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 10B

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 10B

Other Exercises

Question 1.
In an A.P., ten times of its tenth term is equal to thirty times of its 30th term. Find its 40th term.
Solution:
In an A.P.
10T10 = 30T30
We know that,
If m times of mth term = n times of nth term
Then its (m + n)th term = 0
∴ 10T10 = 30T30
Then T10+30 = 0 or T40 = 0

Question 2.
How many two-digit numbers are divisible by 3 ?
Solution:
Two digits numbers are 10 to 99
and two digits numbers which are divisible
by 3 are 12, 15, 18, 21,….99
Here a = 12, d = 3 and l = 99
Now, Tn = l = a + (n – 1 )d
99 = 12 + (n – 1) x 3
=> 99 – 12 = 3(n – 1)
=> 87 = 3(n – 1)
=> \(\\ \frac { 87 }{ 3 } \) = n – 1
=> n – 1 = 29
n = 29 + 1 = 30
Number of two digit number divisible by 3 = 30

Question 3.
Which term of A.P. 5, 15, 25, will be 130 more than its 31st term?
Solution:
A.P. is 5, 15, 25,….
Let Tn = T31 + 130
In A.P. a = 5, d = 15 – 5 = 10
∴ Tn = a + 30d + 130 = 5 + 30 x 10 +130
= 5 + 300 + 130 = 435
∴ n + (n – 1)d = 435
=> 5 + (n – 1) x 10 = 435
=> (n – 1) x 10 = 435 – 5 = 430
n – 1 = \(\\ \frac { 430 }{ 10 } \) = 43
n = 43 + 1 = 44
∴ The required term is 44th.

Question 4.
Find the value of p, if x, 2x + p and 3x + 6 are in A.P.
Solution:
A.P. is x, 2x + p, 3x + 6
2x + p – x = 3x + 6 – 2x + p
x + p = x – p + 6
=>2p = 6
=>p = \(\\ \frac { 6 }{ 2 } \) = 3
Hence p = 3

Question 5.
If the 3rd and the 9th terms of an arithmetic progression are 4 and – 8 respectively, which term of it is zero?
Solution:
In an A.P.
T3 = 4 and T9 = – 8
Let a be the first term and d be the common difference, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B Q5.1

Question 6.
How many three-digit numbers ate divisible by 87 ?
Solution:
Three digits numbers are 100 to 999
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B Q6.1
999 – 42 = 957
Numbers divisible by 87 will be 174,….., 957
Let n be the number of required three digit number.
Here, a = 174 and d = 87
l = 957 = a + (n – l)d
= 174 + (n – 1) x 87
=> (n – 1) x 87 = 957 – 174 = 783
n – 1 = \(\\ \frac { 783 }{ 87 } \) = 9
n = 9 + 1 = 10
There are 10 such numbers.

Question 7.
For what value of n, the nth term of A.P. 63, 65, 67,….. and nth term of A.P. 3, 10, 17,…… are equal to each other?
Solution:
We are given,
nth term of 63, 65, 67, …….
=> nth term of 3, 10, 17,…….
=> In first A.P., a1 = 63, d1 = 2
and in second A.P., a2 = 3, d2 = 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B Q7.1
∴ Required nth term will be 13th

Question 8.
Determine the A.P. whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
T3 = 16, and T7 = T5 + 12
Let a be the first term and d be the common difference
T3 = a + 2d – 16 …(i)
T7 = T5 + 12
a + 6d = a + 4d + 12
=> 6d – 4d = 12 => 2d – 12
=> d = \(\\ \frac { 12 }{ 2 } \) = 6
and in (i),
a + 6 x 2 = 16
=> a + 12 = 16
=>a = 16 – 12 = 4
A.P. will be 4, 10, 16, 22,……

Question 9.
If numbers n – 2, 4n – 1 and 5n + 2 are in A.P., find the value of n and its next two terms.
Solution:
n – 2, 4n – 1 and 5n + 2 are in A.P.
(4n – 1) – (n – 2) = (5n + 2) – (4n – 1)
=> 4n – 1 – n + 2 = 5n + 2 – 4n + 1
=> 4n – n – 5n + 4n = 2 + 1 + 1 – 2
=>2n = 2 =>n = \(\\ \frac { 2 }{ 2 } \) = 1
4n – 1 = 4 x 1 – 1 = 4 – 1 = 3
n – 2 = 1 – 2 = – 1
and 5n + 2 = 5 x 1 + 2 = 5 + 2 = 7
A.P. is – 1, 3, 7, ……
and next 2 terms will 11, 15

Question 10.
Determine the value of k for which k2 + 4k + 8, 2k2 + 3k + 6 and 3k2 + 4k + 4 are in A.P.
Solution:
k2 + 4k+ 8, 2k2 + 3k + 6 and 3k2 + 4k + 4 are in A.P.
(2k2 + 3k + 6) – (k2 + 4k + 8)
= (3k2 + 4k + 4) – (2k2 + 3k + 6)
=> 2k2 + 3k + 6 – k2 – 4k – 8
= 3k2 + 4k + 4 – 2k2 – 3k – 6
=> k2 – k – 2 = k2 + k – 2
=> k2 – k – k2 – k = – 2 + 2
=> – 2k = 0
=> k = 0
Hence k = 0

Question 11.
If a, b and c are in A.P. show that :
(i) 4a, 4b and 4c are in A.P.
(ii) a + 4, b + 4 and c + 4 are in A.P.
Solution:
a, b, c are in A.P.
2b = a + c
(i) 4a, 4b and 4c are in A.P.
If 2(4b) = 4a + 4c
If 8b = 4a + 4c
If 2b = a + c which is given
(ii) a + 4, b + 4 and c + 4 are in A.P.
If 2(b + 4) = a + 4 + c + 4
If 2b + 8 = a + c + 8
If 2b = a + c which is given

Question 12.
An A.P. consists of 57 terms of which 7th term is 13 and the last term is 108. Find the 45th term of this A.P.
Solution:
Number of terms in an A.P. = 57
T7 = 13, l= 108
To find T45
Let a be the first term and d be the common difference
=> a + 6d = 13 …(i)
a + (n – 1 )d= 108
=> a + (57 – 1 )d= 108
=> a + 56d = 108
Subtracting,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B Q12.1

Question 13.
4th term of an A.P. is equal to 3 times its term and 7th term exceeds twice the 3rd term by 1. Find the first term and the common difference.
Solution:
In A.P
T4 = 3 x T1
T7 = 2 x T3 + 1
Let a be the first term and d be the common
difference
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B Q13.1

Question 14.
The sum of the 2nd term and the 7th term of an A.P. is 30. If its 15th term is 1 less than twice of its 8th term, find the A.P.
Solution:
In an A.P.
T2 + T7 = 30
T15 = 2T8 – 1
Let a be the first term and d be the common difference
a + d + a + 6d = 30
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B Q14.1

Question 15.
In an A.P., if mth term is n and nth term is m, show that its rth term is (m + n – r)
Solution:
In an A.P.
Tm = n and Tn = m
Let a be the first term and d be the common difference, then
Tm = a + (m – 1 )d = n
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B Q15.1

Question 16.
Which term of the A.P. 3, 10, 17,…..will be 84 more than its 13th term?
Solution:
A.P. is 3, 10, 17,….
Here, a = 2, d = 10 – 3 = 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10B Q16.1

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

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A

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

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 6A.

Other Exercises

Question 1.
The product of two consecutive integers is 56. Find the integers.
Solution:
Let the first integer = x
Then second integer = x + 1
Now according to the condition given
x (x + 1) = 56
⇒ x² + x – 56 = 0
⇒ x² + 8x – 7x – 56 = 0
⇒ x (x + 8) – 7 (x + 8) = 0
⇒ (x + 8) (x – 7) = 0
Either x + 8 = 0, then x = – 8
or x – 7 = 0, then x = 7
(i) If x = -8, then
first integer = -8
and second integer = – 8 + 1 = – 7
(ii) If x = 7, then
first integer = 7
and second integer = 7 + 1 = 8
Integers are 7, 8 or -8, -7

Question 2.
The sum of the squares of two consecutive natural numbers is 41. Find the numbers.
Solution:
Let the first natural number = x
Then second natural number = x + 1
Now according to the condition given,
(x)² + (x + 1)² = 41
⇒ x² + x² + 2x + 1 = 41
⇒ 2x² + 2x + 1 – 41 = 0
⇒ 2x² + 2x – 40 = 0
⇒ x² + x – 20 = 0 (Dividing by 2)
⇒ x² + 5x – 4x – 20 = 0
⇒ x (x + 5) – 4 (x + 5) = 0
⇒ (x + 5) (x – 4) = 0
Either x + 5 = 0 then x = – 5 But it is not a natural number
or x – 4 = 0, Then x = 4
Numbers are 4 and 5

Question 3.
Find the two natural numbers which differ by 5 and the sum of whose squares is 97.
Solution:
Let the first natural number = x
Then second natural number = x + 5
Now according to the given condition,
(x)² + (x + 5)² = 97
⇒ x² + x² + 10x + 25 – 97 = 0
⇒ 2x² + 10x – 72 = 0
⇒ x² + 5x – 36 = 0 (Dividing by 2)
⇒ x² + 9x – 4x – 36 = 0
⇒ x (x + 9) – 4 (x + 9) = 0
⇒ (x + 9) (x – 4) = 0
Either x + 9 = 0, then x = – 9 But it is not a natural number
or x – 4 = 0, then x = 4
First number = 4
and second number = 4 + 5 = 9

Question 4.
The sum of a number and its reciprocal is 4.25. Find the number.
Solution:
Let the number = x
Now according to the condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q4.1
⇒ 4x (x – 4) – 1 (x – 4) = 0
⇒ (x – 4) (4x – 1) = 0
Either x – 4 = 0, then x = 4
or 4x – 1 = 0, 4x = 1 then x = \(\frac { 1 }{ 4 }\)
Number is 4 or \(\frac { 1 }{ 4 }\)

Question 5.
Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is \(\frac { 7 }{ 10 }\)
Solution:
Let the first natural number = x
Then second natural number = x + 3
Now according to the given condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q5.1
But it is not a natural number
or x – 2 = 0, then x = 2
First number = 2
and second number = 2 + 3 = 5

Question 6.
Divide 15 into two parts such that the sum of their reciprocals is \(\frac { 3 }{ 10 }\).
Solution:
Let first part = x
Then second part = 15 – x (sum = 15)
Now according to the given condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q6.1
⇒ x (x – 5) – 10 (x – 5) = 0
⇒ (x – 5) (x – 10) = 0
Either x – 5 = 0, then x = 5
or x – 10 = 0, then x = 10
If x = 5, then first part = 15 – 5 = 10
If x = 10, then second part = 15 – 10 = 5
Parts are 5, 10

Question 7.
The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the numbers.
Solution:
Let x be the larger number and y be the smaller number, then
According to the conditions x2 = 18 ….(i)
and x² + y² = 208 ….(ii)
⇒ 18y + y² = 208 [From (i)]
⇒ y² + 18y – 208 = 0
⇒ y² + 26y – 8y – 208 = 0
⇒ y (y + 26) – 8 (y + 26) = 0
⇒ (y + 26) (y – 8) = 0
Either y + 26 = 0, then y = -26
But it is not possible as it is not positive
or y – 8 = 0, then y = 8
Then x² = 18y ⇒ y² = 18 x 8 ⇒x² = 144 = (12)² ⇒ x = 12
Number are 12, 8

Question 8.
The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.
Solution:
Let first even number = 2x
and second even number = 2x + 2
Now according to the given condition,
(2x)² + (2x + 2)² = 52
⇒ 4x² + 4x² + 8x + 4 = 52
⇒ 4x² + 4x² + 8x + 4 – 52 = 0
⇒ 8x² + 8x – 48 = 0
⇒ x² + x – 6 = 0 (Dividing by 8)
⇒ x² + 3x – 2x – 6 = 0
⇒ x (x + 3) – 2 (x + 3) = 0
⇒ (x + 3) (x – 2) = 0
Either x + 3 = 0, then x = – 3 But it is not possible because it is not positive.
or x – 2 = 0, then x = 2
First even number = 2 x 2 = 4
Second number = 4 + 2 = 6
Hence numbers are 4, 6

Question 9.
Find two consecutive positive odd numbers, the sum of whose squares is 74.
Solution:
Let first odd number = 2x -1
Second odd number = 2x – 1 + 2 = 2x + 1
Now according to the given condition,
(2x – 1)² + (2x + 1)² = 74
⇒ 4x² – 4x + 1 + 4x² + 4x + 1 = 74
⇒ 8x² + 2 – 74 = 0
⇒ 8x² – 72 = 0
⇒ x² – 9 = 0 (Dividing by 8)
⇒ (x + 3) (x – 3) = 0
Either x + 3 = 0, then x = -3 But it is not possible because it is not positive.
or x – 3 = 0, then x = 3
First odd number = 2x – 1 = 2 x 3 – 1 = 6 – 1 = 5
and second odd number = 5 + 2 = 7
Number are 5, 7

Question 10.
The denominator of a positive fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2.9; find the fraction.
Solution:
Let numerator of a fraction = x
Then denominator = 2x + 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q10.2

Question 11.
Three positive numbers are in the ratio \(\frac { 1 }{ 2 }\) : \(\frac { 1 }{ 3 }\) : \(\frac { 1 }{ 4 }\) Find the numbers; if the sum of their squares is 244.
Solution:
Multiply each ratio by L.C.M. of de-nominators i.e. by 12.
We get, 6 : 4 : 3
Let first positive number = 6x
Then second number = 4x
and third number = 3x
According to the given condition,
(6x)² + (4x)² + (3x)² = 244
⇒ 36x² + 16x² + 9x² = 244
⇒ 61x² = 244
⇒ x² – 4 = 0
⇒ (x + 2) (x – 2) = 0
Either x + 2 = 0, then x = -2 But it is not possible because it is not positive
or x – 2 = 0, then x = 2
First number = 6x = 6 x 2 = 12
Second number = 4x = 4 x 2 = 8
and third number = 3x = 3 x 2 = 6
Numbers are 12, 8, 6

Question 12.
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Solution:
Let first part = x
Then second part = 20 – x (Sum = 20)
Now, according to the given condition,
3(x)² – (20 – x) = 10
⇒ 3x² – 20 + x – 10 = 0
⇒ 3x² + x – 30 = 0
⇒ 3x² + 10x – 9x – 30 = 0
⇒ x (3x + 10) – 3 (3x + 10) = 0
⇒ (3x + 10) (x – 3) = 0
Either 3x + 10 = 0, then 3x = -10 ⇒ x = \(\frac { -10 }{ 3 }\)
But it is not possible.
or x – 3 = 0 then x = 3
Then first part = 3
and second part = 20 – 3 = 17

Question 13.
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Assume the middle number to be x and form a quadratic equation satisfying the above statement Hence ; find the three numbers.
Solution:
Let middle number = x
Then, first number = x – 1
and third number = x + 1
Now according to the condition,
(x)²= [(x + 1)² – (x – 1)²] + 60
⇒ x² = [x² + 2x + 1 – x² + 2x – 1 ] + 60
⇒ x² = 4x + 60
⇒ x² – 4x – 60 = 0
⇒ x² – 10x + 6x – 60 = 0
⇒ x (x – 10) + 6 (x – 10) = 0
⇒ (x – 10) (x + 6) = 0
Either x – 10 = 0 then x = 10
or x + 6 = 0 then x = -6. But it is not a natural number.
Middle number = 10
First number = 10 – 1 = 9
and third number = 10 + 1 = 11
Hence numbers are 9, 10, 11

Question 14.
Out of three consecutive positive integers, the middle number is p. If three times the square of the largest is greater than the sum of the squares of the other two numbers by 67; calculate the value of p.
Solution:
Middle number = p
then smallest number = p – 1
and greatest number = p + 1
Now, according to the condition.
3 (p + 1)² – (p – 1)² – p² = 67
⇒ 3 (p² + 2p + 1) – (p² – 2p + 1) – p² = 67
⇒ 3p² + 6p + 3 – p² + 2p – 1 – p² – 67 = 0
⇒ p² + 8p + 2 – 67 = 0
⇒ p² + 8p – 65 = 0
⇒ p² + 13p – 5p – 65 = 0
⇒ p (p + 13) – 5 (p + 13) = 0
⇒ (p + 13) (p – 5) = 0
Either p + 13 = 0, then p = -13 But it is not possible.
or p – 5 = 0, then p = 5
p = 5

Question 15.
A can do a piece of work in ‘x’ days and B can do the same work in (x + 16) days. If both working together can do it in 15 days. Calculate ‘x’.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q15.2

Question 16.
One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.
Solution:
Let first pipe can fill the cistern in = x hrs.
Second pipe will fill the cistern in = x – 3 hrs.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q16.1
But it is not possible.
x = 15
First pipe can fill in 15 hrs.
and second pipe in 15 – 3 = 12 hrs.

Question 17.
A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking A as the smaller part of the two parts. Find the number. (2010)
Solution:
Let larger part = y
and smaller part = x
x² + y² = 20 ….(i)
and y² = 8x ….(ii)
Substituting the value of y² in „
x² + 8x = 20
⇒ x² + 8x – 20 = 0
⇒ x² + 10x – 1x – 20 = 0 {-20 = 10 x (-2), 8 = 10 – 2}
⇒ x (x + 10) – 2(x + 10) = 0
⇒ (x + 10) (x – 2) = 0
Either x + 10 = 0, then x = – 10 which is not possible because it is negative
or x – 2 = 0, then x = 2
Smaller number = 2
and larger number = y² = 8x = 8 x 2 = 16
⇒ y² = 16 = (4)²
⇒ y = 4
Number = x + y = 2 + 4 = 6

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

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

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 24A.

Other Exercises

Question 1.
Find the mean of following set of numbers:
(i) 6, 9, 11, 12 and 7
(ii) 11, 14, 23, 26, 10, 12, 18 and 6.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q1.1

Question 2.
Marks obtained (in mathematics) by a students are given below :
60, 67, 52, 76, 50, 51, 74, 45 and 56
(a) Find the arithmetic mean
(b) If marks of each student be increased by 4;
what will be the new value of arithmetic mean.
Solution:
(a) Hence x = 9
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q2.1
(b) If marks of each students be increased by 4 then new mean will be = 59 + 4 = 63

Question 3.
Find the mean of natural numbers from 3 to 12.
Solution:
Numbers betweeen 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Here n = 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q3.1

Question 4.
(a) Find the means of 7, 11, 6, 5 and 6. (b) If each number given in (a) is diminished by 2; find the new value of mean.
Solution:
(a) The mean of 7, 11, 6, 5 and 6.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q4.1
(b) If we subtract 2 from each number, then the mean will be 7 – 2 = 5

Question 5.
If the mean of 6, 4, 7, a and 10 is 8. Find the value of ‘a’.
Solution:
No. of terms = 5
Mean = 8
∴ Sum of number (Σxi) = 5 x 8 = 40 …(i)
But Σxi = 6 + 4 + 7 + a+10 = 27 + a ….(ii)
From (i) and (ii)
27 + a = 40 ⇒ a = 40 – 27
∴ a = 13

Question 6.
The mean of the number 6, y, 7, x and 14 is 8. Express y in terms of x.
Solution:
No. of terms = 5 and
mean = 8
∴ Sum of numbers (Σxi) = 5 x 8 = 40 ….(i)
But sum of numbers given = 6 + y + 7 + x + 14
= 21 + y + x + ….(ii)
From (i) and (ii)
27 + y + x = 40
⇒ y = 40 – 27 – x
⇒ y= 13 – x

Question 7.
The ages of 40 students are given in the following table :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q7.1
Find the arithmetic mean.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q7.2

Question 8.
If 69.5 is the mean of 72, 70, x, 62, 50, 71, 90, 64, 58 and 82, find the value of x.
Solution:
No. of terms = 10
Mean = 69.5
∴ Sum of numbers = 69,5 x 10 = 695 ….(i)
But sum of given number = 72 + 70 + x + 62 + 50 + 71 + 90 + 64 + 58 + 82 = 619+x ….(ii)
From (i) and (ii)
619 + x = 695
⇒ x = 695 – 619 = 76

Question 9.
The following table gives the heights of plants in centimeter. If the mean height of plants is 60.95 cm; find the value of ‘f’.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q9.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q9.3

Question 10.
From the data given below, calculate the mean wage, correct to the nearest rupee.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q10.1
(i) If the number of workers in each category is doubled, what would be the new mean wage? [1995]
(ii) If the wages per day in each category are increased by 60%; what is the new mean wage?
(iii) If the number of workers in each category is doubled and the wages per day per worker are reduced by 40%. What would be the new mean wage?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q10.2
(i) Mean remains the same if the number of workers in each catagory is doubled.
∴ Mean = 80.
(ii) Mean will be increased by 60% if the wages per day per worker is increased by 60%.
∴ New mean = 80 x \(\frac { 160 }{ 100 }\)= 128
(iii) No change in the mean if the number of worker is doubled but if wages per worker is reduced by 40%, then
New mean = 80 x \(\frac { 60 }{ 100 }\)= 48

Question 11.
The contents of 100 match boxes were checked to determine the number of matches they contained.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q11.1
(i) Calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean up to exactly 39 matches. [1997]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q11.2
(ii) In the second case,
New mean = 39 matches
∴ Total contents = 39 x 100 = 3900
But total no of matches already given = 3813
∴ Number of new matches to be added = 3900 – 3813 = 87

Question 12.
If the mean of the following distribution is 3, find the value of p.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q12.1
Solution:
Mean = 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q12.3

Question 13.
In the following table, Σf= 200 and mean = 73. Find the missing frequencies f1 and f2.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q13.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q13.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q13.3

Question 14.
Find the arithmetic mean (correct to the nearest whole-number) by using step-deviation method.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q14.1
Solution:
Let the Assumed mean = 30
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q14.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q14.3

Question 15.
Find the mean (correct to one place of decimal) by using short-cut method.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q15.1
Solution:
Let the Assumed mean A = 45
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q15.3

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

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

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 10A

Other Exercises

Question 1.
Which of the following sequences are in arithmetic progression?
(i) 2, 6, 10, 14, …….
(ii) 15, 12, 9, 6,…….
(iii) 5, 9, 12, 18, ……
(iv) \(\frac { 1 }{ 2 } ,\frac { 1 }{ 3 } ,\frac { 1 }{ 4 } ,\frac { 1 }{ 5 } \),…..
Solution:
(i) 2, 6, 10, 14,
Here, d = 6 – 2 = 4
10 – 6 = 4
14 – 10 = 4
∴ In each case (d) is same.
∴It is an arithmetic progression.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q1.2
∴d is not the same
∴It is not an arithmetic progression

Question 2.
The nth term of a sequence is (2n – 3), find its fifteenth term.
Solution:
Tn = 2n – 3
T15 = 2 x 15 – 3
= 30 – 3
= 27

Question 3.
If the pth term of an A.P. is (2p + 3); find the A.P.
Solution:
Tp = 2p + 3
T1 = 2 x 1 + 3 = 2 + 3 = 5
T2 = 2 x 2 + 3 = 4 + 3 = 7
T3 = 2 x 3 + 3 = 6 + 3 = 9
∴A.P. is 5, 7, 9,…..

Question 4.
Find the 24th term of the sequence : 12, 10, 8, 6,…….
Solution:
A.P. is 12, 10, 8, 6,……
Here a = 12, d = 10 – 12 = – 2
Tn = a + (n – 1)d
T24 = 12 + (24 – 1) x ( – 2)
= 12 + 23 x ( – 2)
= 12 – 46
= – 34

Question 5.
Find the 30th term of the sequence :
\(\\ \frac { 1 }{ 2 } \), 1, \(\\ \frac { 3 }{ 2 } \),….
Solution:
\(\\ \frac { 1 }{ 2 } \), 1, \(\\ \frac { 3 }{ 2 } \),….
Here a = \(\\ \frac { 1 }{ 2 } \), d = \(1- \frac { 1 }{ 2 } \)
\(\\ \frac { 1 }{ 2 } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q5.1

Question 6.
Find the 100th term of the sequence :
√3, 2√3, 3√3,….
Solution:
√3, 2√3, 3√3,….
Here a = √3, d = 2√3 – √3 = √3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q6.1

Question 7.
Find the 50th term of the sequence :
\(\\ \frac { 1 }{ n } \), \(\\ \frac { n+1 }{ n } \), \(\\ \frac { 2n+1 }{ n } \),……
Solution:
\(\\ \frac { 1 }{ n } \), \(\\ \frac { n+1 }{ n } \), \(\\ \frac { 2n+1 }{ n } \),……
=>\(\\ \frac { 1 }{ n } \), \(1+ \frac { 1 }{ n } \), \(2+ \frac { 1 }{ n } \),…..
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q7.1

Question 8.
Is 402 a term of the sequence :
8, 13, 18, 23,…?
Solution:
In sequence 8, 13, 18, 23,…..
a = 8 and d = 13 – 8 = 5
Let 402 be the nth term, then
402 = a + (n – 1)d = 8 + (n – 1) x 5
\(\\ \frac { 402-8 }{ 5 } \) = n – 1
=> \(\\ \frac { 394 }{ 5 } \) = n – 1
394 is not exactly divisible by 5,
∴ 402 is not its term.

Question 9.
Find the common difference and 99th term of the arithmetic progression :
\(7 \frac { 3 }{ 4 } \), \(9 \frac { 1 }{ 2 } \), \(11 \frac { 1 }{ 4 } \),…..
Solution:
\(7 \frac { 3 }{ 4 } \), \(9 \frac { 1 }{ 2 } \), \(11 \frac { 1 }{ 4 } \),…..
here a = \(7 \frac { 3 }{ 4 } \)
= \(\\ \frac { 31 }{ 4 } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q9.2

Question 10.
How many terms are there in the series:
(i) 4, 7, 10, 13,…..148 ?
(ii) 0.5, 0.53, 0.56,…..1.1 ?
(iii) \(\\ \frac { 3 }{ 4 } \), 1, \(1 \frac { 1 }{ 4 } \),….3 ?
Solution:
(i) 4, 7, 10, 13,…..148 ?
Here a = 4, d = 7 – 4
= 3
Let 148 be the nth term, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q10.2

Question 11.
Which term of the A.P. 1 + 4 + 7 + 10 +……is 52 ?
Solution:
A.P. is 1, 4, 7, 10,……is 52
Here a = 1, d = 4 – 1 = 3
Let 52 be the nth term, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q11.1

Question 12.
If 5th and 6th terms of an A.P. are respectively 6 and 5, find the 11th term of the A.P.
Solution:
5th term (T5) = 6
=> 6 = a + 4d
and T6 = 5
=> 5 = a + 5d
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q12.1

Question 13.
If tn represents nth term of an A.P., t2 + t5 – t3 = 10 and t2 + t9 = 17, find its first term and its common difference.
Solution:
Let first term of an A.P. be a
and common difference = d
Then Tn is the nth term
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q13.1

Question 14.
Find the 10th term from the end of the A.P. 4, 9, 14,…. 254.
Solution:
We know that rth term from the end
= (n – r + 1)th from the beginning
Here, a = A, d = 9 – 4 = 5,
nth term = 254
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q14.1

Question 15.
Determine the arithmetic progression whose 3rd term is 5 and 7th term is 9.
Solution:
Let a be the first term and d be the common difference, then
T3 = a + 2d – 5
T7 = a + 6d = 9
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q15.1

Question 16.
Find the 31st term of an A.P. whose 10th term is 38 and 16th term is 74.
Solution:
Let a be the first term and d be the common difference, then
T10 = a + 9d =38
and T16 = a + 15d = 74
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q16.1

Question 17.
Which term of the series :
21, 18, 15, is – 81 ?
Can any term of this series be zero ? If yes, find the number of term.
Solution:
21, 18, 15, 1….. – 81
Here a = 21,d = 18 – 21 = – 3,
nth term = – 81
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q17.1

Question 18.
An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 31st term.
Solution:
Given : An A.P. consists of 60 terms
a = 7
and a60 = 125
=> a + (60 – 1 )d = 125
=> 7 + 59 d = 125
=> 59d= 125 – 7
=> d = \(\\ \frac { 118 }{ 59 } \) = 2
now, a31 = a + (n – 1)d
= 7 + (31 – 1) x 2
= 7 + 30 x 2
= 67

Question 19.
The sum of the 4th and the 8th terms of an A.P. is 24 and the sum of the sixth term and the tenth term is 34. Find the first three terms of the A.P.
Solution:
In an A.P.
T4 + T8 = 24
T6 + T10 =34
Let a be the first term and d be the common difference
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q19.2

Question 20.
If the third term of an A.P. is 5 and the seventh terms is 9, find the 17th term
Solution:
Let the first term of an A.P. = a
and the common difference of the given A.P. = d
As, we know that,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q20.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A Q20.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A 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 5 Quadratic Equations Ex 5D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D.

Other Exercises

Solve each of the following equations :
Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q1.2

Question 2.
(2x + 3)² = 81
Solution:
(2x + 3)2 = 81
⇒ 4x² + 12x + 9 = 81
⇒ 4x² + 12x + 9 – 81 = 0
⇒ 4x² + 12x – 72 = 0
⇒ x² + 3x – 18 = 0 (Dividing by 4)
⇒ x² + 6x – 3x – 18 = 0
⇒ x (x + 6) – 3 (x + 6) = 0
⇒ (x + 6) (x – 3) = 0
Either x + 6 = 0, then x = -6
or x – 3 = 0, then x = 3
x = 3, – 6

Question 3.
a² x² – b² = 0
Solution:
a² x² – b² = 0
⇒ (ax)² – (b)² =0
⇒ (ax + b) (ax – b) =
Either ax + b = 0, then x = \(\frac { -b }{ a }\)
or ax – b = 0. then x = \(\frac { b }{ a }\)
x = \(\frac { b }{ a }\) , \(\frac { -b }{ a }\)

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q4.2

Question 5.
x + \(\frac { 4 }{ x }\) = – 4; x ≠ 0
Solution:
x + \(\frac { 4 }{ x }\) = -4
⇒ x² + 4 = -4x
⇒ x² + 4x + 4 = 0
⇒ (x + 2)² = 0
⇒ x + 2 = 0
⇒ x = – 2

Question 6.
2x4 – 5x² + 3 = 0
Solution:
2x4 – 5x² + 3 = 0
⇒ 2(x²)² – 5x² + 3 = 0
⇒ 2(x²)² – 3x² – 2x² + 3 = 0
⇒ 2x4 – 3x² – 2x² + 3 = 0
⇒ x² (2x² – 3) – 1 (2x² – 3) = 0
⇒ (2x² – 3) (x² – 1) = 0
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q6.1

Question 7.
x4 – 2x² – 3 = 0
Solution:
x4 – 2x² – 3 = 0
⇒ (x²)² – 2x² – 3 = 0
⇒ (x²)² – 3x² + x² – 3 = 0
⇒ x² (x² – 3) + 1 (x² -3) = 0
⇒ (x² – 3) (x² + 1) = 0
Either x² – 3 = 0, then x² = 3 ⇒ x = √3
or x² + 1 = 0, then x² = – 1 In this case roots are not real
x = ±√3 or √3 , – √3

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q8.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q8.3

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.4

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q10.3

Question 11.
(x² + 5x + 4)(x² + 5x + 6) = 120
Solution:
Let x² + 5x + 4 = y then x² + 5x + 6 = y + 2
Now (x² + 5x + 4) (x² + 5x + 6) = 120
⇒ y (y + 2) – 120 = 0
⇒ y² + 2y – 120 = 0
⇒ y² + 12y – 10y – 120 = 0
⇒ y (y + 12) – 10 (y + 12) = 0
⇒ (y + 12) (y – 10) = 0
Either y + 12 = 0, then y = – 12
or y – 10 = 0, then y = 10
(i) when y = -12, then x² + 5x + 4 = -12
⇒ x² + 5x + 4 + 12 = 0
⇒ x² + 5x + 16 = 0
Here a = 1, b = 5, c = 16
D = b² – 4ac = (5)² – 4 x 1 x 16 = 25 – 64 = -39
D < 0, then roots are not real
(ii) When y = 10, then x² + 5x + 4 = 10
⇒ x² + 5x + 4 – 10 = 0
⇒ x² + 5x – 6 = 0
⇒ x² + 6x – x – 6 = 0
⇒ x (x + 6) – 1 (x + 6) = 0
⇒ (x + 6) (x – 1) = 0
Either x + 6 = 0, then x = – 6
or x – 1 = 0, then x = 1
x = 1, -6

Question 12.
Solve each of the following equations, giving answer upto two decimal places:
(i) x² – 5x – 10 = 0 [2005]
(ii) 3x² – x – 7 = 0 [2004]
Solution:
(i) Given Equation is : x² – 5x – 10 = 0
On comparing with, ax² + bx + c = 0
a = 1, b = -5 , c = -10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q12.3

Question 13.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q13.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q13.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q13.3

Question 14.
Solve:
(i) x² – 11x – 12 = 0; when x ∈ N
(ii) x² – 4x – 12 = 0; when x ∈ I
(iii) 2x² – 9x + 10 = 0; when x ∈ Q.
Solution:
(i) x² – 11x – 12 = 0
⇒ x² – 12x + x – 12 = 0
⇒ x (x – 12) + 1 (x – 12) = 0
⇒ (x – 12) (x + 1) = 0
Either x – 12 = 0, then x = 12
or x + 1 = 0, then x = -1
x ∈ N
x = 12
(ii) x² – 4x – 12 = 0
⇒ x² – 6x + 2x – 12 = 0
⇒ x (x – 6) + 2 (x – 6)=0
⇒ (x – 6) (x + 2) = 0
Either x – 6 = 0, then x = 6
or x + 2 = 0, then x = -2
x ∈ I
x = 6, -2
(iii) 2x² – 9x + 10 = 0
⇒ 2x² – 4x – 5x + 10 = 0
⇒ 2x (x – 2) – 5 (x – 2) = 0
⇒ (x – 2) (2x – 5) = 0
Either x – 2 = 0, then x = 2
or 2x – 5 = 0, then 2x = 5 ⇒ x = \(\frac { 1 }{ 2 }\)
x ∈ Q
x = 2, \(\frac { 5 }{ 2 }\) or 2, 2.5

Question 15.
Solve: (a + b)² x² – (a + b) x – 6 = 0, a + b ≠ 0.
Solution:
(a + b)² x² – (a + b) x – 6 = 0
Let (a + b) x = y, then y² – y – 6 = 0
⇒ y² – 3y + 2y – 6 = 0
⇒ y (y – 3) + 2 (y – 3) = 0
⇒ (y – 3) (y + 2) = 0
Either y – 3 = 0, then y = 3
or y + 2 = 0, then y = – 2
(i) If y = 3, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q15.1

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q16.1
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q16.3
Either x + p = 0, then x = -p
or x + q = 0, then x = -q
Hence x = -p, -q

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q17.1
Solution:
(i) x (x + 1) + (x + 2) (x + 3) = 42
⇒ x² + x + x² + 3x + 2x + 6 – 42 = 0
⇒ 2x² + 6x – 36 = 0
⇒ x² + 3x – 18 = 0
⇒ x² + 6x – 3x – 18 = 0
⇒ x (x + 6) – 3(x + 6) = 0
⇒ (x + 6) (x – 3) = 0
Either x + 6 = 0, then x = -6
or x – 3 = 0, then x = 3
Hence x = 3, -6
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q17.2

Question 18.
For each equation, given below, find the value of ‘m’ so that the equation has equal roots. Also, find the solution of each equation:
(i) (m – 3) x² – 4x + 1 = 0
(ii) 3x² + 12x + (m + 7) = 0
(iii) x² – (m + 2) x + (m + 5) = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q18.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q18.3

Question 19.
Without solving the following quadratic equation, find the value of ‘p’ for which the roots are equal. px² – 4x + 3 = 0.
Solution:
px² – 4x + 3 = 0 …..(i)
Compare (i) with ax² + bx + c = 0
Here a = p, b = -4, c = 3
D = b² – 4ac = (-4)² – 4.p.(3) = 16 – 12p
As roots are equal, D = 0
16 – 12p = 0
⇒ \(\frac { 16 }{ 12 }\) = p
⇒ p = \(\frac { 4 }{ 3 }\)

Question 20.
Without solving the following quadratic equation, find the value of m for which the given equation has real and equal roots : x² + 2 (m – 1) x + (m + 5) = 0.
Solution:
x² + 2 (m – 1) x + (m + 5) = 0.
Here, a = 1, b = 2 (m – 1), c = m + 5
So, discriminant, D = b² – 4ac
= 4(m – 1)² – 4 x 1 (m + 5)
= 4m² + 4 – 8m – 4m – 20
= 4m² – 12m – 16
For real and equal roots D = 0
So, 4m² – 12m – 16 = 0
⇒ m² – 3m – 4 = 0 (Dividingby4)
⇒ m² – 4m + m – 4 = 0
⇒ m (m – 4) + 1 (m – 4) = 0
⇒ (m – 4) (m + 1) = 0
⇒ m = 4 or m = -1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D 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.