
AC be the ladder, which is placed against the wall. If foot is 2.4 m away from the wall i.e. CB = 2.4m1.
Let AB =x m.
In right ∆ ABC,
tan θ = \(\frac { AB }{ BC }\) ⇒ tan 68° = \(\frac { X }{ 2.4 }\)
∴ x = 2.4 x tan 68° = 2.4 x 2.4751
= 5.94 m

Question 4.
Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30° and 38° respectively. Find the distance between them, if the height of the tower is 50 m.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q4.1
Two persons A and B are standing on the opposite side of the tower TR and height of tower TR = 50 m and angles of elevation with A and B are 30° and 38° respectively. Let AR = x and RB = y
Now in right ∆TAR,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q4.2

Question 5.
A kite is attached to a string. Find the length of the string, when the height of the kite is 60 m. and the string makes an angle 30° with the ground.
Solution:
Let KT be the height of kite and PK is the string which makes an angle of 30° with the ground.
∴ KT = 60 m
Let KP = xm.
Now in right ∆PKT,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q5.1

Question 6.
A boy 1.6m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45° (ii) 60°. Find the height of the tower in each case.
Solution:
(i) Let AB be the tower and MN be the boy who is 20m away from the foot of the tower.
Let AB = x and angle of elevation = 45°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q6.3

Question 7.
The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground; and the distance from the root to the point where the top of the tree touches the ground, is 15m. What was the height of the tree before it was broken ?
Solution:
Let AB be the tree which was broken at the point C which makes an angle of elevation of 45°, with the ground at a distance of 15m.
BD = 15m
AC = CD
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q7.2

Question 8.
The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60° ?
Solution:
Distance of a point from the tower = 80 m
Angle of elevation = 30°
In second case the elevation of lower = 60°
In first case,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q8.3

Question 9.
At a particular time, when the sun’s altitude is 30°, the length of the shadow of’C vertical tower is 45 m. Calculate :
(i) height of the tower.
(ii) the length of the shadow of the same tower, when the sun’s altitude is (a) 45° (b) 60°.
Solution:
Shadow of the tower = 45 m and angle of elevation = 30°
Let AB be the lower and BC is its shadow.
∴ CB = 45 m.
Now in right ∆ABC,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q9.1
(ii) In second case,
(a) Angle of elevation = 45°
and height of tower = 25.98 m or 15\(\sqrt { 3 }\) m
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q9.2
(b) Angle of elevation = 60°
and height of tower = 25.98 m or 15\(\sqrt { 3 }\) m.
Let shadow of the tower DB = xm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q9.3

Question 10.
Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32°24′ with the pole and when it is turned to rest against another pole, it makes angle 32°24′ with the road. Calculate the width of the road.
Solution:
Two poles AB and CD which are at the either end of a road BD. A ladder 30 m long subtends an angle of 32° 24′ with the first pole AB and 32°24′ with the road when it is turned to rest against the second pole CD.
Now in right ∆ABE.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q10.1

Question 11.
Two climbers are at points A and Bona vertical cliff face. To an observer C, 40 m from the foot of the cliff on the level ground, A is at an elevation of 48° and B of 57°. What is the distance between the climbers ?
Solution:
A and B are two climbers on the cliff and ob-server is at C, 40 m from the foot of the cliff while the angles of elevations of each climber is 48° and 57° respectively.
In right ∆ACD,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q11.2

Question 12.
A man stands 9m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28° and the angle of depression of the bottom of the pole is 13°. Calculate the height of the pole.
Solution:
Let PL is the pole and MN is the man The angle of elevation of the top of the pole = 28°
arid the angle of depression of the bottom of the pole =13°
Man is 9 m away from the pole,
i.e. MQ = 9 m
Now in right ∆PMQ,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q12.1

Question 13.
From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate to the nearest metre the distance of the buoy from the foot of the cliff.
Solution:
Let CD be the cliff and CD = 92m, B is the buoy,
then from C ,
the angle of depression is 20°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A Q13.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A 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 11 Geometric Progression Ex 11D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D

Other Exercises

Question 1.
Find the sum of G.P. :
(i) 1 + 3 + 9 + 27 +….to 12 terms.
(ii) 0.3 + 0.03 + 0.003 + 0.0003 +….to 8 terms.
(iii) \(1-\frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } ….to\quad 9\quad terms\)
(iv) \(1-\frac { 1 }{ 3 } +\frac { 1 }{ { 3 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 3 } } ….to\quad n\quad terms \)
(v) \(\frac { x+y }{ x-y } +1+\frac { x-y }{ x+y } +….upto\quad n\quad terms\)
(vi) \(\sqrt { 3 } +\frac { 1 }{ \sqrt { 3 } } +\frac { 1 }{ 3\sqrt { 3 } } +….to\quad n\quad terms\)
Solution:
(i) 1 + 3 + 9 + 27 +….to 12 terms.
Here a = 1, r = 3 and n = 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.6
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q1.8

Question 2.
How many terms of the geometric progression 1 + 4 + 16 + 64 +…. must be added to get sum equal to 5461 ?
Solution:
Sn = 5461 and G.P. is
1 + 4 + 16 + 64 +…..
Here, a = 1, r = 4 (r > 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q2.1

Question 3.
The first term of a G.P. is 27 and its 8th term is \(\\ \frac { 1 }{ 81 } \). Find the sum of its first 10 terms.
Solution:
First term of a G.P (a) = 27
T8 = \(\\ \frac { 1 }{ 81 } \), n = 10
a = 27
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q3.2

Question 4.
A boy spends Rs 10 on first day, Rs 20 on second day, Rs 40 on third day and so on. Find how much, in all, will he spend in 12 days?
Solution:
A boy spends Rs 10 on first day,
Rs 20 on second day
Rs 40 on third day and so on
G.P. is 10 + 20 + 40 +…. 12 terms
Here a = 10, r = 2 and n = 12 (r > 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q4.2

Question 5.
The 4th and the 7th terms of a G.P. are \(\\ \frac { 1 }{ 27 } \) and \(\\ \frac { 1 }{ 729 } \) respectively. Find the sum of n terms of this G.P.
Solution:
In a G.P.
T4 = \(\\ \frac { 1 }{ 27 } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q5.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q5.3

Question 6.
A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728 ; find its first term.
Solution:
In a G.P.
Common ratio (r) = 3
Last term (l) = 486
Sum of its terms (Sn) = 728
Let a be the first term, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q6.1

Question 7.
Find the sum of G.P. : 3, 6, 12, …… 1536.
Solution:
G.P. is 3, 6, 12,….1536
Here a = 3, r = \(\\ \frac { 6 }{ 3 } \) = 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q7.2

Question 8.
How many terms of the series 2 + 6 + 18 +…. must be taken to make the sum equal to 728 ?
Solution:
G.P. is 2 + 6 + 18 +….
Here a = 2, r = \(\\ \frac { 6 }{ 2 } \) = 3, Sn = 728
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q8.1

Question 9.
In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125 : 152. Find its common ratio.
Solution:
In a G.P.
Sum of first 3 terms : Sum of 6 terms = 125 : 152
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q9.2

Question 10.
Find how many terms of G.P.\(\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 } \)… must be added to get the sum equal to \(\\ \frac { 55 }{ 72 } \) ?
Solution:
\(\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 } \)…
Let n terms be added
Now,Sn = \(\\ \frac { 55 }{ 72 } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q10.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q10.4

Question 11.
If the sum of 1 + 2 + 22 +…..+ 2n – 1 is 255, find the value of n.
Solution:
1 + 2 + 22 +…..+ 2n – 1 = 255
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q11.2

Question 12.
Find the geometric mean between :
(i) \(\\ \frac { 4 }{ 9 } \) and \(\\ \frac { 9 }{ 4 } \)
(ii) 14 and \(\\ \frac { 7 }{ 32 } \)
(iii) 2a and 8a3
Solution:
(i) G.M between \(\\ \frac { 4 }{ 9 } \) and \(\\ \frac { 9 }{ 4 } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q12.2

Question 13.
The sum of three numbers in G.P. is \(\\ \frac { 39 }{ 10 } \) and their product is 1. Find the numbers.
Solution:
Sum of three numbers in G.P. = \(\\ \frac { 39 }{ 10 } \)
and their product = 1
Let number be \(\\ \frac { a }{ r } \), a, ar, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q13.2

Question 14.
The first term of a G.P. is – 3 and the square of the second term is equal to its 4th term. Find its 7th term.
Solution:
In G.P.
T1 = – 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q14.1

Question 15.
Find the 5th term of the G.P. \(\\ \frac { 5 }{ 2 } \), 1,…..
Solution:
Given G.P is \(\\ \frac { 5 }{ 2 } \), 1,…..
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q15.1

Question 16.
The first two terms of a G.P. are 125 and 25 respectively. Find the 5th and the 6th terms of the G.P.
Solution:
Given, First term = a = 125….(i)
and Second term = ar = 25…..(ii)
Now, Divide eq. (ii) by eq (i), we get
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q16.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q16.2

Question 17.
Find the sum of the sequence \(– \frac { 1 }{ 3 } \), 1, – 3, 9,….upto 8 terms.
Solution:
Here, First Term, a = \(– \frac { 1 }{ 3 } \)…(i)
and Second Term, ar = 1 …(ii)
Dividing eq. (i) by eq. (ii), we get
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q17.1

Question 18.
The first term of a G.P. in 27. If the 8th term be \(\\ \frac { 1 }{ 81 } \), what will be the sum of 10 terms ?
Solution:
Given, First term (a) = 27, n = 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q18.2

Question 19.
Find a G.P. for which the sum of first two terms is – 4 and the fifth term is 4 times the third term.
Solution:
Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D Q19.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 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 7 Ratio and Proportion Ex 7A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A.

Other Exercises

Question 1.
If a : b = 5 : 3; find: \(\frac { 5a – 3b }{ 5a + 3b }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q1.1

Question 2.
If x : y = 4 : 7; find the value of (3x + 2y) : (5x + y).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q2.1

Question 3.
If a : b = 3 : 8, find the value of \(\frac { 4a + 3b }{ 6a – b }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q3.1

Question 4.
If (a – b): (a + b) = 1 : 11, find the ratio (5a + 4b + 15) : (5a – 4b + 3).
Solution:
(a – b) : (a + b) = 1 : 11
Let a – b = x, then a + b = 11x
Adding we get, 2a = 12x ⇒ a = 6x
Subtracting, -2b = -10x ⇒ b = 5x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q4.1

Question 5.
Find the number which bears the same ratio to \(\frac { 7 }{ 33 }\) that \(\frac { 8 }{ 21 }\) does to \(\frac { 4 }{ 9 }\).
Solution:
Let x be the required number, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q5.1

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q6.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q6.2

Question 7.
Find \(\frac { x }{ y }\) ; when x² + 6y² = 5xy.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q7.2

Question 8.
If the ratio between 8 and 11 is the same as the ratio of 2x – y to x + 2y, find the value of \(\frac { 7x }{ 9y }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q8.1

Question 9.
Divide ₹ 290 into A, B and C such that A is \(\frac { 2 }{ 5 }\) of B and B : C = 4 : 3.
Solution:
Total amount = ₹ 1290
A = \(\frac { 2 }{ 5 }\) B and B : C = 4 : 3
⇒ A : B = 2 : 5 and B : C = 4 : 3
LCM of 5, 4 = 20
A : B = 2 x 4 : 5 x 4 = 8 : 20
and B : C = 4 x 5 : 3 x 5 = 20 : 15
A : B : C = 8 : 20 : 15
Sum of ratios = 8 + 20 + 15 = 43
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q9.1

Question 10.
A school has 630 students. The ratio of the number of boys to the number of girls is 3 : 2. This ratio changes to 7 : 5 after the admission of 90 new students. Find the number of newly admitted boys.
Solution:
Number of students = 630
Ratio in boys and girls = 3 : 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q10.1

Question 11.
What quantity must be subtracted from each term of the ratio 9 : 17, to make it equal to 1 : 3?
Solution:
Let x be subtracted from each term such that
\(\frac { 9 – x }{ 17 – x }\) = \(\frac { 1 }{ 3 }\)
⇒ 17 – x = 27 – 3x
⇒ -x + 3x = 21 – 17
⇒ 2x = 10
⇒ x = 5

Question 12.
The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each save Rs. 80 every month, find their monthly pocket money. [2012]
Solution:
Let the monthly pocket money of Ravi and Sanjeev be 5x and 7x respectively.
Also, let their expenditure be 3y and 5y respectively. So,
5x – 3y = 80 …(i)
and 7x – 5y = 80 …(ii)
Multiplying (i) by 7 and (ii) by 5 and subtracting, we get
4y = 160 ⇒ y = 40
From (i), 5x = 80 + 3 x 40 = 200 ⇒ x = 40
So, monthly pocket money of Ravi = Rs. 5 x 40 = Rs. 200
and monthly pocket money of Sanjeev = Rs. 7 x 40 = Rs. 280

Question 13.
The work done by (x – 2) men in (4x + 1) days and the work done by (4x + 1) men in (2x – 3) days are in the ratio of 3 : 8. Find the value of x.
Solution:
(x – 2) men can do a work in = (4x +1) days
1 man can do a work in = (4x +1) (x-2) days ….(i)
Again (4x + 1) men can do work in = 2x – 3 days
1 man can do the work in = (2x – 3) (4x + 1) days ….(ii)
From (i) and (ii), we get
(4x + 1) (x – 2) : (2x – 3) (4x + 1) = 3 : 8
⇒ (4x + 1) (x – 2) x 8 = (2x – 3) (4x + 1) x 3
⇒ 8 (4x² – 8x + x – 2) – 3 (8x² + 2x – 12x – 3)
⇒ 32x² – 64x + 8x – 16 = 24x² + 6x – 36x – 9
⇒ 32x² – 64x + 8x – 24x² – 6x + 36x – 16 + 9 = 0
⇒ 8x² – 70x + 44x – 7 = 0
⇒ 8x² – 26x – 7 = 0
⇒ 8x² – 28x + 2x – 7 = 0
⇒ 4x ( 2x – 7) + 1 (2x – 7) = 0
⇒ (2x – 7) (4x + 1) = 0
Either 2x – 7 = 0, then x = \(\frac { 7 }{ 2 }\)
or 4x + 1 = 0, then x = \(\frac { -1 }{ 4 }\) but it is not possible.
x = \(\frac { 7 }{ 2 }\) or 3.5 Ans.

Question 14.
The bus fare between two cities is increased in the ratio 7 : 9. Find the increase in the fare, if :
(i) the original fare is Rs. 245 ;
(ii) the increased fare is Rs. 207.
Solution:
The increase in bus fare between two cities is in the ratio = 7 : 9.
(i) If the original fare is Rs. 245
then increase fare = Rs. 245 x \(\frac { 9 }{ 7 }\) = Rs. 315
Increase = Rs. 315 – Rs. 245 = Rs. 70
(ii) The increased fare is Rs. 207
Original fare = Rs. \(\frac { 207 x 7 }{ 9 }\) = Rs. 161
Increase = Rs. 207 – Rs. 161 = Rs. 46

Question 15.
By increasing the cost of entry ticket to a fair in the ratio 10 : 13; the number of visitors to the fair has decreased in the ratio 6 : 5. In what ratio has the total collection increased or decreased ?
Solution:
Increase in the entry tickets = 10 : 13.
But decrease in visitors = 6 : 5
Let original price of per ticket = Rs. 10
Then increased in price will be = Rs. 13
Collection in first case = Rs. 10 x 6 = Rs. 60
and collection in second case = Rs. 13 x 5 = Rs. 65
Hence increase in collection will be = Rs. 60 : Rs. 65 = 12 : 13

Question 16.
In a basket, the ratio between the number of oranges and the number of apples is 7 : 13. If 8 oranges and 11 apples are eaten, the ratio between the number of oranges and the number of apples becomes 1 : 2. Find the original number of oranges and the original number of apples in the basket.
Solution:
The ratio in number of oranges and apples = 7 : 13
Let number of oranges = 7x
Then number of apples = 13x
When 8 oranges and 11 apples are eaten, then
According to the sum,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q16.1
Original number of oranges = 7x = 7 x 5 = 35
and number of apples = 13x = 13 x 5 = 65

Question 17.
In a mixture of 126 kg of milk and water, milk and water are in ratio 5 : 2. How much water must be added to the mixture to make this ratio 3 : 2?
Solution:
Mixture of milk and water = 126 kg
Ratio in milk and water = 5 : 2
Quantity of milk = \(\frac { 126 }{ 5 + 2 }\) x 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q17.1

Question 18.
(a) If A : B = 3 : 4 and B : C = 6 : 7, find:
(i) A : B : C
(ii) A : C
(b) If A : B = 2 : 5 and A : C = 3 : 4, find: A : B : C
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q18.2
A : B : C = 6 : 15 : 8

Question 19.
(i) If 3A = 4B = 6C ; find A : B : C.
(ii) If 2a = 3b and 4b = 5c, find a : c
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q19.2

Question 20.
Find the compound ratio of:
(i) 2 : 3, 9 : 14 and 14 : 27.
(ii) 2a : 3b, mn : x² and x : n.
(iii) √2 : 1, 3 : √5 and √20 = 9.
Solution:
(i) Compound ratio of 2 : 3, 9 : 14 and 14 : 27
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q20.1

Question 21.
Find the duplicate ratio of:
(i) 3 : 4
(ii) 3√3 : 2√5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q21.1

Question 22.
Find triplicate ratio of:
(i) 1 : 3
(ii) \(\frac { m }{ 2 }\) : \(\frac { n }{ 3 }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q22.1

Question 23.
Find sub-duplicate ratio of:
(i) 9 : 16
(ii) (x – y)4 : (x + y)6
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q23.1

Question 24.
Find sub-triplicate ratio of:
(i) 64 : 27
(ii) x3 : 125y3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q24.1

Question 25.
Find the reciprocal ratio of :
(i) 5 : 8
(ii) \(\frac { x }{ 3 }\) : \(\frac { y }{ 7 }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q25.1

Question 26.
If (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.
Solution:
(x + 3) : (4x + 1) is the duplicate ratio of 3 : 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q26.1

Question 27.
If m : n is the duplicate ratio of m + x : n + x; show that x² = mn.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q27.1

Question 28.
If (3x – 9) : (5x + 4) is the triplicate ratio of 3 : 4, find the value of x.
Solution:
(3x – 9) : (5x + 4) is the triplicate ratio of 3 : 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q28.1
⇒ 27 (5x + 4) = 64 (3x – 9)
⇒ 135x + 108 = 192x – 576
⇒ 192x – 135x = 108 + 576
⇒ 57x = 684
⇒ x = 12

Question 29.
Find the ratio compounded of the reciprocal ratio of 15 : 28, the sub-duplicate ratio of 36 : 49 and the triplicate ratio of 5 : 4.
Solution:
Reciprocal ratio of 15 : 28
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q29.1

Question 30.
(a) If r² = pq, show that p : q is the duplicate ratio of (p + r) : (q + r).
(b) If (p – x) : (q – x) be the duplicate ratio of p : q then show that : \(\frac { 1 }{ p }\) + \(\frac { 1 }{ q }\) = \(\frac { 1 }{ x }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q30.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A Q30.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A 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 23 Graphical Representation Ex 23

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation (Histograms, Frequency Polygon and Ogives) Ex 23

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 23 Graphical Representation (Histograms, Frequency Polygon and Ogives) Ex 23.

Question 1.
Draw histograms for the following distributions :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q1.4

Question 2.
Draw a cumulative frequency curve (ogive) for each of the following distributions
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q2.2

Question 3.
Draw an ogive for each of the following distributions :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q3.2
Now plotting the points (10, 8), (20, 25), (30, 38), (40, 50) and (50, 67) on the graph and join them with free hand to obtain an ogive as shown in the graph.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q3.3
Now plot the points (10, 0), (20, 17), (30, 32), (40, 37), (50, 53), (60, 58) and (70, 65) on the graph and join them with free hand to get an ogive as shown in the graph.

Question 4.
Construct a frequency distribution table for the numbers given below, using the class intervals 21-30,31-40…. etc.
75, 67, 57, 50, 26, 33, 44, 58, 67,75, 78, 43, 41, 31, 21, 32, 40, 62, 54, 69, 48, 47,51,38, 39,43,61, 63, 68, 53, 56, 49, 59, 37, 40, 68, 23, 28, 36 and 47.
Use the table obtained to draw:
(i) a histogram
(ii) an ogive
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q4.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q4.3
(ii) Ogive :
We plot the points (30, 4), (40, 13), (50, 22), (60, 29); (70, 37) and (80, 40) on the graph and join them in free hand to obtain an ogive.

Question 5.
(a) Use the information given in the adjoining histogram to c construct a f frequency table.
(b)Use this table to construct an ogive.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q5.1
Solution:
From the histogram given, the required frequency table will be as given below.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q5.2
Plot the points (12, 9), (16, 25), (20, 47), (24, 65), (28, 77), (32, 81) on the graph and join them with free hand to get an ogive as shown.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q5.3

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q6.1
(a) From the distribution given above, construct a frequency table, (b) Use the table obtained in part (a) to draw :
(i) a histogram
(ii) an ogive.
Solution:
Difference in consecutive class marks. = 17.5 – 12.5 = 5
∴ first class interval will be : 10-15 and so on
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q6.2
(ii) We plot the point (15. 12), (20. 29), (25. 51), (30, 78). (35, 108), (40. 129) and (45, 145) on the graph and join them in free hand to obtain the ogive.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q6.3

Question 7.
Use graph paper for this question.
The table given beiow shows the monthly wages of some factory workers.
(i) Using the table calculate the cumulative frequencies of workers.
(ii) Draw the cumulative frequency curve.
Use 2 cm = Rs. 500, starting the origin at Rs. 6500 on x-axis, and 2 cm = 10 workers on the y-axis
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q7.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q7.3
(ii) We plot the points (7000, 10), (7500, 28), (8000, 50), (8500, 75), (9000, 92) (9500, 102) and (10000, 110) on the graph and join them in free hand to obtain an ogive.

Question 8.
The following table shows the distribution of the heights of a group of factory workers :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q8.1
(i) Determine the cumulative frequencies.
(ii) Draw the ‘less than’ cumulative frequency curve on a graph paper. Use 2 cm = 5cm height on one axis and 2 cm = 10 workers on the other.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q8.3
We plot points (155, 6), (160, 18), (165, 36),
(170, 56), (175, 69), (180, 77) and (185, 83) on the graph and join them in free hand to obtain an ogive.

Question 9.
Construct a frequency distribution table for each of the following distributions :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23 Q9.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation Ex 23  are helpful to you.

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

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C

Other Exercises

Question 1.
Find the seventh term from the end of the series :
√2, 2, 2√2,……32
Solution:
√2, 2, 2√2,……32
Here a = √2
r = \(\frac { 2 }{ \surd 2 } =\surd 2\)
and l =32
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q1.1

Question 2.
Find the third term from the end of the GP.
\(\frac { 2 }{ 27 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 3 } ,….162\)
Solution:
G.P is \(\frac { 2 }{ 27 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 3 } ,….162\)
a = \(\\ \frac { 2 }{ 27 } \)
r = \(\frac { 2 }{ 9 } \div \frac { 2 }{ 27 } \)
= \(\frac { 2 }{ 9 } \times \frac { 27 }{ 2 } \)
= 3
l = 162
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q2.1

Question 3.
For the G.P. \(\frac { 1 }{ 27 } ,\frac { 1 }{ 9 } ,\frac { 1 }{ 3 } …..81\)
find the product of fourth term from the beginning and the fourth term from the end.
Solution:
\(\frac { 1 }{ 27 } ,\frac { 1 }{ 9 } ,\frac { 1 }{ 3 } …..81\)
a = \(\\ \frac { 2 }{ 27 } \)
r = \(\frac { 1 }{ 9 } \div \frac { 1 }{ 27 } \)
= \(\frac { 1 }{ 9 } \times \frac { 27 }{ 1 } \)
= 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q3.1

Question 4.
If for a G.P., pth, qth and rth terms are a, b and c respectively ;
prove that :
{q – r) log a + (r – p) log b + (p – q) log c = 0
Solution:
In a G.P
Tp = a,
Tq = b,
Tr = c
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q4.2

Question 5.
If a, b and c in G.P., prove that : log an, log bn and log cn are in A.P.
Solution:
a, b, c are in G.P.
Let A and R be the first term and common ratio respectively.
Therefore,
a = A
b = AR
c = AR2
log a = log A
log b = log AR = log A + log R
log c = log AR2 = log A + 2log R
log a, log b and log c are in A.P.
If 2log b = log a + log c
If 2[logA + logR] = log A + log A + 2log R
If 2log A + 2log R = 2log A + 2log R
which is true.
Hence log a, log b and log c are in A.P.

Question 6.
If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.
Solution:
Let a, b, c are in G.P.
Then b2 = ac …(i)
Now ax, bx + cx will be in G.P. if (bx)2 = ax.cx
=> (bx)2 = ax.cx
=>(b2)= (ac)x
Hence ax, bx, cx are in G.P. (∴ b2 = ac)
Hence proved.

Question 7.
If a, b and c are in A.P. a, x, b are in G.P. whereas b, y and c are also in G.P. Show that : x2, b2, y2 are in A.P.
Solution:
2 b = a + c _(i)
a, x, b are in G.P.
x2 = ab _(ii)
and b, y, c in G.P.
y2 = bc _(iii)
Now x2 + y2 = ab + bc
= b(a + c)
= b x 2b [from(i)]
= 2 b2
Hence x2, b2, y2 are in G.P.

Question 8.
If a, b, c are in G.P. and a, x, b, y, c are in A.P., prove that :
(i)\(\frac { 1 }{ x } +\frac { 1 }{ y } =\frac { 2 }{ b } \)
(ii)\(\frac { a }{ x } +\frac { c }{ y } =2\)
Solution:
a, b, c are in G.P.
b2 = ac
a, x, b, y, c are in A.P.
2x = a + b and 2y = b + c
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q8.2

Question 9.
If a, b and c are in A.P. and also in G.P., show that: a = b = c.
Solution:
a, b, c are in A.R
2 b = a + c ….(i)
Again, a, b, c are in G.P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C Q9.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C 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 6 Solving Problems Ex 6E

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

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E.

Other Exercises

Question 1.
The distance by road between two towns A and B is 216 km., and by rail it is 208 km. A car travels at a speed of x km/hr. and the train travels at a speed which is 16 km/hr faster than the car. Calculate:
(i) the time taken by the car to reach town B from A, in terms of x ;
(ii) the time taken by the train, to reach town B from A, in terms of x.
(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x, and solve it.
(iv) Hence, find the speed of the train. [1998]
Solution:
Distance between two stations by road = 216 km and by rail = 208 km.
Speed of car = x km/hr.
and speed of train = (x + 16) km/ hr.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q1.1
⇒ 8x + 3456 = 2×2 + 32x
⇒ 2x² + 32x – 8x – 3456 = 0
⇒ 2x² + 24x – 3456 = 0
⇒ x² + 12x – 1728 = 0 (Dividing by 2)
⇒ x² + 48x – 36x – 1728 = 0
⇒ x (x + 48) – 36 (x + 48) = 0
⇒ (x + 48 ) (x – 36) = 0
Either x + 48 =0, then x = – 48 which is not possible.
or x – 36 = 0, then x = 36
(iv) Speed of train = x + 16 = 36 + 16 = 52 km/hr.

Question 2.
A trader buys x articles for a total cost of Rs. 600.
(i) Write down the cost of one article in terms of x. If the cost per article were Rs. 5 more, the number of articles that can be bought for Rs. 600 would be four less.
(ii) Write down the equation in x for the above situation and solve it for x. [1999]
Solution:
C.P. of x articles = Rs. 600
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q2.1
⇒ 600x = (x – 4) (600 + 5x) (By cross multiplication)
⇒ 600x = 600x + 5x² – 2400 – 20x
⇒ 5x² – 20x – 2400 = 0
(ii) x² – 4x – 480 = 0
⇒ x² – 24x + 20x – 480 = 0
⇒ x (x – 24) + 20 (x – 24) = 0
⇒ (x – 24) (x + 20) = 0 (Zero Product Rule)
Either x – 24 = 0, then x = 24
or x + 20 = 0, then x = -20 Which is not possible.
Hence no. of articles = 24

Question 3.
A hotel bill for a number of people for overnight stay is Rs. 4,800. If there were 4 people more, the bill each person had to pay would have reduced by Rs. 200. Find the number of people staying overnight. [2000]
Solution:
Amount of the bill = Rs. 4800
Let the number of persons staying overnight = x
Then amount to be paid by each person
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q3.2
⇒ x (x + 12) – 8 (x + 12) = 0
⇒ (x + 12) (x – 8) = 0 (Zero Product Rule)
Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 8 = 0, then x = 8
Hence no. of persons staying overnight = 8

Question 4.
An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression lor the time taken for :
(i) the omvard journey;
(ii) the return journey.
If the return journey took 30 minutes less than the on ward journey, write down an equation in x and find its value. |2002]
Solution:
Distance between A and B = 400 km.
Speed of aeroplane onward journey = x km/hr.
and Speed of aeroplane on return journey = (x + 40) km/hr.
Now time taken for onward journey = \(\frac { 400 }{ x }\) hrs.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q4.2
Which is not possible being negative
or x – 160 = 0, then x = 160
x = 160

Question 5.
Rs. 6,500 were divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs. 30 less. Find the original number of persons.
Solution:
Let original number of persons = x
Amount = Rs. 6,500
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q5.1
⇒ 30x² + 450x – 97500 = 0 (Dividing by 30)
⇒ x² + 15x – 3250 = 0
⇒ x² + 65x – 50x – 3250 = 0
⇒ x (x + 65) – 50 (x + 65) = 0
⇒ (x + 65) (x – 50) = 0
Either x + 65 = 0, then x = -65 which is not possible.
or x – 50 = 0, then x = 50
Original number of persons = 50

Question 6.
A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km. away in time, it has to increase its speed by 250 km./hr. from its usual speed. Find its usual speed.
Solution:
Let the usual speed of plane = x km/hr.
Distance = 1500 km.
Increased speed = (x + 250) km./hr.
Now, according to the condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q6.1
⇒ x² + 250x – 750000 = 0
⇒ x² + 1000x – 750x – 750000 = 0
⇒ x (x + 1000) – 750 (x + 1000) = 0
⇒ (x + 1000) (x – 750) = 0
Either x + 1000 = 0, then x = – 1000 But it is not possible.
or x – 750 = 0, then x = 750
Usual speed of plane = 750 km/hr

Question 7.
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the average speed of each train.
Solution:
Let the speed of first train = x km/hr.
Then speed of second train = (x – 5) km/ hr.
In 2 hours, first train will travel = 2x km.
and second train will travel = 2 (x – 5) km.
According to the condition,
2x + 2 (x – 5) = 50
⇒ 2x + 2x – 10 = 50
⇒ 4x = 50 + 10 = 60
x = 15 km/hr.
and speed of second train = 15 – 5 = 10 km/hr.

Question 8.
The sum S of first n even natural numbers is given by the relation S = n (n + 1). Find n if the sum is 420.
Solution:
S = n (n + 1) and x = 420
⇒ n (n + 1) = 420
⇒ n² + n – 420 = 0
⇒ n² + 21n – 20n – 420 = 0
⇒ n (n + 21) – 20 (n + 21) = 0
⇒ (n + 21) (n – 20) = 0
Either n + 21 =0, then n = -21 which is not possible as it is negative
or n – 20 = 0 then n = 20

Question 9.
The Sum of the ages of a father and his son is 45 years. Five year ago, the product of their ages (in years) was 124. Determine their present ages.
Solution:
Let age of father = x years
Then age of his son = (45 – x) years (sum = 45 years)
5 years ago,
The age of father = (x – 5) years
and age of son = 45 – x – 5 = (40 – x) years
According to the condition,
(x – 5) (40 – x) = 124
⇒ 40x – x² – 200 + 5x = 124
⇒ -x² + 45x – 200 – 124 = 0
⇒ -x² + 45x – 324 = 0
⇒ x² – 45x + 324 = 0
⇒ x² – 36x – 9x + 324 = 0
⇒ x (x – 36) – 9 (x – 36) = 0
⇒ (x – 36) (x – 9) = 0
Either x – 36 = 0, then x = 36
or x – 9 = 0, then x = 9, but it is not possible as age of father cannot be less than his son.
Age of father = 36 years
and age of son = 45 – 36 = 9 years

Question 10.
In an auditorium, seats were arranged in rows and columns. The number of rows w as equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. Find :
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement. [2003]
Solution:
Let the number of rows in the auditorium = x
No. of seats in each row = x
and no. of total seats in the auditorium =
x x x = x²
In second case,
No. of rows = 2x
and no. of seats in each row = x – 10
Then the total seats will = x² + 300
Now, according to the condition,
2x (x – 10) = x² + 300
⇒ 2x² – 20x = x² + 300
⇒ 2x² – x² – 20x – 300 = 0
⇒ x² – 30x + 10x – 300 = 0
⇒ x (x – 30) + 10 (x – 30) = 0
⇒ (x – 30) (x + 10) = 0 (Zero Product Rule)
Either x – 30 = 0, then x = 30
or x + 10 = 0, then x = -10 Which not possible.
(i) No. of rows in original arrangement = 30
(ii) and no. of seats after re-arrangements = x² + 300 = (30)2 + 300 = 900 + 300 = 1200

Question 11.
Mohan takes 16 days less than Manoj to do a piece of work. If both working together can do it in 15 days, how many days will Mohan alone complete the work?
Solution:
Let time taken by Mohan = x days
Time taken by Manoj = (x + 16) days
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q11.1
⇒ 30x + 240 = x² + 16x
⇒ x² + 16x – 30x – 240 = 0
⇒ x² – 14x – 240 = 0
⇒ x² – 24x + 10x – 240 = 0
⇒ x (x – 24) + 10 (x – 24) = 0
⇒ (x + 10) (x – 24) = 0
Either x + 10 = 0, then .x = -10 But it is not possible
or x – 24 = 0, then x = 24
Mohan can do the work in = 24 days

Question 12.
Two years ago. a man’s age was three times the square of his son’s age. In three years time, his age will be four times his son’s age. Find their present ages.
Solution:
2 years ago,
let son’s age = x
Man’s age = 3x
Son’s present age = x + 2
and man’s age = 3x² + 2
and 3 years after,
Son’s age = x + 2 + 3 = x + 5
Man’s age 3x² + 2 + 3 = 3x² + 5
According to condition,
3x² + 5 = 4 (x + 5)
⇒ 3x² + 5 = 4x + 20
⇒ 3x² – 4x + 5 – 20 = 0
⇒ 3x² – 4x – 15 = 0
⇒ 3x² – 9x + 5x – 15 = 0
⇒ 3x (x – 3) + 5 (x – 3) = 0
⇒ (x – 3) (3x + 5) = 0
Either x – 3 = 0, then x = 3
or 3x + 5 = 0, then 3x = -5 ⇒ x = \(\frac { -5 }{ 3 }\)
But it is not possible.
x = 3
Son’s present age = x + 2 = 3 + 2 = 5 years
and man’s present age = 3x² + 2 = 3(3)² + 2 = 27 + 2 = 29 years

Question 13.
In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both. the fraction reduces by \(\frac { 1 }{ 14 }\) Find the fraction.
Solution:
In first case.
Let numerator of a fraction = x
then, its denominator = x + 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q13.2

Question 14.
In a two digit number, the ten’s digit is bigger. The product of the digits is 27 and the difference between two digits is 6. Find the number.
Solution:
Difference of digits = 6
Let one’s digit = x
Then ten’s digit = x + 6
and number = x + 10 (x + 6) = x + 10x + 60 = 11x + 60
But product of digits = 27
x (x + 6) = 27
⇒ x² + 6x – 27 = 0
⇒ x² + 9x – 3x – 27 = 0
⇒ x (x + 9) – 3 (x + 9) = 0
⇒ (x + 9) (x – 3) = 0
Either x + 9 = 0, then x = – 9 But it is not possible
or x – 3 = 0, then x = 3
Number = 11x + 60 = 11 x 3 + 60 = 33 + 60 = 93

Question 15.
Some school children on an excursion by a bus to a picnic spot at a distance of 300 km. While returning, it was raining and the bus had to reduce its speed by 5 km/ hr and it took two hours longer for returning. Find the time taken to return.
Solution:
Distance = 300 km.
Let speed of the bus = x km/hr.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q15.1
⇒ x² – 5x = 750
⇒ x² – 5x – 750 = 0
⇒ x² – 30x + 25x – 750 = 0
⇒ x (x – 30) + 25 (x – 30) = 0
⇒ (x – 30) (x + 25) = 0
Either x – 30 = 0, then x = 30
or x + 25 = 0, then x = -25, but it is not possible as it is negative
Speed of the bus = 30 km/hr
and time taken while returning = \(\frac { 300 }{ x }\) = \(\frac { 300 }{ 25 }\) = 12 hours

Question 16.
Rs. 480 is divided equally among ‘x’ children. If the number of children were 20 more then each would have got Rs. 12 less. Find ‘x’.
Solution:
Total amount = Rs. 480
Number of children = x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q16.1
⇒ x² + 20x – 800 = 0
⇒ x² + 40x – 20x – 800 – 0
⇒ x (x + 40) – 20 (x + 40) = 0
⇒ (x + 40) (x – 20) = 0
Either x + 40 = 0, then x = – 40 which is not possible being negative
or x – 20 = 0 then x = 20
Number of children = 20

Question 17.
A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to covers the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate ‘x’ (2016)
Solution:
Let the original speed be x km/hr
Time taken by the bus with moving at speed x km/h = \(\frac { 240 }{ x }\)
Time taken by the bus with moving at speed (x – 10) km/h = \(\frac { 240 }{ x – 10 }\)
According to the given condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E Q17.1
⇒ x (x – 10) = 10 x 120
⇒ x² – 10x = 1200
⇒ x² – 10x – 1200 = 0
⇒ x² – 40x + 30x – 1200 = 0
⇒ x (x – 40) + 30 (x – 40) = 0
⇒ (x – 40) (x + 30) = 0
⇒ x – 40 = 0 or x + 30 = 0
⇒ x = 40 or x = -30
Since, the speed cannot be negative, the uniform speed is 40 km/h

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E 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 Ex 24E

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

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 24E.

Other Exercises

Question 1.
The following distribution represents the height of 160 students of a school
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q1.1
Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine :
(i) The median height.
(ii) The inter quartile range.
(iii) The number of students whose height is above 172 cm
Solution:
The cumulative frequency table may be prepared as follows :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q1.3
Now, we take height along x-axis and number of students along the y-axis. Now, plot the point (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152) and (180, 160). On the graph paper and join them with free hand.
(i) Here N = 160 ⇒ \(\frac { N }{ 2 }\) = 80
Which is even now take a point A on the y-axis representing 80. Through A draw horizontal line meeting the ogive at B. From B, draw BC ⊥ x-axis, meeting the x-axis at C.
The abscissa of C is 157.5 So, median = 157.5 cm
(ii) Proceeding in the same way as we have done in above, we have, Q1 = 152 and Q3 = 164
So, inter quartile range = Q3 – Q1 = 164 – 152 = 12 cm
(iii) From the ogive, we see that the number of students whose height is less than 172 is 145.
No. of students whose height is above 172 cm = 160 – 145 = 15

Question 2.
The following table gives the weekly wages of workers in a factory.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q2.1
Calculate : (i) the mean, (ii) the modal class, (iii) the number of workers getting weekly w ages below Rs. 80 and (iv) the number of workers get¬ting Rs 65 or more but less than Rs. 85 as weekly wages. [2002]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q2.2
(ii) Modal class 55-60 (It has maximum frequency)
(iii) No. of workers getting wages below Rs. 80 = 60
(iv) No. of worker getting Rs. 65 is more but less than 85 as weekly wages = 37

Question 3.
Draw an ogive for the data given below and from the graph determine :
(i) the median marks,
(ii) the number of students who obtained more than 75% marks ?
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q3.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q3.3
Through 60.5 th marks, draw a line segment parallel to x-axis which meets the curve at A.
From A, draw a line segment perpendicular to, x-axis meeting at B.
∴ B is the median = 43 (approx.)
No. of students who obtained upto 75% marks in the test =111
∴ No. of students who obtained more than 75% = 120- 111 =9

Question 4.
The mean of 1, 7, 5, 3, 4, and 4 is m. The numbers 3,2,4,2,3,3 and p have mean m-1 and median q. Find p and q.
Solution:
Mean of 1,7, 5, 3,4, and 4 = \(\frac { 24 }{ 6 }\) =4
∴ m = 4.
Now mean of 3,2,4,2,3,3 and p = m- 1= 4- 1 = 3
i.e. 17+p = 3xn when n =7
17 + p = 3×7 = 21
⇒ p = 21 – 17 = 4
Median of 3, 2,4,2, 3, 3 and 3 is q
Arranging in ascending order, 2,2, 3,3,3,3,4,4
Mean = 4th terms is 3.
∴ q = 3

P.Q.
The marks of 200 students in a test were recorded as follows :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp1.1
Construct the cumulative frequency table. Draw an ogive and use it to find :
(i) the median and
(ii) the number of students who score more than 35 % marks.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp1.2
Through 100 th scores, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a line segment perpendicular to it which meets at B.
∴ Median = 52.5
No. of students who score more than 35% marks.
= 200 – 28 = 172
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp1.3

Question 5.
In a malaria epidemic, the number of cases diagnosed were as follows :
Date (July) 1 2 3 4 5 6 7 8 9 10 11 12 Number 5 12 20 27 46 30 31 18 11 5 0 1 On what days does the mode, the upper and lower quartiles occur ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q5.1
(i) Mode = 5th July (because it has the maximum frequencies i,e. 46)

Question 6.
The incomes of the parents of 100 students in a class in a certain university are tabulated below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q6.1
(i) Draw a cumulative frequency curve to estimate the median income.
(ii) If 15 % of the students ae given freeships on the basis of the income of their parents, find the annual income of parents, below which the freeships will be awarded.
(iii) Calculate the Arithmetic mean.
Solution:
(Cummulative Frequancy table)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q6.2
(i) No. of terms= 100
∴ Mean = \(\frac { 100 }{ 2 }\) = 50th term
Through 50 mark, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis meeting it at B, B is the median.
∴ B = 17.6 thousands
(ii) Upper quartile = 100 x \(\frac { 3 }{ 4 }\) = 75th term
From the Curve Q3 = 23.2
(iii) Lower-quartile = 100 x \(\frac { 1 }{ 4 }\)=25th term
From the curve Q1 = 12.8
∴ Inter-quartile range = Q3– Q1 = 23.2 – 12.8
= 10.4 thousands
(iv) 15% of 100 students = \(\frac { 100 x 15 }{ 100 }\) = 15
From C.F. 15, draw a horizontal line which intersects the curve at P. From P, draw a perpendicular to x-axis meeting it at 11.2
∴ Freeship to parents = Rs. 11.2 thousands upto
the income of Rs. 11.2 thousands
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q6.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q6.4

Question 7.
The marks of 20 students in a test were as follows : 2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20. Calculate-
(i) the mean
(ii) the median
(iii) the mode. [2002]
Solution:
Arranging in ascending order,
2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 15, 16, 16, 18, 19, 20
No. of terms = 20
Ix = 2 + 6 + 8 + 9+ 10+ 11 + 11 + 12+ 13 + 13 + 14 + 14 + 14 + 15 + 15 + 15 + 16 + 16 + 18 +
19 + 20 = 257
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q7.1
(iii) Mode = 15 (as it has maximum frequency i.e. it has 3)

Question 8.
The marks obtained by 120 students in a Mathematics test are given below:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q8.1
Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate :
(i) the median
(ii) the number of students who obtained more than 75% marks in a test ?
(iii) the number of students who did not pass in the test if the pass percentage was 40. [2002]
(iv) the lower quartile
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q8.3
Lower Quartile (Q1)
∴ \(\frac { N }{ 4 }\)  = \(\frac { 120 }{ 4 }\)  = 30
From a point B (30) on v-axis, draw a line parallel to x- axis meeting the curve at Q and from Q Draw a line parallel to .Y-axis meeting it at 30.
∴ Lower quartile = 30
Through 60.5th marks, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a line segment perpendicular to, x-axis meeting at B.
∴ B is the median = 43 (approx.)
No. of students who obtained upto 75% marks in the test = 110
∴ No. of students who obtained more than 75% = 120- 110 = 10
No. of students who obtained less than 40% marks in the test = 52 (∴ in the graph x = 40, y = 52)

P.Q.
Find the mean for the following frequency distribution: [2003]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp2.2

P.Q.
Draw a histogram and hence estimate the mode for the following frequency distribution: [2003]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp3.1
Solution:
(i) Draw the histogram.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp3.2
(ii) In the highest rectangle which represents the modal class draw two lines points AC and BD intersecting at P.
(iii) From P, draw a perpendicular to x-axis meeting at Q.
(iv) value of Q is the mode which is = 23

P.Q.
For die following set of data find the median :
10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9 and 15.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q4.1

P.Q.
For the following frequency distribution draw a histogram. Hence, calculate the mode.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp5.1
Solution:
Histogram :
(i) Draw a histogram and make the upper corner of the rectangle
(ii) With maximum frequency A and B. Also upper corners of the two other rectangles as C and D which are the next and to maximum rectangle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp5.2
(iii) Join AD and BC which intersect at P.
(iv) From P, draw PM ⊥ X – axis
OM = 13
Hence mode = 13

Question 9.
Using a graph paper, draw an Ogive for the following distribution which shows a record of the weight in kilograms of 200 students.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q9.1
Use your Ogive to estimate the following :
(i) The percentage of students weighing 55 kg or more,
(ii) The weight above which the heaviest 30% of the students fall.
(iii) The number of students who are (a) under-weight and (b) over-weight, if 55.70 kg is considered as standard weight. (2005)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q9.2
Plot the points (45. 5), (50, 22), (55, 44). (60, 89), (65, 140), (70, 171), (75, 191) and (80, 200) on the graph and join them in free hand to get an ogive as shown.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q9.3
(i) From the graph, number of students weighing 55 kg or more = 200 – 44 = 156
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q9.4
∴ Heaviest 60 students in weight = 9 + 20 + 31 = 60 (From the graph, the required weight is 65 kg or more but less than 80 kg)
(iii) Total number of students who are (i) under weight = 47 and (ii) over weight = 152 (∴ Standard weight is 55.70 kg)

P.Q.
Using step deviation method, find the mean of the following distribution.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp6.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp6.2

P.Q.
The daily wages of 80 workers in a building project are given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp7.1
Using graph paper, draw an Ogive for the above distribution.
Use your Ogive to estimate : (i)the median wages of the workers.
(ii) the percentage of workers who earn more than Rs. 75 day.
(iii) the upper quartile wage of the workers.
(iv) the lower quartile wage of the workers.
(v) Inter quartile range.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp7.3
Now plot the points (40, 6), (50, 16), (60, 31), (70, 50), (80, 62), (90, 70) (100, 76). and (110, 80) on the graph and join them with free hand to get an ogive as shown.
(i) Median : \(\frac { N }{ 2 }\) = \(\frac { 80 }{ 2 }\) = 40
From 40 on y-axis, draw a line parallel to x- axis meeting the curve at P. From P, draw PL ⊥ x-axis
Then L is the median which is 65
∴ Median = Rs. 65
(ii) No. of workers earning more than Rs. 75 per day
From 75 on v-axis, draw a perpendicular meeting the curved at Q and from Q( draw a line parallel to x-axis which meet y-axis at B which is 57
∴No. of workers getting more than Rs. 75 per day = 80 – 57 = 23
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp7.4
From 60 on y-axis, draw- a line parallel to x- axis which meets the curve at R. From R, draw a perpendicular on x-axis meeting it at N.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Qp7.5
From 20 on y-axis, draw a line parallel to x- axis meeting the curve at S. From S, draw a perpendicular on x-axis meeting it at T.
T is the lower quartile (Q1) which is 53.5
∴ Q1 = Rs. 53.50
(v) Inter quartile range = Q3 – Q1 = Rs. 78-53.50 = Rs. 24.50

Question 10.
The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q10.2
Mode = Marks with maximum frequency is 6 ∴ Mode = 6

Question 11.
The mean of the following distribution is 52 and the frequency of class interval 30-40 is Find ‘f’ .Find ‘ f ‘.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q11.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q11.2

Question 12.
The monthly income of a group of 320 employees in a company is given below:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q12.1
Draw an ogive of the given distribution on a graph sheet taking 2 cm = Rs. 1000 on one axis and 2 cm= 50 employees on the other axis.
From the graph determine
(i)the median wage
(ii)the number of employees whose income is below Rs. 8500.
(iii)If the salary of a senior employee is above Rs. 11,500, find the number of senior employees in the company.
(iv) the upper quartile.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q12.3
(i)For median wage, Take OP =\(\frac { 320 }{ 2 }\) =160 on y-axis, Draw a line PQ || x-axis and from Q,
draw QM ⊥ x-axis, abcissa of M point is 9400 ⇒ Median = Rs. 9400
(ii) Take OM’ = 8500 on.t-axis. Draw Q’M’|| toy-axis and P’Q’ || X-axis
Where ordinate of P’ is 92.5
There are approximately 93 employees whose monthly wage is below Rs. 8500
(iii) There are approximately 18 employees whose salary is above Rs. 11500.
(iv) Upper quartile
Mark a point A ony-axis on \(\frac { 3N }{ 4 }\) = \(\frac { 3 x 320 }{ 4 }\)= 240 and draw a line AB || X-axis, then draw BB’
⊥ x-axis abscissa of B’ is upper quartile i.e., Rs. 10250.

Question 13.
A Mathematics aptitude test of 50 students was recorded as follows :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q13.1
Draw a histogram for the above data using a graph paper and locate the mode.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q13.2
1. Draw the histogram with given data.
2. Inside the highest rectangle which represents the maximum frequency (or modal class), draw two lines AC and BD diagonally from the upper comer C and D or adjacent rectangle which intersect at K.
3. Draw KL ⊥ X-axis.
Value ofL is the mode which is 82.5 (approx).

Question 14.
Marks obtained by 200 students in an examination are given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q14.1
Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis. Using the graph, determine
(i) The median marks.
(ii) The number of students who failed if minimum marks required to pass is 40.
(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q14.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q14.3
(i) Median is 57.
(ii)44 students failed.
(iii) No. of students who secured grade one = 200 – 188 = 12

Question 15.
Marks obtained by 40 students in a short assessment are given below, where a and ft are two missing data.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q15.1
If the mean of the distribution is 7.2, find a and b.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q15.3

Question 16.
Find the mode and median of the following frequency distribution :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q16.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q16.2

Question 17.
The median of the following observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q17.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q17.2

Question 18.
The numbers 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x. (2014)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q18.1

Question 19.
(Use a graph paper for this question.) The daily pocket expenses of 200 students in a school are given below:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q19.1
Draw a histogram representing the above distribution and estimate the mode from the graph. (2014)
Solution:
Steps of construction :
(i) Draw a line BC = 6.5 cm.
(ii) Centre B and C draw arcs AB = 5.5 cm and AC = 5 cm
(iii) Join AB and AC, ABC is the required triangle,
(iv) Draw the angle bisetors of B and C. Let these bisectors meet at O.
(v) Taking O as centre. Draw a incircle which touches all the sides of the ∆ ABC.
(vi) From O draw a perpendicular to side BC which cut at N.
(vii) Measure ON which is required radius of the incircle.
ON = 1.5 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q19.2

Question 20.
The marks obtained by 100 students in a Mathematics test are given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q20.1
Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm 10 units on both axes).
Use the ogive to estimate the:
(i) median.
(ii) lower quartile.
(iii) number of students who obtained more than 85% marks in the test.
(iv) number of students failed, If the pass percentage was 35. (2014)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q20.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q20.3
N= 100
Median = \(\frac { 100 }{ 2 }\) = 50th term Median = 45
(ii) Lower quartile : (Q1)
N = 100
⇒ \(\frac { N }{ 4 }\) = \(\frac { 100 }{ 4 }\) = 25
∴ Q1 = 32
(iii) Mo. of students with 85% less = 94
∴ More than 85% marks = 100 94 6
(iv) Number of students who did not pass = 30

Question 21.
The marks obtained by 30 students in a eiass assessment of 5 marks is given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q21.1
Calculate the mean, median and mode of the above distribution. (2015)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q21.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q21.3
Which is between 10 and 20
∴ Median = 3
(ii) Mode frequency of 3 is the greatest
∴ Mode = 3

Question 22.
The weight of 50 workers is given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q22.1
Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis and 2 cm = 5 workers along the other axis. Use the ogive drawn to estimate the following:
(i) the upper and lower quartiles.
(ii) if weighing 95 Kg and above is considered overweight, find the number of workers who are overweight. (2015)
Solution:
Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42) (110, 47) and (120, 50) on the graph and join them in order with free hand.
This is the required ogive
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q22.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q22.3
(i) Upper Quartile = \(\frac { 50 x 3 }{ 4 }\) th term = \(\frac { 150 }{ 4 }\) th = -37.5th term
Lower Quartile = \(\frac { 50 }{ 4 }\) th = 12.5th term
Upper quartile is 42 kg and lower quartile is 72 kg.
(ii) 95 kg and above are over weight
∴ No. of over weight students are 50 – 39 = 11 students.

Question 23.
The mean of following number is 68. Find the value of ‘x’. 45, 52, 60, x, 69, 70, 26, 81 and 94. Hence estimate the median. (2016)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q23.1

Question 24.
The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the X-axis and 2 cm = 20 shooters on the Y-axis). (2016)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q24.1
Use your graph to estimate the following :
(i) The median.
(ii) The interquartile range.
(iii) The number of shooters who obtained a score of more than 85%.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q24.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q24.3
Through mark 80 on y-axis, draw a horizontal line which meets the ogive drawn at point Q.
Through Q, draw a vertical line which meets the x-axis at the mark of 43(app.).
∴ Median = 43
(ii) Since the number of terms = 160
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q24.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q24.5
(iii) Since 85% scores = 85% of 100 = 85
Through mark for 85 on x-axis, draw a vertical line which meets the ogive drawn at point B.
Through the point B, draw a horizontal line which meets thej-axis at the mark of 148 = 160- 148= 12
So, the number of shooters who obtained more than 85% score is 12.

Question 25.
The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to :
(i) Frame a frequency distribution table
(ii) To calculate mean
(iii) To determine the Modal class
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q25.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24E Q25.2
(iii) Here the maximum class frequency is 8, and the class corresponding to this frequency is 20-30. So, the modal class is 20-30.

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