Concise Mathematics Class 10 ICSE 2018 Solutions Pdf

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

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 11A

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, which of the following sequence form a G.P. :
(i) 8, 24, 72, 216, ……
(ii) \(\\ \frac { 1 }{ 8 } \),\(\\ \frac { 1 }{ 24 } \),\(\\ \frac { 1 }{ 72 } \),\(\\ \frac { 1 }{ 216 } \)
(iii) 9, 12, 16, 24,…..
Solution:
(i) 8, 24, 72, 216,……
Here, a = 8

Question 2.
Find the 9th term of the series :
1, 4, 16, 64,…….
Solution:
In G.P. 1, 4, 16, 64,….
Here first term (a) = 1
and common ratio (r) = \(\\ \frac { 4 }{ 1 } \) = 4,
T₉ = ar^{n – 1} = 1 x 4^{9 – 1} = 1 x 4⁸ = 4⁸
= 4 x 4 x 4 x 4 x 4 x 4 x 4 x 4
= 65536

Question 3.
Find the seventh term of the G.P. :
1 , √3, 3, 3√3…….
Solution:
G.P. is 1 , √3, 3, 3√3
Here first term (a) = 1

Question 4.
Find the 8th term of the sequence :
\(\\ \frac { 3 }{ 4 } \),\(1 \frac { 1 }{ 2 } \),3……
Solution:
G.P. = \(\\ \frac { 3 }{ 4 } \),\(1 \frac { 1 }{ 2 } \),3…….
= \(\\ \frac { 3 }{ 4 } \),\(\\ \frac { 3 }{ 2 } \),3…….

Question 5.
Find the 10th term of the G.P. :
12, 4, \(1 \frac { 1 }{ 3 } \),……
Solution:
G.P. = 12, 4, \(1 \frac { 1 }{ 3 } \),……..
= 12, 4, \(\\ \frac { 4 }{ 3 } \),…..

Question 6.
Find the nth term of the series :
1, 2, 4, 8 …….
Solution:
1, 2, 4, 8,……
Here, a = 1,r = \(\\ \frac { 2 }{ 1 } \) = 2

Question 7.
Find the next three terms of the sequence :
√5, 5, 5√5…..
Solution:
√5, 5, 5√5……
Here a = √5 and r = \(\frac { 5 }{ \surd 5 }\) = √5

Question 8.
Find the sixth term of the series :
2², 2³, 2⁴,….
Solution:
2², 2³, 2⁴,……
Here, a = 2², r = 2³ ÷ 2² = 2^{3 – 2} = 2¹ = 2

Question 9.
Find the seventh term of the G.P. :
√3 + 1, 1, \(\frac { \surd 3-1 }{ 2 } \),…….
Solution:
√3 + 1, 1, \(\frac { \surd 3-1 }{ 2 } \),…….

Question 10.
Find the G.P. whose first term is 64 and next term is 32.
Solution:
First term of a G.P. (a) = 64
and second term (ar) = 32
G.P. will be 64, 32, 16, 8, 4, 2, 1,…….

Question 11.
Find the next three terms of the series:
\(\\ \frac { 2 }{ 27 } \),\(\\ \frac { 2 }{ 9 } \),\(\\ \frac { 2 }{ 3 } \),…..
Solution:
G.P. is \(\\ \frac { 2 }{ 27 } \),\(\\ \frac { 2 }{ 9 } \),\(\\ \frac { 2 }{ 3 } \),…..
a = \(\\ \frac { 2 }{ 27 } \)

Question 12.
Find the next two terms of the series
2 – 6 + 18 – 54…..
Solution:
G.P. is 2 – 6 + 18 – 54 +………
Here a = 2 and r = \(\\ \frac { -6 }{ 2 } \) = – 3
Next two terms will be
– 54 x ( – 3) = + 162
162 x ( – 3) = – 486
Next two terms are 162 – 486

 

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A 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 (Based on Quadratic Equations) Ex 6D

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 6D.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E

Question 1.
The sum S of n successive odd numbers starting from 3 is given by the relation :
S = n (n + 2). Determine n, if the sum is 168.
Solution:
S = n (n + 2) and S = 168
⇒ n (n + 2) = 168
⇒ n² + 2n – 168 = 0
⇒ n² + 14n – 12n – 168 = 0
⇒ n (n + 14) – 12 (n + 14) = 0
⇒ (n + 14) (n – 12) = 0
Either n + 14 = 0, then n = -14 which is not possible as n is positive.
or n – 12 = 0, then n = 12
Hence n = 12

Question 2.
A stone is thrown vertically downwards and the formula d = 16t² + 4t gives the distance, d metres, that it falls in t seconds. How long does it take to fall 420 metres ?
Solution:
d = 16t² + 4t, d = 420 m
Distance = 420 m.
6t² + 4t = 420
⇒ 16t² + 4t – 420 = 0
⇒ 4t² + t – 105= 0 (Dividing by 4)
⇒ 4t² + 21t – 20t – 105 = 0
⇒ t (4t + 21) – 5 (4t + 21) = 0
⇒ (4t + 21) (t – 5) = 0
Either 4t + 21 = 0, then 4t = -21 ⇒ t = \(\frac { -21 }{ 4 }\)
But it is not possible as time can not be negative.
or t – 5 = 0 , then t = 5
t = 5 seconds

Question 3.
The product of the digits of two digit number is 24. If its unit’s digit exceeds twice its ten’s digit by 2; find the number.
Solution:
Let ten’s digit = x
then unit’s digit = 2x + 2
According to the condition,
x (2x + 2) = 24
⇒ 2x² + 2x – 24 = 0
⇒ x² + x – 12 = 0 (Dividing by 2)
⇒ x² + 4x – 3x – 12 = 0
⇒ x (x + 4) – 3 (x + 4) = 0
⇒ (x + 4) (x – 3) = 0
Either x + 4 = 0, then x = – 4, which is not possible.
x – 3 = 0, then x = 3.
Ten’s digit = 3
and unit’s digit = 3 x 2 + 2 = 6 + 2 = 8
Number = 8 + 10 x 3 = 8 + 30 = 38

Question 4.
The ages of two sisters are 11 years and 14 years. In how many years time will the product of their ages be 304 ?
Solution:
Let the number of years = x
Age of first sister = 11 + x
and of second sister = 14 + x
Now according to the condition,
(11 + x) ( 14 + x) = 304
⇒ 154 + 11x + 14x + x² = 304
⇒ x² + 25x – 150 = 0
⇒ x² + 30x – 5x – 150 = 0
⇒ x (x + 30) – 5 (x + 30 ) = 0
⇒ (x + 30) (x – 5) = 0
Either x + 30 = 0 , then x = -30 But it is not possible as can’t be in negative
or x – 5 = 0 , then x = 5
Number of years = 5

Question 5.
One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.
Solution:
One year ago, let the age of son = x years
and age of his father = 8x.
But present age of father is = (8x + 1) years
8x + 1 = (x + 1)²
⇒ x² + 2x + 1 = 8x + 1
⇒ x² + 2x + 1 – 8x – 1 = 0
⇒ x² – 6x = 0
⇒ x (x – 6) = 0
Either x = 0, which is not possible.
or x – 6 = 0, then x = 6
Present age of father = 8x + 1 = 8 x 6 + 1 = 48 + 1 = 49 years.
and age of son = x + 1 = 6 + 1 = 7 years

Question 6.
The age of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.
Solution:
Let age of son = x
Then age of father will be = 2x²
8 years hence,
age of son = x + 8
and age of father = 2x² + 8
According to the condition,
2x² + 8 = 3 (x + 8) + 4
⇒ 2x² + 8 = 3x + 24 + 4
⇒ 2x² + 8 – 3x – 28 = 0
⇒ 2x² – 3x – 20 = 0
⇒ 2x² – 8x + 5x – 20 = 0
⇒ 2x (x – 4) + 5 (x – 4) = 0
⇒ (x – 4) (2x + 5) = 0
Either x – 4 = 0, then x = 4
or 2x + 5 = 0, then 2x – 5 ⇒ x = \(\frac { -5 }{ 2 }\)
Which is not possible being negative
x = 4
Present age of son = 4 years
and age of father = 2x² = 2 (4)² = 2 x 16 = 32 years

Question 7.
The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return down-stream to the original point in 4 hours 30 minutes, find the speed of the stream.
Solution:
Let the speed of stream = x km/hr.
Distance = 30 km.
Speed of boat in still water = 15 km/hr.

⇒ 9x² – 2025 + 1800
⇒ 9x² – 225 = 0
⇒ x² – 25 = 0
⇒ (x)² – (5)² = 0
⇒ (x + 5) (x – 5) = 0
Either x + 5 = 0, then x = -5 which is not possible.
or x – 5 = 0, then x = 5
Speed of stream = 5 km/hr.

Question 8.
Mr. Mehra sends his servant to the market to buy oranges worth Rs. 15. The servant having eaten three oranges on the way, Mr. Mehra pays 25 paise per orange more than the market price. Taking x to be the number of oranges which Mr. Mehra receives, form a quadratic equation in x. Hence, find the value of x.
Solution:
No. of oranges received by Mr. Mehra = x
No. of oranges eaten by the servant = 3
Total no. of oranges bought = x + 3
Total cost = Rs. 15
Price of one orange = Rs. \(\frac { 15 }{ x + 3 }\)
Now according to the sum,

⇒ x² + 3x = 180
⇒ x² + 3x – 180 = 0
⇒ x² + 15x – 12x – 180 = 0
⇒ x (x + 15) – 12 (x + 15) = 0
⇒ (x + 15) (x – 12) = 0
Either x + 15 = 0, then x = – 15 which is not possible
or x – 12 = 0, then x = 12
x = 12

Question 9.
Rs. 250 is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.
Solution:
Let the number of children = x
Amount to be divided = Rs. 250

⇒ 6250 x 2 = x² + 25x
⇒ x² + 25x – 12500 = 0
⇒ x² + 125x – 100x – 12500 = 0
⇒ x (x + 125) – 100 (x + 125) = 0
⇒ (x + 125) (x – 100) = 0
Either x + 125 = 0 then x = -125 which is not possible.
or x – 100 = 0, then x = 100
No. of children = 100

Question 10.
An employer finds that if he increases the weekly wages of each worker by Rs. 5 and employs five workers less, he increases his weekly wage bill from Rs. 3,150 to Rs. 3,250. Taking the original weekly wage of each worker as Rs. x; obtain an equation in* and then solve it to find the weekly wages of each worker.
Solution:
In first case,
Let weekly wages of each employee = Rs. x
and number of employees = y
and weekly wages = 3150
xy = 3150 ⇒ y = \(\frac { 3150 }{ x }\) ….(i)
In second case,
Weekly wages = x + 5
and number of employees = y – 5
and weekly wages = 3250
(x + 5) (y – 5) = 3250
⇒ xy + 5y – 5x – 25 = 3200

⇒x (x + 70) – 45 (x + 70) = 0
⇒ (x + 70) (x – 45) = 0
Either x + 70 = 0, then x = -70 which is not possible being negative
or x – 45 = 0, then x = 45
Weekly wages per worker = Rs. 45

Question 11.
A trader bought a number of articles for Rs. 1,200. Ten were damaged and he sold each of the remaining articles at Rs. 2 more than what he paid for it, thus getting a profit of Rs. 60 on the whole transaction. Taking the number of articles he bought as x, form an equation in x and solve it.
Solution:
Let number of articles = x
C.P. = Rs. 1200
Profit = Rs. 60
S.P. = Rs. 1200 + 60 = Rs. 1260
No. of articles damaged = 10
Remaining articles = x – 10

⇒ 2x² – 80x – 12000 = 0
⇒ x² – 40x – 6000 = 0 (Dividing by 2)
⇒ x² – 100x + 60x – 6000 = 0
⇒ x (x – 100) + 60 (x – 100) = 0
⇒ (x – 100) (x + 60) = 0
Either x – 100 = 0, then x = 100
or x + 60 = 0, then x = – 60 which is not possible.
Number of articles = 100

Question 12.
The total cost price of a certain number of identical articles is Rs. 4,800. By selling the article at Rs. 100 each, a profit equal to the cost price of 15 articles is made. Find the number of articles bought.
Solution:
Total cost of some articles = Rs. 4800
Let number of articles = x
S.P. of one article = Rs. 100
S.P. of x articles = Rs. 100x
Profit = Cost price of 15 articles

⇒ x² = 48x + 720 (Dividing by 100)
⇒ x² – 48x – 720 = 0
⇒ x² – 60x + 12x – 720 = 0
⇒ x (x – 60) + 12 (x – 60) = 0
⇒ (x – 60) (x + 12) = 0
Either x – 60 = 0, then x = 60
or x + 12 = 0, then x = -12 Which is not possible.
x = 60
Number of articles = 60

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D are helpful to complete your math homework.

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

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

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

Other Exercises

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

Question 1.
The 6th term of an A.P. is 16 and the 14th term is 32. Determine the 36th term.
Solution:
Let the first term and common difference of an A.P. be a and d
As, we know that,

Question 2.
If the third and the 9th terms of an A.P. term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:
Let the first term and common difference of an A.P. be a and d.
As, we know that,
a_n = a + (n – 1 )d
a₃ = a + (3 – 1 )d = a + 2d
Similarly,

Question 3.
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:
Number of terms in an A.P. = 50
T₃= 12, l = 106
To find T₂₉
Let a be the first term and d be the common difference
=> a + 2d = 12 …(i)

Question 4.
Find the arithmetic mean of :
(i) – 5 and 41
(ii) 3x – 2y and 3x + 2y
(iii) (m + n)² and (m – n)²
Solution:
(i) Arithmetic mean between – 5 and 41
= \(\\ \frac { -5+41 }{ 2 } \)
= \(\\ \frac { 36 }{ 2 } \)
= 18

Question 5.
Find the sum of first 10 terms of the A.P. 4 + 6 + 8 +…..
Solution:
A.P. = 4 + 6 + 8 +…….
Here, a = 4, d = 6 – 4 = 2, n = 10

Question 6.
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 60.
Solution:
Sum of first 20 terms of an A.P. in which
a = 3 and a₂₀ = 60
a₂₀ = a + (20 – 1) x d
60 = 3 + 19 x d
19d = 60 – 3
d = \(\\ \frac { 57 }{ 19 } \)
= 3

Question 7.
How many terms of the series 18 + 15 + 12 +…..when added together will give 45 ?
Solution:
A.P. is 18 + 15 + 12 +…..
Here, a = 18, d = 15 – 18 = – 3
Given : S_n = 45

Question 8.
The nth term of a sequence is 8 – 5n. Show that the sequence is an A.P.
Solution:
Given, a_n = 8 – Sn
a₁ = 8 – 5 x (1) = 8 – 5 = 3
a₂ = 8 – 5 x (2) = 8 – 10 = – 2
a₃ = 8 – 5 x (3) = 8 – 15 = – 7
We see that

Question 9.
The the general term (nth term) and 23rd term of the sequence 3, 1, – 1, – 3,……
Solution:
The progression 3, 1, – 1, – 3,…..is A.P.
with first term (a) = 3 and common difference (d) = 1 – 3 = – 2

Question 10.
Which term of the sequence 3, 8, 13,…..is 78 ?
Solution:
Let 78 be the nth term
a = 3, d = 8 – 3 = 5, a_n = 78, n = ?
a + (n – 1)d = a_n

Question 11.
Is – 150 a term of 11, 8, 5, 2,….. ?
Solution:
11, 8, 5, 2,….1st term, a = 11
Common difference, d = 8 – 11 = – 3
a_n = – 150
=> a + (n – 1 )d = – 150
=> 11 + (n – 1) ( – 3) = – 150

Question 12.
How many two digit numbers are divisible by 3 ?
Solution:
Numbers divisible by 3 are 3, 6, 9, 12,….
Hence, lowest two digit number divisible by 3 = 12
and highest two digit number divisible by 3 = 99

Question 13.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Multiples of 4 between 10 and 250 are
12, 16, 20, 24,……, 248
Here, a = 12, d = 4, l = 248

Question 14.
The sum of the 4th term and the 8th term of an A.P. is 24 and the sum of 6th term and the 10th term is 44. Find the first three terms of the A.P.
Solution:
In an A.P.
T₄ + T₈ = 24
T₆ + T₁₀ = 44
Let a be the first term and d be the common difference

Question 15.
The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
Solution:
Given a₁₄ = 140
we know, a_n = a + (n – 1) x d

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F 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 (Based on Quadratic Equations) Ex 6C

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 6C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E

Question 1.
The speed of an ordinary train is x km/hr and that of an express train is (x + 25) km per hour.
(i) Find the time taken by each train to cover 300 km.
(ii) If the ordinary train takes 2 hrs more than the express train; calculate speed of the express train.
Solution:
Speed of an ordinary train = x km/ hr
and speed of an express train = (x + 25) km/hr.
(i) Time taken by ordinary train to cover 300 km = \(\frac { 300 }{ x }\) hr
(ii) Time taken by express train to cover 300 km = \(\frac { 300 }{ x + 25 }\) hr

⇒ 7500 = 2x² + 50x
⇒ 2x² + 50x – 7500 = 0
⇒ x² + 25x – 3750 = 0 (Dividing by 2)
⇒ x² + 75x – 50x – 3750 = 0
⇒ x (x + 75) -50 (x + 75) = 0
⇒ (x + 75) (x – 50) = 0
Either x + 75 = 0 then x = -75 But it is not possible,
or x – 50 = 0 then x = 50
Speed of an ordinary train = 50 km/ hr
and speed of express train = 50 + 25 = 75 km/ hr

Question 2.
If the speed of a car is increased by 10 km per hr, it takes 18 minutes less to cover a distance of 36 km. Find the speed of the car.
Solution:
Let the speed of ear = x km/hr
time taken to cover 36 km = \(\frac { 36 }{ x }\) hr
In second case,
Speed of car = (x + 10) km/hr

⇒ 3x² + 30 x = 3600
⇒ 3x² + 30x – 3600 = 0 .
⇒ x² + 10x – 1200 = 0 (Dividing by 3)
⇒ x² + 40x – 30x – 1200 = 0
⇒ x (x + 40) – 30 (x + 40) = 0
⇒ (x + 40) (x – 30) = 0
Either x + 40 = 0 then x = – 40 But it is not possible,
or x – 30 = 0, then x = 30
Speed of car = 30 km/hr.

Question 3.
If the speed of an aeroplane is reduced by 40 km/per hr, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.
Solution:
Let the speed of aeroplane = x km/hr.

⇒ x² – 40x = 144000
⇒ x² – 40x – 144000 = 0
⇒ x² – 400x + 360x – 144000 = 0
⇒ x (x – 400) + 360 (x – 400) = 0
Either x – 400 = 0, then x = 400
or x + 360 = 0, then x = – 360, But it is not possible.
x = 400
Hence speed of aeroplane = 400 km/hr.

Question 4.
A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
Solution:
Let the original speed of the car be x km/h.

⇒ 1200x + 14400 – 1200x = 5x² + 60x
⇒ 14400 = 5x² + 60x
⇒ x² + 12x – 2880 = 0 (Dividing by 5)
⇒ x² + 60x – 48x – 2880 = 0
⇒ x (x + 60) – 48 (x + 60) = 0
⇒ (x + 60) (x – 48) = 0
⇒ x = 48 or x = -60
⇒ x = 48 (Rejecting x = -60, being speed)
Hence, original speed of the car = 48 km/h.

Question 5.
A girl goes to her friend’s house, which is at a distance of 12 km. She covers half of the distance at a speed of ‘x’ km/hr. and the remaining distance at a speed of (x + 2) km/hr. If she takes 2 hrs. 30 minutes to cover the whole distance; find ‘x’
Solution:
Distance = 12 km.
Speed for the first half distance = x km/hr.
and for the second half distance = (x + 2) km/hr.
Total time taken = 2 hrs. 30 minutes.

But it is not possible.
Speed for first half distance (x) = 4 km/hr

Question 6.
A car made a run of 390 km in ‘x’ hours. If the speed had been 4 km per hour more, it would have taken 2 hours less for the journey. Find ‘x’.
Solution:
Distance = 390 km
time = x hours

⇒ (x – 15) (x + 13) = 0
Either x – 15 = 0, then x = 15
or (x + 13) = 0 then x = -13 Which is not possible.
Value of x = 15

Question 7.
A goods train leaves a station at 6 p.m., followed by an express train which leaves at 8 p.m. and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the trains remain constant between the two stations; calculate their speeds.
Solution:
Departure of goods train = 6 p.m.
and departure of express train = 8 p.m.
Speed of express train is more than goods trains by 20 km/hr
Total distance = 1040 km
Let speed of goods train = x km/hr
Then speed of express train = (x + 20) km/hr
Difference of time taken = 8 p.m. – 6 p.m. + 36 minutes = 2 hours 36 minutes

⇒ x – (x + 100) – 80 (x + 100) = 0
⇒ (x + 100) (x – 80) = 0
Either x + 100 = 0, then x = -100 but it is not possible being negative
or x – 80 = 0, then x = 80
Speed of goods train = 80 km/hr
and speed of express train = 80 + 20 = 100 km/hr
⇒ 13x² + 507x – 35100 = 0
⇒ x² + 39x – 2700 = 0 (Dividing by 13)
⇒ x² + 75x – 36x – 2700 = 0
⇒ x (x + 75) – 36(x + 75) = 0
⇒ (x + 75) (x – 36) = 0
Either x + 75 = 0, then x = – 75 Which is not possible,
or x – 36 = 0 then x = 36
Speed of goods train = 36 km/hr
and speed of express train = 36 + 39 = 75 km/hr.

Question 8.
A man bought an article for Rs. x and sold it for Rs. 16. If his loss was x percent, find the cost price of the article.
Solution:
C.P. of article = Rs. x
S.R = Rs. 16
Loss = C.P. – S.P. = Rs. (x – 16)

⇒ x² – 100x + 1600 = 0
⇒ x² – 20x – 80x + 1600 = 0
⇒ x (x – 20) – 80 (x – 20) = 0
⇒ (x – 20) (x – 80) = 0
Either x – 20 = 0, then x = 20
or x – 80 = 0, then x = 80
Cost Price = Rs. 20 or Rs. 80

Question 9.
A trader bought an article for Rs. x and sold it for Rs. 52, thereby making a profit of (x – 10) per cent on his outlay. Calculate the cost price.
Solution:
Let C.P. = Rs. x
S.R = Rs. 52
Profit = S.P – C.P. = Rs. 52 – x

x² – 10x = 5200 – 100x
⇒ x² – 10x + 100x – 5200 = 0
⇒ x² + 90x – 5200 = 0
⇒ x2 + 130x – 40x – 5200 = 0
⇒ x (x + 130) – 40(x + 130) = 0
⇒ (x + 130) (x – 40 ) = 0
Either x + 130 = 0 , then x = – 130 which is not possible.
or x – 40 = 0 then x = 40
Cost price = Rs. 40

Question 10.
By selling a chair for Rs. 75, Mohan gained as much per cent as its cost. Calculate the cost of the chair.
Solution:
Let. C.P of chair = Rs. x
Profit = x %
S.P. = Rs. 75
Total profit = Rs. (75 – x)

⇒ x² = 7500 – 100x
⇒ x² + 100x – 7500 = 0
⇒ x²+ 150x – 50 – 7500 = 0
⇒ x (x + 150) – 50 (x + 150) = 0
⇒ (x + 150) (x – 50) = 0
Either x + 150 = 0, then x = -150 which is not possible
or x – 50 = 0, Then x = 50
Cost price of chair = Rs. 50

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C are helpful to complete your math homework.

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

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

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

Other Exercises

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

Question 1.
Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km h^-1. The second car goes at a speed of 8 km h^-1 in the first hour and thereafter increasing the speed by 0.5 km h^-1 each succeeding hour. After how many hours will the two cars meet ?
Solution:
Speed of first car = 10 km/hr
Speed of second car = 8 km/hr in first hour

Question 2.
A sum of Rs 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs 20 less than its preceding prize; find the value of each of the prizes.
Solution:
Total amount (S_n) = Rs 700
Cost of each prize is Rs 20 less than its preceding price or d = – 20
d = – 20 and n = 7
\({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \)

Question 3.
An article can be bought by paying Rs 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs 3,000 and every other installment is Rs 100 less than the previous one, find :
(i) amount of installment paid in the 9th month
(ii) total amount paid in the installment scheme.
Solution:
Total price of an article = Rs 28000
No. of installments (n) = 12
First installment (a) = RS 3000
d = Rs 100

Question 4.
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find :
(i) the production in the first year.
(ii) the production in the 10th year.
(iii) the total production in 7 years.
Solution:
A manufacture of TV sets, he produces
No. of units in 3rd year = 600
No. of units in 7th year = 700
Let a be the first term and d be the common difference, then

Question 5.
Mrs. Gupta repays her total loan of Rs 1.18,000 by paying installments every month. If the installment for the first month is Rs 1,000 and it increases by RS 100 every month, what amount will she pay as the 30th installment of loan? What amount of loan she still has to pay after the 30th installment?
Solution:
Total loan to be paid by Mrs. Gupta = Rs 118000
Installment for the first month (a) = Rs 1000

Question 6.
In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be five times of the class to which the respective section belongs. If there are 1 to 10 classes in the school and each class has three sections, find how many trees were planted by the students?
Solution:
Number of classes = 10
Number of sections of each class = 3
Total number of sections = 10 x 3 = 30
Each section plant tree = 5 times of the class
Each section of 1st class will plant = 1 x 15 = 15

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