ICSE Class 10

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

• 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 Ex 7B
• 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 Ex 7D

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

Question 2.
Find the third proportional to :
(i) 2\(\frac { 2 }{ 3 }\) and 4
(ii) a – b and a² – b²
Solution:

Question 3.
Find the mean proportional between :
(i) 6 + 3√3 and 8 – 4√3
(ii) a – b and a³ – a²b.
Solution:

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

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.

Question 7.

Solution:

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:

Solution:

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

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

Question 15.

Solution:

Question 16.
If p : q = r : s ; then show that: mp + nq : q = mr + ns : s.
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

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

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

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

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.

∴ 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

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

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:

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,

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,

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°

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

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,

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,

(ii) In second case,
(a) Angle of elevation = 45°
and height of tower = 25.98 m or 15\(\sqrt { 3 }\) m

(b) Angle of elevation = 60°
and height of tower = 25.98 m or 15\(\sqrt { 3 }\) m.
Let shadow of the tower DB = xm

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.

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,

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,

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°

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

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

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B
• 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 11D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions

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

Question 2.
How many terms of the geometric progression 1 + 4 + 16 + 64 +…. must be added to get sum equal to 5461 ?
Solution:
S_n = 5461 and G.P. is
1 + 4 + 16 + 64 +…..
Here, a = 1, r = 4 (r > 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

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)

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 = \(\\ \frac { 1 }{ 27 } \)

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 (S_n) = 728
Let a be the first term, then

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

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, S_n = 728

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

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,S_n = \(\\ \frac { 55 }{ 72 } \)

Question 11.
If the sum of 1 + 2 + 2² +…..+ 2^{n – 1} is 255, find the value of n.
Solution:
1 + 2 + 2² +…..+ 2^{n – 1} = 255

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 \(\\ \frac { 9 }{ 4 } \)

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

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

Question 15.
Find the 5th term of the G.P. \(\\ \frac { 5 }{ 2 } \), 1,…..
Solution:
Given G.P is \(\\ \frac { 5 }{ 2 } \), 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

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

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

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,

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 (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

• 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 Ex 7B
• 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 Ex 7D

Question 1.
If a : b = 5 : 3; find: \(\frac { 5a – 3b }{ 5a + 3b }\)
Solution:

Question 2.
If x : y = 4 : 7; find the value of (3x + 2y) : (5x + y).
Solution:

Question 3.
If a : b = 3 : 8, find the value of \(\frac { 4a + 3b }{ 6a – b }\)
Solution:

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

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

Question 6.

Solution:

Question 7.
Find \(\frac { x }{ y }\) ; when x² + 6y² = 5xy.
Solution:

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:

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

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

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,

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

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:


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:

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

Question 21.
Find the duplicate ratio of:
(i) 3 : 4
(ii) 3√3 : 2√5
Solution:

Question 22.
Find triplicate ratio of:
(i) 1 : 3
(ii) \(\frac { m }{ 2 }\) : \(\frac { n }{ 3 }\)
Solution:

Question 23.
Find sub-duplicate ratio of:
(i) 9 : 16
(ii) (x – y)⁴ : (x + y)⁶
Solution:

Question 24.
Find sub-triplicate ratio of:
(i) 64 : 27
(ii) x³ : 125y³
Solution:

Question 25.
Find the reciprocal ratio of :
(i) 5 : 8
(ii) \(\frac { x }{ 3 }\) : \(\frac { y }{ 7 }\)
Solution:

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

Question 27.
If m : n is the duplicate ratio of m + x : n + x; show that x² = mn.
Solution:

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

⇒ 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

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:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A are helpful to complete your math homework.

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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 :

Solution:

Question 2.
Draw a cumulative frequency curve (ogive) for each of the following distributions

Solution:

Question 3.
Draw an ogive for each of the following distributions :

Solution:

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.

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:



(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.

Solution:
From the histogram given, the required frequency table will be as given below.

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.

Question 6.

(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

(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.

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

Solution:


(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 :

(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:


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 :

Solution:

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.